CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Enthalpy
Question Bank - Set 2
Liberty University
Question 1
Question
Calculate the change in enthalpy (∆H) for the following reaction at 298 K:
2C(graphite)+3H2(g)→C2H6(g)
Given the following enthalpy values:
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.7 kJ/mol
Solution
Step 1: Write the balanced chemical equation and the standard enthalpy of
formation for each compound involved in the reaction.
2C(graphite)+3H2(g)→C2H6(g)
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.7 kJ/mol
Step 2: Calculate the change in enthalpy using the equation:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given values into the formula and calculate ∆H.
∆H= [(−84.7) kJ/mol] −[(2 ×0) + (3 ×0)] kJ/mol
∆H=−84.7 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H=−84.7 kJ/mol
at 298 K.
Question 2
Question
A reaction is carried out in a bomb calorimeter, and the temperature of the
calorimeter increases by 5
°
C. If the heat capacity of the calorimeter is 20 J/
°
C,
and the reaction releases 5000 J of heat, calculate the change in enthalpy (∆H)
of the reaction.
Solution
Step 1: Calculate the heat absorbed by the calorimeter using the formula:
qcalorimeter =C·∆T
where: C= heat capacity of the calorimeter = 20 J/
°
C, ∆T= change in
temperature = 5
°
C.
Therefore,
qcalorimeter = 20 J/
°
C·5
°
C = 100 J
Step 2: Since the calorimeter is absorbing the heat released by the reaction,
the heat absorbed by the reaction is equal in magnitude but opposite in sign:
qreaction =−qcalorimeter =−100 J
Step 3: The heat absorbed by the reaction is related to the change in enthalpy
of the reaction through the equation:
∆H=qreaction
Therefore, the change in enthalpy of the reaction is:
∆H=−100 J
Question 3
Question
A reaction takes place in a bomb calorimeter and releases 1250 kJ of heat. If
the volume of water in the calorimeter is 500 mL and its initial temperature
was 25
°
C, what is the final temperature of the water? Assume no heat is lost
to the surroundings. The specific heat of water is 4.18 J/g
°
C.
2
Solution
Step 1: Calculate the heat capacity of the water.
Q=mc∆T
1250000 = (500 g)(4.18 J/g
°
C)(Tf−25)
Tf−25 = 1250000
(500)(4.18)
Tf=1250000
(500)(4.18) + 25
Tf≈89.95
°
C
So, the final temperature of the water is approximately 89.95
°
C.
Question 4
Question
Calculate the change in enthalpy (∆H) when 2.5 moles of water (H2O) liquid at
25◦C is converted into steam at 100◦C. Given that the specific heat capacity of
water is 4.18 J/(g·◦C), the molar heat of vaporization of water is 40.79 kJ/mol,
and the molar mass of water is 18.02 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 25◦C
to 100◦C.
The specific heat capacity of water is 4.18 J/(g·◦C), so the heat required is
given by the equation:
q=m×c×∆T
Where: - mis the mass of water in grams - cis the specific heat capacity of
water - ∆Tis the change in temperature
Given that there are 2.5 moles of water and the molar mass of water is 18.02
g/mol, the mass of water is calculated as:
Mass of water = 2.5 moles ×18.02 g/mol = 45.05 g
Substitute the values into the equation:
q= 45.05 g ×4.18 J/(g ·◦C) ×(100 −25)◦C
q= 45.05 ×4.18 ×75
q= 14141.775 J = 14.14 kJ
3
Step 2: Calculate the heat required to convert water at 100◦C to steam at
100◦C.
The molar heat of vaporization of water is 40.79 kJ/mol. Since we have 2.5
moles of water, the heat required for vaporization is:
q= 2.5 moles ×40.79 kJ/mol = 101.975 kJ
Step 3: Calculate the total change in enthalpy using the heat values calcu-
lated in steps 1 and 2.
∆H=qraising temp +qvaporization
∆H= 14.14 kJ + 101.975 kJ
∆H= 116.115 kJ
Therefore, the change in enthalpy (∆H) when 2.5 moles of water liquid at
25◦C is converted into steam at 100◦C is 116.115 kJ.
Question 5
Question
Calculate the change in enthalpy for the reaction
2C(s)+2H2(g)→C2H4(g)
given the following enthalpy values:
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H4(g)) = 52.3 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation is:
2C(s)+2H2(g)→C2H4(g)
Step 2: Calculate the change in enthalpy using the standard enthalpies of
formation. The change in enthalpy, H◦
rxn, can be calculated using the equation:
H◦
rxn =nH◦
f, products −(mH◦
f, reactants)
where nand mare the stoichiometric coefficients in the balanced equation, and
H◦
frepresents the standard enthalpy of formation.
4
Given that:
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H4(g)) = 52.3 kJ/mol
Substitute the values into the equation:
H◦
rxn = (1 ·52.3 kJ/mol) −[(2 ·0 kJ/mol) + (2 ·0 kJ/mol)]
H◦
rxn = 52.3 kJ/mol
Therefore, the change in enthalpy for the reaction is 52.3 kJ/mol .
Question 6
Question
A reaction is known to have ∆H◦=−92.2 kJ and ∆S◦=−145.1 J/K. Calculate
the temperature at which the reaction becomes spontaneous.
Solution
Step 1: Write down the equation for Gibb’s free energy change (∆G):
∆G= ∆H−T·∆S
Step 2: At equilibrium, ∆G= 0. Therefore, we have:
0 = −92.2 kJ −T·(−0.1451 kJ/K)
Step 3: Solve for the temperature T:
T=92.2 kJ
0.1451 kJ/K
T≈635.13 K
Therefore, the reaction becomes spontaneous at approximately 635.13 Kelvin.
Question 7
Question
A reaction occurring in a bomb calorimeter releases 1500 J of heat while doing
300 J of work on the surroundings. Calculate the change in enthalpy (∆H) for
the reaction.
5
Solution
Step 1: Recall that the change in enthalpy (∆H) is given by the equation:
∆H= ∆E+P∆V
where ∆Eis the change in internal energy and P∆Vis the work done on (or
by) the system.
Step 2: We can calculate the change in internal energy (∆E) using the first
law of thermodynamics:
∆E=q+w
where qis the heat absorbed (positive) or released (negative) by the system and
wis the work done on (positive) or by (negative) the system.
Step 3: In this case, the heat released by the reaction is -1500 J (negative
because it is released) and the work done on the surroundings is 300 J (positive
because it is done on the surroundings). Therefore, we have:
∆E=−1500 J + 300 J
∆E=−1200 J
Step 4: Now we can calculate the change in enthalpy (∆H) by adding the
change in internal energy to the work done on the system:
∆H= ∆E+P∆V
∆H=−1200 J + 300 J
∆H=−900 J
Therefore, the change in enthalpy (∆H) for the reaction is -900 J.
Question 8
Question
A gas at a constant pressure of 1 atm is heated from 25◦C to 100◦C. During
this process, the gas absorbs 150 J of heat and expands to a volume of 5 L.
Calculate the change in enthalpy (∆H) for the gas.
Solution
Step 1: Calculate the change in temperature: Given: Initial temperature,
Tinitial = 25◦C = 25 + 273 = 298 K Final temperature, Tfinal = 100◦C =
100 + 273 = 373 K
Change in temperature: ∆T=Tfinal−Tinitial ∆T= 373K−298K∆T= 75K
Step 2: Calculate the heat absorbed by the gas: Given: Heat absorbed, q =
150 J
6
Step 3: Calculate the change in enthalpy using the equation: ∆H=q+P·
∆V
Step 4: Calculate the change in volume: Given: Initial volume, Vinitial = 0
(not needed in this case, as ∆Vwill be equal to the final volume since the gas
expands) Final volume, Vfinal = 5 L
Change in volume: ∆V=Vfinal −Vinitial ∆V= 5 L
Step 5: Plug in the values into the equation for change in enthalpy: ∆H=
q+P·∆V∆H= 150 J + (1 atm) ·5 L ·(101.3 J/L
·
atm) ∆H= 150 J + 506.5 J
∆H= 656.5 J
Therefore, the change in enthalpy (∆H) for the gas is 656.5 J.
Question 9
Question
Calculate the change in enthalpy (∆H) when 15.0 g of water at 25
°
C is converted
to steam at 150
°
C. The specific heat capacity of water is 4.18 J/g
°
C, the molar
enthalpy of vaporization of water is 40.7 kJ/mol, and the molar mass of water
is 18.0 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 25
°
C
to 100
°
C. The heat required can be calculated using the formula:
q=mc∆T
where: q= heat energy, m= mass of the substance, c= specific heat capacity,
and ∆T= change in temperature. Given that m= 15.0 g, c= 4.18 J/g
°
C, and
∆T= 100
°
C, we can substitute these values into the formula to find q.
Step 2: Calculate the heat required to convert water at 100
°
C to steam at
100
°
C (phase change). The heat required for this phase change is equal to the
molar enthalpy of vaporization of water, which is 40.7 kJ/mol. First, we need
to find the number of moles of water in 15.0 g.
Moles = Mass (g)
Molar mass (g/mol)
Moles = 15.0 g
18.0 g/mol
Moles ≈0.833 mol
Next, we calculate the heat required for the phase change:
q=n·∆Hvap
q= 0.833 mol ×40.7×103J/mol
q≈33.91 ×103J
7
Step 3: Calculate the heat required to raise the temperature of steam from
100
°
C to 150
°
C. Using the formula q=mc∆Twith m= 15.0 g, c= 2.01 J/g
°
C
(specific heat capacity of steam), and ∆T= 50
°
C. Substitute these values into
the formula to find q.
Step 4: Calculate the total change in enthalpy (∆H).
∆H=q1+q2+q3
Substitute the values of q1,q2, and q3into the formula to calculate ∆H.
Question 10
Question
A sample of gas undergoes a process at constant pressure where it absorbs 150
J of heat and does 50 J of work on the surroundings. If the change in internal
energy of the gas is 100 J, what is the change in enthalpy of the gas during the
process?
Solution
Step 1: Recall the definition of enthalpy: Enthalpy is defined as H=U+P V ,
where His enthalpy, Uis internal energy, Pis pressure, and Vis volume.
Step 2: The change in enthalpy can be calculated using the equation:
∆H= ∆U+P∆V
Step 3: Given that ∆U= 100 J, and the process is at constant pressure, we
can rewrite P∆Vas w, the work done by the gas on the surroundings. Thus,
∆H= ∆U+w.
Step 4: Plugging in the values, we have:
∆H= 100 J + (−50 J)
Step 5: Simplifying, we find:
∆H= 50 J
Step 6: Therefore, the change in enthalpy of the gas during the process is
50 J .
Question 11
Question
A reaction is carried out in a bomb calorimeter, and it is found that the tem-
perature of the calorimeter increases by 5.82
°
C. If the heat capacity of the
calorimeter is 247 J/
°
C, calculate the enthalpy change (∆H) for the reaction in
kJ/mol.
8
Solution
Step 1: Calculate the heat absorbed by the bomb calorimeter using the formula:
q=C·∆T
where: q= heat absorbed by the calorimeter, C= heat capacity of the calorime-
ter, ∆T= change in temperature.
Plugging in the values, we get:
q= 247 J/
°
C×5.82
°
C = 1439.74 J
Step 2: Convert the heat absorbed by the calorimeter to kilojoules (1 kJ =
1000 J):
q= 1439.74 J ×1 kJ
1000 J = 1.44 kJ
Step 3: Calculate the moles of substance reacting using the molar heat
capacity formula:
∆H=q
n
where: ∆H= enthalpy change, q= heat absorbed by the calorimeter, n=
moles of substance.
Step 4: Rearrange the formula to solve for ∆H:
∆H=q
n
n=q
∆H
Step 5: The molar heat capacity of the reaction is then given by:
∆H=1.44 kJ
n
Step 6: Finally, divide the heat absorbed by the calorimeter by the moles of
substance to determine the enthalpy change in kJ/mol.
Question 12
Question
A reaction takes place in a bomb calorimeter, and the temperature of the water
in the bomb calorimeter changes from 25
°
C to 30
°
C. If the heat capacity of the
bomb calorimeter is 50 J/
°
C, calculate the change in enthalpy of the reaction.
Assume that the heat released from the reaction is only absorbed by the water
in the calorimeter, and the bomb calorimeter is insulated with its surroundings.
9
Solution
Step 1: Calculate the heat absorbed by the water in the bomb calorimeter using
the formula:
q=C·∆T
where: q= heat absorbed by the water in the calorimeter, C= 50 J/
°
C =
heat capacity of the bomb calorimeter, and ∆T= 30C−25C= 5C= change
in temperature.
Substitute the values into the formula:
q= 50 J/
°
C·5C= 250 J
Therefore, the water in the bomb calorimeter absorbs 250 J of heat.
Step 2: Since the reaction is taking place in the bomb calorimeter, the heat
absorbed by the water is equal to the heat released by the reaction according
to the first law of thermodynamics. This heat is given by the formula:
∆H=−q
where: ∆H= change in enthalpy of the reaction and q= 250 J (negative
because the reaction releases heat).
Therefore, the change in enthalpy of the reaction is:
∆H=−250 J
Hence, the change in enthalpy of the reaction is -250 J.
Question 13
Question
A sample of an ideal gas is taken through a cyclic process consisting of two
isobaric and two isochoric processes. The initial temperature and pressure of
the gas are T1and P1, and the final temperature and pressure are T2and P2,
respectively. Calculate the change in enthalpy for the gas during this process.
Solution
Step 1: Recall that for an ideal gas, the change in enthalpy is given by ∆H=
Cp·∆T, where Cpis the specific heat at constant pressure and ∆Tis the change
in temperature.
Step 2: For an isobaric process, Cp=5
2Rwhere Ris the gas constant. Thus,
the change in enthalpy for the first isobaric process is ∆H1=5
2R·(T2−T1).
Step 3: For an isochoric process, ∆H= ∆U, where ∆Uis the change in
internal energy. Since there is no work done in an isochoric process, ∆U=
nCv∆T, where Cvis the specific heat at constant volume.
10
Step 4: For an ideal gas, Cv=3
2R. Therefore, the change in enthalpy for
the first isochoric process is ∆H2=3
2R·(T2−T1).
Step 5: The total change in enthalpy for the cyclic process is then ∆H=
∆H1+ ∆H2=5
2R+3
2R(T2−T1) = 4R(T2−T1).
Question 14
Question
A reaction has an enthalpy change of -285 kJ/mol. If 2.00 moles of the reactant
are consumed in the reaction at constant pressure, what is the enthalpy change
for the reaction in kJ?
Solution
Step 1: Start with the given information: The enthalpy change for the reaction
is -285 kJ/mol. The number of moles of the reactant consumed is 2.00 mol.
