CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Enthalpy
Question Bank - Set 1
Liberty University
Question 1
Question
Calculate the change in enthalpy (∆H) when 2.50 moles of hydrogen gas re-
act with excess oxygen gas to produce water vapor according to the following
balanced chemical equation:
2H2(g)+ O2(g)→2H2O(g)
Given: - ∆H◦
ffor H2(g) = 0 kJ/mol - ∆H◦
ffor O2(g) = 0 kJ/mol - ∆H◦
ffor
H2O(g)=−286 kJ/mol
Solution
Step 1: Calculate the change in enthalpy (∆H) for the given reaction using the
standard enthalpies of formation: ∆H=Pn∆H◦
f(products)−Pm∆H◦
f(reactants)
Given: - ∆H◦
ffor H2(g) = 0 kJ/mol - ∆H◦
ffor O2(g) = 0 kJ/mol - ∆H◦
ffor
H2O(g)=−286 kJ/mol
Substitute the values into the formula: ∆H= 2(−286 kJ/mol)−[2(0 kJ/mol)+
0 kJ/mol]
∆H=−572 kJ/mol
Step 2: Calculate the total change in enthalpy when 2.50 moles of hydrogen
gas react:
∆Htotal = 2.50 mol ×(−572 kJ/mol)
∆Htotal =−1430 kJ
Therefore, the change in enthalpy when 2.50 moles of hydrogen gas react is
-1430 kJ.
Question 2
Question
Given that the enthalpy change for the reaction 2 H2(g) + O2(g) −−→ 2 H2O(g)
is −483.6 kJ, calculate the enthalpy change for the reaction 2 H2O(g) −−→
2 H2(g) + O2(g).
Solution
Step 1: Write the enthalpy change for the second reaction in terms of the
enthalpy change for the first reaction. Let xrepresent the enthalpy change
for the second reaction. The enthalpy change for the second reaction is the
negative of the enthalpy change for the first reaction, since the second reaction
is the reverse of the first reaction.
x=−(−483.6)
x= 483.6 kJ
Therefore, the enthalpy change for the reaction 2 H2O(g) −−→ 2 H2(g) +
O2(g) is 483.6 kJ.
Question 3
Question
Calculate the change in enthalpy (∆H) when 1 mol of liquid water at 25◦C
is heated to form 1 mol of water vapor at 100◦C. Given that the specific heat
capacities of liquid water and steam are 4.18 J/g
°
C and 2.03 J/g
°
C, respectively,
and the heat of vaporization of water is 40.79 kJ/mol.
Solution
Step 1: Calculate the heat required to heat liquid water from 25◦C to 100◦C.
Use the formula: q=mc∆T, where - qis the heat required, - mis the mass
of the substance, - cis the specific heat capacity of the substance, and - ∆Tis
the change in temperature.
Given that 1 mol of water has a mass of 18 g (since the molar mass of water
is 18 g/mol), we can calculate the heat required as follows:
q= (18 g)(4.18 J/g
°
C)(100 −25)◦C
q= 18 mol ×4.18 J/g
°
C×75C
q= 5613 J/mol
Step 2: Calculate the heat required to vaporize the water at 100◦C.
2
The heat required to vaporize water is the heat of vaporization, which is
40.79 kJ/mol.
Step 3: Calculate the total change in enthalpy.
The change in enthalpy (∆H) is the sum of the heat required to heat the
water from 25◦C to 100◦C and the heat required to vaporize the water. Let’s
calculate it:
∆H= 5613 J/mol + 40.79 ×103J/mol
∆H= 46.403 ×103J/mol
∆H= 46.403 kJ/mol
Therefore, the change in enthalpy when 1 mol of liquid water at 25◦C is
heated to form 1 mol of water vapor at 100◦C is 46.403 kJ/mol.
Question 4
Question
Given the enthalpy of formation values for the following reactions:
Reaction 1: 2H2(g)+O2(g)→2H2O(l) ∆H0
f=−572 kJ/mol
Reaction 2: C(s)+O2(g)→CO2(g) ∆H0
f=−394 kJ/mol
Reaction 3: 2H2O(l)→2H2(g)+O2(g) ∆H0
f= 476 kJ/mol
Calculate the standard enthalpy change for the reaction:
C(s)+O2(g)→CO2(g)
Solution
Step 1: Write the target reaction as the sum of the given reactions, adjusting for
stoichiometry and direction if necessary. Step 2: Calculate the overall enthalpy
change for the target reaction using the enthalpy of formation values. Step 3:
Substitute the given enthalpy of formation values into the equation and solve
for the overall enthalpy change.
Step 1: To obtain the target reaction, we need to reverse Reaction 2 and
halve Reaction 1. This will cancel out H2O on both sides, resulting in the target
reaction:
C(s)+O2(g)→CO2(g)
Now, the reaction will be:
C(s) + 2H2(g) + 1
2O2(g)→CO2(g) + 2H2(g)
3
Step 2: The overall enthalpy change for the target reaction can be calculated
using the enthalpy of formation values:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given enthalpy of formation values into the equation:
∆H=−394 kJ
mol−0kJ
mol + 0 kJ
mol +1
2×−572 kJ
mol
∆H=−394 kJ
mol + 286 kJ
mol
∆H=−108 kJ
mol
Therefore, the standard enthalpy change for the reaction is ∆H=−108 kJ/mol.
Question 5
Question
Calculate the change in enthalpy (∆H) when 2 moles of methane gas react with
excess oxygen gas to form carbon dioxide gas and water vapor according to the
following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpies of formation: H◦
f(CH4) = −74.8
kJ/mol H◦
f(CO2) = −393.5 kJ/mol H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of reaction (∆H◦
rxn) using the standard
enthalpies of formation.
The standard enthalpy change for the reaction is given by:
∆H◦
rxn =XνfH◦
f(products) −XνfH◦
f(reactants)
where νfis the stoichiometric coefficient.
Substitute the given standard enthalpies of formation values into the equa-
tion:
∆H◦
rxn = [1 ∗(−393.5) + 2 ∗(−285.8)] −[1 ∗(−74.8) + 2 ∗0]
∆H◦
rxn = (−393.5−571.6) −(−74.8)
∆H◦
rxn =−965.1 + 74.8
∆H◦
rxn =−890.3 kJ
4
Therefore, the standard enthalpy change for the reaction is ∆H◦
rxn =−890.3
kJ.
Step 2: Calculate the actual change in enthalpy (∆H) for 2 moles of methane
gas.
Since the balanced chemical equation shows the reaction of 1 mole of methane,
we need to scale the enthalpy value accordingly:
∆H=2 moles
1 mole ∗∆H◦
rxn
∆H= 2 ∗(−890.3)
∆H=−1780.6 kJ
Therefore, the change in enthalpy for the reaction of 2 moles of methane gas
is ∆H=−1780.6 kJ.
Question 6
Question
Given the reaction:
2A(g) + 3B(g)→C(g) + D(g)
with the following enthalpy changes:
∆Hrxn =−580 kJ/mol
∆Hf C =−390 kJ/mol
∆Hf D =−125 kJ/mol
Calculate the standard enthalpy change of formation for compound A if the
standard enthalpy change of formation for compound B is -200 kJ/mol.
Solution
Step 1: Calculate the total enthalpy of formation for the products C and D.
The total enthalpy of formation for the products is given by:
∆Hf C+D = ∆Hf C + ∆Hf D
∆Hf C+D =−390 kJ/mol + (−125 kJ/mol)
∆Hf C+D =−515 kJ/mol
Step 2: Calculate the total enthalpy change for the reactants A and B. The
total enthalpy change for the reactants is given by:
∆Hrxn =Xνproducts∆Hf products −Xνreactants∆Hf reactants
5
∆Hrxn = (∆Hf C+D)−(2 ×∆Hf A + 3 ×∆Hf B)
−580 kJ/mol = −515 kJ/mol −2×∆Hf A −3×(−200 kJ/mol)
Step 3: Solve for the enthalpy change of formation of compound A.
∆Hf A =−580 kJ/mol −(−515 kJ/mol) + 3 ×(−200 kJ/mol)
2
∆Hf A =−407.5 kJ/mol
Therefore, the standard enthalpy change of formation for compound A is
-407.5 kJ/mol.
Question 7
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of ammonia gas (NH3)
react with excess oxygen gas (O2) according to the following balanced chemical
equation:
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
Given that ∆Hfor the reaction is -940 kJ.
Solution
Step 1: Determine the moles of ammonia reacting Given: Number of moles of
ammonia, n= 5.00 moles.
Step 2: Determine the molar ratio between NH3and ∆HFrom the balanced
chemical equation, the ratio of NH3to ∆His 4:940.
Step 3: Calculate the change in enthalpy (∆H)
∆H=940 kJ
4 mol ×5.00 mol = −2350 kJ
Therefore, the change in enthalpy (∆H) for the reaction when 5.00 moles of
ammonia gas react is -2350 kJ.
Question 8
Question
Given the following reaction at 25
°
C:
2A(g)+3B(g)→C(g)+2D(g)
6
If the standard enthalpies of formation (∆H◦
f) are as follows:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −100 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.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [(∆H◦
f(C) + 2∆H◦
f(D)) −(2∆H◦
f(A) + 3∆H◦
f(B))]
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion.
∆H◦= [(−400 kJ/mol+2(−100 kJ/mol))−(2(−200 kJ/mol)+3(−300 kJ/mol))]
Step 3: Perform the calculations to find the standard enthalpy change.
