CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Enthalpy
Question Bank - Set 4
Liberty University
Question 1
Question
Calculate the change in enthalpy for a reaction where 2 moles of methane (CH4)
are combusted completely with 4 moles of oxygen (O2) to form carbon dioxide
(CO2) and water (H2O) according to the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the standard enthalpies of formation (∆H◦
f) for the compounds in-
volved are:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpies of formation of the compounds involved.
The standard enthalpy change for the reaction can be calculated using the
formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Given:
∆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 is:
∆H◦=1×∆H◦
f(CO2)+2×∆H◦
f(H2O)−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) = −965.1 + 74.8 = −890.3 kJ/mol
Therefore, the change in enthalpy for the given reaction is ∆H◦=−890.3 kJ/mol.
Question 2
Question
The enthalpy of formation of ammonia gas (NH3) is -45.9 kJ/mol. Calculate
the enthalpy change when 3.00 mol of ammonia gas is formed.
Solution
Step 1: Write the balanced chemical equation for the formation of ammonia
gas:
N2(g) + 3H2(g)→2NH3(g)
Step 2: Determine the number of moles of ammonia gas being formed: Given
that 3.00 mol of ammonia gas is formed.
Step 3: Calculate the enthalpy change for the formation of 1 mole of am-
monia gas: From the balanced chemical equation, 2 moles of ammonia gas are
formed when the enthalpy change is -45.9 kJ. Thus, for the formation of 1 mole
of NH3:
Enthalpy change = −45.9 kJ
2 mol =−22.95 kJ/mol
Step 4: Calculate the enthalpy change for the formation of 3.00 mol of
ammonia gas:
Enthalpy change = −22.95 kJ/mol ×3.00 mol = −68.85 kJ
Therefore, the enthalpy change when 3.00 mol of ammonia gas is formed is
-68.85 kJ.
Question 3
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l)
If the enthalpy change for the reaction is -484 kJ, calculate the enthalpy
change when 4 moles of water are formed.
2
Solution
Step 1: Calculate the enthalpy change for the formation of 1 mole of water.
Given that the enthalpy change for the reaction is -484 kJ and the reaction
forms 2 moles of water, the enthalpy change for the formation of 1 mole of
water is:
−484 kJ/2 = −242 kJ
Step 2: Calculate the enthalpy change for the formation of 4 moles of water.
Since the enthalpy change for the formation of 1 mole of water is -242 kJ, the
enthalpy change for the formation of 4 moles of water is:
−242 kJ/mole ×4 moles = −968 kJ
Therefore, the enthalpy change when 4 moles of water are formed is -968 kJ.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2(g)+O2(g)→2H2O(l)
given the following standard enthalpies of formation: ∆H◦
f(H2(g)) = 0
kJ/mol, ∆H◦
f(O2(g)) = 0 kJ/mol, ∆H◦
f(H2O(l)) = −285.8 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and identify the standard en-
thalpies of formation.
2H2(g)+O2(g)→2H2O(l)
Given:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation. We can use the formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the values:
∆H◦=2×∆H◦
f(H2O(l))−2×∆H◦
f(H2(g)) + ∆H◦
f(O2(g))
3
∆H◦= (2 ×(−285.8)) −(2 ×0 + 0)
∆H◦=−571.6 kJ/mol
Answer: The standard enthalpy change for the given reaction is -571.6
kJ/mol.
Question 5
Question
Calculate the change in enthalpy when 2 moles of water at 100◦C and 1 atm
condenses to liquid water at the same temperature. The enthalpy of vaporization
of water at 100◦C is 40.79 kJ/mol.
Solution
Step 1: Determine the change in enthalpy from vapor to liquid: Given that
the enthalpy of vaporization of water at 100◦C is 40.79 kJ/mol, the change in
enthalpy when 2 moles of water condenses to liquid form can be calculated as
follows:
∆H= 40.79 kJ/mol ×2 mol = 81.58 kJ
Therefore, the change in enthalpy when 2 moles of water condenses to liquid
form is 81.58 kJ.
Question 6
Question
A reaction is carried out in a bomb calorimeter. The initial temperature of the
system is 25
°
C and the final temperature after the reaction is 45
°
C. If the heat
capacity of the calorimeter is 150 J/
°
C, and the reaction released 8000 J of heat,
calculate the change in enthalpy (∆H) for the reaction.
Solution
Step 1: Calculate the heat absorbed by the calorimeter
Heat absorbed by calorimeter = Heat released by reaction
Heat absorbed by calorimeter = 8000 J
Step 2: Calculate the temperature change in the calorimeter
∆T=Tf−Ti= 45◦C−25◦C = 20◦C
4
Step 3: Calculate the heat absorbed by the calorimeter
Q=C×∆T
8000 = 150 ×20
Step 4: Calculate the change in enthalpy for the reaction
∆H=Q−Qcalorimeter
∆H= 8000 −3000
∆H= 5000 J
Therefore, the change in enthalpy for the reaction is ∆H= 5000 J.
Question 7
Question
A reaction is carried out in a bomb calorimeter, where the heat capacity of
the calorimeter is known to be 120 J/
°
C. During the reaction, 3.50 grams of a
compound are completely burned, raising the temperature of the calorimeter and
its contents by 4.50
°
C. Calculate the enthalpy change (∆H) for the combustion
of the compound in kJ/mol.
Given: Molar mass of the compound = 102.10 g/mol
Solution
Step 1: Calculate the heat absorbed by the calorimeter and its contents (qcalorimeter).
qcalorimeter =C×∆T
where Cis the heat capacity of the calorimeter and ∆Tis the temperature
change.
qcalorimeter = 120 J/
°
C×4.50
°
C
qcalorimeter = 540 J
Step 2: Calculate the molar quantity of the compound burned.
Moles of compound = Mass
Molar mass =3.50 g
102.10 g/mol
Moles of compound ≈0.0343 mol
Step 3: Calculate the enthalpy change for the combustion reaction.
∆H=−qcalorimeter
moles of compound
∆H=−540 J
0.0343 mol
∆H≈ −15765.77 J/mol ≈ −15.77 kJ/mol
Therefore, the enthalpy change for the combustion of the compound is ap-
proximately -15.77 kJ/mol.
5
Question 8
Question
A reaction at constant pressure has a change in enthalpy of −349 kJ. If 2.50
moles of the reactant are consumed, what is the change in enthalpy per mole of
reactant?
Solution
Step 1: Recall that the change in enthalpy (∆H) per mole of reactant can be
calculated using the formula:
∆Hper mole =∆H
moles of reactant
Step 2: Given that the change in enthalpy (∆H) is −349 kJ and the moles
of reactant is 2.50 mol, we can substitute these values into the formula:
∆Hper mole =−349 kJ
2.50 mol
Step 3: Calculate the change in enthalpy per mole of reactant:
∆Hper mole =−349
2.50 =−139.6 kJ/mol
Therefore, the change in enthalpy per mole of reactant is −139.6 kJ/mol.
Question 9
Question
Calculate the enthalpy change (∆H) for the reaction:
2H2(g)+O2(g)→H2O(l) given the following information:
∆H◦
ffor H2(g) = 0 kJ/mol
∆H◦
ffor O2(g) = 0 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate ∆Hfor the reaction using the standard enthalpies of
formation:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
6
Step 3: Substitute the given values into the equation:
∆H= [2(−285.8 kJ/mol)] −[2(0 kJ/mol) + 0 kJ/mol]
Step 4: Solve for ∆H:
∆H=−571.6 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H=−571.6 kJ/mol.
Question 10
Question
Calculate the change in enthalpy (∆H) when 50.0 g of water at 25
°
C is heated
to steam at 100
°
C. 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
18.015 g/mol.
Solution
Step 1: Calculate the heat required to heat the water from 25
°
C to 100
°
C. The
formula to calculate the heat required is:
q=mc∆T
Where: - qis the heat energy, - mis the mass of the substance, - cis the specific
heat capacity of the substance, - ∆Tis the change in temperature.
Substitute the values:
q= (50.0 g)(4.18 J/g
°
C)(100 −25)C
q= (50.0)(4.18)(75) J
Step 2: Calculate the heat required to vaporize the water. The heat required
to vaporize the water can be calculated using the formula:
q=n·∆Hvap
Where: - nis the number of moles of water, - ∆Hvap is the molar heat of
vaporization.
First, calculate the number of moles of water:
n=50.0 g
18.015 g/mol
Step 3: Convert the heat of vaporization to joules.
40.79 kJ/mol = 40.79 ×103J/mol
7
Now, calculate the heat required to vaporize the water:
q=50.0
18.015×(40.79 ×103) J
Step 4: Calculate the total change in enthalpy. The total change in enthalpy
is the sum of the heat required to raise the temperature and the heat required
to vaporize the water:
∆H=qheating +qvaporization
Question 11
Question
Given the reaction:
2A(g)+3B(g)→C(g)+4D(g)
where ∆H◦=−1258 kJ, calculate the enthalpy change when 1.5 moles of A
reacts with 2.0 moles of B.
Solution
Step 1: Calculate the moles of A and B reacting.
Moles of A = 1.5 mol
Moles of B = 2.0 mol
Step 2: Determine the limiting reactant.
Moles of A required for reaction = 2
2= 1.0 mol
Moles of B required for reaction = 3
2= 1.5 mol
Since A has fewer moles than required, A is the limiting reactant.
Step 3: Calculate the moles of C and D formed.
Since 2 moles of A produces 1 mole of C,
Moles of C formed = 1
2×1.5=0.75 mol
Since 2 moles of A produces 4 moles of D,
Moles of D formed = 2 ×1.5=3.0 mol
Step 4: Calculate the enthalpy change for the reaction.
∆H◦=−1258 kJ
For the reaction of 1 mole of A:
Enthalpy change = −1258
2=−629 kJ
For the reaction of 1.5 moles of A:
Enthalpy change = −629 ×1.5 = −943.5 kJ
Therefore, the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B
is −943.5 kJ .
8
Question 12
Question
A reaction that releases 400 kJ of heat causes a decrease in temperature of 0.5
°
C
in a closed container water at 25
°
C. Assuming the heat capacity of the container
is negligible and that the density of water is 1 g/cm3, calculate the enthalpy
change for the reaction.
Solution
Step 1: Calculate the mass of water affected by the heat exchange. Given that
the density of water is 1 g/cm3, the mass of water affected by the heat exchange
is:
Volume of water = Density ×Volume
Volume = Mass
Density
Volume = 1 g
1 g/cm3= 1 cm3
The volume of water affected by the heat exchange is 1 cm3, which is equivalent
to 1 gram.
Step 2: Calculate the change in temperature of the water. Given that the
temperature change is 0.5
°
C, the heat exchanged is:
q=mc∆T
q= (1 g)(4.18 J/g
°
C)(0.5C)=2.09 J
Step 3: Convert the heat exchanged to kilojoules. To convert Joules to
kilojoules, divide by 1000:
q=2.09 J
1000 = 0.00209 kJ
Step 4: Determine the enthalpy change for the reaction. The enthalpy change
for the reaction is equal in magnitude but opposite in sign to the heat exchanged:
∆H=−q=−0.00209 kJ = -2.09 kJ
Therefore, the enthalpy change for the reaction is -2.09 kJ.
Question 13
Question
For a reaction at constant pressure, the enthalpy change (∆H) is given by the
equation ∆H= 50 −2T, where Tis the temperature in Kelvin. Determine the
temperature at which the reaction becomes exothermic.
9
Solution
Step 1: To find the temperature at which the reaction becomes exothermic, set
∆Hto zero and solve for T.
∆H= 50 −2T= 0
50 = 2T
T= 25 K
Step 2: To determine whether the reaction is exothermic or endothermic at
T= 25 K, substitute T= 25 into the equation ∆H= 50 −2T.
∆H= 50 −2(25) = 50 −50 = 0
Since ∆H= 0 at T= 25 K, the reaction becomes exothermic at this tem-
perature.
Question 14
Question
A gas initially at 300 K and 1 atm pressure is expanded isothermally and re-
versibly to three times its initial volume. Calculate the change in enthalpy for
this process, given that the molar heat capacity at constant pressure, Cp, is 30
J/mol K.
Solution
Step 1: Calculate the work done during the expansion. Given that the process
is isothermal and reversible, the work done is −nRT ln(Vf/Vi), where nis the
number of moles of gas, Ris the ideal gas constant, Tis the temperature, Vfis
the final volume, and Viis the initial volume.
Step 2: Calculate the heat transfer. Since the process is isothermal, the heat
transfer is equal to the work done. Therefore, q=−w.
