CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Thermochemistry
Question Bank - Set 3
Liberty University
Question 1
Question
Given the balanced chemical equation:
2CH3OH(l)+3O2(g)→2CO2(g)+4H2O(l)
If the standard enthalpy of formation of liquid methanol (CH3OH) is -238.7
kJ/mol and the standard enthalpy of formation of liquid water (H2O) is -285.8
kJ/mol, calculate the standard enthalpy change for the combustion of 1 mole of
methanol.
Solution
Step 1: Calculate the standard enthalpy change for the combustion of 1 mole of
methanol using the enthalpy of formation data provided. The standard enthalpy
change for the reaction is given by:
∆H◦=XνfH◦
products −XνfH◦
reactants
Where νfis the stoichiometric coefficient of each species and H◦is the standard
enthalpy of formation. For the combustion of 1 mole of methanol, the reactants
are 2 moles of CH3OH and 3 moles of O2, while the products are 2 moles of
CO2and 4 moles of H2O.
Step 2: Substitute the given standard enthalpy of formation data into the
equation.
∆H◦= 2(−393.5) + 4(−285.8) −[2(−238.7) + 3(0)]
Step 3: Calculate the standard enthalpy change for the combustion of 1 mole
of methanol.
∆H◦=−784.6−1143.2−(−477.4) = −1450.4 kJ/mol
Therefore, the standard enthalpy change for the combustion of 1 mole of
methanol is -1450.4 kJ/mol.
Question 2
Question
Given the following balanced chemical equation for the combustion of ethylene
gas (C2H4):
C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)
and the standard enthalpies of formation, ∆H◦
f, for the substances involved in
the reaction:
∆H◦
f(C2H4) = 52.3 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the combustion of 5.00
moles of ethylene gas.
Solution
Step 1: Calculate ∆H◦for the reaction using the standard enthalpies of forma-
tion.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
= [2 ×∆H◦
f(CO2)+2×∆H◦
f(H2O)] −[∆H◦
f(C2H4)+3×∆H◦
f(O2)]
= [2 ×(−393.5 kJ/mol) + 2 ×(−285.8 kJ/mol)] −[52.3 kJ/mol + 3 ×0 kJ/mol]
= [−787 kJ/mol −571.6 kJ/mol] −52.3 kJ/mol
=−1358.6 kJ/mol
Step 2: Calculate ∆H◦for the combustion of 5.00 moles of ethylene gas.
∆H◦(5.00 moles) = (5.00 moles) ×∆H◦
= (5.00 moles) ×(−1358.6 kJ/mol)
=−6793 kJ
Therefore, the standard enthalpy change for the combustion of 5.00 moles
of ethylene gas is -6793 kJ.
2
Question 3
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2CH3OH(l) + 3O2(g)→2CO2(g) + 4H2O(l)
given the following standard enthalpies of formation:
∆H◦
f(CH3OH) = −238.6 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.
∆H◦
rxn =Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients in the balanced chemical
equation.
Step 2: Substitute the given values into the equation and solve for ∆H◦
rxn.
∆H◦
rxn = [2(−393.5 kJ/mol) + 4(−285.8 kJ/mol)] −[2(−238.6 kJ/mol) + 3(0)]
∆H◦
rxn = [−787.0 kJ/mol −1143.2 kJ/mol]
∆H◦
rxn =−1930.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−1930.2 kJ/mol.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(graphite)+3H2(g)→C2H2(g)
given the following standard enthalpy of formation values: ∆H◦
f:C(graphite) =
0 kJ/mol, H2(g) = 0 kJ/mol, and C2H2(g) = 227 kJ/mol.
3
Solution
Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate
the standard enthalpy change (∆H◦) using the standard enthalpy of formation
values. - ∆H◦=Pn∆H◦
f(products) −Pm∆H◦
f(reactants) - Substitute the
given values into the equation and solve for ∆H◦.
Question 5
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction using
the given standard enthalpies of formation:
2C(s) + 3H2(g)→C2H6(g)
Given:
Standard enthalpy of formation, ∆H◦
f(C2H6(g)) = −84.68 kJ/mol
Standard enthalpy of formation, ∆H◦
f(C(s)) = 0 kJ/mol
Standard enthalpy of formation, ∆H◦
f(H2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical reaction and identify the given enthalpies
of formation.
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the formula:
∆H◦=X∆H◦
fproducts −X∆H◦
freactants
Step 3: Calculate the standard enthalpy change:
∆H◦=1×∆H◦
f(C2H6(g))−2×∆H◦
f(C(s)) + 3 ×∆H◦
f(H2(g))
= (1 × −84.68) −(2 ×0+3×0)
=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
−84.68 kJ/mol.
Question 6
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l)
4
Calculate the standard enthalpy change (∆H◦) for the reaction using the fol-
lowing data:
Substance ∆H◦
f(kJ/mol)
H2(g) 0
O2(g) 0
H2O(l) -285.8
Solution
Step 1: Write the balanced chemical equation for the reaction and denote the
standard enthalpy of formation for each substance:
2H2(g)+O2(g)→2H2O(l)
H2(g):∆H◦
f= 0 kJ/mol
O2(g):∆H◦
f= 0 kJ/mol
H2O(l):∆H◦
f=−285.8 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦) for the given reac-
tion using the standard enthalpy of formation values:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is -571.6
kJ/mol.
Question 7
Question
During an experiment, 4.50 grams of propane gas (C3H8) is burned in excess
oxygen gas to produce carbon dioxide gas and water vapor. The heat released by
the reaction is found to be -2500 kJ. Calculate the molar enthalpy of combustion
of propane gas. (Molar masses: C= 12.01 g/mol, H= 1.008 g/mol)
Solution
Step 1: Write the balanced chemical equation for the combustion of propane
gas. Step 2: Calculate the number of moles of propane used in the reaction.
Step 3: Calculate the heat released per mole of propane. Step 4: Determine the
molar enthalpy of combustion of propane gas.
5
Step 1: The balanced chemical equation for the combustion of propane gas
is:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Step 2: Calculate the number of moles of propane used:
Molar mass of C3H8= (3 ×12.01 g/mol) + (8 ×1.008 g/mol) = 44.11 g/mol
Number of moles of C3H8=4.50 g
44.11 g/mol ≈0.102 mol
Step 3: Calculate the heat released per mole of propane:
Heat released by reaction = −2500 kJ
Heat released per mole of propane = −2500 kJ
0.102 mol ≈ −24509.80 kJ/mol
Step 4: Determine the molar enthalpy of combustion of propane gas:
Molar enthalpy of combustion of propane = −24509.80 kJ/mol
Therefore, the molar enthalpy of combustion of propane gas is approximately
-24509.80 kJ/mol.
Question 8
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2Mg(s) + O2(g)→2M gO(s)
given the following standard enthalpy of formation values: Mg(s): −601.7
kJ/mol O2(g): 0 kJ/mol M gO(s): −601.6 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard enthalpy
of formation values for each substance. The balanced chemical equation is:
2Mg(s) + O2(g)→2M gO(s)
Given:
∆H◦
f(Mg) = −601.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
∆H◦
f(MgO) = −601.6 kJ/mol
6
Step 2: Determine the standard enthalpy change for the reaction using stan-
dard enthalpy of formation values. The standard enthalpy change for the re-
action can be calculated by the sum of the standard enthalpies of formation
of the products minus the sum of the standard enthalpies of formation of the
reactants:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given values into the equation:
∆H◦= 2(∆H◦
f(MgO)) −[2(∆H◦
f(Mg)) + ∆H◦
f(O2)]
∆H◦= 2(−601.6) −[2(−601.7) + 0]
∆H◦=−1203.2 + 1203.4
∆H◦= 0.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is 0.2 kJ/mol.
Question 9
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with ∆H◦=−571.6 kJ/mol, calculate the heat released when 5.00 g of H2reacts
with excess O2.