Step 2: Calculate the enthalpy change for the reaction in kJ. Given that
the enthalpy change is -285 kJ/mol, we can use this value to find the enthalpy
change for the reaction when 2.00 moles of the reactant are consumed:
Enthalpy change for the reaction = Enthalpy change per mole×Number of moles consumed
Enthalpy change for the reaction = −285 kJ/mol ×2.00 mol
Step 3: Perform the calculation.
Enthalpy change for the reaction = −285 kJ/mol ×2.00 mol = −570 kJ
Therefore, when 2.00 moles of the reactant are consumed in the reaction,
the enthalpy change for the reaction is -570 kJ.
Question 15
Question
Calculate the change in enthalpy when 5 moles of nitrogen gas at 200 K and 1
atm pressure are heated at constant pressure to 400 K. Assume that nitrogen
gas behaves ideally.
Solution
Step 1: The change in enthalpy can be calculated using the formula:
∆H= ∆U+P∆V
11
Step 2: First, we calculate the change in internal energy (∆U) using the
formula:
∆U=nCv∆T
where nis the number of moles, Cvis the molar heat capacity at constant
volume, and ∆Tis the change in temperature.
Step 3: Since nitrogen gas is diatomic, the molar heat capacity at constant
volume is Cv=5
2Rwhere Ris the gas constant.
Step 4: Substituting the values into the formula, we get:
∆U= 5 ×5
2R×(400 −200)
Step 5: Next, we calculate the work done by the gas, which is equal to P∆V.
Since the process is at constant pressure, we have:
∆V=nR∆T
Step 6: Substituting the values into the formula, we get:
∆V= 5 ×R×(400 −200)
Step 7: Now we can calculate the work done:
P∆V= 1 ×(5R×200)
Step 8: Finally, we can calculate the change in enthalpy by adding the change
in internal energy and the work done:
∆H= ∆U+P∆V
∆H= (5 ×5
2R×200) + (5R×200)
Question 16
Question
Given the following data for the reaction:
∆H◦
ffor CO2= -393.5 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction:
C(graphite) + 2H2O→CO2+ 2H2O
12
Solution
Step 1: Write the balanced chemical equation for the reaction and determine the
∆H◦value for the reaction. The balanced chemical equation for the reaction is:
C(graphite) + 2H2O→CO2+ 2H2O
To find the ∆H◦for the reaction, we use the following formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the ∆H◦value for the reaction. Given data: ∆H◦
ffor
CO2= -393.5 kJ/mol ∆H◦
ffor H2O(l) = -285.8 kJ/mol ∆H◦
ffor C(graphite)
= 0 kJ/mol
Plugging in the values into the formula:
∆H◦= [1(−393.5 kJ/mol) + 2(−285.8 kJ/mol)] −[1(0 kJ/mol) + 2(0 kJ/mol)]
∆H◦= [−393.5 kJ/mol −571.6 kJ/mol]
∆H◦=−965.1 kJ/mol
Therefore, the standard enthalpy change, ∆H◦, for the reaction:
C(graphite) + 2H2O→CO2+ 2H2O
is -965.1 kJ/mol.
Question 17
Question
A certain chemical reaction is known to have a standard enthalpy change of -383
kJ mol−1. If 2.50 moles of the reactant are consumed in the reaction, calculate
the amount of heat absorbed or released.
Solution
Step 1: Recall the formula relating enthalpy change (∆H) to heat absorbed or
released (q) and amount of substance (n):
q= ∆H×n
Step 2: Substitute the known values into the formula:
q=−383 kJ mol−1×2.50 mol
Step 3: Perform the calculation:
q=−383 ×2.50 = −957.5 kJ
Step 4: Thus, the amount of heat absorbed or released during the reaction
is -957.5 kJ. Since the enthalpy change is negative, it indicates that the reaction
releases heat.
13
Question 18
Question
A reaction is carried out in a bomb calorimeter and releases 150 kJ of heat.
If the volume of the bomb calorimeter is 500 mL and the pressure inside the
calorimeter remains constant at 1 atm, calculate the change in enthalpy (∆H)
of the reaction in kJ/mol. Assume ideal gas behavior.
Solution
Step 1: Recall that ∆H=qp, where qpis the heat exchanged at constant
pressure during a chemical reaction.
Step 2: Convert the volume of the bomb calorimeter from mL to L: V=
500 mL ×1 L
1000 mL = 0.5 L
Step 3: Since the pressure remains constant and the reaction takes place in a
bomb calorimeter, the heat exchanged is equal to the change in enthalpy (∆H).
Step 4: The change in enthalpy of the reaction can be calculated using the
formula: ∆H=q
n
Step 5: Given that the heat exchanged (q) is 150 kJ and the number of moles
(n) is unknown, we have: ∆H=150 kJ
n
Step 6: To find the number of moles, we need to convert the volume at STP
to number of moles of gas. At STP, 22.4 L of any gas contains 1 mole.
Step 7: Assuming ideal gas behavior, the volume of 0.5 L at 1 atm pressure
contains nmoles: n=0.5 L
22.4 L/mol
Step 8: Substitute the value of nback into the formula for ∆Hto find the
change in enthalpy of the reaction: ∆H=150 kJ
0.5 L/22.4 L/mol = 150 kJ ×22.4 L/mol
0.5 L
Step 9: ∆H= 150 ×22.4 = 3360 kJ/mol
Therefore, the change in enthalpy of the reaction is 3360 kJ/mol.
Question 19
Question
A reaction is taking place in a closed system where the enthalpy change (∆H)
is given by the equation:
∆H= 50 −2T+ 0.05T2
where ∆His in kJ/mol and Tis in Kelvin. Determine the temperature at which
the reaction is exothermic.
Solution
Step 1: To determine the temperature at which the reaction is exothermic, we
need to find the point where ∆Hbecomes negative. Step 2: Set ∆Hto be less
14
than zero and solve for T:
50 −2T+ 0.05T2<0
0.05T2−2T+ 50 <0
Step 3: This is a quadratic inequality. To solve it, we find the roots of the
corresponding quadratic equation:
0.05T2−2T+ 50 = 0
T2−40T+ 1000 = 0
Step 4: Use the quadratic formula to find the roots:
T=−(−40) ±p(−40)2−4(1)(1000)
2(1)
T=40 ±√1600 −4000
2
T=40 ±√−2400
2
Step 5: Since the square root of a negative number is not real, the inequality
has no real solutions. Therefore, the reaction is never exothermic.
Question 20
Question
Given the reaction:
2C(graphite)+3H2(g)→C2H6(g)
If the enthalpy change for this reaction is -84 kJ/mol, calculate the enthalpy
change when 6.00 g of carbon reacts with excess hydrogen. (Molar mass of
carbon = 12.01 g/mol, molar mass of hydrogen = 1.008 g/mol)
Solution
Step 1: Calculate the number of moles of carbon in 6.00 g. Step 2: Determine
the molar ratio between carbon and the given reaction. Step 3: Use the enthalpy
change for the reaction to calculate the enthalpy change when 6.00 g of carbon
reacts.
Step 1: Calculate the number of moles of carbon in 6.00 g. Given: Mass of
carbon = 6.00 g Molar mass of carbon = 12.01 g/mol
Number of moles of carbon:
Moles = Mass
Molar mass =6.00 g
12.01 g/mol
15
Moles = 0.499 mol
Step 2: Determine the molar ratio between carbon and the given reaction.
From the balanced equation: 2 moles of carbon react to produce 1 mole of C2H6
Step 3: Use the enthalpy change for the reaction to calculate the enthalpy
change when 6.00 g of carbon reacts. Given enthalpy change for the reaction:
-84 kJ/mol
Since 2 moles of carbon are required to produce 1 mole of C2H6, the enthalpy
change for the reaction involving 2 moles of carbon is -84 kJ. For 0.499 moles
of carbon (half the amount):
(0.499 mol)(−84 kJ/mol) = −41.92 kJ
Therefore, the enthalpy change when 6.00 g of carbon reacts with excess
hydrogen is -41.92 kJ.
Question 21
Question
Calculate the change in enthalpy (∆H) for a reaction where 2 moles of ethylene
(C2H4) are combusted with excess oxygen to form carbon dioxide (CO2) and
water vapor (H2O). The enthalpies of formation (∆H◦
f) for C2H4, CO2, and
H2O are -52.26 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Write the balanced chemical equation for the combustion of 2 moles of
ethylene:
C2H4+ 3O2→2CO2+ 2H2O
Step 2: Calculate the change in enthalpy (∆H) using the equation:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given enthalpies of formation into the equation:
∆H= [2(−393.5 kJ/mol) + 2(−241.8 kJ/mol)] −[1(−52.26 kJ/mol)]
Step 4: Perform the calculations to find ∆H:
∆H= [−787 + (−483.6)] −[−52.26]
∆H=−1270.6 + 52.26
∆H=−1218.34 kJ
Answer: The change in enthalpy for the reaction is ∆H=−1218.34 kJ.
16
Question 22
Question
A reaction takes place in a closed system at constant pressure, where the en-
thalpy change (∆H) of the reaction is −286 kJ. If the reaction releases 672 J
of heat to the surroundings, calculate the work done by the system during the
reaction.
Solution
Step 1: Recall the relationship between the enthalpy change, heat absorbed or
released, and work done for a reaction taking place at constant pressure:
∆H=q+w
where ∆His the enthalpy change, qis the heat absorbed or released by the
system, and wis the work done by or on the system.
Step 2: Given that ∆H=−286 kJ and q=−672 J (since the reaction
releases heat to the surroundings), we can substitute these values into the equa-
tion:
−286 kJ = −672 J + w
Step 3: Convert all units to the same unit. Since 1 kJ = 1000 J, we need to
convert −286 kJ to J:
−286 kJ = −286 ×1000 J = −286000 J
Step 4: Substitute −286000 J for ∆Hand −672 J for qinto the equation
and solve for w:
−286000 J = −672 J + w
w=−286000 J + 672 J
w=−285328 J
Step 5: Therefore, the work done by the system during the reaction is
−285328 J.
Question 23
Question
A gas undergoes a process where its enthalpy change is given by the equation:
∆H= 30T2−20T
where ∆His in kJ/mol and Tis in Kelvin. Determine the heat transfer when
the temperature changes from 300 K to 500 K.
17
Solution
Step 1: To find the heat transfer, we need to calculate the change in enthalpy
for the given temperature range by integrating the enthalpy change equation
with respect to temperature.
∆H=ZTf
Ti
(30T2−20T)dT
Step 2: Integrating the equation, we get
∆H=10T3−10T2500
300
∆H=10(500)3−10(500)2−10(300)3−10(300)2
Step 3: Calculating the values, we get
∆H= (10 ×125000) −(10 ×2500) −(10 ×27000) + (10 ×900)
∆H= 1250000 −25000 −270000 + 9000
∆H= 959000 kJ/mol
Therefore, the heat transfer when the temperature changes from 300 K to
500 K is 959000 kJ/mol.
Question 24
Question
Given the reaction:
2H2(g) + O2(g) →2H2O(g)
with ∆H=−484 kJ/mol, calculate the enthalpy change when 4 moles of water
are produced.
Solution
Step 1: Determine the moles of water produced in the reaction. Since the
reaction produces 2 moles of water for every 1 mole of oxygen, we can calculate
the moles of water produced as follows:
4 moles of water = 4 moles ×2 moles of H2O
1 mole of O2
= 8 moles of H2O
Step 2: Use the given enthalpy change to calculate the total enthalpy change
for 4 moles of water. Since 2 moles of water are produced for every 484 kJ of
energy released, the enthalpy change for 8 moles of water produced will be:
8 moles of H2O×484 kJ
2 moles of H2O= 1936 kJ
Therefore, the enthalpy change when 4 moles of water are produced in this
reaction is 1936 kJ.
18
Question 25
Question
Calculate the change in enthalpy for a reaction where 2 moles of methane (CH4)
react with 4 moles of oxygen (O2) to produce carbon dioxide (CO2) and water
(H2O) according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the standard enthalpies of formation are ∆H◦
f(CH4) = −74.8 kJ/mol,
∆H◦
f(CO2) = −393.5 kJ/mol, and ∆H◦
f(H2O) = −285.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy of the reaction using the given standard
enthalpies of formation. The standard enthalpy change for a reaction is equal
to the sum of the standard enthalpies of formation of the products minus the
sum of the standard enthalpies of formation of the reactants.
Given standard enthalpies of formation: ∆H◦
f(CH4) = −74.8 kJ/mol, ∆H◦
f(CO2) =
−393.5 kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol.
The standard enthalpy change for the reaction can be calculated as:
∆H◦=Σ∆H◦
f(products)−Σ∆H◦
f(reactants)
Given reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Substitute the values:
∆H◦=∆H◦
f(CO2) + 2∆H◦
f(H2O)−∆H◦
f(CH4)+2·0
∆H◦= (−393.5 kJ/mol + 2 ∗(−285.8) kJ/mol) −(−74.8 kJ/mol)
∆H◦= (−393.5 kJ/mol −571.6 kJ/mol) + 74.8 kJ/mol
∆H◦=−965.1 kJ/mol + 74.8 kJ/mol
∆H◦=−890.3 kJ/mol
Therefore, the change in enthalpy for the reaction is -890.3 kJ/mol.
Question 26
Question
A reaction is carried out in a bomb calorimeter, which is also referred to as a
constant volume calorimeter. The initial temperature inside the calorimeter is
25
°
C and the final temperature after the reaction is 35
°
C. The heat capacity of
the calorimeter is 30 J/
°
C. If the reaction released 500 J of heat, calculate the
enthalpy change (∆H) for the reaction.
19
Solution
Step 1: Calculate the change in temperature (∆T) using the initial and final
temperatures:
∆T=Tf−Ti= 35◦C−25◦C = 10◦C
Step 2: Calculate the total heat absorbed by the calorimeter and its contents:
Q=C·∆T
Q= 30 J/
°
C·10◦C = 300 J
Step 3: Since the reaction released 500 J of heat, the total heat absorbed by
the calorimeter and its contents is equal to the negative of the heat released by
the reaction:
Qreaction =−300 J −500 J
Qreaction =−800 J
Step 4: The enthalpy change (∆H) for the reaction is equal to the total heat
absorbed by the calorimeter and its contents:
∆H=Q=−800 J
Therefore, the enthalpy change (∆H) for the reaction is -800 J.
Question 27
Question
Calculate the change in enthalpy (∆H) when 25.0 g of water vapor at 100.0
°
C
condenses to liquid water at 100.0
°
C. The molar enthalpy of vaporization for
water is 40.79 kJ/mol.
Solution
Step 1: Determine the number of moles of water vapor. Given: Mass of water
vapor = 25.0 g Molar mass of water (HO) = 18.015 g/mol
Number of moles of water vapor = Mass
Molar mass =25.0 g
18.015 g/mol ≈1.387 mol
Step 2: Calculate the change in enthalpy. The change in enthalpy (∆H) can
be calculated using the equation:
∆H= moles ×molar enthalpy of vaporization
Substitute the values:
∆H= 1.387 mol ×40.79 kJ/mol = 56.62 kJ
Therefore, the change in enthalpy when 25.0 g of water vapor at 100.0
°
C
condenses to liquid water at 100.0
°
C is 56.62 kJ.