Step 3: Calculate the change in enthalpy. The change in enthalpy is given
by ∆H= ∆U+P∆V, where ∆Uis the change in internal energy, Pis the
pressure, and ∆Vis the change in volume. For an isothermal process, ∆U= 0,
so ∆H=P∆V.
Step 4: Substitute values and calculate. Given: - Initial temperature, Ti=
300 K - Initial pressure, Pi= 1 atm - Final volume, Vf= 3Vi- Molar heat
capacity at constant pressure, Cp= 30 J/mol K - Ideal gas constant, R= 8.314
J/mol K
From the ideal gas law, P V =nRT , we can find nand R. Then, we can
calculate the change in enthalpy using the formula ∆H=P∆V.
Finally, substitute the known values to find the change in enthalpy.
10
Question 15
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with the following enthalpy changes:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
calculate the standard enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy of formation values provided.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
∆H◦=2·∆H◦
f(H2O(l))−2·∆H◦
f(H2(g)) + ∆H◦
f(O2(g))
∆H◦= 2(−285.8 kJ/mol) −2(0 kJ/mol) −0 kJ/mol
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is −571.6 kJ/mol .
Question 16
Question
Given the following reaction:
2A(g)+3B(g)−→ C(g) + D(g)
If the standard enthalpy of formation for each substance is as follows:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −500 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction.
11
Solution
To calculate the standard enthalpy change for the reaction, we can use Hess’s
Law, which states that the total enthalpy change for a reaction is the same
regardless of the number of steps in the reaction or the pathway taken.
Step 1: Write the given reaction in terms of standard enthalpies
of formation.
The given reaction is:
2A(g)+3B(g)−→ C(g) + D(g)
The standard enthalpy change for this reaction can be expressed as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the standard enthalpy change for the reaction.
Given that:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −500 kJ/mol
The standard enthalpy change can be calculated as follows:
∆H◦= [(∆H◦
f(C)+∆H◦
f(D)) −(∆H◦
f(2A)+∆H◦
f(3B))]
∆H◦= [(−400 kJ/mol+(−500 kJ/mol))−((2×−200 kJ/mol)+(3×−300 kJ/mol))]
∆H◦= [−900 kJ/mol −(−200 kJ/mol + −900 kJ/mol)]
∆H◦=−900 kJ/mol + 1100 kJ/mol
∆H◦= 200 kJ/mol
Question 17
Question
A reaction has an enthalpy change of ∆H=−285.8 kJ. If 4.96 moles of the
reactant undergo the reaction, what is the total heat absorbed or released during
the reaction?
12
Solution
Step 1: Determine the total heat absorbed or released per mole of the reactant.
Given: ∆H=−285.8 kJ
We know that the total heat absorbed or released during the reaction can
be calculated as:
Total heat = ∆H×moles of reactant
Step 2: Calculate the total heat absorbed or released.
Plugging in the values:
Total heat = −285.8 kJ ×4.96 moles
Total heat = −1416.688 kJ
Therefore, the total heat absorbed or released during the reaction is -1416.688
kJ.
Question 18
Question
Calculate the change in enthalpy (∆H) when 35.0 g of liquid water at 25◦C is
converted to steam at 100◦C. The specific heat capacity of water is 4.18 J/(g·
°
C),
the molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar mass
of water is 18.015 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of the water from
25◦C to 100◦C using the formula:
q1=m×c×∆T
where - mis the mass of water (35.0 g), - cis the specific heat capacity of water
(4.18 J/(g·
°
C)), - ∆Tis the change in temperature (100◦C - 25◦C).
Step 2: Convert the heat calculated in Step 1 to kilojoules (kJ) by dividing
by 1000.
Step 3: Calculate the moles of water in 35.0 g using the molar mass of water.
Step 4: Calculate the heat required to convert the water to steam at 100◦C
using the formula:
q2=n×∆Hvap
where - nis the number of moles of water calculated in Step 3, - ∆Hvap is the
molar enthalpy of vaporization of water (40.79 kJ/mol).
Step 5: Add the heats calculated in Step 2 and Step 4 to find the total heat
change, ∆H.
13
∆H=q1+q2
Question 19
Question
Given the balanced chemical equation for the combustion of ethane:
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
Calculate the standard enthalpy change (∆H◦) for this reaction. Given the
following standard enthalpy of formation values:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.51 kJ/mol
∆H◦
f(H2O) = −285.83 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of formation for the products using the
given values.
∆H◦
f(2CO2) = 2(−393.51 kJ/mol) = −787.02 kJ/mol
∆H◦
f(3H2O) = 3(−285.83 kJ/mol) = −857.49 kJ/mol
Step 2: Calculate the standard enthalpy of formation for the reactants using
the given value.
∆H◦
f(C2H6) = −84.68 kJ/mol
Step 3: Calculate the net standard enthalpy change for the reaction using
the formula:
∆H◦= (X∆H◦
fproducts) −(X∆H◦
freactants)
∆H◦= [(−787.02 kJ/mol) + (−857.49 kJ/mol)] −(−84.68 kJ/mol)
∆H◦=−1644.51 kJ/mol + 84.68 kJ/mol = −1559.83 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the combustion of ethane
is -1559.83 kJ/mol.
14
Question 20
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas (N2)
react according to the following balanced chemical equation:
N2(g)+3H2(g)→2NH3(g)
Given the enthalpies of formation (∆H◦
f) for N2(g), H2(g), and NH3(g) are
0 kJ/mol, 0 kJ/mol, and -46.2 kJ/mol, respectively.
Solution
Step 1: Write the given balanced chemical equation and the enthalpy change
for the reaction.
N2(g)+3H2(g)→2NH3(g)
The enthalpy change for the reaction is given by:
∆H= Σ∆H◦
f(products) −Σ∆H◦
f(reactants)
Step 2: Calculate the enthalpy change for the reaction. The enthalpies of
formation are:
∆H◦
f(NH3) = −46.2 kJ/mol
∆H◦
f(N2) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Substitute the values into the equation for ∆H:
∆H=2×∆H◦
f(NH3)−∆H◦
f(N2)+3×∆H◦
f(H2)
∆H= (2 × −46.2 kJ/mol) −(0 + 3 ×0 kJ/mol)
∆H=−92.4 kJ/mol
Step 3: Calculate the change in enthalpy when 5.00 moles of N2react. Since
the enthalpy change is given per mole, we need to calculate the total enthalpy
change for 5 moles of N2:
Change in enthalpy = moles ×∆H
Change in enthalpy = 5.00 moles × −92.4 kJ/mol
Change in enthalpy = −462 kJ
Therefore, the change in enthalpy when 5.00 moles of N2react is -462 kJ.
15
Question 21
Question
A reaction is thought to have a certain enthalpy change. By performing a
series of measurements, it is found that the reaction has an enthalpy change of
−289.2 kJ when carried out at 25
°
C. Assuming that the heat capacities of all
substances are constant over the temperature range in question, calculate the
enthalpy change for the reaction at 55
°
C.
Solution
Step 1: Determine the heat capacity change (∆Cp) of the reaction.
∆Hreaction = ∆H25C
reaction +Z55C
25C
∆CpdT
∆Hreaction =−289.2 kJ + Z55C
25C
∆CpdT
Step 2: Since the heat capacities are constant, we can consider ∆Cpas a
constant value, say Cp.
∆Hreaction =−289.2 kJ + CpZ55C
25C
dT
∆Hreaction =−289.2 kJ + Cp·(55 −25) K
∆Hreaction =−289.2 kJ + 30Cp
Step 3: Given that the enthalpy change of the reaction at 25
°
C is -289.2 kJ:
−289.2 = −289.2 + 30Cp
0 = 30Cp=⇒Cp= 0
Step 4: The assumption that the heat capacities are constant is incorrect. We
must therefore consider the correct dependence of heat capacity on temperature
for the substances involved.
Question 22
Question
A reaction is carried out in a bomb calorimeter, where 1.50 mol of a gas is burned
at 1.00 atm pressure in a vessel with a volume of 10.0 L at a temperature of
298 K. The calorimeter itself has a heat capacity of 750 J/K. If the temperature
rises to 316 K during the reaction, calculate the change in enthalpy (∆H) for
the reaction.
Given: Gas constant, R= 8.314 J/(mol
·
K)
16
Solution
Step 1: Calculate the work done by the gas during the reaction. The work done
by the gas is given by the equation:
W=−Pext∆V
Where: Pext = external pressure = 1.00 atm = 1.013×105Pa ∆V= change
in volume = 10.0 L Convert 10.0 L to m
³
:
10.0L= 10.0×10−3m3= 0.0100m3
Now, calculate the work done:
W=−(1.013 ×105P a)(0.0100m3) = −1013J
Step 2: Calculate the heat absorbed by the calorimeter. The heat absorbed
by the calorimeter is given by the equation:
qcal =C∆T
Where: C= heat capacity of the calorimeter = 750 J/K ∆T= change in
temperature = 316 K - 298 K = 18 K
Now, calculate the heat absorbed by the calorimeter:
qcal = (750J/K)(18K) = 13500J
Step 3: Calculate the heat of the reaction. The heat absorbed by the reaction
is given by the equation:
qrxn =qcal −W
Now, calculate the heat of the reaction:
qrxn = 13500J−(−1013J) = 14513J
Step 4: Calculate the change in enthalpy. Since ∆H=qrxn under constant
pressure conditions, the change in enthalpy for the reaction is ∆H= 14513 J.
Therefore, the change in enthalpy for the reaction is 14513 J.
Question 23
Question
A reaction is known to have an enthalpy change of ∆H=−250 kJ/mol. If the
reaction is exothermic, in which direction will the enthalpy change (∆H) be
when the reaction is reversed?
17
Solution
1. When a reaction is reversed, the sign of the enthalpy change is also reversed.
Therefore, if the original reaction has ∆H=−250 kJ/mol, the reversed reaction
will have ∆H= +250 kJ/mol.
2. This change in sign occurs because when a reaction is reversed, the
products become the reactants and vice versa. As a result, the direction of heat
flow (exothermic or endothermic) also changes.
Thus, the enthalpy change for the reversed reaction will be +250 kJ/mol.
Question 24
Question
An ideal gas undergoes a process at constant pressure. The initial temperature
of the gas is 300 K and the final temperature is 400 K. Calculate the change in
enthalpy of the gas during the process.
Solution
Step 1: Recall the definition of enthalpy in terms of internal energy and pressure:
∆H= ∆U+P∆V
Step 2: Since the process is at constant pressure, ∆U=Q−Wwhere Q
is the heat added to the system and Wis the work done by the gas on its
surroundings.
Step 3: We can rewrite the expression for enthalpy change as:
∆H= (Q−W) + P∆V
Step 4: The work done by the gas can be expressed as W=P∆V. Substi-
tuting this into the equation gives:
∆H=Q
Step 5: To calculate the heat added to the system, we can use the formula:
Q=nCp∆T
where nis the number of moles of gas and Cpis the molar heat capacity at
constant pressure.
Step 6: First, we need to find the number of moles of the gas. We can use
the ideal gas law:
P V =nRT
Solving for n, we have:
n=P V
RT
18
Step 7: Substituting ninto the formula for heat gives:
Q=P V
RT Cp∆T
Step 8: Substituting the given values P= constant, V= constant, R=
constant, T1= 300 K, T2= 400 K and Cp= specific heat at constant pressure
into the equation gives the change in enthalpy.
Question 25
Question
A reaction between hydrogen gas and oxygen gas to form water vapor has an
enthalpy change of -483.6 kJ. If 8.00 moles of hydrogen gas react, how much
heat is absorbed or released?
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g)→2H2O(g)
Step 2: Determine the moles of oxygen gas reacting, based on the stoichiom-
etry of the balanced equation. Since the balanced equation states that 1 mole
of oxygen gas reacts with 2 moles of hydrogen gas, the moles of oxygen gas
required will be half of the moles of hydrogen gas:
moles of O2=1
2×8.00 moles = 4.00 moles
Step 3: Calculate the heat absorbed or released using the enthalpy change
and the moles of the limiting reactant. Given enthalpy change: ∆H=−483.6 kJ
Since the equation is for 2 moles of hydrogen gas reacting, the heat change for
1 mole of hydrogen gas will be half of the given value:
Heat change for 1 mole of H2=−483.6 kJ
2=−241.8 kJ/mole
Therefore, the heat change for 8.00 moles of hydrogen gas will be:
Heat change = −241.8 kJ/mole ×8.00 moles = −1934.4 kJ
So, 1934.4 kJ of heat is released in this reaction.