Solution
Step 1: Calculate the number of moles of H2in 5.00 g.
Molar mass of H2= 2.016 g/mol
Moles of H2=5.00 g
2.016 g/mol
Moles of H2= 2.48 mol
Step 2: Use the stoichiometry of the reaction to determine the heat released
when 2.48 moles of H2react.
Moles of water produced = 2.48 mol ×2 mol H2O(l)
2 mol H2
Moles of water produced = 2.48 mol
Step 3: Calculate the heat released when 2.48 moles of H2react.
Heat released = 2.48 mol ×(−571.6 kJ/mol)
Heat released = −1416.9 kJ
Answer: The heat released when 5.00 g of H2reacts with excess O2is
-1416.9 kJ.
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Question 10
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2O2(l)→2H2O(l)+O2(g)
given the following standard enthalpies of formation:
∆H◦
f(H2O2(l)) = −196.1 kJ/mol,∆H◦
f(H2O(l)) = −285.8 kJ/mol,and ∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and determine the ∆H◦for the
reaction based on standard enthalpies of formation. The reaction provided is:
2H2O2(l)→2H2O(l)+O2(g)
The standard enthalpy change for the reaction can be calculated using the
standard enthalpies of formation of the products and reactants:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of products and reactants,
respectively.
Step 2: Substitute the given values into the equation. We have:
∆H◦= [2(−285.8) + 0] −[2(−196.1)]
∆H◦= [−571.6] −[−392.2]
∆H◦=−179.4 kJ
Therefore, the standard enthalpy change for the reaction is ∆H◦=−179.4 kJ.
Question 11
Question
For the reaction
2C5H12(g) + 16O2(g)→10CO2(g) + 12H2O(g)
the standard enthalpy change (∆H◦) is −3500 kJ. Calculate the enthalpy change
when 5.00 g of C5H12 is burned at constant pressure.
8
Solution
Step 1: Calculate the molar quantity of C5H12. Given: Mass of C5H12 = 5.00
g, Molar mass of C5H12 = 5(12.01) + 12(1.01) = 72.15 g/mol.
Number of moles of C5H12:
moles = mass
molar mass =5.00 g
72.15 g/mol = 0.0693 mol
Step 2: Use stoichiometry to calculate the enthalpy change. From the bal-
anced chemical equation, we see that for 2 moles of C5H12 reacted, the enthalpy
change is −3500 kJ. Therefore, for 0.0693 moles of C5H12:
Enthalpy change = −3500 kJ
2 mol ×0.0693 mol = −121 kJ
Therefore, the enthalpy change when 5.00 g of C5H12 is burned at constant
pressure is −121 kJ.
Question 12
Question
Calculate the enthalpy change for the reaction: 2C(graphite)+3H2(g)→C2H6(g)
given the following bond enthalpies: C−Cbond: 347 kJ/mol, C=Cbond:
614 kJ/mol, C−Hbond: 413 kJ/mol, and H−Hbond: 432 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and calculate the total bond
energy of the bonds broken and formed:
Bonds broken: 2(C−C) + 3(H−H) = 2(347) + 3(432) = 694 + 1296 =
1990 kJ
Bond formed: 1(C−C) + 6(C−H) = 1(347) + 6(413) = 347 + 2478 =
2825 kJ
Step 2: Calculate the overall enthalpy change:
∆H= Bonds broken −Bonds formed
∆H= 1990 kJ −2825 kJ = −835 kJ
Step 3: Therefore, the enthalpy change for the reaction is −835 kJ .
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Question 13
Question
Calculate the enthalpy change (∆H) for the combustion of ethylene gas (C2H4)
according to the equation:
C2H4(g) + 3O2(g)→2CO2(g) + 2H2O(l)
Given the standard enthalpies of formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(C2H4) = +52.3 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the combustion of ethylene
gas by using the standard enthalpies of formation of the products and reactants.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
= [2 ×∆H◦
f(CO2)+2×∆H◦
f(H2O)] −[∆H◦
f(C2H4)+3×∆H◦
f(O2)]
= [2 ×(−393.5) + 2 ×(−285.8)] −[52.3+3×0]
= [−787.0−571.6] −52.3
=−1358.6−52.3
=−1410.9 kJ/mol
Therefore, the standard enthalpy change for the combustion of ethylene gas
is ∆H=−1410.9 kJ/mol.
Question 14
Question
Calculate the standard enthalpy change for the following reaction at 25
°
C:
2C(graphite)+3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.68 kJ/mol
10
Solution
Step 1: Write the balanced chemical equation for the reaction and determine
the change in enthalpy.
2C(graphite)+3H2(g)→C2H6(g)
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Step 2: Calculate the change in enthalpy.
∆H= [1 ×∆H◦
f(C2H6(g)) −(2 ×∆H◦
f(C(graphite)) + 3 ×∆H◦
f(H2(g))]
= [(1 × −84.68) −(2 ×0+3×0)] kJ/mol
=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 25
°
C is
-84.68 kJ/mol.
Question 15
Question
A 2.50 g sample of ethylene glycol (C2H6O2) is burned in a bomb calorimeter.
The temperature of the calorimeter increases by 6.50
°
C. If the heat capacity
of the calorimeter is 12.59 kJ/
°
C, calculate the molar heat of combustion of
ethylene glycol. The molar mass of ethylene glycol is 62.07 g/mol.
Solution
Step 1: Calculate the heat released by the calorimeter.
Heat released by the calorimeter = C×∆T
= 12.59 kJ/
°
C×6.50C
= 81.935 kJ
Step 2: Calculate the moles of ethylene glycol burned.
Moles of ethylene glycol = 2.50 g
62.07 g/mol
= 0.04032 mol
Step 3: Calculate the molar heat of combustion of ethylene glycol.
Molar heat of combustion = Heat released by the calorimeter
Moles of ethylene glycol
=81.935 kJ
0.04032 mol
= 2033.12 kJ/mol
Therefore, the molar heat of combustion of ethylene glycol is 2033.12 kJ/mol.
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Question 16
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + 3H2(g)→C2H4(g)
given the following bond dissociation energies:
C-C : 347 kJ/mol
C=C : 614 kJ/mol
H-H : 436 kJ/mol
C-H : 413 kJ/mol
Solution
Step 1: Calculate the total bond dissociation energy for the reactants.
2(C-C) + 3(H-H) + 6(C-H) = 2(347 kJ/mol) + 3(436 kJ/mol) + 6(413 kJ/mol)
= 694 kJ/mol + 1308 kJ/mol + 2478 kJ/mol
= 4480 kJ/mol
Step 2: Calculate the total bond dissociation energy for the products.
C=C + 4(C-H) = 614 kJ/mol + 4(413 kJ/mol)
= 614 kJ/mol + 1652 kJ/mol
= 2266 kJ/mol
Step 3: Calculate the change in bond dissociation energy.
∆H= Energy of bonds broken −Energy of bonds formed
∆H= (4480 kJ/mol) −(2266 kJ/mol)
∆H= 2214 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H= 2214 kJ/mol.
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2C(s)+3H2(g)→C2H6(g)
Given the following standard enthalpy of formation values: C(s) : 0 kJ/mol,
H2(g) : 0 kJ/mol, C2H6(g) : −84.68 kJ/mol.
12
Solution
Step 1: Write the given chemical equation and the formation reaction for
C2H6(g):
Given equation: 2C(s)+3H2(g)→C2H6(g)
Formation reaction for C2H6(g) : 2C(s)+3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change for the formation of C2H6(g):
∆H◦
formation =X∆H◦
products −X∆H◦
reactants
∆H◦
formation =−84.68 kJ/mol −(2(0 kJ/mol) + 3(0 kJ/mol))
∆H◦
formation =−84.68 kJ/mol
Step 3: Remember that the enthalpy change for the reaction is the same as
the enthalpy change for the formation of C2H6(g), but with opposite sign.