20
Question 28
Question
A reaction is carried out at constant pressure, and the enthalpy change (∆H)
for the reaction is measured to be −425 kJ. If the reaction releases 450 kJ of
heat to the surroundings, what is the change in internal energy (∆U) for the
reaction? Assume that the only kind of work done is pressure-volume work.
Solution
Step 1: Recall the relationship between enthalpy change (∆H), change in inter-
nal energy (∆U), and heat exchanged (q) under constant-pressure conditions:
∆H= ∆U+P∆V
where Pis the constant pressure and ∆Vis the change in volume.
Step 2: Since the reaction is carried out at constant pressure, ∆U= ∆H−
P∆V.
Step 3: We are given that ∆H=−425 kJ and q=−450 kJ (since heat is
released to the surroundings).
Step 4: Since the only kind of work done is pressure-volume work, q=
∆U−P∆V= ∆U−w, where wis the work done. Here, w=−P∆V, so
q= ∆U+P∆V.
Step 5: Substitute the given values into the equation:
−450 kJ = ∆U−425 kJ
Step 6: Solve for ∆U:
∆U=−450 kJ + 425 kJ = −25 kJ
Therefore, the change in internal energy for the reaction is ∆U=−25 kJ.
Question 29
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas (CH4)
is burned in excess oxygen gas (O2) according to the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the standard enthalpies of formation are:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
21
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
for the reaction is:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Step 2: Determine the enthalpy change for the reaction using the standard
enthalpies of formation. The change in enthalpy (∆H) for the reaction can be
calculated using the formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given standard enthalpies of formation:
∆H= [1 ×∆H◦
f(CO2)+2×∆H◦
f(H2O(l))] −[1 ×∆H◦
f(CH4)+2×∆H◦
f(O2)]
∆H= [1 × −393.5+2× −285.8] −[1 × −74.8+2×0]
∆H= [−393.5−571.6] −[−74.8]
∆H=−965.1 + 74.8
∆H=−890.3 kJ
Therefore, the change in enthalpy (∆H) when 5.00 moles of methane gas is
burned in excess oxygen gas is -890.3 kJ.
Question 30
Question
Calculate the change in enthalpy (∆H) when 25.0 g of ice at -18.0
°
C is melted
and heated to form water vapor at 125.0
°
C. The specific heat capacity of ice is
2.09 J/g
°
C, the heat of fusion for ice is 333 J/g, the specific heat capacity of
water is 4.18 J/g
°
C, and the heat of vaporization of water is 2260 J/g.
Solution
Step 1: Calculate the heat required to melt the ice.
qice =m·∆Hfusion
where mis the mass of ice and ∆Hfusion is the heat of fusion for ice.
qice = 25.0 g ×333 J/g = 8325 J
22
Step 2: Calculate the heat required to heat the water from -18.0
°
C to 0
°
C.
qice→water =m×cice ×∆T
where cice is the specific heat capacity of ice and ∆Tis the temperature change.
qice→water = 25.0 g ×2.09 J/g
°
C×(0 −(−18.0))
°
C = 943.5 J
Step 3: Calculate the heat required to heat the water from 0
°
C to 100
°
C.
qwater =m×cwater ×∆T
where cwater is the specific heat capacity of water and ∆Tis the temperature
change.
qwater = 25.0 g ×4.18 J/g
°
C×(100 −0)
°
C = 10450 J
Step 4: Calculate the heat required to vaporize the water.
qwater→vapor =m×∆Hvaporization
where ∆Hvaporization is the heat of vaporization of water.
qwater→vapor = 25.0 g ×2260 J/g = 56500 J
Step 5: Calculate the total heat required by adding all the heats calculated
in the previous steps.
qtotal =qice+qice→water+qwater+qwater→vapor = 8325 J+943.5 J+10450 J+56500 J = 8618.5 J
Step 6: Calculate the change in enthalpy using the total heat required.
∆H=qtotal
Therefore, the change in enthalpy (∆H) when 25.0 g of ice at -18.0
°
C is melted
and heated to form water vapor at 125.0
°
C is 8618.5 J.
Question 31
Question
Calculate the change in enthalpy (∆H) for a reaction in which 2 moles of
methane (CH4) are combusted to form carbon dioxide (CO2) and water va-
por (H2O), given the following standard enthalpy of formation values:
CH4:−74.8kJ/mol
CO2:−393.5kJ/mol
H2O:−241.8kJ/mol
23
Solution
Step 1: Write the balanced chemical equation for the combustion of methane:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Step 2: Calculate the standard enthalpy change for the reaction using the
standard enthalpies of formation:
∆H=XνfHf(products)−XνfHf(reactants)
Step 3: Substitute the given standard enthalpy of formation values into the
equation:
∆H= [1(−393.5kJ/mol) + 2(−241.8kJ/mol)] −[1(−74.8kJ/mol)]
Step 4: Perform the calculations:
∆H= [−393.5kJ/mol −483.6kJ/mol] + 74.8kJ/mol
∆H=−877.1kJ/mol + 74.8kJ/mol
∆H=−802.3kJ/mol
Therefore, the change in enthalpy (∆H) for the combustion of 2 moles of
methane is -802.3 kJ.
Question 32
Question
A gas is compressed isothermally at 27
°
C from an initial pressure of 2 atm to a
final pressure of 8 atm. If the enthalpy change during the process is -42 kJ/mol,
calculate the change in molar enthalpy for the gas.
Solution
Step 1: Calculate the initial molar enthalpy using the ideal gas law.
P V =nRT
Given that the initial pressure P1= 2 atm, the final pressure P2= 8 atm, the
temperature T= 27C= 300K, and the gas constant R= 0.0821 atm L mol−1
K−1.
We can rearrange the ideal gas law to solve for the initial molar quantity:
n=P V
RT
24
Substitute the values to find n:
n=(2 atm)V
(0.0821 atm L mol−1K−1)(300 K)
n=2V
24.63 =V
12.315
Step 2: Calculate the final molar enthalpy using the ideal gas law as well.
Given that the final pressure is P2= 8 atm, we can use the ideal gas law to find
nas before:
n=(8 atm)V
(0.0821 atm L mol−1K−1)(300 K)
n=8V
24.63 =4V
12.315
Step 3: Calculate the change in molar enthalpy. The change in molar en-
thalpy, ∆H, is given as -42 kJ/mol. We can express this change in terms of the
initial and final molar quantities:
∆H=Hf−Hi=4V
12.315−V
12.315=3V
12.315 =−42 kJ/mol
Solving for V:3V
12.315 =−42
3V=−517.245
V=−172.415 L
Step 4: Calculate the change in molar enthalpy for the gas. The change in
molar enthalpy, ∆H, is given as -42 kJ/mol. We can express this change in
terms of the initial and final molar quantities:
∆H=Hf−Hi=4V
12.315−V
12.315=3V
12.315 =−42 kJ/mol
Substitute the value of Vto find ∆H:
∆H=3(−172.415)
12.315
∆H=−42 kJ/mol
Therefore, the change in molar enthalpy for the gas during the isothermal
compression process is -42 kJ/mol.
25
Question 33
Question
Given the following reaction at constant pressure:
2H2(g)+O2(g)→2H2O(g)
If the standard enthalpy change of formation of water vapor is -241.8 kJ/mol,
calculate the standard enthalpy change of the reaction.
Solution
Step 1: Write down the balanced chemical equation for the reaction. Step 2: Use
the standard enthalpy of formation values to calculate the standard enthalpy
change of the reaction.
Step 1: The balanced chemical equation for the reaction is:
2H2(g)+O2(g)→2H2O(g)
Step 2: The standard enthalpy change of the reaction can be calculated
using the standard enthalpy of formation values. The standard enthalpy of
formation of water vapor is given as -241.8 kJ/mol.
The standard enthalpy change of the reaction can be calculated as:
∆H◦=X∆H◦
products −X∆H◦
reactants
Substitute the values into the equation:
∆H◦= 2 ×0 kJ/mol −(2 ×(-241.8 kJ/mol))
∆H◦= 483.6 kJ/mol
Therefore, the standard enthalpy change of the reaction is 483.6 kJ/mol.
Question 34
Question
Given the reaction:
2H2O2(l)→2H2O(l)+O2(g)
Calculate the standard enthalpy change for this reaction (∆H◦) given the
following information:
∆H◦
ffor H2O2(l) = -196.1 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
∆H◦
ffor O2(g) = 0 kJ/mol
26
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2O2(l)→2H2O(l)+O2(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation for the given compounds.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the values into the equation.
∆H◦= [2∆H◦
f(H2O(l)) + ∆H◦
f(O2(g))] −[2∆H◦
f(H2O2(l))]
Step 4: Substitute the given values.
∆H◦= [2(−285.8) + 0] −[2(−196.1)]
Step 5: Calculate the final answer.
∆H◦= (−571.6) −(−392.2)
∆H◦=−179.4 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -179.4
kJ/mol.
Question 35
Question
A sample of methane gas at 27
°
C and 1 atm is burned to heat 1 kg of water at
20
°
C. The enthalpy change for the combustion of methane at 25
°
C and 1 atm
is -890 kJ/mol. Calculate the final temperature of the water assuming all the
heat released in the combustion is transferred to the water. Take the specific
heat capacity of water as 4.18 J/(g
°
C) and the molar mass of methane as 16.05
g/mol.
Solution
Step 1: Calculate the heat released by the combustion of 1 mol of methane.
The enthalpy change for the combustion of methane is -890 kJ/mol, which is
equivalent to -890,000 J/mol. Since the molar mass of methane is 16.05 g/mol,
we need to calculate the heat released per gram of methane.
Heat released per gram of methane = −890,000 J/mol
16.05 g/mol =−55517.24 J/g
27
Step 3: Substitute the given values into the formula and calculate ∆H.
∆H= [(−84.7) kJ/mol] −[(2 ×0) + (3 ×0)] kJ/mol
∆H=−84.7 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H=−84.7 kJ/mol
at 298 K.
Question 2
Question
A reaction is carried out in a bomb calorimeter, and the temperature of the
calorimeter increases by 5
°
C. If the heat capacity of the calorimeter is 20 J/
°
C,
and the reaction releases 5000 J of heat, calculate the change in enthalpy (∆H)
of the reaction.
Solution
Step 1: Calculate the heat absorbed by the calorimeter using the formula:
qcalorimeter =C·∆T
where: C= heat capacity of the calorimeter = 20 J/
°
C, ∆T= change in
temperature = 5
°
C.
Therefore,
qcalorimeter = 20 J/
°
C·5
°
C = 100 J
Step 2: Since the calorimeter is absorbing the heat released by the reaction,
the heat absorbed by the reaction is equal in magnitude but opposite in sign:
qreaction =−qcalorimeter =−100 J
Step 3: The heat absorbed by the reaction is related to the change in enthalpy
of the reaction through the equation:
∆H=qreaction
Therefore, the change in enthalpy of the reaction is:
∆H=−100 J
Question 3
Question
A reaction takes place in a bomb calorimeter and releases 1250 kJ of heat. If
the volume of water in the calorimeter is 500 mL and its initial temperature
was 25
°
C, what is the final temperature of the water? Assume no heat is lost
to the surroundings. The specific heat of water is 4.18 J/g
°
C.
2
Solution
Step 1: Calculate the heat capacity of the water.
Q=mc∆T
1250000 = (500 g)(4.18 J/g
°
C)(Tf−25)
Tf−25 = 1250000
(500)(4.18)
Tf=1250000
(500)(4.18) + 25
Tf≈89.95
°
C
So, the final temperature of the water is approximately 89.95
°
C.
Question 4
Question
Calculate the change in enthalpy (∆H) when 2.5 moles of water (H2O) liquid at
25◦C is converted into steam at 100◦C. Given that the specific heat capacity of
water is 4.18 J/(g·◦C), the molar heat of vaporization of water is 40.79 kJ/mol,
and the molar mass of water is 18.02 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 25◦C
to 100◦C.
The specific heat capacity of water is 4.18 J/(g·◦C), so the heat required is
given by the equation:
q=m×c×∆T
Where: - mis the mass of water in grams - cis the specific heat capacity of
water - ∆Tis the change in temperature
Given that there are 2.5 moles of water and the molar mass of water is 18.02
g/mol, the mass of water is calculated as:
Mass of water = 2.5 moles ×18.02 g/mol = 45.05 g
Substitute the values into the equation:
q= 45.05 g ×4.18 J/(g ·◦C) ×(100 −25)◦C
q= 45.05 ×4.18 ×75
q= 14141.775 J = 14.14 kJ
3
Step 2: Calculate the heat required to convert water at 100◦C to steam at
100◦C.
The molar heat of vaporization of water is 40.79 kJ/mol. Since we have 2.5
moles of water, the heat required for vaporization is:
q= 2.5 moles ×40.79 kJ/mol = 101.975 kJ
Step 3: Calculate the total change in enthalpy using the heat values calcu-
lated in steps 1 and 2.
∆H=qraising temp +qvaporization
∆H= 14.14 kJ + 101.975 kJ
∆H= 116.115 kJ
Therefore, the change in enthalpy (∆H) when 2.5 moles of water liquid at
25◦C is converted into steam at 100◦C is 116.115 kJ.
Question 5
Question
Calculate the change in enthalpy for the reaction
2C(s)+2H2(g)→C2H4(g)
given the following enthalpy values:
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H4(g)) = 52.3 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation is:
2C(s)+2H2(g)→C2H4(g)
Step 2: Calculate the change in enthalpy using the standard enthalpies of
formation. The change in enthalpy, H◦
rxn, can be calculated using the equation:
H◦
rxn =nH◦
f, products −(mH◦
f, reactants)
where nand mare the stoichiometric coefficients in the balanced equation, and
H◦
frepresents the standard enthalpy of formation.
4
Given that:
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H4(g)) = 52.3 kJ/mol
Substitute the values into the equation:
H◦
rxn = (1 ·52.3 kJ/mol) −[(2 ·0 kJ/mol) + (2 ·0 kJ/mol)]
H◦
rxn = 52.3 kJ/mol
Therefore, the change in enthalpy for the reaction is 52.3 kJ/mol .
Question 6
Question
A reaction is known to have ∆H◦=−92.2 kJ and ∆S◦=−145.1 J/K. Calculate
the temperature at which the reaction becomes spontaneous.
Solution
Step 1: Write down the equation for Gibb’s free energy change (∆G):
∆G= ∆H−T·∆S
Step 2: At equilibrium, ∆G= 0. Therefore, we have:
0 = −92.2 kJ −T·(−0.1451 kJ/K)
Step 3: Solve for the temperature T:
T=92.2 kJ
0.1451 kJ/K
T≈635.13 K
Therefore, the reaction becomes spontaneous at approximately 635.13 Kelvin.