Question 26
Question
A reaction has an enthalpy change of ∆H=−185 kJ/mol. If 2.50 moles of the
reaction takes place, calculate the total enthalpy change for the reaction.
19
Solution
Step 1: Determine the total enthalpy change for the reaction per mole. Given:
∆H=−185 kJ/mol
Step 2: Calculate the total enthalpy change for the reaction when 2.50 moles
of the reaction takes place. Total Enthalpy Change = ∆H×Number of moles
Total Enthalpy Change = −185 kJ/mol ×2.50 mol Total Enthalpy Change =
−185 kJ/mol ×2.50 = −462.5 kJ
Therefore, the total enthalpy change for the reaction when 2.50 moles of the
reaction takes place is −462.5kJ.
Question 27
Question
Calculate the change in enthalpy when 2 moles of water vapor at 100
°
C and 1
atm pressure is condensed to liquid water at the same temperature and pressure.
Given that the heat of vaporization of water is 40.79 kJ/mol.
Solution
Step 1: Determine the change in enthalpy during the phase change from vapor
to liquid. For 2 moles of water vapor being condensed to liquid water, the
change in enthalpy can be calculated using the heat of vaporization formula:
Change in Enthalpy = moles ×heat of vaporization
Change in Enthalpy = 2 moles ×40.79 kJ/mol
Change in Enthalpy = 81.58 kJ
Therefore, the change in enthalpy during the phase change is 81.58 kJ.
Step 2: Account for the temperature change. Since the temperature remains
constant during the phase change, there is no additional change in enthalpy due
to the temperature change.
Step 3: Calculate the total change in enthalpy. The total change in enthalpy
is the sum of the change in enthalpy during the phase change and any changes
due to temperature:
Total Change in Enthalpy = Change in Enthalpy (phase change)+Change in Enthalpy (temperature change)
Total Change in Enthalpy = 81.58 kJ + 0 kJ
Total Change in Enthalpy = 81.58 kJ
20
Therefore, the total change in enthalpy when 2 moles of water vapor at 100
°
C
and 1 atm pressure is condensed to liquid water at the same temperature and
pressure is 81.58 kJ.
Question 28
Question
A gas initially at a pressure of 10 atm and a volume of 5 L is expanded at
constant temperature to a final volume of 25 L. If the enthalpy change during
the process is 1500 J, calculate the final pressure of the gas.
Solution
Step 1: Calculate the initial number of moles of gas using the ideal gas law,
P V =nRT .
n=P V
RT
Given P= 10 atm, V= 5 L, R= 0.0821 L atm mol−1K−1, and Tis constant
for the process.
Substitute the given values to find n.
n=(10 atm)(5 L)
(0.0821 L atm mol−1K−1)(T)=50
0.0821T=610
T
Step 2: Calculate the final pressure of the gas using the enthalpy change and
the number of moles.
∆H=nCp∆T
Given ∆H= 1500 J and n= 610/T , where Tis the gas temperature for the
process, and Cpis the molar heat capacity at constant pressure.
Since the process is at constant temperature, ∆T= 0 and nCp∆T= 0.
Thus, the enthalpy change during the process does not depend on the molar
heat capacity at constant pressure.
So, ∆H=nCp∆T= 0.
Since ∆H=nCp∆T= 0, nCp∆T= 1500 J implies nCp= 1500 J.
Substitute n= 610/T into nCp= 1500 J and simplify.
610Cp
T= 1500
Step 3: Calculate the final pressure.
P V =nRT
At the final state, PfinalVfinal =nRT . Given Vfinal = 25 L, n= 610/T , and
R= 0.0821 L atm mol−1K−1.
21
Substitute the values to find the final pressure.
Pfinal ·25 = 610
T·0.0821 ·T
25Pfinal = 50.31Pfinal =50.31
25
Therefore, the final pressure of the gas is 2.012 atm.
Question 29
Question
A certain reaction has an enthalpy change of ∆H=−128 kJ/mol at 25
°
C. If
the reaction is exothermic, calculate the standard enthalpy change at 298 K.
Solution
Step 1: Calculate the change in enthalpy due to the change in temperature from
25
°
C to 298 K.
∆H298 = ∆H25 +Z298
25
CpdT
where Cpis the molar heat capacity at constant pressure.
Step 2: Use the heat capacity values and the relationship Cp=dH
dT pto
calculate the integral.
Z298
25
CpdT =Z298
25 dH
dT p
dT
=Z298
25 128
298 −25dT
Step 3: Perform the integration and calculate the new enthalpy change at
298 K.
= 128 ×ln 298
25 kJ/mol
∆H298 =−126.39 kJ/mol
Therefore, the standard enthalpy change at 298 K is ∆H298 =−126.39
kJ/mol.
22
Question 30
Question
Calculate the change in enthalpy for the reaction
2H2(g) + O2(g)→2H2O(l)
given the following bond energies: H−H= 436 kJ/mol, O=O= 498 kJ/mol,
H−O= 463 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Step 2: Calculate the total energy released when the new bonds are formed in
the products. Step 3: Find the difference between the energy required to break
the bonds in the reactants and the energy released when the new bonds are
formed in the products.
Step 1: The total energy required to break the bonds in the reactants is:
2×(H−H)+1×(O=O) = 2 ×436 + 1 ×498 = 1370 kJ/mol
Step 2: The total energy released when the new bonds are formed in the
products is:
2×(H−O) = 2 ×463 = 926 kJ/mol
Step 3: The change in enthalpy for the reaction is given by:
∆H= (Energy required to break the bonds)−(Energy released when new bonds are formed)
∆H= 1370 −926 = 444 kJ/mol
Therefore, the change in enthalpy for the reaction is 444 kJ/mol.
Question 31
Question
Given the following reaction involving the combustion of ethane gas:
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
If the standard enthalpies of formation are −84.7 kJ/mol for CO2and −285.8 kJ/mol
for H2O, and the heating value of ethane gas is −1558.0 kJ/mol, calculate the
standard enthalpy of formation for ethane.
23
Solution
Step 1: Calculate the standard enthalpy change of the reaction using the given
standard enthalpies of formation.
∆H◦
rxn =Xν∆H◦
f,products −Xν∆H◦
f,reactants
Given data:
∆H◦
f(CO2) = −84.7 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
rxn =−1558.0 kJ/mol
∆H◦
rxn = 2(−84.7) + 3(−285.8) −(−1558.0)
∆H◦
rxn =−169.4 kJ/mol
Step 2: Write the standard enthalpy change of the reaction using the en-
thalpy of formation of ethane.
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
∆H◦
rxn = 2∆H◦
f(CO2) + 3∆H◦
f(H2O)−∆H◦
f(C2H6)
Step 3: Solve for the standard enthalpy of formation of ethane.
−169.4 = 2(−84.7) + 3(−285.8) −∆H◦
f(C2H6)
−169.4 = −169.4−∆H◦
f(C2H6)
∆H◦
f(C2H6) = 0 kJ/mol
Therefore, the standard enthalpy of formation of ethane is 0 kJ/mol.
Question 32
Question
A sample of an ideal gas undergoes a process where the pressure changes from
2 atm to 6 atm while the volume changes from 8 L to 2 L. If the initial and
final temperatures are 300 K and 500 K respectively, calculate the change in
enthalpy of the gas during this process.
24
Solution
Step 1: Calculate the initial and final internal energies of the gas using the first
law of thermodynamics:
∆U=Q−W
where ∆Uis the change in internal energy, Q is the heat added to the gas, and
W is the work done by the gas.
Given that the gas undergoes an isobaric process, the work done can be
calculated as:
W=−Pext ·∆V
Since the gas is ideal, we can use the ideal gas law to express Pext:
W=−Pext ·∆V=−nRT ln Vf
Vi
Substitute the given values:
W=−nRT ln 2
8=−nRT ln 1
4=−nRT ln(0.25)
∆U=Q−W=Q+nRT ln(0.25)
∆U=nCV∆T
where CVis the molar heat capacity at constant volume. Since the gas is ideal,
CV=f
2Rwhere f is the degrees of freedom (f=3 for monoatomic gas).
Substitute and rearrange to solve for Q:
Q= ∆U−nR∆Tln(0.25) = n(f
2R)∆T−nR∆Tln(0.25)
Q=nR∆T(f
2−ln(0.25)) = nR∆T(3
2−ln(0.25))
Step 2: Calculate the change in enthalpy, ∆H, of the gas using the equation:
∆H= ∆U+ ∆(P V )
Given that the process is isobaric, substitute the change in internal energy
and pressure-volume work:
∆H=Q+W+P∆V
Substitute the expressions for Q and W calculated earlier:
∆H=nR∆T(3
2−ln(0.25)) + nR∆Tln(0.25) + Pext∆V
25
Given that P∆V=nR∆Tfor an isobaric process:
∆H=nR∆T(3
2−ln(0.25)) + nR∆T+nR∆T
∆H=nR∆T(5
2−ln(0.25))
Now, substitute the given values for n, R, and ∆Tto find the change in
enthalpy.
Question 33
Question
Calculate the standard enthalpy change for the reaction below at 298 K:
2CH4(g)+4O2(g)→2CO2(g)+4H2O(l)
Given the standard enthalpies of formation (∆H◦
f) for CH4(g), CO2(g), and
H2O(l) are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where n and m are the stoichiometric coefficients of the products and reactants,
respectively.
Step 2: Substitute the given values and calculate:
∆H◦= 2×∆H◦
f(CO2(g))+4×∆H◦
f(H2O(l))−[2×∆H◦
f(CH4(g))+4×∆H◦
f(O2(g))]
Step 3: Plug in the values and calculate:
∆H◦= 2 ×(−393.5) + 4 ×(−285.8) −[2 ×(−74.8) + 4 ×0]
∆H◦=−787.0+(−1143.2) −(−149.6)
∆H◦=−787.0−1143.2 + 149.6
∆H◦=−1780.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 298 K is
-1780.6 kJ/mol.
26
Question 34
Question
Calculate the change in enthalpy (∆H) when 10.0 moles of steam at 100◦C is
cooled to water at 25◦C. Given that the heat capacity of steam is 2.08 J/g·K,
the heat capacity of water is 4.18 J/g·K, and the heat of vaporization for water
is 40.79 kJ/mol.
Solution
Step 1: Calculate the heat required to cool the steam to 100◦C to water at
100◦C. The heat capacity of steam is 2.08 J/g·K. Given that 1 mole of water
has a molar mass of 18 g/mol and 1 mole of steam has a molar mass of 18 g/mol
+ 18 g/mol = 36 g/mol. Therefore, the heat capacity of steam is 2.08 J/g·K *
36 g/mol = 74.88 J/mol·K.
The change in temperature is 100◦C - 25◦C = 75◦C = 75 K. Therefore, the
heat required to cool 10.0 moles of steam to water at 100◦C is: 10.0 mol ×
75 K ×74.88 J/mol ·K = 56160 J.
Step 2: Calculate the heat required to condense the water at 100◦C to water
at 25◦C. The heat of vaporization for water is 40.79 kJ/mol = 40790 J/mol.
Therefore, the heat required to condense 10.0 moles of steam at 100◦C to water
at 100◦C is: 10.0 mol ×40790 J/mol = 407900 J.
Step 3: Add the heats from Step 1 and Step 2 to find the total change
in enthalpy. ∆H= 56160 J + 407900 J = 464060 J. Therefore, the change in
enthalpy when 10.0 moles of steam at 100◦C is cooled to water at 25◦C is 464060
J.
Question 35
Question
An experiment involves the mixing of 150 g of water at 25
°
C with 300 g of water
at 85
°
C in a thermally insulated container. Calculate the final temperature of
the mixture. Assume specific heat capacity of water is 4.18 J/g
°
C and neglect
any heat losses to the surroundings.
Solution
Step 1: Calculate the heat lost by the hot water and the heat gained by the
cold water.
The heat lost by the hot water is given by:
qhot =mhot ·c·∆T
27
Question 8
Question
A reaction at constant pressure has a change in enthalpy of −349 kJ. If 2.50
moles of the reactant are consumed, what is the change in enthalpy per mole of
reactant?
Solution
Step 1: Recall that the change in enthalpy (∆H) per mole of reactant can be
calculated using the formula:
∆Hper mole =∆H
moles of reactant
Step 2: Given that the change in enthalpy (∆H) is −349 kJ and the moles
of reactant is 2.50 mol, we can substitute these values into the formula:
∆Hper mole =−349 kJ
2.50 mol
Step 3: Calculate the change in enthalpy per mole of reactant:
∆Hper mole =−349
2.50 =−139.6 kJ/mol
Therefore, the change in enthalpy per mole of reactant is −139.6 kJ/mol.