∆H◦
reaction =−∆H◦
formation
∆H◦
reaction =−(−84.68 kJ/mol)
∆H◦
reaction = 84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 25◦C is
84.68 kJ/mol.
Question 18
Question
Given the standard enthalpy of formation for water gas (H2O(g)) is ∆H◦
f=
−241.8 kJ/mol, calculate the standard enthalpy change (∆H◦) for the reaction
below:
2H2(g) + O2(g) →2H2O(g)
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g) →2H2O(g)
Step 2: Determine the standard enthalpy change (∆H◦) for the reaction
using the standard enthalpy of formation values:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
13
Step 3: Calculate the change in enthalpy based on the given standard en-
thalpy of formation values:
∆H◦= 2 ×∆H◦
f(H2O(g)) −[2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
Step 4: Substitute the known values and calculate:
∆H◦= 2 ×(−241.8 kJ/mol) −[2 ×0 + 0] = −483.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−483.6
kJ/mol.
Question 19
Question
Calculate the enthalpy change (∆H) for the following reaction:
2 C(graphite) + 3 H2(g) →CH4(g)
Given the following enthalpy changes:
∆H◦
ffor CH4(g) = −74.81 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
∆H◦
ffor H2(g) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the relevant enthalpy
changes.
2 C(graphite) + 3 H2(g) →CH4(g)
Given:
∆H◦
ffor CH4(g) = −74.81 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
∆H◦
ffor H2(g) = 0 kJ/mol
Step 2: Calculate the enthalpy change for the reaction using standard en-
thalpies of formation.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H= ∆H◦
f(CH4)−(∆H◦
f(C) + 3 ×∆H◦
f(H2))
∆H=−74.81 kJ/mol −(0 + 3 ×0) kJ/mol
∆H=−74.81 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is -74.81 kJ/mol.
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Question 20
Question
Calculate the enthalpy change (∆H) for the following reaction at 298 K:
2H2(g) + O2(g) →2H2O(l)
Given the following standard enthalpies of formation at 298 K: ∆H◦
f(H2O) =
−285.8 kJ/mol, ∆H◦
f(H2) = 0 kJ/mol, ∆H◦
f(O2) = 0 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and identify the given values.
The balanced chemical equation is:
2H2(g) + O2(g) →2H2O(l)
Given: ∆H◦
f(H2O) = −285.8 kJ/mol ∆H◦
f(H2) = 0 kJ/mol ∆H◦
f(O2) = 0
kJ/mol
Step 2: Calculate the change in enthalpy (∆H) using the standard enthalpies
of formation.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Substitute the given values:
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0]
∆H◦=−571.6 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is −571.6 kJ/mol .
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
given the following standard enthalpy of formation values: ∆H◦
f(C2H6) =
−84.7 kJ/mol, ∆H◦
f(CO2) = −393.5 kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol,
∆H◦
f(O2) = 0 kJ/mol.
15
Solution
Step 1: First, write the balanced chemical equation for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given standard enthalpy of formation values into the
equation:
∆H◦= 4 ×∆H◦
f(CO2)+6×∆H◦
f(H2O)−2×∆H◦
f(C2H6)−7×∆H◦
f(O2)
Step 4: Calculate the standard enthalpy change:
∆H◦= 4(−393.5) + 6(−285.8) −2(−84.7) −7(0)
∆H◦=−1574 −1714.8 + 169.4−0
∆H◦=−3119.4 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−3119.4 kJ/mol.
Question 22
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at 298
K:
2 C(graphite) + 3 H2(g) →C2H6(g)
Given the following standard enthalpies of formation at 298 K:
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Solution
Step 1: Write the given balanced chemical equation with the standard enthalpies
of formation:
2 C(graphite) + 3 H2(g) →C2H6(g)
16
Given standard enthalpies of formation:
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦=∆H◦
f(C2H6)−2·∆H◦
f(C(graphite)) + 3 ·∆H◦
f(H2)
∆H◦= [−84.68 kJ/mol] −[2 ·0+3·0]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol .
Question 23
Question
The combustion of ethanol (C2H5OH) releases 1367 kJ/mol. Calculate the
enthalpy change for the reaction C2H5OH(g)→C2H4(g) + H2O(g).
Solution
Step 1: Write the balanced chemical equation for the reaction and identify the
given enthalpy change. The balanced chemical equation for the reaction is:
C2H5OH(g)→C2H4(g) + H2O(g)
Given: ∆H=−1367 kJ/mol for the reaction C2H5OH(g)+3O2(g)→2CO2(g)+
3H2O(g)
Step 2: Apply Hess’s Law to find the enthalpy change for the target reaction.
To find the enthalpy change for the target reaction, we will use the enthalpies
of formation of the reactants and products. The enthalpy change for the target
reaction can be found by manipulating the given reaction to add up to the target
reaction. First, we reverse the given reaction:
2CO2(g)+3H2O(g)→C2H5OH(g)+3O2(g)
Step 3: Find the enthalpy change for the reversed reaction. Since ∆His for
the forward reaction, we will negate the given ∆Hvalue to find the enthalpy
change for the reversed reaction:
∆Hrev =−(−1367) kJ/mol = 1367 kJ/mol
17
Step 4: Manipulate the equations to obtain the target reaction. To use
the reversed reaction to obtain the target reaction, we need to cancel out the
substances that are not part of the target reaction.
2CO2(g)+3H2O(g)→C2H5OH(g)+3O2(g)
2CO2(g)+3H2O(g)→C2H4(g) + H2O(g)
Step 5: Calculate the enthalpy change for the target reaction. The enthalpy
change for the target reaction is equal to the negative of the enthalpy change
for the reversed reaction, as the target reaction is the reverse of the reversed
reaction. Therefore, the enthalpy change for the target reaction is:
∆Htarget =−1367 kJ/mol
So, the enthalpy change for the reaction C2H5OH(g)→C2H4(g) + H2O(g)
is −1367 kJ/mol.
Question 24
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2C6H12O6(s) + 17O2(g)→12CO2(g) + 12H2O(l)
Given the standard enthalpies of formation: C6H12O6(s) = −1273.3 kJ/mol,
CO2(g) = −393.5 kJ/mol, H2O(l) = −285.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy change for the reaction based on the
standard enthalpies of formation.
The standard enthalpy change can be calculated using the equation:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
where nand mare the stoichiometric coefficients of the products and reactants,
respectively.
Given: C6H12O6(s) = −1273.3 kJ/mol, CO2(g) = −393.5 kJ/mol, H2O(l) =
−285.8 kJ/mol.
The standard enthalpy change for the reaction is:
∆H◦= 12(−393.5) + 12(−285.8) −2(−1273.3)
Step 2: Solve for ∆H◦.
∆H◦=−4722 + (−3429.6) + 2546.6
18
∆H◦=−4205.0 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction at 25◦C is
-4205.0 kJ/mol.
Question 25
Question
Given the following reaction:
2H2(g)+ O2(g)→2H2O(l)
the standard enthalpy change of formation for water (∆H◦
f) is -285.8 kJ/mol.
Calculate the standard enthalpy change of the reaction in kJ.
Solution
Step 1: Write down the balanced chemical equation with the given enthalpy
change of formation.
2H2(g) + O2(g) →2H2O(l) ∆H◦
f=−285.8 kJ/mol
Step 2: Calculate the standard enthalpy change of the reaction using the
enthalpy change of formation values.
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = 2 ×∆H◦
f(H2O(l)) −[2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
∆H◦
rxn = 2 ×(−285.8) −[2 ×0 + 0] = −571.6 kJ
Therefore, the standard enthalpy change of the reaction is −571.6 kJ.