∆H◦= [(−400 kJ/mol −200 kJ/mol) −(−400 kJ/mol −900 kJ/mol)]
∆H◦= [−600 kJ/mol −(−1300 kJ/mol)]
∆H◦=−600 kJ/mol + 1300 kJ/mol
∆H◦= 700 kJ/mol
Therefore, the standard enthalpy change for the reaction is 700 kJ/mol .
Question 9
Question
Calculate the change in enthalpy (∆H) when 15.0 g of ice at -10.0
°
C is converted
to liquid water at 50.0
°
C. The specific heat capacity of ice is 2.09 J/g
°
C, the
heat of fusion for ice is 333 J/g, the specific heat capacity of water is 4.18 J/g
°
C,
and the heat of vaporization for water is 2260 J/g. Assume no heat is lost to
the surroundings.
7
Solution
Step 1: Calculate the heat required to raise the temperature of the ice from
-10.0
°
C to 0
°
C.
q=m×c×∆T
q= 15.0 g ×2.09 J/g
°
C×(0C−(−10.0C))
q= 313.5 J
Step 2: Calculate the heat required to melt the ice at 0
°
C.
q=m×∆Hfusion
q= 15.0 g ×333 J/g
q= 4995 J
Step 3: Calculate the heat required to raise the temperature of the water
from 0
°
C to 50.0
°
C.
q=m×c×∆T
q= 15.0 g ×4.18 J/g
°
C×(50.0C−0C)
q= 3135 J
Step 4: Calculate the total heat required for the process.
qtotal =q1+q2+q3
qtotal = 313.5 J + 4995 J + 3135 J
qtotal = 8443.5 J
Step 5: Calculate the change in enthalpy (∆H).
∆H=qtotal
∆H= 8443.5 J
Therefore, the change in enthalpy (∆H) when 15.0 g of ice at -10.0
°
C is
converted to liquid water at 50.0
°
C is 8443.5 J.
Question 10
Question
Given the reaction:
2A+ 3B→C+ 4D
where ∆H=−232 kJ, calculate the enthalpy change when 4.50 mol of A reacts
completely with 2.00 mol of B. Assume the reaction takes place under standard
conditions.
8
Solution
Step 1: Calculate the moles of limiting reagent Let’s first determine the limiting
reagent between A and B to find out which one will be completely consumed.
The stoichiometric ratio between A and B is 2:3. Therefore, we need to compare
the moles of A and B: For A: 4.50 mol A For B: 2.00 mol B
Calculating the moles of B needed to react with 4.50 mol of A:
4.50 mol A
2 mol A ×3 mol B = 6.75 mol B
Since we only have 2.00 mol of B, B is the limiting reagent.
Step 2: Calculate the moles of products formed From the given reaction, we
can determine the moles of C and D formed when 2.00 mol of B is used: Using
the stoichiometry of the reaction: For B: 2.00 mol B For C: 1 mol C For D: 4
mol D
Step 3: Calculate the change in enthalpy Given: ∆H=−232 kJ for the
reaction. The enthalpy change can be calculated using the formula:
∆Htotal = ∆H×moles of products formed
stoichiometric coefficient of limiting reactant
Plugging in the values:
∆Htotal =−232 kJ ×2.00 mol of products
3=−154.67 kJ
Therefore, the enthalpy change when 4.50 mol of A reacts completely with
2.00 mol of B is −154.67 kJ.
Question 11
Question
Calculate the change in enthalpy (∆H) when 2 moles of nitrogen gas reacts
with 5 moles of hydrogen gas to form ammonia gas according to the following
reaction:
N2(g)+3H2(g)→2NH3(g)
Given the enthalpies of formation are: Hf(N2) = 0 kJ/mol, Hf(H2) = 0 kJ/mol,
Hf(NH3) = −46.1 kJ/mol.
Solution
Step 1: Calculate the enthalpy change for the reaction using the enthalpies of
formation. The standard enthalpy change for the reaction can be calculated
using the formula:
∆H=Xn·Hf(products) −Xn·Hf(reactants)
9
where nis the stoichiometric coefficient and Hfis the standard enthalpy of
formation.
Given: n(N2) = 2 moles, n(H2) = 3 moles, n(NH3) = 2 moles.
Plugging in the values:
∆H= [2 ·Hf(NH3)] −[2 ·Hf(N2)+3·Hf(H2)]
∆H= [2 ·(−46.1 kJ/mol)] −[2 ·0+3·0]
∆H=−92.2 kJ −0 kJ
∆H=−92.2 kJ
Therefore, the change in enthalpy for the reaction is ∆H=−92.2 kJ.
Question 12
Question
A reaction is known to have an enthalpy change of ∆H=−185 kJ/mol. If
the reaction is spontaneous at low temperatures, will it be spontaneous at high
temperatures as well? Justify your answer.
Solution
Step 1: Recall that the spontaneity of a reaction can be determined by the sign
of the Gibbs free energy change (∆G) using the equation:
∆G= ∆H−T∆S
where ∆His the enthalpy change, ∆Sis the entropy change, and Tis the
temperature in Kelvin.
Step 2: At low temperatures, the reaction is spontaneous, meaning ∆G < 0.
Given that ∆H=−185 kJ/mol, we can rewrite the equation as:
∆G=−185 −T∆S < 0
Step 3: To determine if the reaction remains spontaneous at high tempera-
tures, we need to consider the effect of temperature on the spontaneity of the
reaction. As temperature increases, the T∆Sterm in the equation becomes
more significant.
Step 4: If the increase in T∆Sis greater than the decrease in ∆H, it is
possible that the reaction will no longer be spontaneous at high temperatures.
Step 5: Therefore, whether the reaction remains spontaneous at high tem-
peratures depends on the magnitude of the change in entropy with temperature.
If the increase in entropy with temperature compensates for the decrease in en-
thalpy, the reaction could become non-spontaneous.
Step 6: In conclusion, the spontaneity of the reaction at high temperatures
cannot be definitively determined without knowing the entropy change (∆S)
and the specific temperature. An increase in temperature may lead to a change
in spontaneity depending on the entropy change of the system.
10
Question 13
Question
Calculate the change in enthalpy (∆H) for the reaction below at 25◦C:
2H2(g)+O2(g)→2H2O(l)
Given the following standard enthalpy of formation values:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(l)) = −286 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction and identify the
formation reactions involved.
The balanced chemical equation is:
2H2(g)+O2(g)→2H2O(l)
The formation reactions involved are:
H2(g) + 1
2O2(g)→H2O(l) ∆H1=−286 kJ/mol
Step 2: Calculate the change in enthalpy (∆H) for the overall reaction using
Hess’s Law.
Since we have two moles of water in the desired reaction, we need to multiply
the formation reaction for water by 2 and subtract the formation reactions for
hydrogen and oxygen:
∆H= 2 ×∆H1−(0 + 0) = 2 ×(−286) = −572 kJ/mol
Therefore, the change in enthalpy (∆H) for the given reaction is -572 kJ/mol
at 25◦C.
Question 14
Question
Calculate the change in enthalpy (∆H) when 10.0 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 heat of vaporization of water is 2260 J/g, and the specific heat capacity of
steam is 1.84 J/g
°
C.
11
Solution
Step 1: Calculate the heat required to raise the temperature of water from
25
°
C to 100
°
C using the specific heat capacity of water. Step 2: Calculate
the heat required to convert the water at 100
°
C to steam at 100
°
C using the
heat of vaporization of water. Step 3: Calculate the heat required to raise the
temperature of steam from 100
°
C to 100
°
C using the specific heat capacity of
steam. Step 4: Add up the heats calculated in steps 1, 2, and 3 to find the total
heat absorbed, which is equal to the change in enthalpy.
Question 15
Question
Calculate the change in enthalpy (∆H) when 2 moles of methane gas (CH4)
react with excess oxygen gas to produce carbon dioxide (CO2) and water (H2O)
according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the standard enthalpies of formation (∆H◦
f) for the compounds in-
volved in this reaction:
∆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 ∆Hfor the reaction using the given standard enthalpies
of formation.
The standard enthalpy change for the reaction can be calculated using the
formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Plugging in the values:
∆H=2×∆H◦
f(H2O)+∆H◦
f(CO2)−∆H◦
f(CH4)+2×∆H◦
f(O2)
∆H= [2 ×(−285.8) + (−393.5)] −[(−74.8) + 2 ×0]
∆H= (−571.6−393.5) −(−74.8)
∆H=−965.1 + 74.8
∆H=−890.3 kJ/mol
Therefore, the change in enthalpy for the given reaction is -890.3 kJ.
12
Question 16
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + 3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction. The given reac-
tion is:
2C(graphite) + 3H2(g)→C2H6(g)
Step 2: Determine the change in enthalpy using the standard enthalpies
of formation. The standard enthalpy change (∆H◦) for the reaction can be
calculated using the equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Given:
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
Substitute the values into the equation:
∆H◦= [−84.68] −[2(0) + 3(0)]
∆H◦=−84.68 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is −84.68 kJ/mol.
Question 17
Question
Calculate the change in enthalpy (∆H) when 50.0 g of water at 30.0
°
C is
converted to steam at 100.0
°
C. The specific heat capacity of water is 4.18
J/(g
°
C), the heat of vaporization for water is 2260 J/g, and the specific heat
capacity of steam is 2.01 J/(g
°
C).
13
Solution
Step 1: Calculate the heat required to heat the water from 30.0
°
C to 100.0
°
C.
The formula to calculate the heat required is given by:
q=mc∆T
where: q= heat energy, m= mass of the substance, c= specific heat capacity
of the substance, ∆T= change in temperature.