Question 9
Question
Calculate the enthalpy change (∆H) for the reaction:
2H2(g)+O2(g)→H2O(l) given the following information:
∆H◦
ffor H2(g) = 0 kJ/mol
∆H◦
ffor O2(g) = 0 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate ∆Hfor the reaction using the standard enthalpies of
formation:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
6
Step 3: Substitute the given values into the equation:
∆H= [2(−285.8 kJ/mol)] −[2(0 kJ/mol) + 0 kJ/mol]
Step 4: Solve for ∆H:
∆H=−571.6 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H=−571.6 kJ/mol.
Question 10
Question
Calculate the change in enthalpy (∆H) when 50.0 g of water at 25
°
C is heated
to steam at 100
°
C. 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
18.015 g/mol.
Solution
Step 1: Calculate the heat required to heat the water from 25
°
C to 100
°
C. The
formula to calculate the heat required is:
q=mc∆T
Where: - qis the heat energy, - mis the mass of the substance, - cis the specific
heat capacity of the substance, - ∆Tis the change in temperature.
Substitute the values:
q= (50.0 g)(4.18 J/g
°
C)(100 −25)C
q= (50.0)(4.18)(75) J
Step 2: Calculate the heat required to vaporize the water. The heat required
to vaporize the water can be calculated using the formula:
q=n·∆Hvap
Where: - nis the number of moles of water, - ∆Hvap is the molar heat of
vaporization.
First, calculate the number of moles of water:
n=50.0 g
18.015 g/mol
Step 3: Convert the heat of vaporization to joules.
40.79 kJ/mol = 40.79 ×103J/mol
7
Now, calculate the heat required to vaporize the water:
q=50.0
18.015×(40.79 ×103) J
Step 4: Calculate the total change in enthalpy. The total change in enthalpy
is the sum of the heat required to raise the temperature and the heat required
to vaporize the water:
∆H=qheating +qvaporization
Question 11
Question
Given the reaction:
2A(g)+3B(g)→C(g)+4D(g)
where ∆H◦=−1258 kJ, calculate the enthalpy change when 1.5 moles of A
reacts with 2.0 moles of B.
Solution
Step 1: Calculate the moles of A and B reacting.
Moles of A = 1.5 mol
Moles of B = 2.0 mol
Step 2: Determine the limiting reactant.
Moles of A required for reaction = 2
2= 1.0 mol
Moles of B required for reaction = 3
2= 1.5 mol
Since A has fewer moles than required, A is the limiting reactant.
Step 3: Calculate the moles of C and D formed.
Since 2 moles of A produces 1 mole of C,
Moles of C formed = 1
2×1.5=0.75 mol
Since 2 moles of A produces 4 moles of D,
Moles of D formed = 2 ×1.5=3.0 mol
Step 4: Calculate the enthalpy change for the reaction.
∆H◦=−1258 kJ
For the reaction of 1 mole of A:
Enthalpy change = −1258
2=−629 kJ
For the reaction of 1.5 moles of A:
Enthalpy change = −629 ×1.5 = −943.5 kJ
Therefore, the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B
is −943.5 kJ .
8
Question 12
Question
A reaction that releases 400 kJ of heat causes a decrease in temperature of 0.5
°
C
in a closed container water at 25
°
C. Assuming the heat capacity of the container
is negligible and that the density of water is 1 g/cm3, calculate the enthalpy
change for the reaction.
Solution
Step 1: Calculate the mass of water affected by the heat exchange. Given that
the density of water is 1 g/cm3, the mass of water affected by the heat exchange
is:
Volume of water = Density ×Volume
Volume = Mass
Density
Volume = 1 g
1 g/cm3= 1 cm3
The volume of water affected by the heat exchange is 1 cm3, which is equivalent
to 1 gram.
Step 2: Calculate the change in temperature of the water. Given that the
temperature change is 0.5
°
C, the heat exchanged is:
q=mc∆T
q= (1 g)(4.18 J/g
°
C)(0.5C)=2.09 J
Step 3: Convert the heat exchanged to kilojoules. To convert Joules to
kilojoules, divide by 1000:
q=2.09 J
1000 = 0.00209 kJ
Step 4: Determine the enthalpy change for the reaction. The enthalpy change
for the reaction is equal in magnitude but opposite in sign to the heat exchanged:
∆H=−q=−0.00209 kJ = -2.09 kJ
Therefore, the enthalpy change for the reaction is -2.09 kJ.
Question 13
Question
For a reaction at constant pressure, the enthalpy change (∆H) is given by the
equation ∆H= 50 −2T, where Tis the temperature in Kelvin. Determine the
temperature at which the reaction becomes exothermic.
9
Solution
Step 1: To find the temperature at which the reaction becomes exothermic, set
∆Hto zero and solve for T.
∆H= 50 −2T= 0
50 = 2T
T= 25 K
Step 2: To determine whether the reaction is exothermic or endothermic at
T= 25 K, substitute T= 25 into the equation ∆H= 50 −2T.
∆H= 50 −2(25) = 50 −50 = 0
Since ∆H= 0 at T= 25 K, the reaction becomes exothermic at this tem-
perature.
Question 14
Question
A gas initially at 300 K and 1 atm pressure is expanded isothermally and re-
versibly to three times its initial volume. Calculate the change in enthalpy for
this process, given that the molar heat capacity at constant pressure, Cp, is 30
J/mol K.
Solution
Step 1: Calculate the work done during the expansion. Given that the process
is isothermal and reversible, the work done is −nRT ln(Vf/Vi), where nis the
number of moles of gas, Ris the ideal gas constant, Tis the temperature, Vfis
the final volume, and Viis the initial volume.
Step 2: Calculate the heat transfer. Since the process is isothermal, the heat
transfer is equal to the work done. Therefore, q=−w.
Step 3: Calculate the change in enthalpy. The change in enthalpy is given
by ∆H= ∆U+P∆V, where ∆Uis the change in internal energy, Pis the
pressure, and ∆Vis the change in volume. For an isothermal process, ∆U= 0,
so ∆H=P∆V.
Step 4: Substitute values and calculate. Given: - Initial temperature, Ti=
300 K - Initial pressure, Pi= 1 atm - Final volume, Vf= 3Vi- Molar heat
capacity at constant pressure, Cp= 30 J/mol K - Ideal gas constant, R= 8.314
J/mol K
From the ideal gas law, P V =nRT , we can find nand R. Then, we can
calculate the change in enthalpy using the formula ∆H=P∆V.
Finally, substitute the known values to find the change in enthalpy.
10
Question 15
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with the following enthalpy changes:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
calculate the standard enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy of formation values provided.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
∆H◦=2·∆H◦
f(H2O(l))−2·∆H◦
f(H2(g)) + ∆H◦
f(O2(g))
∆H◦= 2(−285.8 kJ/mol) −2(0 kJ/mol) −0 kJ/mol
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is −571.6 kJ/mol .
Question 16
Question
Given the following reaction:
2A(g)+3B(g)−→ C(g) + D(g)
If the standard enthalpy of formation for each substance is as follows:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −500 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction.
11
Solution
To calculate the standard enthalpy change for the reaction, we can use Hess’s
Law, which states that the total enthalpy change for a reaction is the same
regardless of the number of steps in the reaction or the pathway taken.
Step 1: Write the given reaction in terms of standard enthalpies
of formation.
The given reaction is:
2A(g)+3B(g)−→ C(g) + D(g)
The standard enthalpy change for this reaction can be expressed as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the standard enthalpy change for the reaction.
Given that:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −500 kJ/mol
The standard enthalpy change can be calculated as follows:
∆H◦= [(∆H◦
f(C)+∆H◦
f(D)) −(∆H◦
f(2A)+∆H◦
f(3B))]
∆H◦= [(−400 kJ/mol+(−500 kJ/mol))−((2×−200 kJ/mol)+(3×−300 kJ/mol))]
∆H◦= [−900 kJ/mol −(−200 kJ/mol + −900 kJ/mol)]
∆H◦=−900 kJ/mol + 1100 kJ/mol
∆H◦= 200 kJ/mol
Question 17
Question
A reaction has an enthalpy change of ∆H=−285.8 kJ. If 4.96 moles of the
reactant undergo the reaction, what is the total heat absorbed or released during
the reaction?
12
Solution
Step 1: Determine the total heat absorbed or released per mole of the reactant.
Given: ∆H=−285.8 kJ
We know that the total heat absorbed or released during the reaction can
be calculated as:
Total heat = ∆H×moles of reactant
Step 2: Calculate the total heat absorbed or released.
Plugging in the values:
Total heat = −285.8 kJ ×4.96 moles
Total heat = −1416.688 kJ
Therefore, the total heat absorbed or released during the reaction is -1416.688
kJ.
Question 18
Question
Calculate the change in enthalpy (∆H) when 35.0 g of liquid water at 25◦C is
converted to steam at 100◦C. The specific heat capacity of water is 4.18 J/(g·
°
C),
the molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar mass
of water is 18.015 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of the water from
25◦C to 100◦C using the formula:
q1=m×c×∆T
where - mis the mass of water (35.0 g), - cis the specific heat capacity of water
(4.18 J/(g·
°
C)), - ∆Tis the change in temperature (100◦C - 25◦C).
Step 2: Convert the heat calculated in Step 1 to kilojoules (kJ) by dividing
by 1000.
Step 3: Calculate the moles of water in 35.0 g using the molar mass of water.
Step 4: Calculate the heat required to convert the water to steam at 100◦C
using the formula:
q2=n×∆Hvap
where - nis the number of moles of water calculated in Step 3, - ∆Hvap is the
molar enthalpy of vaporization of water (40.79 kJ/mol).
Step 5: Add the heats calculated in Step 2 and Step 4 to find the total heat
change, ∆H.
13
∆H=q1+q2
Question 19
Question
Given the balanced chemical equation for the combustion of ethane:
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
Calculate the standard enthalpy change (∆H◦) for this reaction. Given the
following standard enthalpy of formation values:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.51 kJ/mol
∆H◦
f(H2O) = −285.83 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of formation for the products using the
given values.
∆H◦
f(2CO2) = 2(−393.51 kJ/mol) = −787.02 kJ/mol
∆H◦
f(3H2O) = 3(−285.83 kJ/mol) = −857.49 kJ/mol
Step 2: Calculate the standard enthalpy of formation for the reactants using
the given value.
∆H◦
f(C2H6) = −84.68 kJ/mol
Step 3: Calculate the net standard enthalpy change for the reaction using
the formula:
∆H◦= (X∆H◦
fproducts) −(X∆H◦
freactants)
∆H◦= [(−787.02 kJ/mol) + (−857.49 kJ/mol)] −(−84.68 kJ/mol)
∆H◦=−1644.51 kJ/mol + 84.68 kJ/mol = −1559.83 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the combustion of ethane
is -1559.83 kJ/mol.
14
Question 20
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas (N2)
react according to the following balanced chemical equation:
N2(g)+3H2(g)→2NH3(g)
Given the enthalpies of formation (∆H◦
f) for N2(g), H2(g), and NH3(g) are
0 kJ/mol, 0 kJ/mol, and -46.2 kJ/mol, respectively.
Solution
Step 1: Write the given balanced chemical equation and the enthalpy change
for the reaction.
N2(g)+3H2(g)→2NH3(g)
The enthalpy change for the reaction is given by:
∆H= Σ∆H◦
f(products) −Σ∆H◦
f(reactants)
Step 2: Calculate the enthalpy change for the reaction. The enthalpies of
formation are:
∆H◦
f(NH3) = −46.2 kJ/mol
∆H◦
f(N2) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Substitute the values into the equation for ∆H:
∆H=2×∆H◦
f(NH3)−∆H◦
f(N2)+3×∆H◦
f(H2)
∆H= (2 × −46.2 kJ/mol) −(0 + 3 ×0 kJ/mol)
∆H=−92.4 kJ/mol
Step 3: Calculate the change in enthalpy when 5.00 moles of N2react. Since
the enthalpy change is given per mole, we need to calculate the total enthalpy
change for 5 moles of N2:
Change in enthalpy = moles ×∆H
Change in enthalpy = 5.00 moles × −92.4 kJ/mol
Change in enthalpy = −462 kJ
Therefore, the change in enthalpy when 5.00 moles of N2react is -462 kJ.