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of acety-
lene (C2H2(g)) to form carbon dioxide (CO2(g)) and water (H2O(l)) given the
following standard enthalpies of formation:
∆H◦
f(C2H2(g)) = 226.7 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
19
Solution
Step 1: Write the balanced chemical equation for the combustion of acetylene:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(l)
Step 2: Calculate the standard enthalpy change using the given standard
enthalpies of formation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [4∆H◦
f(CO2(g)) + 2∆H◦
f(H2O(l))] −[2∆H◦
f(C2H2(g))]
∆H◦= [4(−393.5) + 2(−285.8)] −[2(226.7)]
∆H◦= [−1574 −571.6] −453.4
∆H◦=−2145.6−453.4
∆H◦=−2599.0 kJ/mol
Therefore, the standard enthalpy change for the combustion of acetylene to
form carbon dioxide and water is −2599.0 kJ/mol.
Question 27
Question
For the reaction:
2A(g)+3B(g)→4C(g)+2D(g)
the standard enthalpy change (∆rH◦) is −1820 kJ/mol at 298 K. Calculate
the standard enthalpy of formation of compound A if the standard enthalpy of
formation of compound B is −410 kJ/mol, compound C is −790 kJ/mol, and
compound D is −560 kJ/mol.
Solution
Step 1: Determine the standard enthalpy of formation of compound A using
Hess’s law. Step 2: Apply Hess’s law to the enthalpy of the reaction to solve
for ∆H◦
fof compound A.
Step 1: Write the standard enthalpy change of the reaction in terms of the
standard enthalpies of formation of the compounds involved:
2∆H◦
f(A) + 3∆H◦
f(B) = 4∆H◦
f(C) + 2∆H◦
f(D)
Step 2: Substitute the values provided into the equation:
2∆H◦
f(A) + 3(−410 kJ/mol) = 4(−790 kJ/mol) + 2(−560 kJ/mol)
20
2∆H◦
f(A)−1230 kJ/mol = −3160 kJ/mol
2∆H◦
f(A) = −1930 kJ/mol
∆H◦
f(A) = −965 kJ/mol
Therefore, the standard enthalpy of formation of compound A is −965
kJ/mol.
Question 28
Question
Given the following thermochemical equation:
2A(g)+3B(g)→C(g)
The enthalpy change for this reaction is ∆H=−150 kJ. If 5.00 moles of A
and 7.00 moles of B react to form 2.00 moles of C under constant pressure, what
is the enthalpy change for the reaction when 1.00 mole of C is formed under the
same conditions?
Solution
Step 1: Find the enthalpy change for the reaction when 1.00 mole of C is formed.
Given:
2A(g)+3B(g)→C(g)
∆H=−150 kJ
We want to find the enthalpy change for the reaction when 1.00 mole of C is
formed. Using stoichiometry, we can relate the enthalpy changes for the given
reaction:
Moles of C formed = 2
Moles of C formed = 2 ×∆H
2= 2 × −150 kJ = −300 kJ
Therefore, the enthalpy change for the reaction when 1.00 mole of C is formed
is −300 kJ .
21
Question 29
Question
Calculate the enthalpy change (∆H) for the reaction below at 25◦C given the
following information:
2H2(g) + O2(g)→2H2O(l)
∆H◦
f(kJ/mol): H2(g) = 0; O2(g) = 0; H2O(l) = -286
Solution
Step 1: Write the balanced equation for the reaction and determine the ∆Hof
the reaction using the standard enthalpies of formation (∆H◦
f) of the compounds
involved.
2H2(g) + O2(g)→2H2O(l)
∆H=P∆H◦
f(products) −P∆H◦
f(reactants)
Step 2: Substitute the given ∆H◦
fvalues into the equation and solve.
∆H= [2 ·∆H◦
f(H2O(l))] −[2 ·∆H◦
f(H2(g)) + 1 ·∆H◦
f(O2(g))]
∆H= [2 ·(−286)] −[2 ·0+1·0] = −572 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction at 25◦C is -572 kJ/mol.
Question 30
Question
Given the following reaction, calculate the heat of reaction at 298 K using
standard enthalpies of formation:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.83 kJ/mol
22
Question 3
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2CH3OH(l) + 3O2(g)→2CO2(g) + 4H2O(l)
given the following standard enthalpies of formation:
∆H◦
f(CH3OH) = −238.6 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.
∆H◦
rxn =Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients in the balanced chemical
equation.
Step 2: Substitute the given values into the equation and solve for ∆H◦
rxn.
∆H◦
rxn = [2(−393.5 kJ/mol) + 4(−285.8 kJ/mol)] −[2(−238.6 kJ/mol) + 3(0)]
∆H◦
rxn = [−787.0 kJ/mol −1143.2 kJ/mol]
∆H◦
rxn =−1930.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−1930.2 kJ/mol.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(graphite)+3H2(g)→C2H2(g)
given the following standard enthalpy of formation values: ∆H◦
f:C(graphite) =
0 kJ/mol, H2(g) = 0 kJ/mol, and C2H2(g) = 227 kJ/mol.
3
Solution
Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate
the standard enthalpy change (∆H◦) using the standard enthalpy of formation
values. - ∆H◦=Pn∆H◦
f(products) −Pm∆H◦
f(reactants) - Substitute the
given values into the equation and solve for ∆H◦.
Question 5
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction using
the given standard enthalpies of formation:
2C(s) + 3H2(g)→C2H6(g)
Given:
Standard enthalpy of formation, ∆H◦
f(C2H6(g)) = −84.68 kJ/mol
Standard enthalpy of formation, ∆H◦
f(C(s)) = 0 kJ/mol
Standard enthalpy of formation, ∆H◦
f(H2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical reaction and identify the given enthalpies
of formation.
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the formula:
∆H◦=X∆H◦
fproducts −X∆H◦
freactants
Step 3: Calculate the standard enthalpy change:
∆H◦=1×∆H◦
f(C2H6(g))−2×∆H◦
f(C(s)) + 3 ×∆H◦
f(H2(g))
= (1 × −84.68) −(2 ×0+3×0)
=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
−84.68 kJ/mol.
Question 6
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l)
4
Calculate the standard enthalpy change (∆H◦) for the reaction using the fol-
lowing data:
Substance ∆H◦
f(kJ/mol)
H2(g) 0
O2(g) 0
H2O(l) -285.8
Solution
Step 1: Write the balanced chemical equation for the reaction and denote the
standard enthalpy of formation for each substance:
2H2(g)+O2(g)→2H2O(l)
H2(g):∆H◦
f= 0 kJ/mol
O2(g):∆H◦
f= 0 kJ/mol
H2O(l):∆H◦
f=−285.8 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦) for the given reac-
tion using the standard enthalpy of formation values:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is -571.6
kJ/mol.
Question 7
Question
During an experiment, 4.50 grams of propane gas (C3H8) is burned in excess
oxygen gas to produce carbon dioxide gas and water vapor. The heat released by
the reaction is found to be -2500 kJ. Calculate the molar enthalpy of combustion
of propane gas. (Molar masses: C= 12.01 g/mol, H= 1.008 g/mol)
Solution
Step 1: Write the balanced chemical equation for the combustion of propane
gas. Step 2: Calculate the number of moles of propane used in the reaction.
Step 3: Calculate the heat released per mole of propane. Step 4: Determine the
molar enthalpy of combustion of propane gas.
5
Step 1: The balanced chemical equation for the combustion of propane gas
is:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Step 2: Calculate the number of moles of propane used:
Molar mass of C3H8= (3 ×12.01 g/mol) + (8 ×1.008 g/mol) = 44.11 g/mol
Number of moles of C3H8=4.50 g
44.11 g/mol ≈0.102 mol
Step 3: Calculate the heat released per mole of propane:
Heat released by reaction = −2500 kJ
Heat released per mole of propane = −2500 kJ
0.102 mol ≈ −24509.80 kJ/mol
Step 4: Determine the molar enthalpy of combustion of propane gas:
Molar enthalpy of combustion of propane = −24509.80 kJ/mol
Therefore, the molar enthalpy of combustion of propane gas is approximately
-24509.80 kJ/mol.