Substitute the given values into the formula:
q= (50.0 g)(4.18 J/(g
°
C))(100.0−30.0)
°
C
q= (50.0)(4.18)(70)
q= 14690 J
Step 2: Calculate the heat required to convert water at 100.0
°
C to steam
at 100.0
°
C. The formula to calculate the heat required is given by:
q=mL
where: q= heat energy, m= mass of the substance, L= heat of vaporization.
Substitute the given values into the formula:
q= (50.0 g)(2260 J/g) = 113000 J
Step 3: Calculate the heat required to heat the steam from 100.0
°
C to steam
at 100.0
°
C. The formula to calculate the heat required is given by:
q=mc∆T
Substitute the given values into the formula:
q= (50.0 g)(2.01 J/(g
°
C))(100.0−100.0)
°
C = 0 J
Step 4: Calculate the total heat required:
qtotal =q1+q2+q3= 14690 + 113000 + 0 = 127690 J
Step 5: Calculate the change in enthalpy:
∆H=qtotal
∆H= 127690 J
Therefore, the change in enthalpy when 50.0 g of water at 30.0
°
C is converted
to steam at 100.0
°
C is 127690 J.
14
Question 18
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas react
with 15.00 moles of oxygen gas to form 10.00 moles of nitrogen dioxide gas. The
reaction is carried out at constant pressure and 298 K.
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
equation for the reaction is:
2N2(g) + 4O2(g)→4NO2(g)
Step 2: Determine the enthalpies of formation for all reactants and products.
The enthalpy change for the reaction is given by:
∆H=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
where ∆H◦
fis the standard enthalpy of formation.
Given: ∆H◦
f(N2) = 0 kJ/mol ∆H◦
f(O2) = 0 kJ/mol ∆H◦
f(NO2) = 34.0
kJ/mol
Step 3: Calculate the change in enthalpy using the enthalpies of formation.
Substitute the values into the equation:
∆H= (4 ×34.0 kJ/mol) −(2 ×0 kJ/mol + 4 ×0 kJ/mol)
∆H= 136.0 kJ/mol
Step 4: Calculate the change in enthalpy for the given number of moles of
reactants. Since the reaction produces 4 moles of nitrogen dioxide for every 2
moles of nitrogen reacted:
∆H= (136.0 kJ/mol) ×10.00
2= 680 kJ
Therefore, the change in enthalpy for the reaction of 5.00 moles of nitrogen
with 15.00 moles of oxygen to produce 10.00 moles of nitrogen dioxide is 680
kJ.
Question 19
Question
A reaction has a standard enthalpy change of -92 kJ/mol. Calculate the maxi-
mum amount of heat (in kJ) that could be released by this reaction if 5.0 moles
of the reactant are consumed.
15
Solution
Given: ∆H=−92 kJ/mol, n= 5.0 mol
We can use the formula:
Maximum heat released = ∆H×n
Step 1: Substitute the given values into the formula.
Maximum heat released = −92 kJ/mol ×5.0 mol
Step 2: Calculate the maximum heat released.
Maximum heat released = −92 kJ/mol ×5.0 mol = −460 kJ
Therefore, the maximum amount of heat that could be released by this
reaction if 5.0 moles of the reactant are consumed is 460 kJ.
Question 20
Question
Calculate the change in enthalpy (∆H) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
given the following enthalpy values:
C2H6(g)→2CO2(g)+3H2O(g) ∆H=−3120 kJ/mol
2CO2(g)+3H2O(g)→2C2H6(g)+7/2O2(g) ∆H= +3530 kJ/mol
Solution
Step 1: Calculate ∆Hfor the given reaction using the enthalpy values provided.
∆H=X∆Hproducts −X∆Hreactants
Step 2: First, let’s find ∆Hproducts.
∆Hproducts = 4(0) + 6(−241.8) = −1450.8 kJ/mol
Step 3: Now, let’s find ∆Hreactants.
∆Hreactants =−3120 + 3530 = 410 kJ/mol
Step 4: Finally, calculate the change in enthalpy for the reaction.
∆H= ∆Hproducts −∆Hreactants =−1450.8−410 = −1860.8 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H=−1860.8 kJ/mol.
16
Question 21
Question
Calculate the change in enthalpy when 10 grams of ice at -10
°
C is heated to form
steam at 110
°
C. Given the specific heat capacities: ice = 2.09 J/g
°
C, water =
4.18 J/g
°
C, steam = 2.03 J/g
°
C, heat of fusion for ice = 333.55 J/g, and heat
of vaporization for water = 2260 J/g.
Solution
Step 1: Calculate the heat required to heat the ice to 0
°
C and melt it.
Heat required to heat ice to 0
°
C: q=m×cice ×∆T
q= 10 g ×2.09 J/g
°
C×(0 −(−10))
°
C
q= 10 g ×2.09 J/g
°
C×10
°
C = 209 J
Heat required to melt ice at 0
°
C: q=m×heat of fusion
q= 10 g ×333.55 J/g = 3335.5 J
Total heat required: qtotal = 209 J + 3335.5 J = 3544.5 J
Step 2: Calculate the heat required to heat the water from 0
°
C to 100
°
C and
vaporize it.
Heat required to heat water to 100
°
C: q=m×cwater ×∆T
q= 10 g ×4.18 J/g
°
C×(100 −0)
°
C = 4180 J
Heat required to vaporize water at 100
°
C: q=m×heat of vaporization
q= 10 g ×2260 J/g = 22600 J
Total heat required: qtotal = 4180 J + 22600 J = 26780 J
Step 3: Calculate the heat required to heat the steam from 100
°
C to 110
°
C.
Heat required to heat steam to 110
°
C: q=m×csteam ×∆T
q= 10 g ×2.03 J/g
°
C×(110 −100)
°
C = 203 J
Step 4: Calculate the total change in enthalpy.
Total change in enthalpy: ∆H=qtotal = 3544.5 J + 26780 J + 203 J =
30327.5 J
17
Question 22
Question
A reaction at constant pressure releases 250 kJ of heat and does 150 kJ of work
on the surroundings. Determine the change in enthalpy for the system.
Solution
Step 1: The change in enthalpy (∆H) for the system can be calculated using
the formula:
∆H= ∆U+P∆V
where ∆Uis the change in internal energy, Pis the pressure, and ∆Vis the
change in volume.
Step 2: Given that the reaction releases 250 kJ of heat, this means ∆U=
−250 kJ (as heat is being released from the system). Given that work is done
on the surroundings, work is negative for the system, so W=−150 kJ.
Step 3: From the first law of thermodynamics, we have:
∆U=q+W
where qis the heat exchanged in the system
Step 4: Rearranging the equation, we get:
q= ∆U−W
Step 5: Substituting the given values, we find:
q=−250 kJ −(−150 kJ) = −250 kJ + 150 kJ = −100 kJ
Step 6: Now, we can determine the change in enthalpy using the formula:
∆H= ∆U+P∆V=q+W
∆H=−100 kJ −150 kJ = −250 kJ
Step 7: Therefore, the change in enthalpy for the system is ∆H=−250 kJ.
Question 23
Question
A reaction is carried out in a bomb calorimeter, and it is found that 1.50 kJ of
heat is released. The temperature of the calorimeter and its contents increases
by 2.50
°
C. Calculate the enthalpy change (∆H) for the reaction, assuming the
specific heat capacity of the calorimeter and its contents is 4.18 J/g
°
C.
18
Solution
Step 1: Calculate the heat absorbed by the calorimeter and its contents. The
heat absorbed (q) is given by the equation:
q=mc∆T
where: - m= mass of the calorimeter and its contents - c= specific heat capacity
of the calorimeter and its contents - ∆T= change in temperature
Given that c= 4.18 J/g
°
C, ∆T= 2.50
°
C, and q=−1.50 kJ: Converting q
to joules: −1.50 kJ = −1.50 ×103J Plugging the values into the formula:
−1.50 ×103=m×4.18 ×2.50
Solving for m:
m=−1.50 ×103
4.18 ×2.50
m≈ −1436.12 g
Step 2: Calculate the enthalpy change (∆H) of the reaction. The enthalpy
change is related to the heat absorbed or released by the reaction and the moles
of reactant or product involved in the reaction. The formula to calculate the
enthalpy change is:
∆H=q
n
where: - q= heat absorbed or released - n= moles of reactant or product
Since the heat released was −1.50 kJ (or −1.50 ×103J) and assuming 1
mole of reaction occurred:
∆H=−1.50 ×103
1
∆H=−1.50 ×103J/mol
Therefore, the enthalpy change for the reaction is −1.50 kJ/mol.
Question 24
Question
A sample of water at 25
°
C is heated until it reaches a temperature of 75
°
C.
During this process, the water absorbs 200 kJ of heat. Calculate the change in
enthalpy (∆H) for this process.
19
Solution
Step 1: Determine the mass of water. Given that the specific heat capacity of
water is 4.18 J/g
°
C, we can use the formula:
q=mc∆T
Where: q= 200 kJ = 200000 J m= mass of water c= 4.18 J/g
°
C ∆T=
75C−25C= 50C
Solving for m:
200000 = m×4.18 ×50
m=200000
209
m≈956.93 g = 0.957 kg
Step 2: Calculate the change in enthalpy. The change in enthalpy (∆H) can
be calculated using the formula:
∆H=q=mc∆T
Substitute the known values:
∆H= 0.957 ×4.18 ×50
∆H≈200.292 kJ
Therefore, the change in enthalpy for this process is approximately 200.292
kJ.
Question 25
Question
A chemical reaction at constant pressure involves the combustion of methane
gas. The enthalpy change for this reaction is -890 kJ/mol. If 3.00 moles of
methane gas are combusted, what is the total amount of heat absorbed or
released by the reaction?