Question 7
Question
A reaction occurring in a bomb calorimeter releases 1500 J of heat while doing
300 J of work on the surroundings. Calculate the change in enthalpy (∆H) for
the reaction.
5
Solution
Step 1: Recall that the change in enthalpy (∆H) is given by the equation:
∆H= ∆E+P∆V
where ∆Eis the change in internal energy and P∆Vis the work done on (or
by) the system.
Step 2: We can calculate the change in internal energy (∆E) using the first
law of thermodynamics:
∆E=q+w
where qis the heat absorbed (positive) or released (negative) by the system and
wis the work done on (positive) or by (negative) the system.
Step 3: In this case, the heat released by the reaction is -1500 J (negative
because it is released) and the work done on the surroundings is 300 J (positive
because it is done on the surroundings). Therefore, we have:
∆E=−1500 J + 300 J
∆E=−1200 J
Step 4: Now we can calculate the change in enthalpy (∆H) by adding the
change in internal energy to the work done on the system:
∆H= ∆E+P∆V
∆H=−1200 J + 300 J
∆H=−900 J
Therefore, the change in enthalpy (∆H) for the reaction is -900 J.
Question 8
Question
A gas at a constant pressure of 1 atm is heated from 25◦C to 100◦C. During
this process, the gas absorbs 150 J of heat and expands to a volume of 5 L.
Calculate the change in enthalpy (∆H) for the gas.
Solution
Step 1: Calculate the change in temperature: Given: Initial temperature,
Tinitial = 25◦C = 25 + 273 = 298 K Final temperature, Tfinal = 100◦C =
100 + 273 = 373 K
Change in temperature: ∆T=Tfinal−Tinitial ∆T= 373K−298K∆T= 75K
Step 2: Calculate the heat absorbed by the gas: Given: Heat absorbed, q =
150 J
6
Step 3: Calculate the change in enthalpy using the equation: ∆H=q+P·
∆V
Step 4: Calculate the change in volume: Given: Initial volume, Vinitial = 0
(not needed in this case, as ∆Vwill be equal to the final volume since the gas
expands) Final volume, Vfinal = 5 L
Change in volume: ∆V=Vfinal −Vinitial ∆V= 5 L
Step 5: Plug in the values into the equation for change in enthalpy: ∆H=
q+P·∆V∆H= 150 J + (1 atm) ·5 L ·(101.3 J/L
·
atm) ∆H= 150 J + 506.5 J
∆H= 656.5 J
Therefore, the change in enthalpy (∆H) for the gas is 656.5 J.
Question 9
Question
Calculate the change in enthalpy (∆H) when 15.0 g of water at 25
°
C is converted
to steam at 150
°
C. The specific heat capacity of water is 4.18 J/g
°
C, the molar
enthalpy of vaporization of water is 40.7 kJ/mol, and the molar mass of water
is 18.0 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 25
°
C
to 100
°
C. The heat required can be calculated using the formula:
q=mc∆T
where: q= heat energy, m= mass of the substance, c= specific heat capacity,
and ∆T= change in temperature. Given that m= 15.0 g, c= 4.18 J/g
°
C, and
∆T= 100
°
C, we can substitute these values into the formula to find q.
Step 2: Calculate the heat required to convert water at 100
°
C to steam at
100
°
C (phase change). The heat required for this phase change is equal to the
molar enthalpy of vaporization of water, which is 40.7 kJ/mol. First, we need
to find the number of moles of water in 15.0 g.
Moles = Mass (g)
Molar mass (g/mol)
Moles = 15.0 g
18.0 g/mol
Moles ≈0.833 mol
Next, we calculate the heat required for the phase change:
q=n·∆Hvap
q= 0.833 mol ×40.7×103J/mol
q≈33.91 ×103J
7
Step 3: Calculate the heat required to raise the temperature of steam from
100
°
C to 150
°
C. Using the formula q=mc∆Twith m= 15.0 g, c= 2.01 J/g
°
C
(specific heat capacity of steam), and ∆T= 50
°
C. Substitute these values into
the formula to find q.
Step 4: Calculate the total change in enthalpy (∆H).
∆H=q1+q2+q3
Substitute the values of q1,q2, and q3into the formula to calculate ∆H.
Question 10
Question
A sample of gas undergoes a process at constant pressure where it absorbs 150
J of heat and does 50 J of work on the surroundings. If the change in internal
energy of the gas is 100 J, what is the change in enthalpy of the gas during the
process?
Solution
Step 1: Recall the definition of enthalpy: Enthalpy is defined as H=U+P V ,
where His enthalpy, Uis internal energy, Pis pressure, and Vis volume.
Step 2: The change in enthalpy can be calculated using the equation:
∆H= ∆U+P∆V
Step 3: Given that ∆U= 100 J, and the process is at constant pressure, we
can rewrite P∆Vas w, the work done by the gas on the surroundings. Thus,
∆H= ∆U+w.
Step 4: Plugging in the values, we have:
∆H= 100 J + (−50 J)
Step 5: Simplifying, we find:
∆H= 50 J
Step 6: Therefore, the change in enthalpy of the gas during the process is
50 J .
Question 11
Question
A reaction is carried out in a bomb calorimeter, and it is found that the tem-
perature of the calorimeter increases by 5.82
°
C. If the heat capacity of the
calorimeter is 247 J/
°
C, calculate the enthalpy change (∆H) for the reaction in
kJ/mol.
8
Solution
Step 1: Calculate the heat absorbed by the bomb calorimeter using the formula:
q=C·∆T
where: q= heat absorbed by the calorimeter, C= heat capacity of the calorime-
ter, ∆T= change in temperature.
Plugging in the values, we get:
q= 247 J/
°
C×5.82
°
C = 1439.74 J
Step 2: Convert the heat absorbed by the calorimeter to kilojoules (1 kJ =
1000 J):
q= 1439.74 J ×1 kJ
1000 J = 1.44 kJ
Step 3: Calculate the moles of substance reacting using the molar heat
capacity formula:
∆H=q
n
where: ∆H= enthalpy change, q= heat absorbed by the calorimeter, n=
moles of substance.
Step 4: Rearrange the formula to solve for ∆H:
∆H=q
n
n=q
∆H
Step 5: The molar heat capacity of the reaction is then given by:
∆H=1.44 kJ
n
Step 6: Finally, divide the heat absorbed by the calorimeter by the moles of
substance to determine the enthalpy change in kJ/mol.
Question 12
Question
A reaction takes place in a bomb calorimeter, and the temperature of the water
in the bomb calorimeter changes from 25
°
C to 30
°
C. If the heat capacity of the
bomb calorimeter is 50 J/
°
C, calculate the change in enthalpy of the reaction.
Assume that the heat released from the reaction is only absorbed by the water
in the calorimeter, and the bomb calorimeter is insulated with its surroundings.
9
Solution
Step 1: Calculate the heat absorbed by the water in the bomb calorimeter using
the formula:
q=C·∆T
where: q= heat absorbed by the water in the calorimeter, C= 50 J/
°
C =
heat capacity of the bomb calorimeter, and ∆T= 30C−25C= 5C= change
in temperature.
Substitute the values into the formula:
q= 50 J/
°
C·5C= 250 J
Therefore, the water in the bomb calorimeter absorbs 250 J of heat.
Step 2: Since the reaction is taking place in the bomb calorimeter, the heat
absorbed by the water is equal to the heat released by the reaction according
to the first law of thermodynamics. This heat is given by the formula:
∆H=−q
where: ∆H= change in enthalpy of the reaction and q= 250 J (negative
because the reaction releases heat).
Therefore, the change in enthalpy of the reaction is:
∆H=−250 J
Hence, the change in enthalpy of the reaction is -250 J.
Question 13
Question
A sample of an ideal gas is taken through a cyclic process consisting of two
isobaric and two isochoric processes. The initial temperature and pressure of
the gas are T1and P1, and the final temperature and pressure are T2and P2,
respectively. Calculate the change in enthalpy for the gas during this process.
Solution
Step 1: Recall that for an ideal gas, the change in enthalpy is given by ∆H=
Cp·∆T, where Cpis the specific heat at constant pressure and ∆Tis the change
in temperature.
Step 2: For an isobaric process, Cp=5
2Rwhere Ris the gas constant. Thus,
the change in enthalpy for the first isobaric process is ∆H1=5
2R·(T2−T1).
Step 3: For an isochoric process, ∆H= ∆U, where ∆Uis the change in
internal energy. Since there is no work done in an isochoric process, ∆U=
nCv∆T, where Cvis the specific heat at constant volume.
10
Step 4: For an ideal gas, Cv=3
2R. Therefore, the change in enthalpy for
the first isochoric process is ∆H2=3
2R·(T2−T1).
Step 5: The total change in enthalpy for the cyclic process is then ∆H=
∆H1+ ∆H2=5
2R+3
2R(T2−T1) = 4R(T2−T1).
Question 14
Question
A reaction has an enthalpy change of -285 kJ/mol. If 2.00 moles of the reactant
are consumed in the reaction at constant pressure, what is the enthalpy change
for the reaction in kJ?
Solution
Step 1: Start with the given information: The enthalpy change for the reaction
is -285 kJ/mol. The number of moles of the reactant consumed is 2.00 mol.
Step 2: Calculate the enthalpy change for the reaction in kJ. Given that
the enthalpy change is -285 kJ/mol, we can use this value to find the enthalpy
change for the reaction when 2.00 moles of the reactant are consumed:
Enthalpy change for the reaction = Enthalpy change per mole×Number of moles consumed
Enthalpy change for the reaction = −285 kJ/mol ×2.00 mol
Step 3: Perform the calculation.
Enthalpy change for the reaction = −285 kJ/mol ×2.00 mol = −570 kJ
Therefore, when 2.00 moles of the reactant are consumed in the reaction,
the enthalpy change for the reaction is -570 kJ.
Question 15
Question
Calculate the change in enthalpy when 5 moles of nitrogen gas at 200 K and 1
atm pressure are heated at constant pressure to 400 K. Assume that nitrogen
gas behaves ideally.
Solution
Step 1: The change in enthalpy can be calculated using the formula:
∆H= ∆U+P∆V
11
Step 2: First, we calculate the change in internal energy (∆U) using the
formula:
∆U=nCv∆T
where nis the number of moles, Cvis the molar heat capacity at constant
volume, and ∆Tis the change in temperature.
Step 3: Since nitrogen gas is diatomic, the molar heat capacity at constant
volume is Cv=5
2Rwhere Ris the gas constant.
Step 4: Substituting the values into the formula, we get:
∆U= 5 ×5
2R×(400 −200)
Step 5: Next, we calculate the work done by the gas, which is equal to P∆V.
Since the process is at constant pressure, we have:
∆V=nR∆T
Step 6: Substituting the values into the formula, we get:
∆V= 5 ×R×(400 −200)
Step 7: Now we can calculate the work done:
P∆V= 1 ×(5R×200)
Step 8: Finally, we can calculate the change in enthalpy by adding the change
in internal energy and the work done:
∆H= ∆U+P∆V
∆H= (5 ×5
2R×200) + (5R×200)
Question 16
Question
Given the following data for the reaction:
∆H◦
ffor CO2= -393.5 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction:
C(graphite) + 2H2O→CO2+ 2H2O
12
Solution
Step 1: Write the balanced chemical equation for the reaction and determine the
∆H◦value for the reaction. The balanced chemical equation for the reaction is:
C(graphite) + 2H2O→CO2+ 2H2O
To find the ∆H◦for the reaction, we use the following formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the ∆H◦value for the reaction. Given data: ∆H◦
ffor
CO2= -393.5 kJ/mol ∆H◦
ffor H2O(l) = -285.8 kJ/mol ∆H◦
ffor C(graphite)
= 0 kJ/mol
Plugging in the values into the formula:
∆H◦= [1(−393.5 kJ/mol) + 2(−285.8 kJ/mol)] −[1(0 kJ/mol) + 2(0 kJ/mol)]
∆H◦= [−393.5 kJ/mol −571.6 kJ/mol]
∆H◦=−965.1 kJ/mol
Therefore, the standard enthalpy change, ∆H◦, for the reaction:
C(graphite) + 2H2O→CO2+ 2H2O
is -965.1 kJ/mol.
Question 17
Question
A certain chemical reaction is known to have a standard enthalpy change of -383
kJ mol−1. If 2.50 moles of the reactant are consumed in the reaction, calculate
the amount of heat absorbed or released.
Solution
Step 1: Recall the formula relating enthalpy change (∆H) to heat absorbed or
released (q) and amount of substance (n):
q= ∆H×n
Step 2: Substitute the known values into the formula:
q=−383 kJ mol−1×2.50 mol
Step 3: Perform the calculation:
q=−383 ×2.50 = −957.5 kJ
Step 4: Thus, the amount of heat absorbed or released during the reaction
is -957.5 kJ. Since the enthalpy change is negative, it indicates that the reaction
releases heat.
13
Question 18
Question
A reaction is carried out in a bomb calorimeter and releases 150 kJ of heat.
If the volume of the bomb calorimeter is 500 mL and the pressure inside the
calorimeter remains constant at 1 atm, calculate the change in enthalpy (∆H)
of the reaction in kJ/mol. Assume ideal gas behavior.
Solution
Step 1: Recall that ∆H=qp, where qpis the heat exchanged at constant
pressure during a chemical reaction.
Step 2: Convert the volume of the bomb calorimeter from mL to L: V=
500 mL ×1 L
1000 mL = 0.5 L
Step 3: Since the pressure remains constant and the reaction takes place in a
bomb calorimeter, the heat exchanged is equal to the change in enthalpy (∆H).
Step 4: The change in enthalpy of the reaction can be calculated using the
formula: ∆H=q
n
Step 5: Given that the heat exchanged (q) is 150 kJ and the number of moles
(n) is unknown, we have: ∆H=150 kJ
n
Step 6: To find the number of moles, we need to convert the volume at STP
to number of moles of gas. At STP, 22.4 L of any gas contains 1 mole.
Step 7: Assuming ideal gas behavior, the volume of 0.5 L at 1 atm pressure
contains nmoles: n=0.5 L
22.4 L/mol
Step 8: Substitute the value of nback into the formula for ∆Hto find the
change in enthalpy of the reaction: ∆H=150 kJ
0.5 L/22.4 L/mol = 150 kJ ×22.4 L/mol
0.5 L
Step 9: ∆H= 150 ×22.4 = 3360 kJ/mol
Therefore, the change in enthalpy of the reaction is 3360 kJ/mol.
Question 19
Question
A reaction is taking place in a closed system where the enthalpy change (∆H)
is given by the equation:
∆H= 50 −2T+ 0.05T2
where ∆His in kJ/mol and Tis in Kelvin. Determine the temperature at which
the reaction is exothermic.