Question 8
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2Mg(s) + O2(g)→2M gO(s)
given the following standard enthalpy of formation values: Mg(s): −601.7
kJ/mol O2(g): 0 kJ/mol M gO(s): −601.6 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard enthalpy
of formation values for each substance. The balanced chemical equation is:
2Mg(s) + O2(g)→2M gO(s)
Given:
∆H◦
f(Mg) = −601.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
∆H◦
f(MgO) = −601.6 kJ/mol
6
Step 2: Determine the standard enthalpy change for the reaction using stan-
dard enthalpy of formation values. The standard enthalpy change for the re-
action can be calculated by the sum of the standard enthalpies of formation
of the products minus the sum of the standard enthalpies of formation of the
reactants:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given values into the equation:
∆H◦= 2(∆H◦
f(MgO)) −[2(∆H◦
f(Mg)) + ∆H◦
f(O2)]
∆H◦= 2(−601.6) −[2(−601.7) + 0]
∆H◦=−1203.2 + 1203.4
∆H◦= 0.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is 0.2 kJ/mol.
Question 9
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with ∆H◦=−571.6 kJ/mol, calculate the heat released when 5.00 g of H2reacts
with excess O2.
Solution
Step 1: Calculate the number of moles of H2in 5.00 g.
Molar mass of H2= 2.016 g/mol
Moles of H2=5.00 g
2.016 g/mol
Moles of H2= 2.48 mol
Step 2: Use the stoichiometry of the reaction to determine the heat released
when 2.48 moles of H2react.
Moles of water produced = 2.48 mol ×2 mol H2O(l)
2 mol H2
Moles of water produced = 2.48 mol
Step 3: Calculate the heat released when 2.48 moles of H2react.
Heat released = 2.48 mol ×(−571.6 kJ/mol)
Heat released = −1416.9 kJ
Answer: The heat released when 5.00 g of H2reacts with excess O2is
-1416.9 kJ.
7
Question 10
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2O2(l)→2H2O(l)+O2(g)
given the following standard enthalpies of formation:
∆H◦
f(H2O2(l)) = −196.1 kJ/mol,∆H◦
f(H2O(l)) = −285.8 kJ/mol,and ∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and determine the ∆H◦for the
reaction based on standard enthalpies of formation. The reaction provided is:
2H2O2(l)→2H2O(l)+O2(g)
The standard enthalpy change for the reaction can be calculated using the
standard enthalpies of formation of the products and reactants:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of products and reactants,
respectively.
Step 2: Substitute the given values into the equation. We have:
∆H◦= [2(−285.8) + 0] −[2(−196.1)]
∆H◦= [−571.6] −[−392.2]
∆H◦=−179.4 kJ
Therefore, the standard enthalpy change for the reaction is ∆H◦=−179.4 kJ.
Question 11
Question
For the reaction
2C5H12(g) + 16O2(g)→10CO2(g) + 12H2O(g)
the standard enthalpy change (∆H◦) is −3500 kJ. Calculate the enthalpy change
when 5.00 g of C5H12 is burned at constant pressure.
8
Solution
Step 1: Calculate the molar quantity of C5H12. Given: Mass of C5H12 = 5.00
g, Molar mass of C5H12 = 5(12.01) + 12(1.01) = 72.15 g/mol.
Number of moles of C5H12:
moles = mass
molar mass =5.00 g
72.15 g/mol = 0.0693 mol
Step 2: Use stoichiometry to calculate the enthalpy change. From the bal-
anced chemical equation, we see that for 2 moles of C5H12 reacted, the enthalpy
change is −3500 kJ. Therefore, for 0.0693 moles of C5H12:
Enthalpy change = −3500 kJ
2 mol ×0.0693 mol = −121 kJ
Therefore, the enthalpy change when 5.00 g of C5H12 is burned at constant
pressure is −121 kJ.
Question 12
Question
Calculate the enthalpy change for the reaction: 2C(graphite)+3H2(g)→C2H6(g)
given the following bond enthalpies: C−Cbond: 347 kJ/mol, C=Cbond:
614 kJ/mol, C−Hbond: 413 kJ/mol, and H−Hbond: 432 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and calculate the total bond
energy of the bonds broken and formed:
Bonds broken: 2(C−C) + 3(H−H) = 2(347) + 3(432) = 694 + 1296 =
1990 kJ
Bond formed: 1(C−C) + 6(C−H) = 1(347) + 6(413) = 347 + 2478 =
2825 kJ
Step 2: Calculate the overall enthalpy change:
∆H= Bonds broken −Bonds formed
∆H= 1990 kJ −2825 kJ = −835 kJ
Step 3: Therefore, the enthalpy change for the reaction is −835 kJ .
9
Question 13
Question
Calculate the enthalpy change (∆H) for the combustion of ethylene gas (C2H4)
according to the equation:
C2H4(g) + 3O2(g)→2CO2(g) + 2H2O(l)
Given the standard enthalpies of formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(C2H4) = +52.3 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the combustion of ethylene
gas by using the standard enthalpies of formation of the products and reactants.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
= [2 ×∆H◦
f(CO2)+2×∆H◦
f(H2O)] −[∆H◦
f(C2H4)+3×∆H◦
f(O2)]
= [2 ×(−393.5) + 2 ×(−285.8)] −[52.3+3×0]
= [−787.0−571.6] −52.3
=−1358.6−52.3
=−1410.9 kJ/mol
Therefore, the standard enthalpy change for the combustion of ethylene gas
is ∆H=−1410.9 kJ/mol.
Question 14
Question
Calculate the standard enthalpy change for the following reaction at 25
°
C:
2C(graphite)+3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.68 kJ/mol
10
Solution
Step 1: Write the balanced chemical equation for the reaction and determine
the change in enthalpy.
2C(graphite)+3H2(g)→C2H6(g)
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Step 2: Calculate the change in enthalpy.
∆H= [1 ×∆H◦
f(C2H6(g)) −(2 ×∆H◦
f(C(graphite)) + 3 ×∆H◦
f(H2(g))]
= [(1 × −84.68) −(2 ×0+3×0)] kJ/mol
=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 25
°
C is
-84.68 kJ/mol.
Question 15
Question
A 2.50 g sample of ethylene glycol (C2H6O2) is burned in a bomb calorimeter.
The temperature of the calorimeter increases by 6.50
°
C. If the heat capacity
of the calorimeter is 12.59 kJ/
°
C, calculate the molar heat of combustion of
ethylene glycol. The molar mass of ethylene glycol is 62.07 g/mol.
Solution
Step 1: Calculate the heat released by the calorimeter.
Heat released by the calorimeter = C×∆T
= 12.59 kJ/
°
C×6.50C
= 81.935 kJ
Step 2: Calculate the moles of ethylene glycol burned.
Moles of ethylene glycol = 2.50 g
62.07 g/mol
= 0.04032 mol
Step 3: Calculate the molar heat of combustion of ethylene glycol.
Molar heat of combustion = Heat released by the calorimeter
Moles of ethylene glycol
=81.935 kJ
0.04032 mol
= 2033.12 kJ/mol
Therefore, the molar heat of combustion of ethylene glycol is 2033.12 kJ/mol.
11
Question 16
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + 3H2(g)→C2H4(g)
given the following bond dissociation energies:
C-C : 347 kJ/mol
C=C : 614 kJ/mol
H-H : 436 kJ/mol
C-H : 413 kJ/mol
Solution
Step 1: Calculate the total bond dissociation energy for the reactants.