Solution
Step 1: Determine the total amount of heat absorbed or released by the reaction.
Given: Enthalpy change for the reaction (∆H) = -890 kJ/mol Number of moles
of methane gas combusted = 3.00 moles
Step 2: Calculate the total heat absorbed or released. The total heat ab-
sorbed or released by the reaction can be calculated using the formula: Total
heat = ∆H×moles of gas
20
Substitute the given values into the formula: Total heat = -890 kJ/mol ×
3.00 moles Total heat = -2670 kJ
Therefore, the total amount of heat absorbed or released by the reaction is
2670 kJ , with the negative sign indicating that heat is released by the reaction.
Question 26
Question
A reaction has an enthalpy change of -346 kJ/mol. If 0.25 mol of the reactant
is consumed, calculate the amount of heat exchanged in kJ.
Solution
Step 1: Determine the heat exchanged for 1 mol of the reactant. Given that
the enthalpy change for the reaction is -346 kJ/mol, it means that 1 mol of the
reactant reacts with the release of 346 kJ of heat.
Therefore, the heat exchanged for 1 mol of the reactant is -346 kJ.
Step 2: Calculate the heat exchanged for 0.25 mol of the reactant. To find
the heat exchanged for 0.25 mol of the reactant, we need to multiply the heat
exchanged for 1 mol by the number of moles consumed.
0.25 mol ×(−346 kJ/mol) = −86.5 kJ
Therefore, when 0.25 mol of the reactant is consumed, the amount of heat
exchanged is -86.5 kJ.
Question 27
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following standard enthalpy of formation values:
∆H◦
f(H2O(l)) = −286 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g)+O2(g)→2H2O(l)
21
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given values into the formula to find ∆H◦.
∆H◦= [2 ·∆H◦
f(H2O(l))] −[2 ·∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
Step 4: Substitute the specific values for the enthalpies of formation and
solve the equation.
∆H◦= [2 ·(−286 kJ/mol)] −[2 ·0 kJ/mol + 0 kJ/mol]
∆H◦=−572 kJ/mol
So, the standard enthalpy change for the given reaction is ∆H◦=−572 kJ/mol.
Question 28
Question
A certain reaction has an enthalpy change of ∆H=−345 kJ. Calculate the
amount of heat (in joules) absorbed or released when 5.00 mol of the reactant
is consumed in the reaction.
Solution
Step 1: First, we need to convert the given enthalpy change from kilojoules to
joules.
Given: ∆H=−345 kJ = −345 ×1000 J = −345000 J
Step 2: Calculate the heat absorbed or released when 1.00 mol of the reactant
is consumed using the given enthalpy change.
Heat for 1.00 mol = ∆H=−345000 J = −345000 J/mole
Step 3: Determine the heat absorbed or released when 5.00 mol of the reac-
tant is consumed by multiplying the heat for 1.00 mol by 5.00.
Heat for 5.00 mol = −345000 J/mole ×5.00 mol = −1725000 J
Therefore, 5.00 mol of the reactant in the reaction would absorb or release
−1725000 J of heat.
22
Question 29
Question
Calculate the change in enthalpy (∆H) when 10.0 grams of water at 25
°
C is
heated to form steam at 100
°
C. The specific heat capacity of water is 4.18 J/g
°
C,
the specific heat capacity of steam is 2.0 J/g
°
C, and the heat of vaporization of
water is 2260 J/g.
Solution
Step 1: Calculate the energy required to heat the water from 25
°
C to 100
°
C.
q=mc∆T
q= (10.0 g)(4.18 J/g
°
C)(100C−25C)
q= 2930 J
Step 2: Calculate the energy required to evaporate the water at 100
°
C.
q=m×heat of vaporization
q= (10.0 g)(2260 J/g)
q= 22600 J
Step 3: Calculate the total energy change.
∆H=qheating +qvaporization
∆H= 2930 J + 22600 J
∆H= 25530 J
Therefore, the change in enthalpy (∆H) when 10.0 grams of water is heated
to form steam is 25530 J.
Question 30
Question
Calculate the change in enthalpy for a reaction in which 2 moles of magnesium
metal react with excess hydrochloric acid to form magnesium chloride and hy-
drogen gas. The enthalpies of formation for magnesium chloride and hydrogen
gas are -641 kJ/mol and 0 kJ/mol, respectively.
23
Solution
Step 1: Write the balanced chemical equation for the reaction.
Mg(s) + 2HCl(aq) →MgCl2(aq)+H2(g)
Step 2: Determine the molar enthalpy change for the reaction using the
enthalpies of formation. The enthalpy change for the reaction can be calculated
using the formula:
∆H=X∆Hproducts −X∆Hreactants
Substitute the values given into the equation:
∆H= [1(−641 kJ/mol) + 1(0 kJ/mol)] −[1(0 kJ/mol) + 2(0 kJ/mol)]
∆H=−641 kJ/mol
Step 3: Calculate the change in enthalpy for the given reaction. Since 2
moles of magnesium is involved in the reaction, the total change in enthalpy
would be:
∆Htotal = 2 ×∆H= 2 ×(−641 kJ/mol)
∆Htotal =−1282 kJ
Therefore, the change in enthalpy for the reaction is -1282 kJ.
Question 31
Question
A reaction at constant pressure has a change in enthalpy of −158 kJ. If the
reaction also has a change in internal energy of −120 kJ, calculate the work
done by the reaction.
Solution
Step 1: Recall the relationship between enthalpy change, internal energy change,
and work done by the reaction. The relationship between the change in enthalpy
(∆H), change in internal energy (∆U), and work done (W) by the reaction is
given by:
∆H= ∆U+P∆V
where Pis the pressure and ∆Vis the change in volume.
Step 2: Given that ∆H=−158 kJ, ∆U=−120 kJ, and the reaction is at
constant pressure, we can rearrange the equation to solve for the work done:
W= ∆H−∆U
24
Step 3: Substitute the given values into the formula to calculate the work
done:
W=−158 kJ −(−120 kJ)
W=−158 kJ + 120 kJ
W=−38 kJ
Therefore, the work done by the reaction is −38 kJ.
Question 32
Question
Determine the change in enthalpy (∆H) for the reaction:
2C(s) + 3H2(g) →C2H6(g)
given the following standard enthalpies of formation:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Solution
Step 1: Write the equation for the reaction in terms of the standard enthalpies
of formation of the reactants and products. Step 2: Calculate the standard en-
thalpy of reaction using the standard enthalpies of formation. Step 3: Interpret
the sign of ∆Hin terms of whether the reaction is exothermic or endothermic.
Step 1: The enthalpy change for the reaction can be expressed as:
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Step 2: Substitute the given values into the formula:
∆H= 1(−84.68 kJ/mol) −2(0 kJ/mol) −3(0 kJ/mol)
=−84.68 kJ/mol
So, the change in enthalpy for the reaction is ∆H=−84.68 kJ/mol.
Step 3: Since the value of ∆His negative, the reaction is exothermic (it
releases heat).
Question 33
Question
A chemical reaction is carried out at constant pressure. The reaction releases
250 kJ of heat and does work of 100 kJ on the surroundings. Determine the
change in enthalpy for the reaction.
25
Solution
Step 1: Recall that the change in enthalpy (∆H) for a reaction is equal to the
heat exchanged at constant pressure (qp) plus the work done on the surround-
ings.
∆H=qp+w
Step 2: Given that the reaction releases 250 kJ of heat (qp=−250 kJ) and
does work of 100 kJ on the surroundings (w=−100 kJ), substitute these values
into the equation.
∆H=−250 kJ −100 kJ
Step 3: Calculate the change in enthalpy.
∆H=−350 kJ
Therefore, the change in enthalpy for the reaction is -350 kJ.
Question 34
Question
A reaction is carried out in a bomb calorimeter. Initially, the temperature of
the calorimeter was 25
°
C. During the reaction, the temperature of the water in
the calorimeter rose to 35
°
C. The bomb calorimeter has a heat capacity of 400
J/
°
C. If the enthalpy change for the reaction is -150 kJ/mol, how many moles
of the reactant are present in the bomb calorimeter?
Solution
Step 1: Calculate the heat absorbed by the bomb calorimeter. The heat ab-
sorbed by the bomb calorimeter is equal to the heat released by the reaction:
q=C·∆T
where: q= heat absorbed by calorimeter C= heat capacity of calorimeter (400
J/
°
C) ∆T= change in temperature (10
°
C) Plugging in the values:
q= 400 J/
°
C·10
°
C = 4000 J
Step 2: Convert heat absorbed to kJ.
q= 4000 J = 4 kJ
Step 3: Calculate the number of moles of reactant. Given: Enthalpy change
(∆H) = -150 kJ/mol ∆H=qfor the reaction
−150 kJ/mol = 4 kJ ×n
26
where: n= number of moles of reactant Solving for n:
n=−150 kJ/mol
4 kJ =−37.5 mol
Since moles cannot be negative, it implies that the sign of the enthalpy change
was chosen incorrectly. The correct sign should be:
n=150 kJ/mol
4 kJ = 37.5 mol
Therefore, there are 37.5 moles of the reactant present in the bomb calorime-
ter.
Question 35
Question
A certain reaction has an enthalpy change of −285 kJ/mol. If this reaction is
carried out at constant pressure and releases 95 kJ of heat, what is the change
in internal energy for the reaction?