Solution
Step 1: To determine the temperature at which the reaction is exothermic, we
need to find the point where ∆Hbecomes negative. Step 2: Set ∆Hto be less
14
than zero and solve for T:
50 −2T+ 0.05T2<0
0.05T2−2T+ 50 <0
Step 3: This is a quadratic inequality. To solve it, we find the roots of the
corresponding quadratic equation:
0.05T2−2T+ 50 = 0
T2−40T+ 1000 = 0
Step 4: Use the quadratic formula to find the roots:
T=−(−40) ±p(−40)2−4(1)(1000)
2(1)
T=40 ±√1600 −4000
2
T=40 ±√−2400
2
Step 5: Since the square root of a negative number is not real, the inequality
has no real solutions. Therefore, the reaction is never exothermic.
Question 20
Question
Given the reaction:
2C(graphite)+3H2(g)→C2H6(g)
If the enthalpy change for this reaction is -84 kJ/mol, calculate the enthalpy
change when 6.00 g of carbon reacts with excess hydrogen. (Molar mass of
carbon = 12.01 g/mol, molar mass of hydrogen = 1.008 g/mol)
Solution
Step 1: Calculate the number of moles of carbon in 6.00 g. Step 2: Determine
the molar ratio between carbon and the given reaction. Step 3: Use the enthalpy
change for the reaction to calculate the enthalpy change when 6.00 g of carbon
reacts.
Step 1: Calculate the number of moles of carbon in 6.00 g. Given: Mass of
carbon = 6.00 g Molar mass of carbon = 12.01 g/mol
Number of moles of carbon:
Moles = Mass
Molar mass =6.00 g
12.01 g/mol
15
Moles = 0.499 mol
Step 2: Determine the molar ratio between carbon and the given reaction.
From the balanced equation: 2 moles of carbon react to produce 1 mole of C2H6
Step 3: Use the enthalpy change for the reaction to calculate the enthalpy
change when 6.00 g of carbon reacts. Given enthalpy change for the reaction:
-84 kJ/mol
Since 2 moles of carbon are required to produce 1 mole of C2H6, the enthalpy
change for the reaction involving 2 moles of carbon is -84 kJ. For 0.499 moles
of carbon (half the amount):
(0.499 mol)(−84 kJ/mol) = −41.92 kJ
Therefore, the enthalpy change when 6.00 g of carbon reacts with excess
hydrogen is -41.92 kJ.
Question 21
Question
Calculate the change in enthalpy (∆H) for a reaction where 2 moles of ethylene
(C2H4) are combusted with excess oxygen to form carbon dioxide (CO2) and
water vapor (H2O). The enthalpies of formation (∆H◦
f) for C2H4, CO2, and
H2O are -52.26 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Write the balanced chemical equation for the combustion of 2 moles of
ethylene:
C2H4+ 3O2→2CO2+ 2H2O
Step 2: Calculate the change in enthalpy (∆H) using the equation:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given enthalpies of formation into the equation:
∆H= [2(−393.5 kJ/mol) + 2(−241.8 kJ/mol)] −[1(−52.26 kJ/mol)]
Step 4: Perform the calculations to find ∆H:
∆H= [−787 + (−483.6)] −[−52.26]
∆H=−1270.6 + 52.26
∆H=−1218.34 kJ
Answer: The change in enthalpy for the reaction is ∆H=−1218.34 kJ.
16
Question 22
Question
A reaction takes place in a closed system at constant pressure, where the en-
thalpy change (∆H) of the reaction is −286 kJ. If the reaction releases 672 J
of heat to the surroundings, calculate the work done by the system during the
reaction.
Solution
Step 1: Recall the relationship between the enthalpy change, heat absorbed or
released, and work done for a reaction taking place at constant pressure:
∆H=q+w
where ∆His the enthalpy change, qis the heat absorbed or released by the
system, and wis the work done by or on the system.
Step 2: Given that ∆H=−286 kJ and q=−672 J (since the reaction
releases heat to the surroundings), we can substitute these values into the equa-
tion:
−286 kJ = −672 J + w
Step 3: Convert all units to the same unit. Since 1 kJ = 1000 J, we need to
convert −286 kJ to J:
−286 kJ = −286 ×1000 J = −286000 J
Step 4: Substitute −286000 J for ∆Hand −672 J for qinto the equation
and solve for w:
−286000 J = −672 J + w
w=−286000 J + 672 J
w=−285328 J
Step 5: Therefore, the work done by the system during the reaction is
−285328 J.
Question 23
Question
A gas undergoes a process where its enthalpy change is given by the equation:
∆H= 30T2−20T
where ∆His in kJ/mol and Tis in Kelvin. Determine the heat transfer when
the temperature changes from 300 K to 500 K.
17
Solution
Step 1: To find the heat transfer, we need to calculate the change in enthalpy
for the given temperature range by integrating the enthalpy change equation
with respect to temperature.
∆H=ZTf
Ti
(30T2−20T)dT
Step 2: Integrating the equation, we get
∆H=10T3−10T2500
300
∆H=10(500)3−10(500)2−10(300)3−10(300)2
Step 3: Calculating the values, we get
∆H= (10 ×125000) −(10 ×2500) −(10 ×27000) + (10 ×900)
∆H= 1250000 −25000 −270000 + 9000
∆H= 959000 kJ/mol
Therefore, the heat transfer when the temperature changes from 300 K to
500 K is 959000 kJ/mol.
Question 24
Question
Given the reaction:
2H2(g) + O2(g) →2H2O(g)
with ∆H=−484 kJ/mol, calculate the enthalpy change when 4 moles of water
are produced.
Solution
Step 1: Determine the moles of water produced in the reaction. Since the
reaction produces 2 moles of water for every 1 mole of oxygen, we can calculate
the moles of water produced as follows:
4 moles of water = 4 moles ×2 moles of H2O
1 mole of O2
= 8 moles of H2O
Step 2: Use the given enthalpy change to calculate the total enthalpy change
for 4 moles of water. Since 2 moles of water are produced for every 484 kJ of
energy released, the enthalpy change for 8 moles of water produced will be:
8 moles of H2O×484 kJ
2 moles of H2O= 1936 kJ
Therefore, the enthalpy change when 4 moles of water are produced in this
reaction is 1936 kJ.
18
Question 25
Question
Calculate the change in enthalpy for a reaction where 2 moles of methane (CH4)
react with 4 moles of oxygen (O2) to produce carbon dioxide (CO2) and water
(H2O) according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the standard enthalpies of formation are ∆H◦
f(CH4) = −74.8 kJ/mol,
∆H◦
f(CO2) = −393.5 kJ/mol, and ∆H◦
f(H2O) = −285.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy of the reaction using the given standard
enthalpies of formation. The standard enthalpy change for a reaction is equal
to the sum of the standard enthalpies of formation of the products minus the
sum of the standard enthalpies of formation of the reactants.
Given standard enthalpies of formation: ∆H◦
f(CH4) = −74.8 kJ/mol, ∆H◦
f(CO2) =
−393.5 kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol.
The standard enthalpy change for the reaction can be calculated as:
∆H◦=Σ∆H◦
f(products)−Σ∆H◦
f(reactants)
Given reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Substitute the values:
∆H◦=∆H◦
f(CO2) + 2∆H◦
f(H2O)−∆H◦
f(CH4)+2·0
∆H◦= (−393.5 kJ/mol + 2 ∗(−285.8) kJ/mol) −(−74.8 kJ/mol)
∆H◦= (−393.5 kJ/mol −571.6 kJ/mol) + 74.8 kJ/mol
∆H◦=−965.1 kJ/mol + 74.8 kJ/mol
∆H◦=−890.3 kJ/mol
Therefore, the change in enthalpy for the reaction is -890.3 kJ/mol.
Question 26
Question
A reaction is carried out in a bomb calorimeter, which is also referred to as a
constant volume calorimeter. The initial temperature inside the calorimeter is
25
°
C and the final temperature after the reaction is 35
°
C. The heat capacity of
the calorimeter is 30 J/
°
C. If the reaction released 500 J of heat, calculate the
enthalpy change (∆H) for the reaction.
19
Solution
Step 1: Calculate the change in temperature (∆T) using the initial and final
temperatures:
∆T=Tf−Ti= 35◦C−25◦C = 10◦C
Step 2: Calculate the total heat absorbed by the calorimeter and its contents:
Q=C·∆T
Q= 30 J/
°
C·10◦C = 300 J
Step 3: Since the reaction released 500 J of heat, the total heat absorbed by
the calorimeter and its contents is equal to the negative of the heat released by
the reaction:
Qreaction =−300 J −500 J
Qreaction =−800 J
Step 4: The enthalpy change (∆H) for the reaction is equal to the total heat
absorbed by the calorimeter and its contents:
∆H=Q=−800 J
Therefore, the enthalpy change (∆H) for the reaction is -800 J.
Question 27
Question
Calculate the change in enthalpy (∆H) when 25.0 g of water vapor at 100.0
°
C
condenses to liquid water at 100.0
°
C. The molar enthalpy of vaporization for
water is 40.79 kJ/mol.
Solution
Step 1: Determine the number of moles of water vapor. Given: Mass of water
vapor = 25.0 g Molar mass of water (HO) = 18.015 g/mol
Number of moles of water vapor = Mass
Molar mass =25.0 g
18.015 g/mol ≈1.387 mol
Step 2: Calculate the change in enthalpy. The change in enthalpy (∆H) can
be calculated using the equation:
∆H= moles ×molar enthalpy of vaporization
Substitute the values:
∆H= 1.387 mol ×40.79 kJ/mol = 56.62 kJ
Therefore, the change in enthalpy when 25.0 g of water vapor at 100.0
°
C
condenses to liquid water at 100.0
°
C is 56.62 kJ.
20
Question 28
Question
A reaction is carried out at constant pressure, and the enthalpy change (∆H)
for the reaction is measured to be −425 kJ. If the reaction releases 450 kJ of
heat to the surroundings, what is the change in internal energy (∆U) for the
reaction? Assume that the only kind of work done is pressure-volume work.
Solution
Step 1: Recall the relationship between enthalpy change (∆H), change in inter-
nal energy (∆U), and heat exchanged (q) under constant-pressure conditions:
∆H= ∆U+P∆V
where Pis the constant pressure and ∆Vis the change in volume.
Step 2: Since the reaction is carried out at constant pressure, ∆U= ∆H−
P∆V.
Step 3: We are given that ∆H=−425 kJ and q=−450 kJ (since heat is
released to the surroundings).
Step 4: Since the only kind of work done is pressure-volume work, q=
∆U−P∆V= ∆U−w, where wis the work done. Here, w=−P∆V, so
q= ∆U+P∆V.
Step 5: Substitute the given values into the equation:
−450 kJ = ∆U−425 kJ
Step 6: Solve for ∆U:
∆U=−450 kJ + 425 kJ = −25 kJ
Therefore, the change in internal energy for the reaction is ∆U=−25 kJ.
Question 29
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas (CH4)
is burned in excess oxygen gas (O2) according to the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the standard enthalpies of formation are:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
21
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
for the reaction is:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Step 2: Determine the enthalpy change for the reaction using the standard
enthalpies of formation. The change in enthalpy (∆H) for the reaction can be
calculated using the formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given standard enthalpies of formation:
∆H= [1 ×∆H◦
f(CO2)+2×∆H◦
f(H2O(l))] −[1 ×∆H◦
f(CH4)+2×∆H◦
f(O2)]
∆H= [1 × −393.5+2× −285.8] −[1 × −74.8+2×0]
∆H= [−393.5−571.6] −[−74.8]
∆H=−965.1 + 74.8
∆H=−890.3 kJ
Therefore, the change in enthalpy (∆H) when 5.00 moles of methane gas is
burned in excess oxygen gas is -890.3 kJ.
Question 30
Question
Calculate the change in enthalpy (∆H) when 25.0 g of ice at -18.0
°
C is melted
and heated to form water vapor at 125.0
°
C. The specific heat capacity of ice is
2.09 J/g
°
C, the heat of fusion for ice is 333 J/g, the specific heat capacity of
water is 4.18 J/g
°
C, and the heat of vaporization of water is 2260 J/g.
Solution
Step 1: Calculate the heat required to melt the ice.
qice =m·∆Hfusion
where mis the mass of ice and ∆Hfusion is the heat of fusion for ice.
qice = 25.0 g ×333 J/g = 8325 J
22
Step 2: Calculate the heat required to heat the water from -18.0
°
C to 0
°
C.
qice→water =m×cice ×∆T
where cice is the specific heat capacity of ice and ∆Tis the temperature change.
qice→water = 25.0 g ×2.09 J/g
°
C×(0 −(−18.0))
°
C = 943.5 J
Step 3: Calculate the heat required to heat the water from 0
°
C to 100
°
C.
qwater =m×cwater ×∆T
where cwater is the specific heat capacity of water and ∆Tis the temperature
change.
qwater = 25.0 g ×4.18 J/g
°
C×(100 −0)
°
C = 10450 J
Step 4: Calculate the heat required to vaporize the water.
qwater→vapor =m×∆Hvaporization
where ∆Hvaporization is the heat of vaporization of water.
qwater→vapor = 25.0 g ×2260 J/g = 56500 J
Step 5: Calculate the total heat required by adding all the heats calculated
in the previous steps.
qtotal =qice+qice→water+qwater+qwater→vapor = 8325 J+943.5 J+10450 J+56500 J = 8618.5 J
Step 6: Calculate the change in enthalpy using the total heat required.
∆H=qtotal
Therefore, the change in enthalpy (∆H) when 25.0 g of ice at -18.0
°
C is melted
and heated to form water vapor at 125.0
°
C is 8618.5 J.
Question 31
Question
Calculate the change in enthalpy (∆H) for a reaction in which 2 moles of
methane (CH4) are combusted to form carbon dioxide (CO2) and water va-
por (H2O), given the following standard enthalpy of formation values:
CH4:−74.8kJ/mol
CO2:−393.5kJ/mol
H2O:−241.8kJ/mol
23
Solution
Step 1: Write the balanced chemical equation for the combustion of methane:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Step 2: Calculate the standard enthalpy change for the reaction using the
standard enthalpies of formation:
∆H=XνfHf(products)−XνfHf(reactants)
Step 3: Substitute the given standard enthalpy of formation values into the
equation:
∆H= [1(−393.5kJ/mol) + 2(−241.8kJ/mol)] −[1(−74.8kJ/mol)]
Step 4: Perform the calculations:
∆H= [−393.5kJ/mol −483.6kJ/mol] + 74.8kJ/mol
∆H=−877.1kJ/mol + 74.8kJ/mol
∆H=−802.3kJ/mol
Therefore, the change in enthalpy (∆H) for the combustion of 2 moles of
methane is -802.3 kJ.
Question 32
Question
A gas is compressed isothermally at 27
°
C from an initial pressure of 2 atm to a
final pressure of 8 atm. If the enthalpy change during the process is -42 kJ/mol,
calculate the change in molar enthalpy for the gas.