2(C-C) + 3(H-H) + 6(C-H) = 2(347 kJ/mol) + 3(436 kJ/mol) + 6(413 kJ/mol)
= 694 kJ/mol + 1308 kJ/mol + 2478 kJ/mol
= 4480 kJ/mol
Step 2: Calculate the total bond dissociation energy for the products.
C=C + 4(C-H) = 614 kJ/mol + 4(413 kJ/mol)
= 614 kJ/mol + 1652 kJ/mol
= 2266 kJ/mol
Step 3: Calculate the change in bond dissociation energy.
∆H= Energy of bonds broken −Energy of bonds formed
∆H= (4480 kJ/mol) −(2266 kJ/mol)
∆H= 2214 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H= 2214 kJ/mol.
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2C(s)+3H2(g)→C2H6(g)
Given the following standard enthalpy of formation values: C(s) : 0 kJ/mol,
H2(g) : 0 kJ/mol, C2H6(g) : −84.68 kJ/mol.
12
Solution
Step 1: Write the given chemical equation and the formation reaction for
C2H6(g):
Given equation: 2C(s)+3H2(g)→C2H6(g)
Formation reaction for C2H6(g) : 2C(s)+3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change for the formation of C2H6(g):
∆H◦
formation =X∆H◦
products −X∆H◦
reactants
∆H◦
formation =−84.68 kJ/mol −(2(0 kJ/mol) + 3(0 kJ/mol))
∆H◦
formation =−84.68 kJ/mol
Step 3: Remember that the enthalpy change for the reaction is the same as
the enthalpy change for the formation of C2H6(g), but with opposite sign.
∆H◦
reaction =−∆H◦
formation
∆H◦
reaction =−(−84.68 kJ/mol)
∆H◦
reaction = 84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 25◦C is
84.68 kJ/mol.
Question 18
Question
Given the standard enthalpy of formation for water gas (H2O(g)) is ∆H◦
f=
−241.8 kJ/mol, calculate the standard enthalpy change (∆H◦) for the reaction
below:
2H2(g) + O2(g) →2H2O(g)
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g) →2H2O(g)
Step 2: Determine the standard enthalpy change (∆H◦) for the reaction
using the standard enthalpy of formation values:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
13
Step 3: Calculate the change in enthalpy based on the given standard en-
thalpy of formation values:
∆H◦= 2 ×∆H◦
f(H2O(g)) −[2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
Step 4: Substitute the known values and calculate:
∆H◦= 2 ×(−241.8 kJ/mol) −[2 ×0 + 0] = −483.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−483.6
kJ/mol.
Question 19
Question
Calculate the enthalpy change (∆H) for the following reaction:
2 C(graphite) + 3 H2(g) →CH4(g)
Given the following enthalpy changes:
∆H◦
ffor CH4(g) = −74.81 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
∆H◦
ffor H2(g) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the relevant enthalpy
changes.
2 C(graphite) + 3 H2(g) →CH4(g)
Given:
∆H◦
ffor CH4(g) = −74.81 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
∆H◦
ffor H2(g) = 0 kJ/mol
Step 2: Calculate the enthalpy change for the reaction using standard en-
thalpies of formation.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H= ∆H◦
f(CH4)−(∆H◦
f(C) + 3 ×∆H◦
f(H2))
∆H=−74.81 kJ/mol −(0 + 3 ×0) kJ/mol
∆H=−74.81 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is -74.81 kJ/mol.
14
Question 20
Question
Calculate the enthalpy change (∆H) for the following reaction at 298 K:
2H2(g) + O2(g) →2H2O(l)
Given the following standard enthalpies of formation at 298 K: ∆H◦
f(H2O) =
−285.8 kJ/mol, ∆H◦
f(H2) = 0 kJ/mol, ∆H◦
f(O2) = 0 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and identify the given values.
The balanced chemical equation is:
2H2(g) + O2(g) →2H2O(l)
Given: ∆H◦
f(H2O) = −285.8 kJ/mol ∆H◦
f(H2) = 0 kJ/mol ∆H◦
f(O2) = 0
kJ/mol
Step 2: Calculate the change in enthalpy (∆H) using the standard enthalpies
of formation.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Substitute the given values:
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0]
∆H◦=−571.6 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is −571.6 kJ/mol .
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
given the following standard enthalpy of formation values: ∆H◦
f(C2H6) =
−84.7 kJ/mol, ∆H◦
f(CO2) = −393.5 kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol,
∆H◦
f(O2) = 0 kJ/mol.
15
Solution
Step 1: First, write the balanced chemical equation for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given standard enthalpy of formation values into the
equation:
∆H◦= 4 ×∆H◦
f(CO2)+6×∆H◦
f(H2O)−2×∆H◦
f(C2H6)−7×∆H◦
f(O2)
Step 4: Calculate the standard enthalpy change:
∆H◦= 4(−393.5) + 6(−285.8) −2(−84.7) −7(0)
∆H◦=−1574 −1714.8 + 169.4−0
∆H◦=−3119.4 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−3119.4 kJ/mol.
Question 22
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at 298
K:
2 C(graphite) + 3 H2(g) →C2H6(g)
Given the following standard enthalpies of formation at 298 K:
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Solution
Step 1: Write the given balanced chemical equation with the standard enthalpies
of formation:
2 C(graphite) + 3 H2(g) →C2H6(g)
16
Given standard enthalpies of formation:
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦=∆H◦
f(C2H6)−2·∆H◦
f(C(graphite)) + 3 ·∆H◦
f(H2)
∆H◦= [−84.68 kJ/mol] −[2 ·0+3·0]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol .
Question 23
Question
The combustion of ethanol (C2H5OH) releases 1367 kJ/mol. Calculate the
enthalpy change for the reaction C2H5OH(g)→C2H4(g) + H2O(g).
Solution
Step 1: Write the balanced chemical equation for the reaction and identify the
given enthalpy change. The balanced chemical equation for the reaction is:
C2H5OH(g)→C2H4(g) + H2O(g)
Given: ∆H=−1367 kJ/mol for the reaction C2H5OH(g)+3O2(g)→2CO2(g)+
3H2O(g)
Step 2: Apply Hess’s Law to find the enthalpy change for the target reaction.
To find the enthalpy change for the target reaction, we will use the enthalpies
of formation of the reactants and products. The enthalpy change for the target
reaction can be found by manipulating the given reaction to add up to the target
reaction. First, we reverse the given reaction:
2CO2(g)+3H2O(g)→C2H5OH(g)+3O2(g)
Step 3: Find the enthalpy change for the reversed reaction. Since ∆His for
the forward reaction, we will negate the given ∆Hvalue to find the enthalpy
change for the reversed reaction:
∆Hrev =−(−1367) kJ/mol = 1367 kJ/mol
17
Step 4: Manipulate the equations to obtain the target reaction. To use
the reversed reaction to obtain the target reaction, we need to cancel out the
substances that are not part of the target reaction.
2CO2(g)+3H2O(g)→C2H5OH(g)+3O2(g)
2CO2(g)+3H2O(g)→C2H4(g) + H2O(g)
Step 5: Calculate the enthalpy change for the target reaction. The enthalpy
change for the target reaction is equal to the negative of the enthalpy change
for the reversed reaction, as the target reaction is the reverse of the reversed
reaction. Therefore, the enthalpy change for the target reaction is:
∆Htarget =−1367 kJ/mol
So, the enthalpy change for the reaction C2H5OH(g)→C2H4(g) + H2O(g)
is −1367 kJ/mol.
Question 24
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2C6H12O6(s) + 17O2(g)→12CO2(g) + 12H2O(l)
Given the standard enthalpies of formation: C6H12O6(s) = −1273.3 kJ/mol,
CO2(g) = −393.5 kJ/mol, H2O(l) = −285.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy change for the reaction based on the
standard enthalpies of formation.