Solution
Step 1: Recall that the relationship between enthalpy change (∆H), heat re-
leased (q), and change in internal energy (∆U) is given by the equation ∆H=
∆U+P∆V, where P∆Vis the work done by the system. Step 2: Since the
reaction is carried out at constant pressure, P∆V= ∆nRT , where ∆nis the
change in moles of gas, Ris the ideal gas constant, and Tis the temperature in
Kelvin. Step 3: In this case, since P∆Vis not given and we are asked to find
∆U, we can simplify the equation to ∆U= ∆H−q. Step 4: Substitute the
given values into the equation: ∆U=−285 kJ/mol −95 kJ. Step 5: Calculate
the change in internal energy: ∆U=−380 kJ. Step 6: Therefore, the change
in internal energy for the reaction is −380 kJ.
27
Question 2
Question
Given that the enthalpy change for the reaction 2 H2(g) + O2(g) −−→ 2 H2O(g)
is −483.6 kJ, calculate the enthalpy change for the reaction 2 H2O(g) −−→
2 H2(g) + O2(g).
Solution
Step 1: Write the enthalpy change for the second reaction in terms of the
enthalpy change for the first reaction. Let xrepresent the enthalpy change
for the second reaction. The enthalpy change for the second reaction is the
negative of the enthalpy change for the first reaction, since the second reaction
is the reverse of the first reaction.
x=−(−483.6)
x= 483.6 kJ
Therefore, the enthalpy change for the reaction 2 H2O(g) −−→ 2 H2(g) +
O2(g) is 483.6 kJ.
Question 3
Question
Calculate the change in enthalpy (∆H) when 1 mol of liquid water at 25◦C
is heated to form 1 mol of water vapor at 100◦C. Given that the specific heat
capacities of liquid water and steam are 4.18 J/g
°
C and 2.03 J/g
°
C, respectively,
and the heat of vaporization of water is 40.79 kJ/mol.
Solution
Step 1: Calculate the heat required to heat liquid water from 25◦C to 100◦C.
Use the formula: q=mc∆T, where - qis the heat required, - mis the mass
of the substance, - cis the specific heat capacity of the substance, and - ∆Tis
the change in temperature.
Given that 1 mol of water has a mass of 18 g (since the molar mass of water
is 18 g/mol), we can calculate the heat required as follows:
q= (18 g)(4.18 J/g
°
C)(100 −25)◦C
q= 18 mol ×4.18 J/g
°
C×75C
q= 5613 J/mol
Step 2: Calculate the heat required to vaporize the water at 100◦C.
2
The heat required to vaporize water is the heat of vaporization, which is
40.79 kJ/mol.
Step 3: Calculate the total change in enthalpy.
The change in enthalpy (∆H) is the sum of the heat required to heat the
water from 25◦C to 100◦C and the heat required to vaporize the water. Let’s
calculate it:
∆H= 5613 J/mol + 40.79 ×103J/mol
∆H= 46.403 ×103J/mol
∆H= 46.403 kJ/mol
Therefore, the change in enthalpy when 1 mol of liquid water at 25◦C is
heated to form 1 mol of water vapor at 100◦C is 46.403 kJ/mol.
Question 4
Question
Given the enthalpy of formation values for the following reactions:
Reaction 1: 2H2(g)+O2(g)→2H2O(l) ∆H0
f=−572 kJ/mol
Reaction 2: C(s)+O2(g)→CO2(g) ∆H0
f=−394 kJ/mol
Reaction 3: 2H2O(l)→2H2(g)+O2(g) ∆H0
f= 476 kJ/mol
Calculate the standard enthalpy change for the reaction:
C(s)+O2(g)→CO2(g)
Solution
Step 1: Write the target reaction as the sum of the given reactions, adjusting for
stoichiometry and direction if necessary. Step 2: Calculate the overall enthalpy
change for the target reaction using the enthalpy of formation values. Step 3:
Substitute the given enthalpy of formation values into the equation and solve
for the overall enthalpy change.
Step 1: To obtain the target reaction, we need to reverse Reaction 2 and
halve Reaction 1. This will cancel out H2O on both sides, resulting in the target
reaction:
C(s)+O2(g)→CO2(g)
Now, the reaction will be:
C(s) + 2H2(g) + 1
2O2(g)→CO2(g) + 2H2(g)
3
Step 2: The overall enthalpy change for the target reaction can be calculated
using the enthalpy of formation values:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given enthalpy of formation values into the equation:
∆H=−394 kJ
mol−0kJ
mol + 0 kJ
mol +1
2×−572 kJ
mol
∆H=−394 kJ
mol + 286 kJ
mol
∆H=−108 kJ
mol
Therefore, the standard enthalpy change for the reaction is ∆H=−108 kJ/mol.
Question 5
Question
Calculate the change in enthalpy (∆H) when 2 moles of methane gas react with
excess oxygen gas to form carbon dioxide gas and water vapor according to the
following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpies of formation: H◦
f(CH4) = −74.8
kJ/mol H◦
f(CO2) = −393.5 kJ/mol H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of reaction (∆H◦
rxn) using the standard
enthalpies of formation.
The standard enthalpy change for the reaction is given by:
∆H◦
rxn =XνfH◦
f(products) −XνfH◦
f(reactants)
where νfis the stoichiometric coefficient.
Substitute the given standard enthalpies of formation values into the equa-
tion:
∆H◦
rxn = [1 ∗(−393.5) + 2 ∗(−285.8)] −[1 ∗(−74.8) + 2 ∗0]
∆H◦
rxn = (−393.5−571.6) −(−74.8)
∆H◦
rxn =−965.1 + 74.8
∆H◦
rxn =−890.3 kJ
4
Therefore, the standard enthalpy change for the reaction is ∆H◦
rxn =−890.3
kJ.
Step 2: Calculate the actual change in enthalpy (∆H) for 2 moles of methane
gas.
Since the balanced chemical equation shows the reaction of 1 mole of methane,
we need to scale the enthalpy value accordingly:
∆H=2 moles
1 mole ∗∆H◦
rxn
∆H= 2 ∗(−890.3)
∆H=−1780.6 kJ
Therefore, the change in enthalpy for the reaction of 2 moles of methane gas
is ∆H=−1780.6 kJ.
Question 6
Question
Given the reaction:
2A(g) + 3B(g)→C(g) + D(g)
with the following enthalpy changes:
∆Hrxn =−580 kJ/mol
∆Hf C =−390 kJ/mol
∆Hf D =−125 kJ/mol
Calculate the standard enthalpy change of formation for compound A if the
standard enthalpy change of formation for compound B is -200 kJ/mol.
Solution
Step 1: Calculate the total enthalpy of formation for the products C and D.
The total enthalpy of formation for the products is given by:
∆Hf C+D = ∆Hf C + ∆Hf D
∆Hf C+D =−390 kJ/mol + (−125 kJ/mol)
∆Hf C+D =−515 kJ/mol
Step 2: Calculate the total enthalpy change for the reactants A and B. The
total enthalpy change for the reactants is given by:
∆Hrxn =Xνproducts∆Hf products −Xνreactants∆Hf reactants
5
∆Hrxn = (∆Hf C+D)−(2 ×∆Hf A + 3 ×∆Hf B)
−580 kJ/mol = −515 kJ/mol −2×∆Hf A −3×(−200 kJ/mol)
Step 3: Solve for the enthalpy change of formation of compound A.
∆Hf A =−580 kJ/mol −(−515 kJ/mol) + 3 ×(−200 kJ/mol)
2
∆Hf A =−407.5 kJ/mol
Therefore, the standard enthalpy change of formation for compound A is
-407.5 kJ/mol.
Question 7
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of ammonia gas (NH3)
react with excess oxygen gas (O2) according to the following balanced chemical
equation:
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
Given that ∆Hfor the reaction is -940 kJ.
Solution
Step 1: Determine the moles of ammonia reacting Given: Number of moles of
ammonia, n= 5.00 moles.
Step 2: Determine the molar ratio between NH3and ∆HFrom the balanced
chemical equation, the ratio of NH3to ∆His 4:940.
Step 3: Calculate the change in enthalpy (∆H)
∆H=940 kJ
4 mol ×5.00 mol = −2350 kJ
Therefore, the change in enthalpy (∆H) for the reaction when 5.00 moles of
ammonia gas react is -2350 kJ.
Question 8
Question
Given the following reaction at 25
°
C:
2A(g)+3B(g)→C(g)+2D(g)
6
If the standard enthalpies of formation (∆H◦
f) are as follows:
∆H◦
f(A) = −200 kJ/mol
∆H◦
f(B) = −300 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −100 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.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [(∆H◦
f(C) + 2∆H◦
f(D)) −(2∆H◦
f(A) + 3∆H◦
f(B))]
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion.
∆H◦= [(−400 kJ/mol+2(−100 kJ/mol))−(2(−200 kJ/mol)+3(−300 kJ/mol))]
Step 3: Perform the calculations to find the standard enthalpy change.
∆H◦= [(−400 kJ/mol −200 kJ/mol) −(−400 kJ/mol −900 kJ/mol)]
∆H◦= [−600 kJ/mol −(−1300 kJ/mol)]
∆H◦=−600 kJ/mol + 1300 kJ/mol
∆H◦= 700 kJ/mol
Therefore, the standard enthalpy change for the reaction is 700 kJ/mol .
Question 9
Question
Calculate the change in enthalpy (∆H) when 15.0 g of ice at -10.0
°
C is converted
to liquid water at 50.0
°
C. The specific heat capacity of ice is 2.09 J/g
°
C, the
heat of fusion for ice is 333 J/g, the specific heat capacity of water is 4.18 J/g
°
C,
and the heat of vaporization for water is 2260 J/g. Assume no heat is lost to
the surroundings.