Solution
Step 1: Calculate the initial molar enthalpy using the ideal gas law.
P V =nRT
Given that the initial pressure P1= 2 atm, the final pressure P2= 8 atm, the
temperature T= 27C= 300K, and the gas constant R= 0.0821 atm L mol−1
K−1.
We can rearrange the ideal gas law to solve for the initial molar quantity:
n=P V
RT
24
Substitute the values to find n:
n=(2 atm)V
(0.0821 atm L mol−1K−1)(300 K)
n=2V
24.63 =V
12.315
Step 2: Calculate the final molar enthalpy using the ideal gas law as well.
Given that the final pressure is P2= 8 atm, we can use the ideal gas law to find
nas before:
n=(8 atm)V
(0.0821 atm L mol−1K−1)(300 K)
n=8V
24.63 =4V
12.315
Step 3: Calculate the change in molar enthalpy. The change in molar en-
thalpy, ∆H, is given as -42 kJ/mol. We can express this change in terms of the
initial and final molar quantities:
∆H=Hf−Hi=4V
12.315−V
12.315=3V
12.315 =−42 kJ/mol
Solving for V:3V
12.315 =−42
3V=−517.245
V=−172.415 L
Step 4: Calculate the change in molar enthalpy for the gas. The change in
molar enthalpy, ∆H, is given as -42 kJ/mol. We can express this change in
terms of the initial and final molar quantities:
∆H=Hf−Hi=4V
12.315−V
12.315=3V
12.315 =−42 kJ/mol
Substitute the value of Vto find ∆H:
∆H=3(−172.415)
12.315
∆H=−42 kJ/mol
Therefore, the change in molar enthalpy for the gas during the isothermal
compression process is -42 kJ/mol.
25
Question 33
Question
Given the following reaction at constant pressure:
2H2(g)+O2(g)→2H2O(g)
If the standard enthalpy change of formation of water vapor is -241.8 kJ/mol,
calculate the standard enthalpy change of the reaction.
Solution
Step 1: Write down the balanced chemical equation for the reaction. Step 2: Use
the standard enthalpy of formation values to calculate the standard enthalpy
change of the reaction.
Step 1: The balanced chemical equation for the reaction is:
2H2(g)+O2(g)→2H2O(g)
Step 2: The standard enthalpy change of the reaction can be calculated
using the standard enthalpy of formation values. The standard enthalpy of
formation of water vapor is given as -241.8 kJ/mol.
The standard enthalpy change of the reaction can be calculated as:
∆H◦=X∆H◦
products −X∆H◦
reactants
Substitute the values into the equation:
∆H◦= 2 ×0 kJ/mol −(2 ×(-241.8 kJ/mol))
∆H◦= 483.6 kJ/mol
Therefore, the standard enthalpy change of the reaction is 483.6 kJ/mol.
Question 34
Question
Given the reaction:
2H2O2(l)→2H2O(l)+O2(g)
Calculate the standard enthalpy change for this reaction (∆H◦) given the
following information:
∆H◦
ffor H2O2(l) = -196.1 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
∆H◦
ffor O2(g) = 0 kJ/mol
26
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2O2(l)→2H2O(l)+O2(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation for the given compounds.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the values into the equation.
∆H◦= [2∆H◦
f(H2O(l)) + ∆H◦
f(O2(g))] −[2∆H◦
f(H2O2(l))]
Step 4: Substitute the given values.
∆H◦= [2(−285.8) + 0] −[2(−196.1)]
Step 5: Calculate the final answer.
∆H◦= (−571.6) −(−392.2)
∆H◦=−179.4 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -179.4
kJ/mol.
Question 35
Question
A sample of methane gas at 27
°
C and 1 atm is burned to heat 1 kg of water at
20
°
C. The enthalpy change for the combustion of methane at 25
°
C and 1 atm
is -890 kJ/mol. Calculate the final temperature of the water assuming all the
heat released in the combustion is transferred to the water. Take the specific
heat capacity of water as 4.18 J/(g
°
C) and the molar mass of methane as 16.05
g/mol.
Solution
Step 1: Calculate the heat released by the combustion of 1 mol of methane.
The enthalpy change for the combustion of methane is -890 kJ/mol, which is
equivalent to -890,000 J/mol. Since the molar mass of methane is 16.05 g/mol,
we need to calculate the heat released per gram of methane.
Heat released per gram of methane = −890,000 J/mol
16.05 g/mol =−55517.24 J/g
27
Step 3: Substitute the given values into the formula and calculate ∆H.
∆H= [(−84.7) kJ/mol] −[(2 ×0) + (3 ×0)] kJ/mol
∆H=−84.7 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H=−84.7 kJ/mol
at 298 K.
Question 2
Question
A reaction is carried out in a bomb calorimeter, and the temperature of the
calorimeter increases by 5
°
C. If the heat capacity of the calorimeter is 20 J/
°
C,
and the reaction releases 5000 J of heat, calculate the change in enthalpy (∆H)
of the reaction.
Solution
Step 1: Calculate the heat absorbed by the calorimeter using the formula:
qcalorimeter =C·∆T
where: C= heat capacity of the calorimeter = 20 J/
°
C, ∆T= change in
temperature = 5
°
C.
Therefore,
qcalorimeter = 20 J/
°
C·5
°
C = 100 J
Step 2: Since the calorimeter is absorbing the heat released by the reaction,
the heat absorbed by the reaction is equal in magnitude but opposite in sign:
qreaction =−qcalorimeter =−100 J
Step 3: The heat absorbed by the reaction is related to the change in enthalpy
of the reaction through the equation:
∆H=qreaction
Therefore, the change in enthalpy of the reaction is:
∆H=−100 J
Question 3
Question
A reaction takes place in a bomb calorimeter and releases 1250 kJ of heat. If
the volume of water in the calorimeter is 500 mL and its initial temperature
was 25
°
C, what is the final temperature of the water? Assume no heat is lost
to the surroundings. The specific heat of water is 4.18 J/g
°
C.
2
Solution
Step 1: Calculate the heat capacity of the water.
Q=mc∆T
1250000 = (500 g)(4.18 J/g
°
C)(Tf−25)
Tf−25 = 1250000
(500)(4.18)
Tf=1250000
(500)(4.18) + 25
Tf≈89.95
°
C
So, the final temperature of the water is approximately 89.95
°
C.
Question 4
Question
Calculate the change in enthalpy (∆H) when 2.5 moles of water (H2O) liquid at
25◦C is converted into steam at 100◦C. Given that the specific heat capacity of
water is 4.18 J/(g·◦C), the molar heat of vaporization of water is 40.79 kJ/mol,
and the molar mass of water is 18.02 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 25◦C
to 100◦C.
The specific heat capacity of water is 4.18 J/(g·◦C), so the heat required is
given by the equation:
q=m×c×∆T
Where: - mis the mass of water in grams - cis the specific heat capacity of
water - ∆Tis the change in temperature
Given that there are 2.5 moles of water and the molar mass of water is 18.02
g/mol, the mass of water is calculated as:
Mass of water = 2.5 moles ×18.02 g/mol = 45.05 g
Substitute the values into the equation:
q= 45.05 g ×4.18 J/(g ·◦C) ×(100 −25)◦C
q= 45.05 ×4.18 ×75
q= 14141.775 J = 14.14 kJ
3
Step 2: Calculate the heat required to convert water at 100◦C to steam at
100◦C.
The molar heat of vaporization of water is 40.79 kJ/mol. Since we have 2.5
moles of water, the heat required for vaporization is:
q= 2.5 moles ×40.79 kJ/mol = 101.975 kJ
Step 3: Calculate the total change in enthalpy using the heat values calcu-
lated in steps 1 and 2.
∆H=qraising temp +qvaporization
∆H= 14.14 kJ + 101.975 kJ
∆H= 116.115 kJ
Therefore, the change in enthalpy (∆H) when 2.5 moles of water liquid at
25◦C is converted into steam at 100◦C is 116.115 kJ.
Question 5
Question
Calculate the change in enthalpy for the reaction
2C(s)+2H2(g)→C2H4(g)
given the following enthalpy values:
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H4(g)) = 52.3 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
chemical equation is:
2C(s)+2H2(g)→C2H4(g)
Step 2: Calculate the change in enthalpy using the standard enthalpies of
formation. The change in enthalpy, H◦
rxn, can be calculated using the equation:
H◦
rxn =nH◦
f, products −(mH◦
f, reactants)
where nand mare the stoichiometric coefficients in the balanced equation, and
H◦
frepresents the standard enthalpy of formation.
4
Given that:
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H4(g)) = 52.3 kJ/mol
Substitute the values into the equation:
H◦
rxn = (1 ·52.3 kJ/mol) −[(2 ·0 kJ/mol) + (2 ·0 kJ/mol)]
H◦
rxn = 52.3 kJ/mol
Therefore, the change in enthalpy for the reaction is 52.3 kJ/mol .
Question 6
Question
A reaction is known to have ∆H◦=−92.2 kJ and ∆S◦=−145.1 J/K. Calculate
the temperature at which the reaction becomes spontaneous.
Solution
Step 1: Write down the equation for Gibb’s free energy change (∆G):
∆G= ∆H−T·∆S
Step 2: At equilibrium, ∆G= 0. Therefore, we have:
0 = −92.2 kJ −T·(−0.1451 kJ/K)
Step 3: Solve for the temperature T:
T=92.2 kJ
0.1451 kJ/K
T≈635.13 K
Therefore, the reaction becomes spontaneous at approximately 635.13 Kelvin.
Question 7
Question
A reaction occurring in a bomb calorimeter releases 1500 J of heat while doing
300 J of work on the surroundings. Calculate the change in enthalpy (∆H) for
the reaction.
5
Solution
Step 1: Recall that the change in enthalpy (∆H) is given by the equation:
∆H= ∆E+P∆V
where ∆Eis the change in internal energy and P∆Vis the work done on (or
by) the system.
Step 2: We can calculate the change in internal energy (∆E) using the first
law of thermodynamics:
∆E=q+w
where qis the heat absorbed (positive) or released (negative) by the system and
wis the work done on (positive) or by (negative) the system.
Step 3: In this case, the heat released by the reaction is -1500 J (negative
because it is released) and the work done on the surroundings is 300 J (positive
because it is done on the surroundings). Therefore, we have:
∆E=−1500 J + 300 J
∆E=−1200 J
Step 4: Now we can calculate the change in enthalpy (∆H) by adding the
change in internal energy to the work done on the system:
∆H= ∆E+P∆V
∆H=−1200 J + 300 J
∆H=−900 J
Therefore, the change in enthalpy (∆H) for the reaction is -900 J.
Question 8
Question
A gas at a constant pressure of 1 atm is heated from 25◦C to 100◦C. During
this process, the gas absorbs 150 J of heat and expands to a volume of 5 L.
Calculate the change in enthalpy (∆H) for the gas.
Solution
Step 1: Calculate the change in temperature: Given: Initial temperature,
Tinitial = 25◦C = 25 + 273 = 298 K Final temperature, Tfinal = 100◦C =
100 + 273 = 373 K
Change in temperature: ∆T=Tfinal−Tinitial ∆T= 373K−298K∆T= 75K
Step 2: Calculate the heat absorbed by the gas: Given: Heat absorbed, q =
150 J
6
Step 3: Calculate the change in enthalpy using the equation: ∆H=q+P·
∆V
Step 4: Calculate the change in volume: Given: Initial volume, Vinitial = 0
(not needed in this case, as ∆Vwill be equal to the final volume since the gas
expands) Final volume, Vfinal = 5 L
Change in volume: ∆V=Vfinal −Vinitial ∆V= 5 L
Step 5: Plug in the values into the equation for change in enthalpy: ∆H=
q+P·∆V∆H= 150 J + (1 atm) ·5 L ·(101.3 J/L
·
atm) ∆H= 150 J + 506.5 J
∆H= 656.5 J
Therefore, the change in enthalpy (∆H) for the gas is 656.5 J.
Question 9
Question
Calculate the change in enthalpy (∆H) when 15.0 g of water at 25
°
C is converted
to steam at 150
°
C. The specific heat capacity of water is 4.18 J/g
°
C, the molar
enthalpy of vaporization of water is 40.7 kJ/mol, and the molar mass of water
is 18.0 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of water from 25
°
C
to 100
°
C. The heat required can be calculated using the formula:
q=mc∆T
where: q= heat energy, m= mass of the substance, c= specific heat capacity,
and ∆T= change in temperature. Given that m= 15.0 g, c= 4.18 J/g
°
C, and
∆T= 100
°
C, we can substitute these values into the formula to find q.
Step 2: Calculate the heat required to convert water at 100
°
C to steam at
100
°
C (phase change). The heat required for this phase change is equal to the
molar enthalpy of vaporization of water, which is 40.7 kJ/mol. First, we need
to find the number of moles of water in 15.0 g.
Moles = Mass (g)
Molar mass (g/mol)
Moles = 15.0 g
18.0 g/mol
Moles ≈0.833 mol
Next, we calculate the heat required for the phase change:
q=n·∆Hvap
q= 0.833 mol ×40.7×103J/mol
q≈33.91 ×103J
7
Step 3: Calculate the heat required to raise the temperature of steam from
100
°
C to 150
°
C. Using the formula q=mc∆Twith m= 15.0 g, c= 2.01 J/g
°
C
(specific heat capacity of steam), and ∆T= 50
°
C. Substitute these values into
the formula to find q.
Step 4: Calculate the total change in enthalpy (∆H).
∆H=q1+q2+q3
Substitute the values of q1,q2, and q3into the formula to calculate ∆H.
Question 10
Question
A sample of gas undergoes a process at constant pressure where it absorbs 150
J of heat and does 50 J of work on the surroundings. If the change in internal
energy of the gas is 100 J, what is the change in enthalpy of the gas during the
process?
Solution
Step 1: Recall the definition of enthalpy: Enthalpy is defined as H=U+P V ,
where His enthalpy, Uis internal energy, Pis pressure, and Vis volume.
Step 2: The change in enthalpy can be calculated using the equation:
∆H= ∆U+P∆V
Step 3: Given that ∆U= 100 J, and the process is at constant pressure, we
can rewrite P∆Vas w, the work done by the gas on the surroundings. Thus,
∆H= ∆U+w.
Step 4: Plugging in the values, we have:
∆H= 100 J + (−50 J)
Step 5: Simplifying, we find:
∆H= 50 J
Step 6: Therefore, the change in enthalpy of the gas during the process is
50 J .
Question 11
Question
A reaction is carried out in a bomb calorimeter, and it is found that the tem-
perature of the calorimeter increases by 5.82
°
C. If the heat capacity of the
calorimeter is 247 J/
°
C, calculate the enthalpy change (∆H) for the reaction in
kJ/mol.