The standard enthalpy change can be calculated using the equation:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
where nand mare the stoichiometric coefficients of the products and reactants,
respectively.
Given: C6H12O6(s) = −1273.3 kJ/mol, CO2(g) = −393.5 kJ/mol, H2O(l) =
−285.8 kJ/mol.
The standard enthalpy change for the reaction is:
∆H◦= 12(−393.5) + 12(−285.8) −2(−1273.3)
Step 2: Solve for ∆H◦.
∆H◦=−4722 + (−3429.6) + 2546.6
18
∆H◦=−4205.0 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction at 25◦C is
-4205.0 kJ/mol.
Question 25
Question
Given the following reaction:
2H2(g)+ O2(g)→2H2O(l)
the standard enthalpy change of formation for water (∆H◦
f) is -285.8 kJ/mol.
Calculate the standard enthalpy change of the reaction in kJ.
Solution
Step 1: Write down the balanced chemical equation with the given enthalpy
change of formation.
2H2(g) + O2(g) →2H2O(l) ∆H◦
f=−285.8 kJ/mol
Step 2: Calculate the standard enthalpy change of the reaction using the
enthalpy change of formation values.
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = 2 ×∆H◦
f(H2O(l)) −[2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
∆H◦
rxn = 2 ×(−285.8) −[2 ×0 + 0] = −571.6 kJ
Therefore, the standard enthalpy change of the reaction is −571.6 kJ.
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of acety-
lene (C2H2(g)) to form carbon dioxide (CO2(g)) and water (H2O(l)) given the
following standard enthalpies of formation:
∆H◦
f(C2H2(g)) = 226.7 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
19
Solution
Step 1: Write the balanced chemical equation for the combustion of acetylene:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(l)
Step 2: Calculate the standard enthalpy change using the given standard
enthalpies of formation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [4∆H◦
f(CO2(g)) + 2∆H◦
f(H2O(l))] −[2∆H◦
f(C2H2(g))]
∆H◦= [4(−393.5) + 2(−285.8)] −[2(226.7)]
∆H◦= [−1574 −571.6] −453.4
∆H◦=−2145.6−453.4
∆H◦=−2599.0 kJ/mol
Therefore, the standard enthalpy change for the combustion of acetylene to
form carbon dioxide and water is −2599.0 kJ/mol.
Question 27
Question
For the reaction:
2A(g)+3B(g)→4C(g)+2D(g)
the standard enthalpy change (∆rH◦) is −1820 kJ/mol at 298 K. Calculate
the standard enthalpy of formation of compound A if the standard enthalpy of
formation of compound B is −410 kJ/mol, compound C is −790 kJ/mol, and
compound D is −560 kJ/mol.
Solution
Step 1: Determine the standard enthalpy of formation of compound A using
Hess’s law. Step 2: Apply Hess’s law to the enthalpy of the reaction to solve
for ∆H◦
fof compound A.
Step 1: Write the standard enthalpy change of the reaction in terms of the
standard enthalpies of formation of the compounds involved:
2∆H◦
f(A) + 3∆H◦
f(B) = 4∆H◦
f(C) + 2∆H◦
f(D)
Step 2: Substitute the values provided into the equation:
2∆H◦
f(A) + 3(−410 kJ/mol) = 4(−790 kJ/mol) + 2(−560 kJ/mol)
20
2∆H◦
f(A)−1230 kJ/mol = −3160 kJ/mol
2∆H◦
f(A) = −1930 kJ/mol
∆H◦
f(A) = −965 kJ/mol
Therefore, the standard enthalpy of formation of compound A is −965
kJ/mol.
Question 28
Question
Given the following thermochemical equation:
2A(g)+3B(g)→C(g)
The enthalpy change for this reaction is ∆H=−150 kJ. If 5.00 moles of A
and 7.00 moles of B react to form 2.00 moles of C under constant pressure, what
is the enthalpy change for the reaction when 1.00 mole of C is formed under the
same conditions?
Solution
Step 1: Find the enthalpy change for the reaction when 1.00 mole of C is formed.
Given:
2A(g)+3B(g)→C(g)
∆H=−150 kJ
We want to find the enthalpy change for the reaction when 1.00 mole of C is
formed. Using stoichiometry, we can relate the enthalpy changes for the given
reaction:
Moles of C formed = 2
Moles of C formed = 2 ×∆H
2= 2 × −150 kJ = −300 kJ
Therefore, the enthalpy change for the reaction when 1.00 mole of C is formed
is −300 kJ .
21
Question 29
Question
Calculate the enthalpy change (∆H) for the reaction below at 25◦C given the
following information:
2H2(g) + O2(g)→2H2O(l)
∆H◦
f(kJ/mol): H2(g) = 0; O2(g) = 0; H2O(l) = -286
Solution
Step 1: Write the balanced equation for the reaction and determine the ∆Hof
the reaction using the standard enthalpies of formation (∆H◦
f) of the compounds
involved.
2H2(g) + O2(g)→2H2O(l)
∆H=P∆H◦
f(products) −P∆H◦
f(reactants)
Step 2: Substitute the given ∆H◦
fvalues into the equation and solve.
∆H= [2 ·∆H◦
f(H2O(l))] −[2 ·∆H◦
f(H2(g)) + 1 ·∆H◦
f(O2(g))]
∆H= [2 ·(−286)] −[2 ·0+1·0] = −572 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction at 25◦C is -572 kJ/mol.
Question 30
Question
Given the following reaction, calculate the heat of reaction at 298 K using
standard enthalpies of formation:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.83 kJ/mol
22
Question 3
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2CH3OH(l) + 3O2(g)→2CO2(g) + 4H2O(l)
given the following standard enthalpies of formation:
∆H◦
f(CH3OH) = −238.6 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.
∆H◦
rxn =Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients in the balanced chemical
equation.
Step 2: Substitute the given values into the equation and solve for ∆H◦
rxn.
∆H◦
rxn = [2(−393.5 kJ/mol) + 4(−285.8 kJ/mol)] −[2(−238.6 kJ/mol) + 3(0)]
∆H◦
rxn = [−787.0 kJ/mol −1143.2 kJ/mol]
∆H◦
rxn =−1930.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−1930.2 kJ/mol.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(graphite)+3H2(g)→C2H2(g)
given the following standard enthalpy of formation values: ∆H◦
f:C(graphite) =
0 kJ/mol, H2(g) = 0 kJ/mol, and C2H2(g) = 227 kJ/mol.
3
Solution
Step 1: Write the balanced chemical equation for the reaction. Step 2: Calculate
the standard enthalpy change (∆H◦) using the standard enthalpy of formation
values. - ∆H◦=Pn∆H◦
f(products) −Pm∆H◦
f(reactants) - Substitute the
given values into the equation and solve for ∆H◦.
Question 5
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction using
the given standard enthalpies of formation:
2C(s) + 3H2(g)→C2H6(g)
Given:
Standard enthalpy of formation, ∆H◦
f(C2H6(g)) = −84.68 kJ/mol
Standard enthalpy of formation, ∆H◦
f(C(s)) = 0 kJ/mol
Standard enthalpy of formation, ∆H◦
f(H2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical reaction and identify the given enthalpies
of formation.
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the formula:
∆H◦=X∆H◦
fproducts −X∆H◦
freactants
Step 3: Calculate the standard enthalpy change:
∆H◦=1×∆H◦
f(C2H6(g))−2×∆H◦
f(C(s)) + 3 ×∆H◦
f(H2(g))
= (1 × −84.68) −(2 ×0+3×0)
=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
−84.68 kJ/mol.