15
Question 21
Question
A reaction is thought to have a certain enthalpy change. By performing a
series of measurements, it is found that the reaction has an enthalpy change of
−289.2 kJ when carried out at 25
°
C. Assuming that the heat capacities of all
substances are constant over the temperature range in question, calculate the
enthalpy change for the reaction at 55
°
C.
Solution
Step 1: Determine the heat capacity change (∆Cp) of the reaction.
∆Hreaction = ∆H25C
reaction +Z55C
25C
∆CpdT
∆Hreaction =−289.2 kJ + Z55C
25C
∆CpdT
Step 2: Since the heat capacities are constant, we can consider ∆Cpas a
constant value, say Cp.
∆Hreaction =−289.2 kJ + CpZ55C
25C
dT
∆Hreaction =−289.2 kJ + Cp·(55 −25) K
∆Hreaction =−289.2 kJ + 30Cp
Step 3: Given that the enthalpy change of the reaction at 25
°
C is -289.2 kJ:
−289.2 = −289.2 + 30Cp
0 = 30Cp=⇒Cp= 0
Step 4: The assumption that the heat capacities are constant is incorrect. We
must therefore consider the correct dependence of heat capacity on temperature
for the substances involved.
Question 22
Question
A reaction is carried out in a bomb calorimeter, where 1.50 mol of a gas is burned
at 1.00 atm pressure in a vessel with a volume of 10.0 L at a temperature of
298 K. The calorimeter itself has a heat capacity of 750 J/K. If the temperature
rises to 316 K during the reaction, calculate the change in enthalpy (∆H) for
the reaction.
Given: Gas constant, R= 8.314 J/(mol
·
K)
16
Solution
Step 1: Calculate the work done by the gas during the reaction. The work done
by the gas is given by the equation:
W=−Pext∆V
Where: Pext = external pressure = 1.00 atm = 1.013×105Pa ∆V= change
in volume = 10.0 L Convert 10.0 L to m
³
:
10.0L= 10.0×10−3m3= 0.0100m3
Now, calculate the work done:
W=−(1.013 ×105P a)(0.0100m3) = −1013J
Step 2: Calculate the heat absorbed by the calorimeter. The heat absorbed
by the calorimeter is given by the equation:
qcal =C∆T
Where: C= heat capacity of the calorimeter = 750 J/K ∆T= change in
temperature = 316 K - 298 K = 18 K
Now, calculate the heat absorbed by the calorimeter:
qcal = (750J/K)(18K) = 13500J
Step 3: Calculate the heat of the reaction. The heat absorbed by the reaction
is given by the equation:
qrxn =qcal −W
Now, calculate the heat of the reaction:
qrxn = 13500J−(−1013J) = 14513J
Step 4: Calculate the change in enthalpy. Since ∆H=qrxn under constant
pressure conditions, the change in enthalpy for the reaction is ∆H= 14513 J.
Therefore, the change in enthalpy for the reaction is 14513 J.
Question 23
Question
A reaction is known to have an enthalpy change of ∆H=−250 kJ/mol. If the
reaction is exothermic, in which direction will the enthalpy change (∆H) be
when the reaction is reversed?
17
Solution
1. When a reaction is reversed, the sign of the enthalpy change is also reversed.
Therefore, if the original reaction has ∆H=−250 kJ/mol, the reversed reaction
will have ∆H= +250 kJ/mol.
2. This change in sign occurs because when a reaction is reversed, the
products become the reactants and vice versa. As a result, the direction of heat
flow (exothermic or endothermic) also changes.
Thus, the enthalpy change for the reversed reaction will be +250 kJ/mol.
Question 24
Question
An ideal gas undergoes a process at constant pressure. The initial temperature
of the gas is 300 K and the final temperature is 400 K. Calculate the change in
enthalpy of the gas during the process.
Solution
Step 1: Recall the definition of enthalpy in terms of internal energy and pressure:
∆H= ∆U+P∆V
Step 2: Since the process is at constant pressure, ∆U=Q−Wwhere Q
is the heat added to the system and Wis the work done by the gas on its
surroundings.
Step 3: We can rewrite the expression for enthalpy change as:
∆H= (Q−W) + P∆V
Step 4: The work done by the gas can be expressed as W=P∆V. Substi-
tuting this into the equation gives:
∆H=Q
Step 5: To calculate the heat added to the system, we can use the formula:
Q=nCp∆T
where nis the number of moles of gas and Cpis the molar heat capacity at
constant pressure.
Step 6: First, we need to find the number of moles of the gas. We can use
the ideal gas law:
P V =nRT
Solving for n, we have:
n=P V
RT
18
Step 7: Substituting ninto the formula for heat gives:
Q=P V
RT Cp∆T
Step 8: Substituting the given values P= constant, V= constant, R=
constant, T1= 300 K, T2= 400 K and Cp= specific heat at constant pressure
into the equation gives the change in enthalpy.
Question 25
Question
A reaction between hydrogen gas and oxygen gas to form water vapor has an
enthalpy change of -483.6 kJ. If 8.00 moles of hydrogen gas react, how much
heat is absorbed or released?
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g)→2H2O(g)
Step 2: Determine the moles of oxygen gas reacting, based on the stoichiom-
etry of the balanced equation. Since the balanced equation states that 1 mole
of oxygen gas reacts with 2 moles of hydrogen gas, the moles of oxygen gas
required will be half of the moles of hydrogen gas:
moles of O2=1
2×8.00 moles = 4.00 moles
Step 3: Calculate the heat absorbed or released using the enthalpy change
and the moles of the limiting reactant. Given enthalpy change: ∆H=−483.6 kJ
Since the equation is for 2 moles of hydrogen gas reacting, the heat change for
1 mole of hydrogen gas will be half of the given value:
Heat change for 1 mole of H2=−483.6 kJ
2=−241.8 kJ/mole
Therefore, the heat change for 8.00 moles of hydrogen gas will be:
Heat change = −241.8 kJ/mole ×8.00 moles = −1934.4 kJ
So, 1934.4 kJ of heat is released in this reaction.
Question 26
Question
A reaction has an enthalpy change of ∆H=−185 kJ/mol. If 2.50 moles of the
reaction takes place, calculate the total enthalpy change for the reaction.
19
Solution
Step 1: Determine the total enthalpy change for the reaction per mole. Given:
∆H=−185 kJ/mol
Step 2: Calculate the total enthalpy change for the reaction when 2.50 moles
of the reaction takes place. Total Enthalpy Change = ∆H×Number of moles
Total Enthalpy Change = −185 kJ/mol ×2.50 mol Total Enthalpy Change =
−185 kJ/mol ×2.50 = −462.5 kJ
Therefore, the total enthalpy change for the reaction when 2.50 moles of the
reaction takes place is −462.5kJ.
Question 27
Question
Calculate the change in enthalpy when 2 moles of water vapor at 100
°
C and 1
atm pressure is condensed to liquid water at the same temperature and pressure.
Given that the heat of vaporization of water is 40.79 kJ/mol.
Solution
Step 1: Determine the change in enthalpy during the phase change from vapor
to liquid. For 2 moles of water vapor being condensed to liquid water, the
change in enthalpy can be calculated using the heat of vaporization formula:
Change in Enthalpy = moles ×heat of vaporization
Change in Enthalpy = 2 moles ×40.79 kJ/mol
Change in Enthalpy = 81.58 kJ
Therefore, the change in enthalpy during the phase change is 81.58 kJ.
Step 2: Account for the temperature change. Since the temperature remains
constant during the phase change, there is no additional change in enthalpy due
to the temperature change.
Step 3: Calculate the total change in enthalpy. The total change in enthalpy
is the sum of the change in enthalpy during the phase change and any changes
due to temperature:
Total Change in Enthalpy = Change in Enthalpy (phase change)+Change in Enthalpy (temperature change)
Total Change in Enthalpy = 81.58 kJ + 0 kJ
Total Change in Enthalpy = 81.58 kJ
20
Therefore, the total change in enthalpy when 2 moles of water vapor at 100
°
C
and 1 atm pressure is condensed to liquid water at the same temperature and
pressure is 81.58 kJ.
Question 28
Question
A gas initially at a pressure of 10 atm and a volume of 5 L is expanded at
constant temperature to a final volume of 25 L. If the enthalpy change during
the process is 1500 J, calculate the final pressure of the gas.
Solution
Step 1: Calculate the initial number of moles of gas using the ideal gas law,
P V =nRT .
n=P V
RT
Given P= 10 atm, V= 5 L, R= 0.0821 L atm mol−1K−1, and Tis constant
for the process.
Substitute the given values to find n.
n=(10 atm)(5 L)
(0.0821 L atm mol−1K−1)(T)=50
0.0821T=610
T
Step 2: Calculate the final pressure of the gas using the enthalpy change and
the number of moles.
∆H=nCp∆T
Given ∆H= 1500 J and n= 610/T , where Tis the gas temperature for the
process, and Cpis the molar heat capacity at constant pressure.
Since the process is at constant temperature, ∆T= 0 and nCp∆T= 0.
Thus, the enthalpy change during the process does not depend on the molar
heat capacity at constant pressure.
So, ∆H=nCp∆T= 0.
Since ∆H=nCp∆T= 0, nCp∆T= 1500 J implies nCp= 1500 J.
Substitute n= 610/T into nCp= 1500 J and simplify.
610Cp
T= 1500
Step 3: Calculate the final pressure.
P V =nRT
At the final state, PfinalVfinal =nRT . Given Vfinal = 25 L, n= 610/T , and
R= 0.0821 L atm mol−1K−1.
21
Substitute the values to find the final pressure.
Pfinal ·25 = 610
T·0.0821 ·T
25Pfinal = 50.31Pfinal =50.31
25
Therefore, the final pressure of the gas is 2.012 atm.
Question 29
Question
A certain reaction has an enthalpy change of ∆H=−128 kJ/mol at 25
°
C. If
the reaction is exothermic, calculate the standard enthalpy change at 298 K.
Solution
Step 1: Calculate the change in enthalpy due to the change in temperature from
25
°
C to 298 K.
∆H298 = ∆H25 +Z298
25
CpdT
where Cpis the molar heat capacity at constant pressure.
Step 2: Use the heat capacity values and the relationship Cp=dH
dT pto
calculate the integral.
Z298
25
CpdT =Z298
25 dH
dT p
dT
=Z298
25 128
298 −25dT
Step 3: Perform the integration and calculate the new enthalpy change at
298 K.
= 128 ×ln 298
25 kJ/mol
∆H298 =−126.39 kJ/mol
Therefore, the standard enthalpy change at 298 K is ∆H298 =−126.39
kJ/mol.
22
Question 30
Question
Calculate the change in enthalpy for the reaction
2H2(g) + O2(g)→2H2O(l)
given the following bond energies: H−H= 436 kJ/mol, O=O= 498 kJ/mol,
H−O= 463 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Step 2: Calculate the total energy released when the new bonds are formed in
the products. Step 3: Find the difference between the energy required to break
the bonds in the reactants and the energy released when the new bonds are
formed in the products.
Step 1: The total energy required to break the bonds in the reactants is:
2×(H−H)+1×(O=O) = 2 ×436 + 1 ×498 = 1370 kJ/mol
Step 2: The total energy released when the new bonds are formed in the
products is:
2×(H−O) = 2 ×463 = 926 kJ/mol
Step 3: The change in enthalpy for the reaction is given by:
∆H= (Energy required to break the bonds)−(Energy released when new bonds are formed)
∆H= 1370 −926 = 444 kJ/mol
Therefore, the change in enthalpy for the reaction is 444 kJ/mol.
Question 31
Question
Given the following reaction involving the combustion of ethane gas:
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
If the standard enthalpies of formation are −84.7 kJ/mol for CO2and −285.8 kJ/mol
for H2O, and the heating value of ethane gas is −1558.0 kJ/mol, calculate the
standard enthalpy of formation for ethane.
23
Solution
Step 1: Calculate the standard enthalpy change of the reaction using the given
standard enthalpies of formation.
∆H◦
rxn =Xν∆H◦
f,products −Xν∆H◦
f,reactants
Given data:
∆H◦
f(CO2) = −84.7 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
rxn =−1558.0 kJ/mol
∆H◦
rxn = 2(−84.7) + 3(−285.8) −(−1558.0)
∆H◦
rxn =−169.4 kJ/mol
Step 2: Write the standard enthalpy change of the reaction using the en-
thalpy of formation of ethane.
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
∆H◦
rxn = 2∆H◦
f(CO2) + 3∆H◦
f(H2O)−∆H◦
f(C2H6)
Step 3: Solve for the standard enthalpy of formation of ethane.