8
Solution
Step 1: Calculate the heat absorbed by the bomb calorimeter using the formula:
q=C·∆T
where: q= heat absorbed by the calorimeter, C= heat capacity of the calorime-
ter, ∆T= change in temperature.
Plugging in the values, we get:
q= 247 J/
°
C×5.82
°
C = 1439.74 J
Step 2: Convert the heat absorbed by the calorimeter to kilojoules (1 kJ =
1000 J):
q= 1439.74 J ×1 kJ
1000 J = 1.44 kJ
Step 3: Calculate the moles of substance reacting using the molar heat
capacity formula:
∆H=q
n
where: ∆H= enthalpy change, q= heat absorbed by the calorimeter, n=
moles of substance.
Step 4: Rearrange the formula to solve for ∆H:
∆H=q
n
n=q
∆H
Step 5: The molar heat capacity of the reaction is then given by:
∆H=1.44 kJ
n
Step 6: Finally, divide the heat absorbed by the calorimeter by the moles of
substance to determine the enthalpy change in kJ/mol.
Question 12
Question
A reaction takes place in a bomb calorimeter, and the temperature of the water
in the bomb calorimeter changes from 25
°
C to 30
°
C. If the heat capacity of the
bomb calorimeter is 50 J/
°
C, calculate the change in enthalpy of the reaction.
Assume that the heat released from the reaction is only absorbed by the water
in the calorimeter, and the bomb calorimeter is insulated with its surroundings.
9
Solution
Step 1: Calculate the heat absorbed by the water in the bomb calorimeter using
the formula:
q=C·∆T
where: q= heat absorbed by the water in the calorimeter, C= 50 J/
°
C =
heat capacity of the bomb calorimeter, and ∆T= 30C−25C= 5C= change
in temperature.
Substitute the values into the formula:
q= 50 J/
°
C·5C= 250 J
Therefore, the water in the bomb calorimeter absorbs 250 J of heat.
Step 2: Since the reaction is taking place in the bomb calorimeter, the heat
absorbed by the water is equal to the heat released by the reaction according
to the first law of thermodynamics. This heat is given by the formula:
∆H=−q
where: ∆H= change in enthalpy of the reaction and q= 250 J (negative
because the reaction releases heat).
Therefore, the change in enthalpy of the reaction is:
∆H=−250 J
Hence, the change in enthalpy of the reaction is -250 J.
Question 13
Question
A sample of an ideal gas is taken through a cyclic process consisting of two
isobaric and two isochoric processes. The initial temperature and pressure of
the gas are T1and P1, and the final temperature and pressure are T2and P2,
respectively. Calculate the change in enthalpy for the gas during this process.
Solution
Step 1: Recall that for an ideal gas, the change in enthalpy is given by ∆H=
Cp·∆T, where Cpis the specific heat at constant pressure and ∆Tis the change
in temperature.
Step 2: For an isobaric process, Cp=5
2Rwhere Ris the gas constant. Thus,
the change in enthalpy for the first isobaric process is ∆H1=5
2R·(T2−T1).
Step 3: For an isochoric process, ∆H= ∆U, where ∆Uis the change in
internal energy. Since there is no work done in an isochoric process, ∆U=
nCv∆T, where Cvis the specific heat at constant volume.
10
Step 4: For an ideal gas, Cv=3
2R. Therefore, the change in enthalpy for
the first isochoric process is ∆H2=3
2R·(T2−T1).
Step 5: The total change in enthalpy for the cyclic process is then ∆H=
∆H1+ ∆H2=5
2R+3
2R(T2−T1) = 4R(T2−T1).
Question 14
Question
A reaction has an enthalpy change of -285 kJ/mol. If 2.00 moles of the reactant
are consumed in the reaction at constant pressure, what is the enthalpy change
for the reaction in kJ?
Solution
Step 1: Start with the given information: The enthalpy change for the reaction
is -285 kJ/mol. The number of moles of the reactant consumed is 2.00 mol.
Step 2: Calculate the enthalpy change for the reaction in kJ. Given that
the enthalpy change is -285 kJ/mol, we can use this value to find the enthalpy
change for the reaction when 2.00 moles of the reactant are consumed:
Enthalpy change for the reaction = Enthalpy change per mole×Number of moles consumed
Enthalpy change for the reaction = −285 kJ/mol ×2.00 mol
Step 3: Perform the calculation.
Enthalpy change for the reaction = −285 kJ/mol ×2.00 mol = −570 kJ
Therefore, when 2.00 moles of the reactant are consumed in the reaction,
the enthalpy change for the reaction is -570 kJ.
Question 15
Question
Calculate the change in enthalpy when 5 moles of nitrogen gas at 200 K and 1
atm pressure are heated at constant pressure to 400 K. Assume that nitrogen
gas behaves ideally.
Solution
Step 1: The change in enthalpy can be calculated using the formula:
∆H= ∆U+P∆V
11
Step 2: First, we calculate the change in internal energy (∆U) using the
formula:
∆U=nCv∆T
where nis the number of moles, Cvis the molar heat capacity at constant
volume, and ∆Tis the change in temperature.
Step 3: Since nitrogen gas is diatomic, the molar heat capacity at constant
volume is Cv=5
2Rwhere Ris the gas constant.
Step 4: Substituting the values into the formula, we get:
∆U= 5 ×5
2R×(400 −200)
Step 5: Next, we calculate the work done by the gas, which is equal to P∆V.
Since the process is at constant pressure, we have:
∆V=nR∆T
Step 6: Substituting the values into the formula, we get:
∆V= 5 ×R×(400 −200)
Step 7: Now we can calculate the work done:
P∆V= 1 ×(5R×200)
Step 8: Finally, we can calculate the change in enthalpy by adding the change
in internal energy and the work done:
∆H= ∆U+P∆V
∆H= (5 ×5
2R×200) + (5R×200)
Question 16
Question
Given the following data for the reaction:
∆H◦
ffor CO2= -393.5 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction:
C(graphite) + 2H2O→CO2+ 2H2O
12
Solution
Step 1: Write the balanced chemical equation for the reaction and determine the
∆H◦value for the reaction. The balanced chemical equation for the reaction is:
C(graphite) + 2H2O→CO2+ 2H2O
To find the ∆H◦for the reaction, we use the following formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the ∆H◦value for the reaction. Given data: ∆H◦
ffor
CO2= -393.5 kJ/mol ∆H◦
ffor H2O(l) = -285.8 kJ/mol ∆H◦
ffor C(graphite)
= 0 kJ/mol
Plugging in the values into the formula:
∆H◦= [1(−393.5 kJ/mol) + 2(−285.8 kJ/mol)] −[1(0 kJ/mol) + 2(0 kJ/mol)]
∆H◦= [−393.5 kJ/mol −571.6 kJ/mol]
∆H◦=−965.1 kJ/mol
Therefore, the standard enthalpy change, ∆H◦, for the reaction:
C(graphite) + 2H2O→CO2+ 2H2O
is -965.1 kJ/mol.
Question 17
Question
A certain chemical reaction is known to have a standard enthalpy change of -383
kJ mol−1. If 2.50 moles of the reactant are consumed in the reaction, calculate
the amount of heat absorbed or released.
Solution
Step 1: Recall the formula relating enthalpy change (∆H) to heat absorbed or
released (q) and amount of substance (n):
q= ∆H×n
Step 2: Substitute the known values into the formula:
q=−383 kJ mol−1×2.50 mol
Step 3: Perform the calculation:
q=−383 ×2.50 = −957.5 kJ
Step 4: Thus, the amount of heat absorbed or released during the reaction
is -957.5 kJ. Since the enthalpy change is negative, it indicates that the reaction
releases heat.
13
Question 18
Question
A reaction is carried out in a bomb calorimeter and releases 150 kJ of heat.
If the volume of the bomb calorimeter is 500 mL and the pressure inside the
calorimeter remains constant at 1 atm, calculate the change in enthalpy (∆H)
of the reaction in kJ/mol. Assume ideal gas behavior.
Solution
Step 1: Recall that ∆H=qp, where qpis the heat exchanged at constant
pressure during a chemical reaction.
Step 2: Convert the volume of the bomb calorimeter from mL to L: V=
500 mL ×1 L
1000 mL = 0.5 L
Step 3: Since the pressure remains constant and the reaction takes place in a
bomb calorimeter, the heat exchanged is equal to the change in enthalpy (∆H).
Step 4: The change in enthalpy of the reaction can be calculated using the
formula: ∆H=q
n
Step 5: Given that the heat exchanged (q) is 150 kJ and the number of moles
(n) is unknown, we have: ∆H=150 kJ
n
Step 6: To find the number of moles, we need to convert the volume at STP
to number of moles of gas. At STP, 22.4 L of any gas contains 1 mole.
Step 7: Assuming ideal gas behavior, the volume of 0.5 L at 1 atm pressure
contains nmoles: n=0.5 L
22.4 L/mol
Step 8: Substitute the value of nback into the formula for ∆Hto find the
change in enthalpy of the reaction: ∆H=150 kJ
0.5 L/22.4 L/mol = 150 kJ ×22.4 L/mol
0.5 L
Step 9: ∆H= 150 ×22.4 = 3360 kJ/mol
Therefore, the change in enthalpy of the reaction is 3360 kJ/mol.
Question 19
Question
A reaction is taking place in a closed system where the enthalpy change (∆H)
is given by the equation:
∆H= 50 −2T+ 0.05T2
where ∆His in kJ/mol and Tis in Kelvin. Determine the temperature at which
the reaction is exothermic.
Solution
Step 1: To determine the temperature at which the reaction is exothermic, we
need to find the point where ∆Hbecomes negative. Step 2: Set ∆Hto be less
14
than zero and solve for T:
50 −2T+ 0.05T2<0
0.05T2−2T+ 50 <0
Step 3: This is a quadratic inequality. To solve it, we find the roots of the
corresponding quadratic equation:
0.05T2−2T+ 50 = 0
T2−40T+ 1000 = 0
Step 4: Use the quadratic formula to find the roots:
T=−(−40) ±p(−40)2−4(1)(1000)
2(1)
T=40 ±√1600 −4000
2
T=40 ±√−2400
2
Step 5: Since the square root of a negative number is not real, the inequality
has no real solutions. Therefore, the reaction is never exothermic.
Question 20
Question
Given the reaction:
2C(graphite)+3H2(g)→C2H6(g)
If the enthalpy change for this reaction is -84 kJ/mol, calculate the enthalpy
change when 6.00 g of carbon reacts with excess hydrogen. (Molar mass of
carbon = 12.01 g/mol, molar mass of hydrogen = 1.008 g/mol)
Solution
Step 1: Calculate the number of moles of carbon in 6.00 g. Step 2: Determine
the molar ratio between carbon and the given reaction. Step 3: Use the enthalpy
change for the reaction to calculate the enthalpy change when 6.00 g of carbon
reacts.
Step 1: Calculate the number of moles of carbon in 6.00 g. Given: Mass of
carbon = 6.00 g Molar mass of carbon = 12.01 g/mol
Number of moles of carbon:
Moles = Mass
Molar mass =6.00 g
12.01 g/mol
15
Moles = 0.499 mol
Step 2: Determine the molar ratio between carbon and the given reaction.
From the balanced equation: 2 moles of carbon react to produce 1 mole of C2H6
Step 3: Use the enthalpy change for the reaction to calculate the enthalpy
change when 6.00 g of carbon reacts. Given enthalpy change for the reaction:
-84 kJ/mol
Since 2 moles of carbon are required to produce 1 mole of C2H6, the enthalpy
change for the reaction involving 2 moles of carbon is -84 kJ. For 0.499 moles
of carbon (half the amount):
(0.499 mol)(−84 kJ/mol) = −41.92 kJ
Therefore, the enthalpy change when 6.00 g of carbon reacts with excess
hydrogen is -41.92 kJ.
Question 21
Question
Calculate the change in enthalpy (∆H) for a reaction where 2 moles of ethylene
(C2H4) are combusted with excess oxygen to form carbon dioxide (CO2) and
water vapor (H2O). The enthalpies of formation (∆H◦
f) for C2H4, CO2, and
H2O are -52.26 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Write the balanced chemical equation for the combustion of 2 moles of
ethylene:
C2H4+ 3O2→2CO2+ 2H2O
Step 2: Calculate the change in enthalpy (∆H) using the equation:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given enthalpies of formation into the equation:
∆H= [2(−393.5 kJ/mol) + 2(−241.8 kJ/mol)] −[1(−52.26 kJ/mol)]
Step 4: Perform the calculations to find ∆H:
∆H= [−787 + (−483.6)] −[−52.26]
∆H=−1270.6 + 52.26
∆H=−1218.34 kJ
Answer: The change in enthalpy for the reaction is ∆H=−1218.34 kJ.
16
Question 22
Question
A reaction takes place in a closed system at constant pressure, where the en-
thalpy change (∆H) of the reaction is −286 kJ. If the reaction releases 672 J
of heat to the surroundings, calculate the work done by the system during the
reaction.
Solution
Step 1: Recall the relationship between the enthalpy change, heat absorbed or
released, and work done for a reaction taking place at constant pressure:
∆H=q+w
where ∆His the enthalpy change, qis the heat absorbed or released by the
system, and wis the work done by or on the system.
Step 2: Given that ∆H=−286 kJ and q=−672 J (since the reaction
releases heat to the surroundings), we can substitute these values into the equa-
tion:
−286 kJ = −672 J + w
Step 3: Convert all units to the same unit. Since 1 kJ = 1000 J, we need to
convert −286 kJ to J:
−286 kJ = −286 ×1000 J = −286000 J
Step 4: Substitute −286000 J for ∆Hand −672 J for qinto the equation
and solve for w:
−286000 J = −672 J + w
w=−286000 J + 672 J
w=−285328 J
Step 5: Therefore, the work done by the system during the reaction is
−285328 J.
Question 23
Question
A gas undergoes a process where its enthalpy change is given by the equation:
∆H= 30T2−20T
where ∆His in kJ/mol and Tis in Kelvin. Determine the heat transfer when
the temperature changes from 300 K to 500 K.
17
Solution
Step 1: To find the heat transfer, we need to calculate the change in enthalpy
for the given temperature range by integrating the enthalpy change equation
with respect to temperature.
∆H=ZTf
Ti
(30T2−20T)dT
Step 2: Integrating the equation, we get
∆H=10T3−10T2500
300
∆H=10(500)3−10(500)2−10(300)3−10(300)2
Step 3: Calculating the values, we get
∆H= (10 ×125000) −(10 ×2500) −(10 ×27000) + (10 ×900)
∆H= 1250000 −25000 −270000 + 9000
∆H= 959000 kJ/mol
Therefore, the heat transfer when the temperature changes from 300 K to
500 K is 959000 kJ/mol.
Question 24
Question
Given the reaction:
2H2(g) + O2(g) →2H2O(g)
with ∆H=−484 kJ/mol, calculate the enthalpy change when 4 moles of water
are produced.