7
Solution
Step 1: Calculate the heat required to raise the temperature of the ice from
-10.0
°
C to 0
°
C.
q=m×c×∆T
q= 15.0 g ×2.09 J/g
°
C×(0C−(−10.0C))
q= 313.5 J
Step 2: Calculate the heat required to melt the ice at 0
°
C.
q=m×∆Hfusion
q= 15.0 g ×333 J/g
q= 4995 J
Step 3: Calculate the heat required to raise the temperature of the water
from 0
°
C to 50.0
°
C.
q=m×c×∆T
q= 15.0 g ×4.18 J/g
°
C×(50.0C−0C)
q= 3135 J
Step 4: Calculate the total heat required for the process.
qtotal =q1+q2+q3
qtotal = 313.5 J + 4995 J + 3135 J
qtotal = 8443.5 J
Step 5: Calculate the change in enthalpy (∆H).
∆H=qtotal
∆H= 8443.5 J
Therefore, the change in enthalpy (∆H) when 15.0 g of ice at -10.0
°
C is
converted to liquid water at 50.0
°
C is 8443.5 J.
Question 10
Question
Given the reaction:
2A+ 3B→C+ 4D
where ∆H=−232 kJ, calculate the enthalpy change when 4.50 mol of A reacts
completely with 2.00 mol of B. Assume the reaction takes place under standard
conditions.
8
Solution
Step 1: Calculate the moles of limiting reagent Let’s first determine the limiting
reagent between A and B to find out which one will be completely consumed.
The stoichiometric ratio between A and B is 2:3. Therefore, we need to compare
the moles of A and B: For A: 4.50 mol A For B: 2.00 mol B
Calculating the moles of B needed to react with 4.50 mol of A:
4.50 mol A
2 mol A ×3 mol B = 6.75 mol B
Since we only have 2.00 mol of B, B is the limiting reagent.
Step 2: Calculate the moles of products formed From the given reaction, we
can determine the moles of C and D formed when 2.00 mol of B is used: Using
the stoichiometry of the reaction: For B: 2.00 mol B For C: 1 mol C For D: 4
mol D
Step 3: Calculate the change in enthalpy Given: ∆H=−232 kJ for the
reaction. The enthalpy change can be calculated using the formula:
∆Htotal = ∆H×moles of products formed
stoichiometric coefficient of limiting reactant
Plugging in the values:
∆Htotal =−232 kJ ×2.00 mol of products
3=−154.67 kJ
Therefore, the enthalpy change when 4.50 mol of A reacts completely with
2.00 mol of B is −154.67 kJ.
Question 11
Question
Calculate the change in enthalpy (∆H) when 2 moles of nitrogen gas reacts
with 5 moles of hydrogen gas to form ammonia gas according to the following
reaction:
N2(g)+3H2(g)→2NH3(g)
Given the enthalpies of formation are: Hf(N2) = 0 kJ/mol, Hf(H2) = 0 kJ/mol,
Hf(NH3) = −46.1 kJ/mol.
Solution
Step 1: Calculate the enthalpy change for the reaction using the enthalpies of
formation. The standard enthalpy change for the reaction can be calculated
using the formula:
∆H=Xn·Hf(products) −Xn·Hf(reactants)
9
where nis the stoichiometric coefficient and Hfis the standard enthalpy of
formation.
Given: n(N2) = 2 moles, n(H2) = 3 moles, n(NH3) = 2 moles.
Plugging in the values:
∆H= [2 ·Hf(NH3)] −[2 ·Hf(N2)+3·Hf(H2)]
∆H= [2 ·(−46.1 kJ/mol)] −[2 ·0+3·0]
∆H=−92.2 kJ −0 kJ
∆H=−92.2 kJ
Therefore, the change in enthalpy for the reaction is ∆H=−92.2 kJ.
Question 12
Question
A reaction is known to have an enthalpy change of ∆H=−185 kJ/mol. If
the reaction is spontaneous at low temperatures, will it be spontaneous at high
temperatures as well? Justify your answer.
Solution
Step 1: Recall that the spontaneity of a reaction can be determined by the sign
of the Gibbs free energy change (∆G) using the equation:
∆G= ∆H−T∆S
where ∆His the enthalpy change, ∆Sis the entropy change, and Tis the
temperature in Kelvin.
Step 2: At low temperatures, the reaction is spontaneous, meaning ∆G < 0.
Given that ∆H=−185 kJ/mol, we can rewrite the equation as:
∆G=−185 −T∆S < 0
Step 3: To determine if the reaction remains spontaneous at high tempera-
tures, we need to consider the effect of temperature on the spontaneity of the
reaction. As temperature increases, the T∆Sterm in the equation becomes
more significant.
Step 4: If the increase in T∆Sis greater than the decrease in ∆H, it is
possible that the reaction will no longer be spontaneous at high temperatures.
Step 5: Therefore, whether the reaction remains spontaneous at high tem-
peratures depends on the magnitude of the change in entropy with temperature.
If the increase in entropy with temperature compensates for the decrease in en-
thalpy, the reaction could become non-spontaneous.
Step 6: In conclusion, the spontaneity of the reaction at high temperatures
cannot be definitively determined without knowing the entropy change (∆S)
and the specific temperature. An increase in temperature may lead to a change
in spontaneity depending on the entropy change of the system.
10
Question 13
Question
Calculate the change in enthalpy (∆H) for the reaction below at 25◦C:
2H2(g)+O2(g)→2H2O(l)
Given the following standard enthalpy of formation values:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(l)) = −286 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction and identify the
formation reactions involved.
The balanced chemical equation is:
2H2(g)+O2(g)→2H2O(l)
The formation reactions involved are:
H2(g) + 1
2O2(g)→H2O(l) ∆H1=−286 kJ/mol
Step 2: Calculate the change in enthalpy (∆H) for the overall reaction using
Hess’s Law.
Since we have two moles of water in the desired reaction, we need to multiply
the formation reaction for water by 2 and subtract the formation reactions for
hydrogen and oxygen:
∆H= 2 ×∆H1−(0 + 0) = 2 ×(−286) = −572 kJ/mol
Therefore, the change in enthalpy (∆H) for the given reaction is -572 kJ/mol
at 25◦C.
Question 14
Question
Calculate the change in enthalpy (∆H) when 10.0 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 heat of vaporization of water is 2260 J/g, and the specific heat capacity of
steam is 1.84 J/g
°
C.
11
Solution
Step 1: Calculate the heat required to raise the temperature of water from
25
°
C to 100
°
C using the specific heat capacity of water. Step 2: Calculate
the heat required to convert the water at 100
°
C to steam at 100
°
C using the
heat of vaporization of water. Step 3: Calculate the heat required to raise the
temperature of steam from 100
°
C to 100
°
C using the specific heat capacity of
steam. Step 4: Add up the heats calculated in steps 1, 2, and 3 to find the total
heat absorbed, which is equal to the change in enthalpy.
Question 15
Question
Calculate the change in enthalpy (∆H) when 2 moles of methane gas (CH4)
react with excess oxygen gas to produce carbon dioxide (CO2) and water (H2O)
according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the standard enthalpies of formation (∆H◦
f) for the compounds in-
volved in this reaction:
∆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 ∆Hfor the reaction using the given standard enthalpies
of formation.
The standard enthalpy change for the reaction can be calculated using the
formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Plugging in the values:
∆H=2×∆H◦
f(H2O)+∆H◦
f(CO2)−∆H◦
f(CH4)+2×∆H◦
f(O2)
∆H= [2 ×(−285.8) + (−393.5)] −[(−74.8) + 2 ×0]
∆H= (−571.6−393.5) −(−74.8)
∆H=−965.1 + 74.8
∆H=−890.3 kJ/mol
Therefore, the change in enthalpy for the given reaction is -890.3 kJ.
12
Question 16
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + 3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction. The given reac-
tion is:
2C(graphite) + 3H2(g)→C2H6(g)
Step 2: Determine the change in enthalpy using the standard enthalpies
of formation. The standard enthalpy change (∆H◦) for the reaction can be
calculated using the equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Given:
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
Substitute the values into the equation:
∆H◦= [−84.68] −[2(0) + 3(0)]
∆H◦=−84.68 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is −84.68 kJ/mol.
Question 17
Question
Calculate the change in enthalpy (∆H) when 50.0 g of water at 30.0
°
C is
converted to steam at 100.0
°
C. The specific heat capacity of water is 4.18
J/(g
°
C), the heat of vaporization for water is 2260 J/g, and the specific heat
capacity of steam is 2.01 J/(g
°
C).
13
Solution
Step 1: Calculate the heat required to heat the water from 30.0
°
C to 100.0
°
C.
The formula to calculate the heat required is given by:
q=mc∆T
where: q= heat energy, m= mass of the substance, c= specific heat capacity
of the substance, ∆T= change in temperature.
Substitute the given values into the formula:
q= (50.0 g)(4.18 J/(g
°
C))(100.0−30.0)
°
C
q= (50.0)(4.18)(70)
q= 14690 J
Step 2: Calculate the heat required to convert water at 100.0
°
C to steam
at 100.0
°
C. The formula to calculate the heat required is given by:
q=mL
where: q= heat energy, m= mass of the substance, L= heat of vaporization.