−169.4 = 2(−84.7) + 3(−285.8) −∆H◦
f(C2H6)
−169.4 = −169.4−∆H◦
f(C2H6)
∆H◦
f(C2H6) = 0 kJ/mol
Therefore, the standard enthalpy of formation of ethane is 0 kJ/mol.
Question 32
Question
A sample of an ideal gas undergoes a process where the pressure changes from
2 atm to 6 atm while the volume changes from 8 L to 2 L. If the initial and
final temperatures are 300 K and 500 K respectively, calculate the change in
enthalpy of the gas during this process.
24
Solution
Step 1: Calculate the initial and final internal energies of the gas using the first
law of thermodynamics:
∆U=Q−W
where ∆Uis the change in internal energy, Q is the heat added to the gas, and
W is the work done by the gas.
Given that the gas undergoes an isobaric process, the work done can be
calculated as:
W=−Pext ·∆V
Since the gas is ideal, we can use the ideal gas law to express Pext:
W=−Pext ·∆V=−nRT ln Vf
Vi
Substitute the given values:
W=−nRT ln 2
8=−nRT ln 1
4=−nRT ln(0.25)
∆U=Q−W=Q+nRT ln(0.25)
∆U=nCV∆T
where CVis the molar heat capacity at constant volume. Since the gas is ideal,
CV=f
2Rwhere f is the degrees of freedom (f=3 for monoatomic gas).
Substitute and rearrange to solve for Q:
Q= ∆U−nR∆Tln(0.25) = n(f
2R)∆T−nR∆Tln(0.25)
Q=nR∆T(f
2−ln(0.25)) = nR∆T(3
2−ln(0.25))
Step 2: Calculate the change in enthalpy, ∆H, of the gas using the equation:
∆H= ∆U+ ∆(P V )
Given that the process is isobaric, substitute the change in internal energy
and pressure-volume work:
∆H=Q+W+P∆V
Substitute the expressions for Q and W calculated earlier:
∆H=nR∆T(3
2−ln(0.25)) + nR∆Tln(0.25) + Pext∆V
25
Given that P∆V=nR∆Tfor an isobaric process:
∆H=nR∆T(3
2−ln(0.25)) + nR∆T+nR∆T
∆H=nR∆T(5
2−ln(0.25))
Now, substitute the given values for n, R, and ∆Tto find the change in
enthalpy.
Question 33
Question
Calculate the standard enthalpy change for the reaction below at 298 K:
2CH4(g)+4O2(g)→2CO2(g)+4H2O(l)
Given the standard enthalpies of formation (∆H◦
f) for CH4(g), CO2(g), and
H2O(l) are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where n and m are the stoichiometric coefficients of the products and reactants,
respectively.
Step 2: Substitute the given values and calculate:
∆H◦= 2×∆H◦
f(CO2(g))+4×∆H◦
f(H2O(l))−[2×∆H◦
f(CH4(g))+4×∆H◦
f(O2(g))]
Step 3: Plug in the values and calculate:
∆H◦= 2 ×(−393.5) + 4 ×(−285.8) −[2 ×(−74.8) + 4 ×0]
∆H◦=−787.0+(−1143.2) −(−149.6)
∆H◦=−787.0−1143.2 + 149.6
∆H◦=−1780.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 298 K is
-1780.6 kJ/mol.
26
Question 34
Question
Calculate the change in enthalpy (∆H) when 10.0 moles of steam at 100◦C is
cooled to water at 25◦C. Given that the heat capacity of steam is 2.08 J/g·K,
the heat capacity of water is 4.18 J/g·K, and the heat of vaporization for water
is 40.79 kJ/mol.
Solution
Step 1: Calculate the heat required to cool the steam to 100◦C to water at
100◦C. The heat capacity of steam is 2.08 J/g·K. Given that 1 mole of water
has a molar mass of 18 g/mol and 1 mole of steam has a molar mass of 18 g/mol
+ 18 g/mol = 36 g/mol. Therefore, the heat capacity of steam is 2.08 J/g·K *
36 g/mol = 74.88 J/mol·K.
The change in temperature is 100◦C - 25◦C = 75◦C = 75 K. Therefore, the
heat required to cool 10.0 moles of steam to water at 100◦C is: 10.0 mol ×
75 K ×74.88 J/mol ·K = 56160 J.
Step 2: Calculate the heat required to condense the water at 100◦C to water
at 25◦C. The heat of vaporization for water is 40.79 kJ/mol = 40790 J/mol.
Therefore, the heat required to condense 10.0 moles of steam at 100◦C to water
at 100◦C is: 10.0 mol ×40790 J/mol = 407900 J.
Step 3: Add the heats from Step 1 and Step 2 to find the total change
in enthalpy. ∆H= 56160 J + 407900 J = 464060 J. Therefore, the change in
enthalpy when 10.0 moles of steam at 100◦C is cooled to water at 25◦C is 464060
J.
Question 35
Question
An experiment involves the mixing of 150 g of water at 25
°
C with 300 g of water
at 85
°
C in a thermally insulated container. Calculate the final temperature of
the mixture. Assume specific heat capacity of water is 4.18 J/g
°
C and neglect
any heat losses to the surroundings.
Solution
Step 1: Calculate the heat lost by the hot water and the heat gained by the
cold water.
The heat lost by the hot water is given by:
qhot =mhot ·c·∆T
27
Question 8
Question
A reaction at constant pressure has a change in enthalpy of −349 kJ. If 2.50
moles of the reactant are consumed, what is the change in enthalpy per mole of
reactant?
Solution
Step 1: Recall that the change in enthalpy (∆H) per mole of reactant can be
calculated using the formula:
∆Hper mole =∆H
moles of reactant
Step 2: Given that the change in enthalpy (∆H) is −349 kJ and the moles
of reactant is 2.50 mol, we can substitute these values into the formula:
∆Hper mole =−349 kJ
2.50 mol
Step 3: Calculate the change in enthalpy per mole of reactant:
∆Hper mole =−349
2.50 =−139.6 kJ/mol
Therefore, the change in enthalpy per mole of reactant is −139.6 kJ/mol.
Question 9
Question
Calculate the enthalpy change (∆H) for the reaction:
2H2(g)+O2(g)→H2O(l) given the following information:
∆H◦
ffor H2(g) = 0 kJ/mol
∆H◦
ffor O2(g) = 0 kJ/mol
∆H◦
ffor H2O(l) = -285.8 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate ∆Hfor the reaction using the standard enthalpies of
formation:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
6
Step 3: Substitute the given values into the equation:
∆H= [2(−285.8 kJ/mol)] −[2(0 kJ/mol) + 0 kJ/mol]
Step 4: Solve for ∆H:
∆H=−571.6 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H=−571.6 kJ/mol.
Question 10
Question
Calculate the change in enthalpy (∆H) when 50.0 g of water at 25
°
C is heated
to steam at 100
°
C. 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
18.015 g/mol.
Solution
Step 1: Calculate the heat required to heat the water from 25
°
C to 100
°
C. The
formula to calculate the heat required is:
q=mc∆T
Where: - qis the heat energy, - mis the mass of the substance, - cis the specific
heat capacity of the substance, - ∆Tis the change in temperature.
Substitute the values:
q= (50.0 g)(4.18 J/g
°
C)(100 −25)C
q= (50.0)(4.18)(75) J
Step 2: Calculate the heat required to vaporize the water. The heat required
to vaporize the water can be calculated using the formula:
q=n·∆Hvap
Where: - nis the number of moles of water, - ∆Hvap is the molar heat of
vaporization.
First, calculate the number of moles of water:
n=50.0 g
18.015 g/mol
Step 3: Convert the heat of vaporization to joules.
40.79 kJ/mol = 40.79 ×103J/mol
7
Now, calculate the heat required to vaporize the water:
q=50.0
18.015×(40.79 ×103) J
Step 4: Calculate the total change in enthalpy. The total change in enthalpy
is the sum of the heat required to raise the temperature and the heat required
to vaporize the water:
∆H=qheating +qvaporization
Question 11
Question
Given the reaction:
2A(g)+3B(g)→C(g)+4D(g)
where ∆H◦=−1258 kJ, calculate the enthalpy change when 1.5 moles of A
reacts with 2.0 moles of B.
Solution
Step 1: Calculate the moles of A and B reacting.
Moles of A = 1.5 mol
Moles of B = 2.0 mol
Step 2: Determine the limiting reactant.
Moles of A required for reaction = 2
2= 1.0 mol
Moles of B required for reaction = 3
2= 1.5 mol
Since A has fewer moles than required, A is the limiting reactant.
Step 3: Calculate the moles of C and D formed.
Since 2 moles of A produces 1 mole of C,
Moles of C formed = 1
2×1.5=0.75 mol
Since 2 moles of A produces 4 moles of D,
Moles of D formed = 2 ×1.5=3.0 mol
Step 4: Calculate the enthalpy change for the reaction.
∆H◦=−1258 kJ
For the reaction of 1 mole of A:
Enthalpy change = −1258
2=−629 kJ
For the reaction of 1.5 moles of A:
Enthalpy change = −629 ×1.5 = −943.5 kJ
Therefore, the enthalpy change when 1.5 moles of A reacts with 2.0 moles of B
is −943.5 kJ .
8
Question 12
Question
A reaction that releases 400 kJ of heat causes a decrease in temperature of 0.5
°
C
in a closed container water at 25
°
C. Assuming the heat capacity of the container
is negligible and that the density of water is 1 g/cm3, calculate the enthalpy
change for the reaction.
Solution
Step 1: Calculate the mass of water affected by the heat exchange. Given that
the density of water is 1 g/cm3, the mass of water affected by the heat exchange
is:
Volume of water = Density ×Volume
Volume = Mass
Density
Volume = 1 g
1 g/cm3= 1 cm3
The volume of water affected by the heat exchange is 1 cm3, which is equivalent
to 1 gram.
Step 2: Calculate the change in temperature of the water. Given that the
temperature change is 0.5
°
C, the heat exchanged is:
q=mc∆T
q= (1 g)(4.18 J/g
°
C)(0.5C)=2.09 J
Step 3: Convert the heat exchanged to kilojoules. To convert Joules to
kilojoules, divide by 1000:
q=2.09 J
1000 = 0.00209 kJ
Step 4: Determine the enthalpy change for the reaction. The enthalpy change
for the reaction is equal in magnitude but opposite in sign to the heat exchanged:
∆H=−q=−0.00209 kJ = -2.09 kJ
Therefore, the enthalpy change for the reaction is -2.09 kJ.
Question 13
Question
For a reaction at constant pressure, the enthalpy change (∆H) is given by the
equation ∆H= 50 −2T, where Tis the temperature in Kelvin. Determine the
temperature at which the reaction becomes exothermic.
9
Solution
Step 1: To find the temperature at which the reaction becomes exothermic, set
∆Hto zero and solve for T.
∆H= 50 −2T= 0
50 = 2T
T= 25 K
Step 2: To determine whether the reaction is exothermic or endothermic at
T= 25 K, substitute T= 25 into the equation ∆H= 50 −2T.
∆H= 50 −2(25) = 50 −50 = 0
Since ∆H= 0 at T= 25 K, the reaction becomes exothermic at this tem-
perature.
Question 14
Question
A gas initially at 300 K and 1 atm pressure is expanded isothermally and re-
versibly to three times its initial volume. Calculate the change in enthalpy for
this process, given that the molar heat capacity at constant pressure, Cp, is 30
J/mol K.
Solution
Step 1: Calculate the work done during the expansion. Given that the process
is isothermal and reversible, the work done is −nRT ln(Vf/Vi), where nis the
number of moles of gas, Ris the ideal gas constant, Tis the temperature, Vfis
the final volume, and Viis the initial volume.
Step 2: Calculate the heat transfer. Since the process is isothermal, the heat
transfer is equal to the work done. Therefore, q=−w.
Step 3: Calculate the change in enthalpy. The change in enthalpy is given
by ∆H= ∆U+P∆V, where ∆Uis the change in internal energy, Pis the
pressure, and ∆Vis the change in volume. For an isothermal process, ∆U= 0,
so ∆H=P∆V.
Step 4: Substitute values and calculate. Given: - Initial temperature, Ti=
300 K - Initial pressure, Pi= 1 atm - Final volume, Vf= 3Vi- Molar heat
capacity at constant pressure, Cp= 30 J/mol K - Ideal gas constant, R= 8.314
J/mol K
From the ideal gas law, P V =nRT , we can find nand R. Then, we can
calculate the change in enthalpy using the formula ∆H=P∆V.
Finally, substitute the known values to find the change in enthalpy.