Solution
Step 1: Determine the moles of water produced in the reaction. Since the
reaction produces 2 moles of water for every 1 mole of oxygen, we can calculate
the moles of water produced as follows:
4 moles of water = 4 moles ×2 moles of H2O
1 mole of O2
= 8 moles of H2O
Step 2: Use the given enthalpy change to calculate the total enthalpy change
for 4 moles of water. Since 2 moles of water are produced for every 484 kJ of
energy released, the enthalpy change for 8 moles of water produced will be:
8 moles of H2O×484 kJ
2 moles of H2O= 1936 kJ
Therefore, the enthalpy change when 4 moles of water are produced in this
reaction is 1936 kJ.
18
Question 25
Question
Calculate the change in enthalpy for a reaction where 2 moles of methane (CH4)
react with 4 moles of oxygen (O2) to produce carbon dioxide (CO2) and water
(H2O) according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the standard enthalpies of formation are ∆H◦
f(CH4) = −74.8 kJ/mol,
∆H◦
f(CO2) = −393.5 kJ/mol, and ∆H◦
f(H2O) = −285.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy of the reaction using the given standard
enthalpies of formation. The standard enthalpy change for a reaction is equal
to the sum of the standard enthalpies of formation of the products minus the
sum of the standard enthalpies of formation of the reactants.
Given standard enthalpies of formation: ∆H◦
f(CH4) = −74.8 kJ/mol, ∆H◦
f(CO2) =
−393.5 kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol.
The standard enthalpy change for the reaction can be calculated as:
∆H◦=Σ∆H◦
f(products)−Σ∆H◦
f(reactants)
Given reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Substitute the values:
∆H◦=∆H◦
f(CO2) + 2∆H◦
f(H2O)−∆H◦
f(CH4)+2·0
∆H◦= (−393.5 kJ/mol + 2 ∗(−285.8) kJ/mol) −(−74.8 kJ/mol)
∆H◦= (−393.5 kJ/mol −571.6 kJ/mol) + 74.8 kJ/mol
∆H◦=−965.1 kJ/mol + 74.8 kJ/mol
∆H◦=−890.3 kJ/mol
Therefore, the change in enthalpy for the reaction is -890.3 kJ/mol.
Question 26
Question
A reaction is carried out in a bomb calorimeter, which is also referred to as a
constant volume calorimeter. The initial temperature inside the calorimeter is
25
°
C and the final temperature after the reaction is 35
°
C. The heat capacity of
the calorimeter is 30 J/
°
C. If the reaction released 500 J of heat, calculate the
enthalpy change (∆H) for the reaction.
19
Solution
Step 1: Calculate the change in temperature (∆T) using the initial and final
temperatures:
∆T=Tf−Ti= 35◦C−25◦C = 10◦C
Step 2: Calculate the total heat absorbed by the calorimeter and its contents:
Q=C·∆T
Q= 30 J/
°
C·10◦C = 300 J
Step 3: Since the reaction released 500 J of heat, the total heat absorbed by
the calorimeter and its contents is equal to the negative of the heat released by
the reaction:
Qreaction =−300 J −500 J
Qreaction =−800 J
Step 4: The enthalpy change (∆H) for the reaction is equal to the total heat
absorbed by the calorimeter and its contents:
∆H=Q=−800 J
Therefore, the enthalpy change (∆H) for the reaction is -800 J.
Question 27
Question
Calculate the change in enthalpy (∆H) when 25.0 g of water vapor at 100.0
°
C
condenses to liquid water at 100.0
°
C. The molar enthalpy of vaporization for
water is 40.79 kJ/mol.
Solution
Step 1: Determine the number of moles of water vapor. Given: Mass of water
vapor = 25.0 g Molar mass of water (HO) = 18.015 g/mol
Number of moles of water vapor = Mass
Molar mass =25.0 g
18.015 g/mol ≈1.387 mol
Step 2: Calculate the change in enthalpy. The change in enthalpy (∆H) can
be calculated using the equation:
∆H= moles ×molar enthalpy of vaporization
Substitute the values:
∆H= 1.387 mol ×40.79 kJ/mol = 56.62 kJ
Therefore, the change in enthalpy when 25.0 g of water vapor at 100.0
°
C
condenses to liquid water at 100.0
°
C is 56.62 kJ.
20
Question 28
Question
A reaction is carried out at constant pressure, and the enthalpy change (∆H)
for the reaction is measured to be −425 kJ. If the reaction releases 450 kJ of
heat to the surroundings, what is the change in internal energy (∆U) for the
reaction? Assume that the only kind of work done is pressure-volume work.
Solution
Step 1: Recall the relationship between enthalpy change (∆H), change in inter-
nal energy (∆U), and heat exchanged (q) under constant-pressure conditions:
∆H= ∆U+P∆V
where Pis the constant pressure and ∆Vis the change in volume.
Step 2: Since the reaction is carried out at constant pressure, ∆U= ∆H−
P∆V.
Step 3: We are given that ∆H=−425 kJ and q=−450 kJ (since heat is
released to the surroundings).
Step 4: Since the only kind of work done is pressure-volume work, q=
∆U−P∆V= ∆U−w, where wis the work done. Here, w=−P∆V, so
q= ∆U+P∆V.
Step 5: Substitute the given values into the equation:
−450 kJ = ∆U−425 kJ
Step 6: Solve for ∆U:
∆U=−450 kJ + 425 kJ = −25 kJ
Therefore, the change in internal energy for the reaction is ∆U=−25 kJ.
Question 29
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas (CH4)
is burned in excess oxygen gas (O2) according to the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the standard enthalpies of formation are:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
21
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
for the reaction is:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Step 2: Determine the enthalpy change for the reaction using the standard
enthalpies of formation. The change in enthalpy (∆H) for the reaction can be
calculated using the formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given standard enthalpies of formation:
∆H= [1 ×∆H◦
f(CO2)+2×∆H◦
f(H2O(l))] −[1 ×∆H◦
f(CH4)+2×∆H◦
f(O2)]
∆H= [1 × −393.5+2× −285.8] −[1 × −74.8+2×0]
∆H= [−393.5−571.6] −[−74.8]
∆H=−965.1 + 74.8
∆H=−890.3 kJ
Therefore, the change in enthalpy (∆H) when 5.00 moles of methane gas is
burned in excess oxygen gas is -890.3 kJ.
Question 30
Question
Calculate the change in enthalpy (∆H) when 25.0 g of ice at -18.0
°
C is melted
and heated to form water vapor at 125.0
°
C. The specific heat capacity of ice is
2.09 J/g
°
C, the heat of fusion for ice is 333 J/g, the specific heat capacity of
water is 4.18 J/g
°
C, and the heat of vaporization of water is 2260 J/g.
Solution
Step 1: Calculate the heat required to melt the ice.
qice =m·∆Hfusion
where mis the mass of ice and ∆Hfusion is the heat of fusion for ice.
qice = 25.0 g ×333 J/g = 8325 J
22
Step 2: Calculate the heat required to heat the water from -18.0
°
C to 0
°
C.
qice→water =m×cice ×∆T
where cice is the specific heat capacity of ice and ∆Tis the temperature change.
qice→water = 25.0 g ×2.09 J/g
°
C×(0 −(−18.0))
°
C = 943.5 J
Step 3: Calculate the heat required to heat the water from 0
°
C to 100
°
C.
qwater =m×cwater ×∆T
where cwater is the specific heat capacity of water and ∆Tis the temperature
change.
qwater = 25.0 g ×4.18 J/g
°
C×(100 −0)
°
C = 10450 J
Step 4: Calculate the heat required to vaporize the water.
qwater→vapor =m×∆Hvaporization
where ∆Hvaporization is the heat of vaporization of water.
qwater→vapor = 25.0 g ×2260 J/g = 56500 J
Step 5: Calculate the total heat required by adding all the heats calculated
in the previous steps.
qtotal =qice+qice→water+qwater+qwater→vapor = 8325 J+943.5 J+10450 J+56500 J = 8618.5 J
Step 6: Calculate the change in enthalpy using the total heat required.
∆H=qtotal
Therefore, the change in enthalpy (∆H) when 25.0 g of ice at -18.0
°
C is melted
and heated to form water vapor at 125.0
°
C is 8618.5 J.
Question 31
Question
Calculate the change in enthalpy (∆H) for a reaction in which 2 moles of
methane (CH4) are combusted to form carbon dioxide (CO2) and water va-
por (H2O), given the following standard enthalpy of formation values:
CH4:−74.8kJ/mol
CO2:−393.5kJ/mol
H2O:−241.8kJ/mol
23
Solution
Step 1: Write the balanced chemical equation for the combustion of methane:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Step 2: Calculate the standard enthalpy change for the reaction using the
standard enthalpies of formation:
∆H=XνfHf(products)−XνfHf(reactants)
Step 3: Substitute the given standard enthalpy of formation values into the
equation:
∆H= [1(−393.5kJ/mol) + 2(−241.8kJ/mol)] −[1(−74.8kJ/mol)]
Step 4: Perform the calculations:
∆H= [−393.5kJ/mol −483.6kJ/mol] + 74.8kJ/mol
∆H=−877.1kJ/mol + 74.8kJ/mol
∆H=−802.3kJ/mol
Therefore, the change in enthalpy (∆H) for the combustion of 2 moles of
methane is -802.3 kJ.
Question 32
Question
A gas is compressed isothermally at 27
°
C from an initial pressure of 2 atm to a
final pressure of 8 atm. If the enthalpy change during the process is -42 kJ/mol,
calculate the change in molar enthalpy for the gas.
Solution
Step 1: Calculate the initial molar enthalpy using the ideal gas law.
P V =nRT
Given that the initial pressure P1= 2 atm, the final pressure P2= 8 atm, the
temperature T= 27C= 300K, and the gas constant R= 0.0821 atm L mol−1
K−1.
We can rearrange the ideal gas law to solve for the initial molar quantity:
n=P V
RT
24
Substitute the values to find n:
n=(2 atm)V
(0.0821 atm L mol−1K−1)(300 K)
n=2V
24.63 =V
12.315
Step 2: Calculate the final molar enthalpy using the ideal gas law as well.
Given that the final pressure is P2= 8 atm, we can use the ideal gas law to find
nas before:
n=(8 atm)V
(0.0821 atm L mol−1K−1)(300 K)
n=8V
24.63 =4V
12.315
Step 3: Calculate the change in molar enthalpy. The change in molar en-
thalpy, ∆H, is given as -42 kJ/mol. We can express this change in terms of the
initial and final molar quantities:
∆H=Hf−Hi=4V
12.315−V
12.315=3V
12.315 =−42 kJ/mol
Solving for V:3V
12.315 =−42
3V=−517.245
V=−172.415 L
Step 4: Calculate the change in molar enthalpy for the gas. The change in
molar enthalpy, ∆H, is given as -42 kJ/mol. We can express this change in
terms of the initial and final molar quantities:
∆H=Hf−Hi=4V
12.315−V
12.315=3V
12.315 =−42 kJ/mol
Substitute the value of Vto find ∆H:
∆H=3(−172.415)
12.315
∆H=−42 kJ/mol
Therefore, the change in molar enthalpy for the gas during the isothermal
compression process is -42 kJ/mol.
25
Question 33
Question
Given the following reaction at constant pressure:
2H2(g)+O2(g)→2H2O(g)
If the standard enthalpy change of formation of water vapor is -241.8 kJ/mol,
calculate the standard enthalpy change of the reaction.
Solution
Step 1: Write down the balanced chemical equation for the reaction. Step 2: Use
the standard enthalpy of formation values to calculate the standard enthalpy
change of the reaction.
Step 1: The balanced chemical equation for the reaction is:
2H2(g)+O2(g)→2H2O(g)
Step 2: The standard enthalpy change of the reaction can be calculated
using the standard enthalpy of formation values. The standard enthalpy of
formation of water vapor is given as -241.8 kJ/mol.
The standard enthalpy change of the reaction can be calculated as:
∆H◦=X∆H◦
products −X∆H◦
reactants
Substitute the values into the equation:
∆H◦= 2 ×0 kJ/mol −(2 ×(-241.8 kJ/mol))
∆H◦= 483.6 kJ/mol
Therefore, the standard enthalpy change of the reaction is 483.6 kJ/mol.
Question 34
Question
Given the reaction:
2H2O2(l)→2H2O(l)+O2(g)
Calculate the standard enthalpy change for this reaction (∆H◦) given the
following information:
∆H◦
ffor H2O2(l) = -196.1 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
∆H◦
ffor O2(g) = 0 kJ/mol
26
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2O2(l)→2H2O(l)+O2(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation for the given compounds.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the values into the equation.
∆H◦= [2∆H◦
f(H2O(l)) + ∆H◦
f(O2(g))] −[2∆H◦
f(H2O2(l))]
Step 4: Substitute the given values.
∆H◦= [2(−285.8) + 0] −[2(−196.1)]
Step 5: Calculate the final answer.
∆H◦= (−571.6) −(−392.2)
∆H◦=−179.4 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -179.4
kJ/mol.
Question 35
Question
A sample of methane gas at 27
°
C and 1 atm is burned to heat 1 kg of water at
20
°
C. The enthalpy change for the combustion of methane at 25
°
C and 1 atm
is -890 kJ/mol. Calculate the final temperature of the water assuming all the
heat released in the combustion is transferred to the water. Take the specific
heat capacity of water as 4.18 J/(g
°
C) and the molar mass of methane as 16.05
g/mol.
Solution
Step 1: Calculate the heat released by the combustion of 1 mol of methane.
The enthalpy change for the combustion of methane is -890 kJ/mol, which is
equivalent to -890,000 J/mol. Since the molar mass of methane is 16.05 g/mol,
we need to calculate the heat released per gram of methane.
Heat released per gram of methane = −890,000 J/mol
16.05 g/mol =−55517.24 J/g
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Step 2: Calculate the heat released by the combustion of methane in this
specific scenario. The heat released by the combustion of methane is propor-
tional to the amount of methane burned. Since we have not been given the
amount of methane burned, we will assume it is 1 g for ease of calculation.
Therefore, the heat released by burning 1 g of methane is -55517.24 J.
Step 3: Calculate the heat absorbed by the water. Given that the specific
heat capacity of water is 4.18 J/(g
°
C) and the initial temperature of the water
is 20
°
C, we can write the heat absorbed by the water as:
Q= mass ×specific heat capacity ×∆T
Q= 1000 g ×4.18 J/(g
°
C) ×(Tf−20C)
−55517.24 J = 1000 g ×4.18 J/(g
°
C) ×(Tf−20C)
Step 4: Solve for the final temperature of the water.
−55517.24 = 4180 ×(Tf−20)
−55517.24 = 4180Tf−83600
4180Tf= 28082.76
Tf≈6.72C
Therefore, the final temperature of the water after all the heat released in
the combustion is transferred to it is approximately 6.72
°
C.
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