Substitute the given values into the formula:
q= (50.0 g)(2260 J/g) = 113000 J
Step 3: Calculate the heat required to heat the steam from 100.0
°
C to steam
at 100.0
°
C. The formula to calculate the heat required is given by:
q=mc∆T
Substitute the given values into the formula:
q= (50.0 g)(2.01 J/(g
°
C))(100.0−100.0)
°
C = 0 J
Step 4: Calculate the total heat required:
qtotal =q1+q2+q3= 14690 + 113000 + 0 = 127690 J
Step 5: Calculate the change in enthalpy:
∆H=qtotal
∆H= 127690 J
Therefore, the change in enthalpy when 50.0 g of water at 30.0
°
C is converted
to steam at 100.0
°
C is 127690 J.
14
Question 18
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of nitrogen gas react
with 15.00 moles of oxygen gas to form 10.00 moles of nitrogen dioxide gas. The
reaction is carried out at constant pressure and 298 K.
Solution
Step 1: Write the balanced chemical equation for the reaction. The balanced
equation for the reaction is:
2N2(g) + 4O2(g)→4NO2(g)
Step 2: Determine the enthalpies of formation for all reactants and products.
The enthalpy change for the reaction is given by:
∆H=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
where ∆H◦
fis the standard enthalpy of formation.
Given: ∆H◦
f(N2) = 0 kJ/mol ∆H◦
f(O2) = 0 kJ/mol ∆H◦
f(NO2) = 34.0
kJ/mol
Step 3: Calculate the change in enthalpy using the enthalpies of formation.
Substitute the values into the equation:
∆H= (4 ×34.0 kJ/mol) −(2 ×0 kJ/mol + 4 ×0 kJ/mol)
∆H= 136.0 kJ/mol
Step 4: Calculate the change in enthalpy for the given number of moles of
reactants. Since the reaction produces 4 moles of nitrogen dioxide for every 2
moles of nitrogen reacted:
∆H= (136.0 kJ/mol) ×10.00
2= 680 kJ
Therefore, the change in enthalpy for the reaction of 5.00 moles of nitrogen
with 15.00 moles of oxygen to produce 10.00 moles of nitrogen dioxide is 680
kJ.
Question 19
Question
A reaction has a standard enthalpy change of -92 kJ/mol. Calculate the maxi-
mum amount of heat (in kJ) that could be released by this reaction if 5.0 moles
of the reactant are consumed.
15
Solution
Given: ∆H=−92 kJ/mol, n= 5.0 mol
We can use the formula:
Maximum heat released = ∆H×n
Step 1: Substitute the given values into the formula.
Maximum heat released = −92 kJ/mol ×5.0 mol
Step 2: Calculate the maximum heat released.
Maximum heat released = −92 kJ/mol ×5.0 mol = −460 kJ
Therefore, the maximum amount of heat that could be released by this
reaction if 5.0 moles of the reactant are consumed is 460 kJ.
Question 20
Question
Calculate the change in enthalpy (∆H) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
given the following enthalpy values:
C2H6(g)→2CO2(g)+3H2O(g) ∆H=−3120 kJ/mol
2CO2(g)+3H2O(g)→2C2H6(g)+7/2O2(g) ∆H= +3530 kJ/mol
Solution
Step 1: Calculate ∆Hfor the given reaction using the enthalpy values provided.
∆H=X∆Hproducts −X∆Hreactants
Step 2: First, let’s find ∆Hproducts.
∆Hproducts = 4(0) + 6(−241.8) = −1450.8 kJ/mol
Step 3: Now, let’s find ∆Hreactants.
∆Hreactants =−3120 + 3530 = 410 kJ/mol
Step 4: Finally, calculate the change in enthalpy for the reaction.
∆H= ∆Hproducts −∆Hreactants =−1450.8−410 = −1860.8 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H=−1860.8 kJ/mol.
16
Question 21
Question
Calculate the change in enthalpy when 10 grams of ice at -10
°
C is heated to form
steam at 110
°
C. Given the specific heat capacities: ice = 2.09 J/g
°
C, water =
4.18 J/g
°
C, steam = 2.03 J/g
°
C, heat of fusion for ice = 333.55 J/g, and heat
of vaporization for water = 2260 J/g.
Solution
Step 1: Calculate the heat required to heat the ice to 0
°
C and melt it.
Heat required to heat ice to 0
°
C: q=m×cice ×∆T
q= 10 g ×2.09 J/g
°
C×(0 −(−10))
°
C
q= 10 g ×2.09 J/g
°
C×10
°
C = 209 J
Heat required to melt ice at 0
°
C: q=m×heat of fusion
q= 10 g ×333.55 J/g = 3335.5 J
Total heat required: qtotal = 209 J + 3335.5 J = 3544.5 J
Step 2: Calculate the heat required to heat the water from 0
°
C to 100
°
C and
vaporize it.
Heat required to heat water to 100
°
C: q=m×cwater ×∆T
q= 10 g ×4.18 J/g
°
C×(100 −0)
°
C = 4180 J
Heat required to vaporize water at 100
°
C: q=m×heat of vaporization
q= 10 g ×2260 J/g = 22600 J
Total heat required: qtotal = 4180 J + 22600 J = 26780 J
Step 3: Calculate the heat required to heat the steam from 100
°
C to 110
°
C.
Heat required to heat steam to 110
°
C: q=m×csteam ×∆T
q= 10 g ×2.03 J/g
°
C×(110 −100)
°
C = 203 J
Step 4: Calculate the total change in enthalpy.
Total change in enthalpy: ∆H=qtotal = 3544.5 J + 26780 J + 203 J =
30327.5 J
17
Question 22
Question
A reaction at constant pressure releases 250 kJ of heat and does 150 kJ of work
on the surroundings. Determine the change in enthalpy for the system.
Solution
Step 1: The change in enthalpy (∆H) for the system can be calculated using
the formula:
∆H= ∆U+P∆V
where ∆Uis the change in internal energy, Pis the pressure, and ∆Vis the
change in volume.
Step 2: Given that the reaction releases 250 kJ of heat, this means ∆U=
−250 kJ (as heat is being released from the system). Given that work is done
on the surroundings, work is negative for the system, so W=−150 kJ.
Step 3: From the first law of thermodynamics, we have:
∆U=q+W
where qis the heat exchanged in the system
Step 4: Rearranging the equation, we get:
q= ∆U−W
Step 5: Substituting the given values, we find:
q=−250 kJ −(−150 kJ) = −250 kJ + 150 kJ = −100 kJ
Step 6: Now, we can determine the change in enthalpy using the formula:
∆H= ∆U+P∆V=q+W
∆H=−100 kJ −150 kJ = −250 kJ
Step 7: Therefore, the change in enthalpy for the system is ∆H=−250 kJ.
Question 23
Question
A reaction is carried out in a bomb calorimeter, and it is found that 1.50 kJ of
heat is released. The temperature of the calorimeter and its contents increases
by 2.50
°
C. Calculate the enthalpy change (∆H) for the reaction, assuming the
specific heat capacity of the calorimeter and its contents is 4.18 J/g
°
C.
18
Solution
Step 1: Calculate the heat absorbed by the calorimeter and its contents. The
heat absorbed (q) is given by the equation:
q=mc∆T
where: - m= mass of the calorimeter and its contents - c= specific heat capacity
of the calorimeter and its contents - ∆T= change in temperature
Given that c= 4.18 J/g
°
C, ∆T= 2.50
°
C, and q=−1.50 kJ: Converting q
to joules: −1.50 kJ = −1.50 ×103J Plugging the values into the formula:
−1.50 ×103=m×4.18 ×2.50
Solving for m:
m=−1.50 ×103
4.18 ×2.50
m≈ −1436.12 g
Step 2: Calculate the enthalpy change (∆H) of the reaction. The enthalpy
change is related to the heat absorbed or released by the reaction and the moles
of reactant or product involved in the reaction. The formula to calculate the
enthalpy change is:
∆H=q
n
where: - q= heat absorbed or released - n= moles of reactant or product
Since the heat released was −1.50 kJ (or −1.50 ×103J) and assuming 1
mole of reaction occurred:
∆H=−1.50 ×103
1
∆H=−1.50 ×103J/mol
Therefore, the enthalpy change for the reaction is −1.50 kJ/mol.
Question 24
Question
A sample of water at 25
°
C is heated until it reaches a temperature of 75
°
C.
During this process, the water absorbs 200 kJ of heat. Calculate the change in
enthalpy (∆H) for this process.
19
Solution
Step 1: Determine the mass of water. Given that the specific heat capacity of
water is 4.18 J/g
°
C, we can use the formula:
q=mc∆T
Where: q= 200 kJ = 200000 J m= mass of water c= 4.18 J/g
°
C ∆T=
75C−25C= 50C
Solving for m:
200000 = m×4.18 ×50
m=200000
209
m≈956.93 g = 0.957 kg
Step 2: Calculate the change in enthalpy. The change in enthalpy (∆H) can
be calculated using the formula:
∆H=q=mc∆T
Substitute the known values:
∆H= 0.957 ×4.18 ×50
∆H≈200.292 kJ
Therefore, the change in enthalpy for this process is approximately 200.292
kJ.
Question 25
Question
A chemical reaction at constant pressure involves the combustion of methane
gas. The enthalpy change for this reaction is -890 kJ/mol. If 3.00 moles of
methane gas are combusted, what is the total amount of heat absorbed or
released by the reaction?
Solution
Step 1: Determine the total amount of heat absorbed or released by the reaction.
Given: Enthalpy change for the reaction (∆H) = -890 kJ/mol Number of moles
of methane gas combusted = 3.00 moles
Step 2: Calculate the total heat absorbed or released. The total heat ab-
sorbed or released by the reaction can be calculated using the formula: Total
heat = ∆H×moles of gas
20
Substitute the given values into the formula: Total heat = -890 kJ/mol ×
3.00 moles Total heat = -2670 kJ
Therefore, the total amount of heat absorbed or released by the reaction is
2670 kJ , with the negative sign indicating that heat is released by the reaction.