10
Question 15
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with the following enthalpy changes:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
calculate the standard enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy of formation values provided.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
∆H◦=2·∆H◦
f(H2O(l))−2·∆H◦
f(H2(g)) + ∆H◦
f(O2(g))
∆H◦= 2(−285.8 kJ/mol) −2(0 kJ/mol) −0 kJ/mol
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is −571.6 kJ/mol .
Question 16
Question
Given the following reaction:
2A(g)+3B(g)−→ C(g) + D(g)
If the standard enthalpy of formation for each substance is as follows:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −500 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction.
11
Solution
To calculate the standard enthalpy change for the reaction, we can use Hess’s
Law, which states that the total enthalpy change for a reaction is the same
regardless of the number of steps in the reaction or the pathway taken.
Step 1: Write the given reaction in terms of standard enthalpies
of formation.
The given reaction is:
2A(g)+3B(g)−→ C(g) + D(g)
The standard enthalpy change for this reaction can be expressed as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the standard enthalpy change for the reaction.
Given that:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −500 kJ/mol
The standard enthalpy change can be calculated as follows:
∆H◦= [(∆H◦
f(C)+∆H◦
f(D)) −(∆H◦
f(2A)+∆H◦
f(3B))]
∆H◦= [(−400 kJ/mol+(−500 kJ/mol))−((2×−200 kJ/mol)+(3×−300 kJ/mol))]
∆H◦= [−900 kJ/mol −(−200 kJ/mol + −900 kJ/mol)]
∆H◦=−900 kJ/mol + 1100 kJ/mol
∆H◦= 200 kJ/mol
Question 17
Question
A reaction has an enthalpy change of ∆H=−285.8 kJ. If 4.96 moles of the
reactant undergo the reaction, what is the total heat absorbed or released during
the reaction?
12
Solution
Step 1: Determine the total heat absorbed or released per mole of the reactant.
Given: ∆H=−285.8 kJ
We know that the total heat absorbed or released during the reaction can
be calculated as:
Total heat = ∆H×moles of reactant
Step 2: Calculate the total heat absorbed or released.
Plugging in the values:
Total heat = −285.8 kJ ×4.96 moles
Total heat = −1416.688 kJ
Therefore, the total heat absorbed or released during the reaction is -1416.688
kJ.
Question 18
Question
Calculate the change in enthalpy (∆H) when 35.0 g of liquid water at 25◦C is
converted to steam at 100◦C. The specific heat capacity of water is 4.18 J/(g·
°
C),
the molar enthalpy of vaporization of water is 40.79 kJ/mol, and the molar mass
of water is 18.015 g/mol.
Solution
Step 1: Calculate the heat required to raise the temperature of the water from
25◦C to 100◦C using the formula:
q1=m×c×∆T
where - mis the mass of water (35.0 g), - cis the specific heat capacity of water
(4.18 J/(g·
°
C)), - ∆Tis the change in temperature (100◦C - 25◦C).
Step 2: Convert the heat calculated in Step 1 to kilojoules (kJ) by dividing
by 1000.
Step 3: Calculate the moles of water in 35.0 g using the molar mass of water.
Step 4: Calculate the heat required to convert the water to steam at 100◦C
using the formula:
q2=n×∆Hvap
where - nis the number of moles of water calculated in Step 3, - ∆Hvap is the
molar enthalpy of vaporization of water (40.79 kJ/mol).
Step 5: Add the heats calculated in Step 2 and Step 4 to find the total heat
change, ∆H.
13
∆H=q1+q2
Question 19
Question
Given the balanced chemical equation for the combustion of ethane:
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
Calculate the standard enthalpy change (∆H◦) for this reaction. Given the
following standard enthalpy of formation values:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.51 kJ/mol
∆H◦
f(H2O) = −285.83 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of formation for the products using the
given values.
∆H◦
f(2CO2) = 2(−393.51 kJ/mol) = −787.02 kJ/mol
∆H◦
f(3H2O) = 3(−285.83 kJ/mol) = −857.49 kJ/mol
Step 2: Calculate the standard enthalpy of formation for the reactants using
the given value.
∆H◦
f(C2H6) = −84.68 kJ/mol
Step 3: Calculate the net standard enthalpy change for the reaction using
the formula:
∆H◦= (X∆H◦
fproducts) −(X∆H◦
freactants)
∆H◦= [(−787.02 kJ/mol) + (−857.49 kJ/mol)] −(−84.68 kJ/mol)
∆H◦=−1644.51 kJ/mol + 84.68 kJ/mol = −1559.83 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the combustion of ethane
is -1559.83 kJ/mol.
14
Question 20
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas (N2)
react according to the following balanced chemical equation:
N2(g)+3H2(g)→2NH3(g)
Given the enthalpies of formation (∆H◦
f) for N2(g), H2(g), and NH3(g) are
0 kJ/mol, 0 kJ/mol, and -46.2 kJ/mol, respectively.
Solution
Step 1: Write the given balanced chemical equation and the enthalpy change
for the reaction.
N2(g)+3H2(g)→2NH3(g)
The enthalpy change for the reaction is given by:
∆H= Σ∆H◦
f(products) −Σ∆H◦
f(reactants)
Step 2: Calculate the enthalpy change for the reaction. The enthalpies of
formation are:
∆H◦
f(NH3) = −46.2 kJ/mol
∆H◦
f(N2) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Substitute the values into the equation for ∆H:
∆H=2×∆H◦
f(NH3)−∆H◦
f(N2)+3×∆H◦
f(H2)
∆H= (2 × −46.2 kJ/mol) −(0 + 3 ×0 kJ/mol)
∆H=−92.4 kJ/mol
Step 3: Calculate the change in enthalpy when 5.00 moles of N2react. Since
the enthalpy change is given per mole, we need to calculate the total enthalpy
change for 5 moles of N2:
Change in enthalpy = moles ×∆H
Change in enthalpy = 5.00 moles × −92.4 kJ/mol
Change in enthalpy = −462 kJ
Therefore, the change in enthalpy when 5.00 moles of N2react is -462 kJ.
15
Question 21
Question
A reaction is thought to have a certain enthalpy change. By performing a
series of measurements, it is found that the reaction has an enthalpy change of
−289.2 kJ when carried out at 25
°
C. Assuming that the heat capacities of all
substances are constant over the temperature range in question, calculate the
enthalpy change for the reaction at 55
°
C.
Solution
Step 1: Determine the heat capacity change (∆Cp) of the reaction.
∆Hreaction = ∆H25C
reaction +Z55C
25C
∆CpdT
∆Hreaction =−289.2 kJ + Z55C
25C
∆CpdT
Step 2: Since the heat capacities are constant, we can consider ∆Cpas a
constant value, say Cp.
∆Hreaction =−289.2 kJ + CpZ55C
25C
dT
∆Hreaction =−289.2 kJ + Cp·(55 −25) K
∆Hreaction =−289.2 kJ + 30Cp
Step 3: Given that the enthalpy change of the reaction at 25
°
C is -289.2 kJ:
−289.2 = −289.2 + 30Cp
0 = 30Cp=⇒Cp= 0
Step 4: The assumption that the heat capacities are constant is incorrect. We
must therefore consider the correct dependence of heat capacity on temperature
for the substances involved.
Question 22
Question
A reaction is carried out in a bomb calorimeter, where 1.50 mol of a gas is burned
at 1.00 atm pressure in a vessel with a volume of 10.0 L at a temperature of
298 K. The calorimeter itself has a heat capacity of 750 J/K. If the temperature
rises to 316 K during the reaction, calculate the change in enthalpy (∆H) for
the reaction.
Given: Gas constant, R= 8.314 J/(mol
·
K)
16
Solution
Step 1: Calculate the work done by the gas during the reaction. The work done
by the gas is given by the equation:
W=−Pext∆V
Where: Pext = external pressure = 1.00 atm = 1.013×105Pa ∆V= change
in volume = 10.0 L Convert 10.0 L to m
³
:
10.0L= 10.0×10−3m3= 0.0100m3
Now, calculate the work done:
W=−(1.013 ×105P a)(0.0100m3) = −1013J
Step 2: Calculate the heat absorbed by the calorimeter. The heat absorbed
by the calorimeter is given by the equation:
qcal =C∆T
Where: C= heat capacity of the calorimeter = 750 J/K ∆T= change in
temperature = 316 K - 298 K = 18 K
Now, calculate the heat absorbed by the calorimeter:
qcal = (750J/K)(18K) = 13500J
Step 3: Calculate the heat of the reaction. The heat absorbed by the reaction
is given by the equation:
qrxn =qcal −W
Now, calculate the heat of the reaction:
qrxn = 13500J−(−1013J) = 14513J
Step 4: Calculate the change in enthalpy. Since ∆H=qrxn under constant
pressure conditions, the change in enthalpy for the reaction is ∆H= 14513 J.
Therefore, the change in enthalpy for the reaction is 14513 J.
Question 23
Question
A reaction is known to have an enthalpy change of ∆H=−250 kJ/mol. If the
reaction is exothermic, in which direction will the enthalpy change (∆H) be
when the reaction is reversed?
17
Solution
1. When a reaction is reversed, the sign of the enthalpy change is also reversed.
Therefore, if the original reaction has ∆H=−250 kJ/mol, the reversed reaction
will have ∆H= +250 kJ/mol.
2. This change in sign occurs because when a reaction is reversed, the
products become the reactants and vice versa. As a result, the direction of heat
flow (exothermic or endothermic) also changes.
Thus, the enthalpy change for the reversed reaction will be +250 kJ/mol.
Question 24
Question
An ideal gas undergoes a process at constant pressure. The initial temperature
of the gas is 300 K and the final temperature is 400 K. Calculate the change in
enthalpy of the gas during the process.
Solution
Step 1: Recall the definition of enthalpy in terms of internal energy and pressure:
∆H= ∆U+P∆V
Step 2: Since the process is at constant pressure, ∆U=Q−Wwhere Q
is the heat added to the system and Wis the work done by the gas on its
surroundings.
Step 3: We can rewrite the expression for enthalpy change as:
∆H= (Q−W) + P∆V
Step 4: The work done by the gas can be expressed as W=P∆V. Substi-
tuting this into the equation gives:
∆H=Q
Step 5: To calculate the heat added to the system, we can use the formula:
Q=nCp∆T
where nis the number of moles of gas and Cpis the molar heat capacity at
constant pressure.
Step 6: First, we need to find the number of moles of the gas. We can use
the ideal gas law:
P V =nRT
Solving for n, we have:
n=P V
RT
18
Step 7: Substituting ninto the formula for heat gives:
Q=P V
RT Cp∆T
Step 8: Substituting the given values P= constant, V= constant, R=
constant, T1= 300 K, T2= 400 K and Cp= specific heat at constant pressure
into the equation gives the change in enthalpy.
Question 25
Question
A reaction between hydrogen gas and oxygen gas to form water vapor has an
enthalpy change of -483.6 kJ. If 8.00 moles of hydrogen gas react, how much
heat is absorbed or released?
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g)→2H2O(g)
Step 2: Determine the moles of oxygen gas reacting, based on the stoichiom-
etry of the balanced equation. Since the balanced equation states that 1 mole
of oxygen gas reacts with 2 moles of hydrogen gas, the moles of oxygen gas
required will be half of the moles of hydrogen gas:
moles of O2=1
2×8.00 moles = 4.00 moles
Step 3: Calculate the heat absorbed or released using the enthalpy change
and the moles of the limiting reactant. Given enthalpy change: ∆H=−483.6 kJ
Since the equation is for 2 moles of hydrogen gas reacting, the heat change for
1 mole of hydrogen gas will be half of the given value:
Heat change for 1 mole of H2=−483.6 kJ
2=−241.8 kJ/mole
Therefore, the heat change for 8.00 moles of hydrogen gas will be:
Heat change = −241.8 kJ/mole ×8.00 moles = −1934.4 kJ
So, 1934.4 kJ of heat is released in this reaction.
Question 26
Question
A reaction has an enthalpy change of ∆H=−185 kJ/mol. If 2.50 moles of the
reaction takes place, calculate the total enthalpy change for the reaction.
19
Solution
Step 1: Determine the total enthalpy change for the reaction per mole. Given:
∆H=−185 kJ/mol
Step 2: Calculate the total enthalpy change for the reaction when 2.50 moles
of the reaction takes place. Total Enthalpy Change = ∆H×Number of moles
Total Enthalpy Change = −185 kJ/mol ×2.50 mol Total Enthalpy Change =
−185 kJ/mol ×2.50 = −462.5 kJ
Therefore, the total enthalpy change for the reaction when 2.50 moles of the
reaction takes place is −462.5kJ.
Question 27
Question
Calculate the change in enthalpy when 2 moles of water vapor at 100
°
C and 1
atm pressure is condensed to liquid water at the same temperature and pressure.
Given that the heat of vaporization of water is 40.79 kJ/mol.
Solution
Step 1: Determine the change in enthalpy during the phase change from vapor
to liquid. For 2 moles of water vapor being condensed to liquid water, the
change in enthalpy can be calculated using the heat of vaporization formula:
Change in Enthalpy = moles ×heat of vaporization
Change in Enthalpy = 2 moles ×40.79 kJ/mol
Change in Enthalpy = 81.58 kJ
Therefore, the change in enthalpy during the phase change is 81.58 kJ.
Step 2: Account for the temperature change. Since the temperature remains
constant during the phase change, there is no additional change in enthalpy due
to the temperature change.
Step 3: Calculate the total change in enthalpy. The total change in enthalpy
is the sum of the change in enthalpy during the phase change and any changes
due to temperature:
Total Change in Enthalpy = Change in Enthalpy (phase change)+Change in Enthalpy (temperature change)
Total Change in Enthalpy = 81.58 kJ + 0 kJ
Total Change in Enthalpy = 81.58 kJ
20
Therefore, the total change in enthalpy when 2 moles of water vapor at 100
°
C
and 1 atm pressure is condensed to liquid water at the same temperature and
pressure is 81.58 kJ.
Question 28
Question
A gas initially at a pressure of 10 atm and a volume of 5 L is expanded at
constant temperature to a final volume of 25 L. If the enthalpy change during
the process is 1500 J, calculate the final pressure of the gas.
Solution
Step 1: Calculate the initial number of moles of gas using the ideal gas law,
P V =nRT .
n=P V
RT
Given P= 10 atm, V= 5 L, R= 0.0821 L atm mol−1K−1, and Tis constant
for the process.
Substitute the given values to find n.
n=(10 atm)(5 L)
(0.0821 L atm mol−1K−1)(T)=50
0.0821T=610
T
Step 2: Calculate the final pressure of the gas using the enthalpy change and
the number of moles.
∆H=nCp∆T
Given ∆H= 1500 J and n= 610/T , where Tis the gas temperature for the
process, and Cpis the molar heat capacity at constant pressure.
Since the process is at constant temperature, ∆T= 0 and nCp∆T= 0.
Thus, the enthalpy change during the process does not depend on the molar
heat capacity at constant pressure.
So, ∆H=nCp∆T= 0.
Since ∆H=nCp∆T= 0, nCp∆T= 1500 J implies nCp= 1500 J.
Substitute n= 610/T into nCp= 1500 J and simplify.
610Cp
T= 1500
Step 3: Calculate the final pressure.
P V =nRT
At the final state, PfinalVfinal =nRT . Given Vfinal = 25 L, n= 610/T , and
R= 0.0821 L atm mol−1K−1.
21
Substitute the values to find the final pressure.
Pfinal ·25 = 610
T·0.0821 ·T
25Pfinal = 50.31Pfinal =50.31
25
Therefore, the final pressure of the gas is 2.012 atm.
Question 29
Question
A certain reaction has an enthalpy change of ∆H=−128 kJ/mol at 25
°
C. If
the reaction is exothermic, calculate the standard enthalpy change at 298 K.
Solution
Step 1: Calculate the change in enthalpy due to the change in temperature from
25
°
C to 298 K.
∆H298 = ∆H25 +Z298
25
CpdT
where Cpis the molar heat capacity at constant pressure.
Step 2: Use the heat capacity values and the relationship Cp=dH
dT pto
calculate the integral.
Z298
25
CpdT =Z298
25 dH
dT p
dT
=Z298
25 128
298 −25dT
Step 3: Perform the integration and calculate the new enthalpy change at
298 K.
= 128 ×ln 298
25 kJ/mol
∆H298 =−126.39 kJ/mol
Therefore, the standard enthalpy change at 298 K is ∆H298 =−126.39
kJ/mol.
22
Question 30
Question
Calculate the change in enthalpy for the reaction
2H2(g) + O2(g)→2H2O(l)
given the following bond energies: H−H= 436 kJ/mol, O=O= 498 kJ/mol,
H−O= 463 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Step 2: Calculate the total energy released when the new bonds are formed in
the products. Step 3: Find the difference between the energy required to break
the bonds in the reactants and the energy released when the new bonds are
formed in the products.
Step 1: The total energy required to break the bonds in the reactants is:
2×(H−H)+1×(O=O) = 2 ×436 + 1 ×498 = 1370 kJ/mol
Step 2: The total energy released when the new bonds are formed in the
products is:
2×(H−O) = 2 ×463 = 926 kJ/mol
Step 3: The change in enthalpy for the reaction is given by:
∆H= (Energy required to break the bonds)−(Energy released when new bonds are formed)
∆H= 1370 −926 = 444 kJ/mol
Therefore, the change in enthalpy for the reaction is 444 kJ/mol.
Question 31
Question
Given the following reaction involving the combustion of ethane gas:
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
If the standard enthalpies of formation are −84.7 kJ/mol for CO2and −285.8 kJ/mol
for H2O, and the heating value of ethane gas is −1558.0 kJ/mol, calculate the
standard enthalpy of formation for ethane.
23
Solution
Step 1: Calculate the standard enthalpy change of the reaction using the given
standard enthalpies of formation.
∆H◦
rxn =Xν∆H◦
f,products −Xν∆H◦
f,reactants
Given data:
∆H◦
f(CO2) = −84.7 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
rxn =−1558.0 kJ/mol
∆H◦
rxn = 2(−84.7) + 3(−285.8) −(−1558.0)
∆H◦
rxn =−169.4 kJ/mol
Step 2: Write the standard enthalpy change of the reaction using the en-
thalpy of formation of ethane.
C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
∆H◦
rxn = 2∆H◦
f(CO2) + 3∆H◦
f(H2O)−∆H◦
f(C2H6)
Step 3: Solve for the standard enthalpy of formation of ethane.
−169.4 = 2(−84.7) + 3(−285.8) −∆H◦
f(C2H6)
−169.4 = −169.4−∆H◦
f(C2H6)
∆H◦
f(C2H6) = 0 kJ/mol
Therefore, the standard enthalpy of formation of ethane is 0 kJ/mol.
Question 32
Question
A sample of an ideal gas undergoes a process where the pressure changes from
2 atm to 6 atm while the volume changes from 8 L to 2 L. If the initial and
final temperatures are 300 K and 500 K respectively, calculate the change in
enthalpy of the gas during this process.
24
Solution
Step 1: Calculate the initial and final internal energies of the gas using the first
law of thermodynamics:
∆U=Q−W
where ∆Uis the change in internal energy, Q is the heat added to the gas, and
W is the work done by the gas.
Given that the gas undergoes an isobaric process, the work done can be
calculated as:
W=−Pext ·∆V
Since the gas is ideal, we can use the ideal gas law to express Pext:
W=−Pext ·∆V=−nRT ln Vf
Vi
Substitute the given values:
W=−nRT ln 2
8=−nRT ln 1
4=−nRT ln(0.25)
∆U=Q−W=Q+nRT ln(0.25)
∆U=nCV∆T
where CVis the molar heat capacity at constant volume. Since the gas is ideal,
CV=f
2Rwhere f is the degrees of freedom (f=3 for monoatomic gas).
Substitute and rearrange to solve for Q:
Q= ∆U−nR∆Tln(0.25) = n(f
2R)∆T−nR∆Tln(0.25)
Q=nR∆T(f
2−ln(0.25)) = nR∆T(3
2−ln(0.25))
Step 2: Calculate the change in enthalpy, ∆H, of the gas using the equation:
∆H= ∆U+ ∆(P V )
Given that the process is isobaric, substitute the change in internal energy
and pressure-volume work:
∆H=Q+W+P∆V
Substitute the expressions for Q and W calculated earlier:
∆H=nR∆T(3
2−ln(0.25)) + nR∆Tln(0.25) + Pext∆V
25
Given that P∆V=nR∆Tfor an isobaric process:
∆H=nR∆T(3
2−ln(0.25)) + nR∆T+nR∆T
∆H=nR∆T(5
2−ln(0.25))
Now, substitute the given values for n, R, and ∆Tto find the change in
enthalpy.
Question 33
Question
Calculate the standard enthalpy change for the reaction below at 298 K:
2CH4(g)+4O2(g)→2CO2(g)+4H2O(l)
Given the standard enthalpies of formation (∆H◦
f) for CH4(g), CO2(g), and
H2O(l) are -74.8 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol, respectively.
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where n and m are the stoichiometric coefficients of the products and reactants,
respectively.
Step 2: Substitute the given values and calculate:
∆H◦= 2×∆H◦
f(CO2(g))+4×∆H◦
f(H2O(l))−[2×∆H◦
f(CH4(g))+4×∆H◦
f(O2(g))]
Step 3: Plug in the values and calculate:
∆H◦= 2 ×(−393.5) + 4 ×(−285.8) −[2 ×(−74.8) + 4 ×0]
∆H◦=−787.0+(−1143.2) −(−149.6)
∆H◦=−787.0−1143.2 + 149.6
∆H◦=−1780.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 298 K is
-1780.6 kJ/mol.
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Question 34
Question
Calculate the change in enthalpy (∆H) when 10.0 moles of steam at 100◦C is
cooled to water at 25◦C. Given that the heat capacity of steam is 2.08 J/g·K,
the heat capacity of water is 4.18 J/g·K, and the heat of vaporization for water
is 40.79 kJ/mol.
Solution
Step 1: Calculate the heat required to cool the steam to 100◦C to water at
100◦C. The heat capacity of steam is 2.08 J/g·K. Given that 1 mole of water
has a molar mass of 18 g/mol and 1 mole of steam has a molar mass of 18 g/mol
+ 18 g/mol = 36 g/mol. Therefore, the heat capacity of steam is 2.08 J/g·K *
36 g/mol = 74.88 J/mol·K.
The change in temperature is 100◦C - 25◦C = 75◦C = 75 K. Therefore, the
heat required to cool 10.0 moles of steam to water at 100◦C is: 10.0 mol ×
75 K ×74.88 J/mol ·K = 56160 J.
Step 2: Calculate the heat required to condense the water at 100◦C to water
at 25◦C. The heat of vaporization for water is 40.79 kJ/mol = 40790 J/mol.
Therefore, the heat required to condense 10.0 moles of steam at 100◦C to water
at 100◦C is: 10.0 mol ×40790 J/mol = 407900 J.
Step 3: Add the heats from Step 1 and Step 2 to find the total change
in enthalpy. ∆H= 56160 J + 407900 J = 464060 J. Therefore, the change in
enthalpy when 10.0 moles of steam at 100◦C is cooled to water at 25◦C is 464060
J.
Question 35
Question
An experiment involves the mixing of 150 g of water at 25
°
C with 300 g of water
at 85
°
C in a thermally insulated container. Calculate the final temperature of
the mixture. Assume specific heat capacity of water is 4.18 J/g
°
C and neglect
any heat losses to the surroundings.
Solution
Step 1: Calculate the heat lost by the hot water and the heat gained by the
cold water.
The heat lost by the hot water is given by:
qhot =mhot ·c·∆T
27
where mhot = 300 g is the mass of hot water, c= 4.18 J/g
°
C is the specific heat
capacity of water, and ∆T= 85 −Tfinal is the decrease in temperature of the
hot water.
The heat gained by the cold water is given by:
qcold =mcold ·c·∆T
where mcold = 150 g is the mass of cold water and ∆T=Tfinal −25 is the
increase in temperature of the cold water.
Since the system is thermally isolated, the heat lost by the hot water is equal
to the heat gained by the cold water:
qhot =qcold
Step 2: Set up and solve the equation based on the heat lost and gained.
mhot ·c·(85 −Tfinal) = mcold ·c·(Tfinal −25)
Substitute in the given values:
300 ·4.18 ·(85 −Tfinal) = 150 ·4.18 ·(Tfinal −25)
Step 3: Solve for the final temperature (Tfinal).
Simplify the equation: 1260 ·(85 −Tfinal) = 630 ·(Tfinal −25)
Expand: 107100 −1260Tfinal = 630Tfinal −15750
Combine like terms: 107100 + 15750 = 630Tfinal + 1260Tfinal
Solve for Tfinal :Tfinal =107100 + 15750
1890 ≈70.37C
Therefore, the final temperature of the mixture is approximately 70.37C.
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