Question 26
Question
A reaction has an enthalpy change of -346 kJ/mol. If 0.25 mol of the reactant
is consumed, calculate the amount of heat exchanged in kJ.
Solution
Step 1: Determine the heat exchanged for 1 mol of the reactant. Given that
the enthalpy change for the reaction is -346 kJ/mol, it means that 1 mol of the
reactant reacts with the release of 346 kJ of heat.
Therefore, the heat exchanged for 1 mol of the reactant is -346 kJ.
Step 2: Calculate the heat exchanged for 0.25 mol of the reactant. To find
the heat exchanged for 0.25 mol of the reactant, we need to multiply the heat
exchanged for 1 mol by the number of moles consumed.
0.25 mol ×(−346 kJ/mol) = −86.5 kJ
Therefore, when 0.25 mol of the reactant is consumed, the amount of heat
exchanged is -86.5 kJ.
Question 27
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following standard enthalpy of formation values:
∆H◦
f(H2O(l)) = −286 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g)+O2(g)→2H2O(l)
21
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given values into the formula to find ∆H◦.
∆H◦= [2 ·∆H◦
f(H2O(l))] −[2 ·∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
Step 4: Substitute the specific values for the enthalpies of formation and
solve the equation.
∆H◦= [2 ·(−286 kJ/mol)] −[2 ·0 kJ/mol + 0 kJ/mol]
∆H◦=−572 kJ/mol
So, the standard enthalpy change for the given reaction is ∆H◦=−572 kJ/mol.
Question 28
Question
A certain reaction has an enthalpy change of ∆H=−345 kJ. Calculate the
amount of heat (in joules) absorbed or released when 5.00 mol of the reactant
is consumed in the reaction.
Solution
Step 1: First, we need to convert the given enthalpy change from kilojoules to
joules.
Given: ∆H=−345 kJ = −345 ×1000 J = −345000 J
Step 2: Calculate the heat absorbed or released when 1.00 mol of the reactant
is consumed using the given enthalpy change.
Heat for 1.00 mol = ∆H=−345000 J = −345000 J/mole
Step 3: Determine the heat absorbed or released when 5.00 mol of the reac-
tant is consumed by multiplying the heat for 1.00 mol by 5.00.
Heat for 5.00 mol = −345000 J/mole ×5.00 mol = −1725000 J
Therefore, 5.00 mol of the reactant in the reaction would absorb or release
−1725000 J of heat.
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Question 29
Question
Calculate the change in enthalpy (∆H) when 10.0 grams of water at 25
°
C is
heated to form steam at 100
°
C. The specific heat capacity of water is 4.18 J/g
°
C,
the specific heat capacity of steam is 2.0 J/g
°
C, and the heat of vaporization of
water is 2260 J/g.
Solution
Step 1: Calculate the energy required to heat the water from 25
°
C to 100
°
C.
q=mc∆T
q= (10.0 g)(4.18 J/g
°
C)(100C−25C)
q= 2930 J
Step 2: Calculate the energy required to evaporate the water at 100
°
C.
q=m×heat of vaporization
q= (10.0 g)(2260 J/g)
q= 22600 J
Step 3: Calculate the total energy change.
∆H=qheating +qvaporization
∆H= 2930 J + 22600 J
∆H= 25530 J
Therefore, the change in enthalpy (∆H) when 10.0 grams of water is heated
to form steam is 25530 J.
Question 30
Question
Calculate the change in enthalpy for a reaction in which 2 moles of magnesium
metal react with excess hydrochloric acid to form magnesium chloride and hy-
drogen gas. The enthalpies of formation for magnesium chloride and hydrogen
gas are -641 kJ/mol and 0 kJ/mol, respectively.
23
Solution
Step 1: Write the balanced chemical equation for the reaction.
Mg(s) + 2HCl(aq) →MgCl2(aq)+H2(g)
Step 2: Determine the molar enthalpy change for the reaction using the
enthalpies of formation. The enthalpy change for the reaction can be calculated
using the formula:
∆H=X∆Hproducts −X∆Hreactants
Substitute the values given into the equation:
∆H= [1(−641 kJ/mol) + 1(0 kJ/mol)] −[1(0 kJ/mol) + 2(0 kJ/mol)]
∆H=−641 kJ/mol
Step 3: Calculate the change in enthalpy for the given reaction. Since 2
moles of magnesium is involved in the reaction, the total change in enthalpy
would be:
∆Htotal = 2 ×∆H= 2 ×(−641 kJ/mol)
∆Htotal =−1282 kJ
Therefore, the change in enthalpy for the reaction is -1282 kJ.
Question 31
Question
A reaction at constant pressure has a change in enthalpy of −158 kJ. If the
reaction also has a change in internal energy of −120 kJ, calculate the work
done by the reaction.
Solution
Step 1: Recall the relationship between enthalpy change, internal energy change,
and work done by the reaction. The relationship between the change in enthalpy
(∆H), change in internal energy (∆U), and work done (W) by the reaction is
given by:
∆H= ∆U+P∆V
where Pis the pressure and ∆Vis the change in volume.
Step 2: Given that ∆H=−158 kJ, ∆U=−120 kJ, and the reaction is at
constant pressure, we can rearrange the equation to solve for the work done:
W= ∆H−∆U
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Step 3: Substitute the given values into the formula to calculate the work
done:
W=−158 kJ −(−120 kJ)
W=−158 kJ + 120 kJ
W=−38 kJ
Therefore, the work done by the reaction is −38 kJ.
Question 32
Question
Determine the change in enthalpy (∆H) for the reaction:
2C(s) + 3H2(g) →C2H6(g)
given the following standard enthalpies of formation:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Solution
Step 1: Write the equation for the reaction in terms of the standard enthalpies
of formation of the reactants and products. Step 2: Calculate the standard en-
thalpy of reaction using the standard enthalpies of formation. Step 3: Interpret
the sign of ∆Hin terms of whether the reaction is exothermic or endothermic.
Step 1: The enthalpy change for the reaction can be expressed as:
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Step 2: Substitute the given values into the formula:
∆H= 1(−84.68 kJ/mol) −2(0 kJ/mol) −3(0 kJ/mol)
=−84.68 kJ/mol
So, the change in enthalpy for the reaction is ∆H=−84.68 kJ/mol.
Step 3: Since the value of ∆His negative, the reaction is exothermic (it
releases heat).
Question 33
Question
A chemical reaction is carried out at constant pressure. The reaction releases
250 kJ of heat and does work of 100 kJ on the surroundings. Determine the
change in enthalpy for the reaction.
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Solution
Step 1: Recall that the change in enthalpy (∆H) for a reaction is equal to the
heat exchanged at constant pressure (qp) plus the work done on the surround-
ings.
∆H=qp+w
Step 2: Given that the reaction releases 250 kJ of heat (qp=−250 kJ) and
does work of 100 kJ on the surroundings (w=−100 kJ), substitute these values
into the equation.
∆H=−250 kJ −100 kJ
Step 3: Calculate the change in enthalpy.
∆H=−350 kJ
Therefore, the change in enthalpy for the reaction is -350 kJ.
Question 34
Question
A reaction is carried out in a bomb calorimeter. Initially, the temperature of
the calorimeter was 25
°
C. During the reaction, the temperature of the water in
the calorimeter rose to 35
°
C. The bomb calorimeter has a heat capacity of 400
J/
°
C. If the enthalpy change for the reaction is -150 kJ/mol, how many moles
of the reactant are present in the bomb calorimeter?
Solution
Step 1: Calculate the heat absorbed by the bomb calorimeter. The heat ab-
sorbed by the bomb calorimeter is equal to the heat released by the reaction:
q=C·∆T
where: q= heat absorbed by calorimeter C= heat capacity of calorimeter (400
J/
°
C) ∆T= change in temperature (10
°
C) Plugging in the values:
q= 400 J/
°
C·10
°
C = 4000 J
Step 2: Convert heat absorbed to kJ.
q= 4000 J = 4 kJ
Step 3: Calculate the number of moles of reactant. Given: Enthalpy change
(∆H) = -150 kJ/mol ∆H=qfor the reaction
−150 kJ/mol = 4 kJ ×n
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where: n= number of moles of reactant Solving for n:
n=−150 kJ/mol
4 kJ =−37.5 mol
Since moles cannot be negative, it implies that the sign of the enthalpy change
was chosen incorrectly. The correct sign should be:
n=150 kJ/mol
4 kJ = 37.5 mol
Therefore, there are 37.5 moles of the reactant present in the bomb calorime-
ter.
Question 35
Question
A certain reaction has an enthalpy change of −285 kJ/mol. If this reaction is
carried out at constant pressure and releases 95 kJ of heat, what is the change
in internal energy for the reaction?
Solution
Step 1: Recall that the relationship between enthalpy change (∆H), heat re-
leased (q), and change in internal energy (∆U) is given by the equation ∆H=
∆U+P∆V, where P∆Vis the work done by the system. Step 2: Since the
reaction is carried out at constant pressure, P∆V= ∆nRT , where ∆nis the
change in moles of gas, Ris the ideal gas constant, and Tis the temperature in
Kelvin. Step 3: In this case, since P∆Vis not given and we are asked to find
∆U, we can simplify the equation to ∆U= ∆H−q. Step 4: Substitute the
given values into the equation: ∆U=−285 kJ/mol −95 kJ. Step 5: Calculate
the change in internal energy: ∆U=−380 kJ. Step 6: Therefore, the change
in internal energy for the reaction is −380 kJ.
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