Question 6
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l)
4
Calculate the standard enthalpy change (∆H◦) for the reaction using the fol-
lowing data:
Substance ∆H◦
f(kJ/mol)
H2(g) 0
O2(g) 0
H2O(l) -285.8
Solution
Step 1: Write the balanced chemical equation for the reaction and denote the
standard enthalpy of formation for each substance:
2H2(g)+O2(g)→2H2O(l)
H2(g):∆H◦
f= 0 kJ/mol
O2(g):∆H◦
f= 0 kJ/mol
H2O(l):∆H◦
f=−285.8 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦) for the given reac-
tion using the standard enthalpy of formation values:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is -571.6
kJ/mol.
Question 7
Question
During an experiment, 4.50 grams of propane gas (C3H8) is burned in excess
oxygen gas to produce carbon dioxide gas and water vapor. The heat released by
the reaction is found to be -2500 kJ. Calculate the molar enthalpy of combustion
of propane gas. (Molar masses: C= 12.01 g/mol, H= 1.008 g/mol)
Solution
Step 1: Write the balanced chemical equation for the combustion of propane
gas. Step 2: Calculate the number of moles of propane used in the reaction.
Step 3: Calculate the heat released per mole of propane. Step 4: Determine the
molar enthalpy of combustion of propane gas.
5
Step 1: The balanced chemical equation for the combustion of propane gas
is:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
Step 2: Calculate the number of moles of propane used:
Molar mass of C3H8= (3 ×12.01 g/mol) + (8 ×1.008 g/mol) = 44.11 g/mol
Number of moles of C3H8=4.50 g
44.11 g/mol ≈0.102 mol
Step 3: Calculate the heat released per mole of propane:
Heat released by reaction = −2500 kJ
Heat released per mole of propane = −2500 kJ
0.102 mol ≈ −24509.80 kJ/mol
Step 4: Determine the molar enthalpy of combustion of propane gas:
Molar enthalpy of combustion of propane = −24509.80 kJ/mol
Therefore, the molar enthalpy of combustion of propane gas is approximately
-24509.80 kJ/mol.
Question 8
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2Mg(s) + O2(g)→2M gO(s)
given the following standard enthalpy of formation values: Mg(s): −601.7
kJ/mol O2(g): 0 kJ/mol M gO(s): −601.6 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard enthalpy
of formation values for each substance. The balanced chemical equation is:
2Mg(s) + O2(g)→2M gO(s)
Given:
∆H◦
f(Mg) = −601.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
∆H◦
f(MgO) = −601.6 kJ/mol
6
Step 2: Determine the standard enthalpy change for the reaction using stan-
dard enthalpy of formation values. The standard enthalpy change for the re-
action can be calculated by the sum of the standard enthalpies of formation
of the products minus the sum of the standard enthalpies of formation of the
reactants:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given values into the equation:
∆H◦= 2(∆H◦
f(MgO)) −[2(∆H◦
f(Mg)) + ∆H◦
f(O2)]
∆H◦= 2(−601.6) −[2(−601.7) + 0]
∆H◦=−1203.2 + 1203.4
∆H◦= 0.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is 0.2 kJ/mol.
Question 9
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with ∆H◦=−571.6 kJ/mol, calculate the heat released when 5.00 g of H2reacts
with excess O2.
Solution
Step 1: Calculate the number of moles of H2in 5.00 g.
Molar mass of H2= 2.016 g/mol
Moles of H2=5.00 g
2.016 g/mol
Moles of H2= 2.48 mol
Step 2: Use the stoichiometry of the reaction to determine the heat released
when 2.48 moles of H2react.
Moles of water produced = 2.48 mol ×2 mol H2O(l)
2 mol H2
Moles of water produced = 2.48 mol
Step 3: Calculate the heat released when 2.48 moles of H2react.
Heat released = 2.48 mol ×(−571.6 kJ/mol)
Heat released = −1416.9 kJ
Answer: The heat released when 5.00 g of H2reacts with excess O2is
-1416.9 kJ.
7
Question 10
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2O2(l)→2H2O(l)+O2(g)
given the following standard enthalpies of formation:
∆H◦
f(H2O2(l)) = −196.1 kJ/mol,∆H◦
f(H2O(l)) = −285.8 kJ/mol,and ∆H◦
f(O2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and determine the ∆H◦for the
reaction based on standard enthalpies of formation. The reaction provided is:
2H2O2(l)→2H2O(l)+O2(g)
The standard enthalpy change for the reaction can be calculated using the
standard enthalpies of formation of the products and reactants:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of products and reactants,
respectively.
Step 2: Substitute the given values into the equation. We have:
∆H◦= [2(−285.8) + 0] −[2(−196.1)]
∆H◦= [−571.6] −[−392.2]
∆H◦=−179.4 kJ
Therefore, the standard enthalpy change for the reaction is ∆H◦=−179.4 kJ.
Question 11
Question
For the reaction
2C5H12(g) + 16O2(g)→10CO2(g) + 12H2O(g)
the standard enthalpy change (∆H◦) is −3500 kJ. Calculate the enthalpy change
when 5.00 g of C5H12 is burned at constant pressure.
8
Solution
Step 1: Calculate the molar quantity of C5H12. Given: Mass of C5H12 = 5.00
g, Molar mass of C5H12 = 5(12.01) + 12(1.01) = 72.15 g/mol.
Number of moles of C5H12:
moles = mass
molar mass =5.00 g
72.15 g/mol = 0.0693 mol
Step 2: Use stoichiometry to calculate the enthalpy change. From the bal-
anced chemical equation, we see that for 2 moles of C5H12 reacted, the enthalpy
change is −3500 kJ. Therefore, for 0.0693 moles of C5H12:
Enthalpy change = −3500 kJ
2 mol ×0.0693 mol = −121 kJ
Therefore, the enthalpy change when 5.00 g of C5H12 is burned at constant
pressure is −121 kJ.
Question 12
Question
Calculate the enthalpy change for the reaction: 2C(graphite)+3H2(g)→C2H6(g)
given the following bond enthalpies: C−Cbond: 347 kJ/mol, C=Cbond:
614 kJ/mol, C−Hbond: 413 kJ/mol, and H−Hbond: 432 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and calculate the total bond
energy of the bonds broken and formed:
Bonds broken: 2(C−C) + 3(H−H) = 2(347) + 3(432) = 694 + 1296 =
1990 kJ
Bond formed: 1(C−C) + 6(C−H) = 1(347) + 6(413) = 347 + 2478 =
2825 kJ
Step 2: Calculate the overall enthalpy change:
∆H= Bonds broken −Bonds formed
∆H= 1990 kJ −2825 kJ = −835 kJ
Step 3: Therefore, the enthalpy change for the reaction is −835 kJ .
9
Question 13
Question
Calculate the enthalpy change (∆H) for the combustion of ethylene gas (C2H4)
according to the equation:
C2H4(g) + 3O2(g)→2CO2(g) + 2H2O(l)
Given the standard enthalpies of formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(C2H4) = +52.3 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the combustion of ethylene
gas by using the standard enthalpies of formation of the products and reactants.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
= [2 ×∆H◦
f(CO2)+2×∆H◦
f(H2O)] −[∆H◦
f(C2H4)+3×∆H◦
f(O2)]
= [2 ×(−393.5) + 2 ×(−285.8)] −[52.3+3×0]
= [−787.0−571.6] −52.3
=−1358.6−52.3
=−1410.9 kJ/mol
Therefore, the standard enthalpy change for the combustion of ethylene gas
is ∆H=−1410.9 kJ/mol.
Question 14
Question
Calculate the standard enthalpy change for the following reaction at 25
°
C:
2C(graphite)+3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.68 kJ/mol
10
Solution
Step 1: Write the balanced chemical equation for the reaction and determine
the change in enthalpy.
2C(graphite)+3H2(g)→C2H6(g)
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Step 2: Calculate the change in enthalpy.
∆H= [1 ×∆H◦
f(C2H6(g)) −(2 ×∆H◦
f(C(graphite)) + 3 ×∆H◦
f(H2(g))]
= [(1 × −84.68) −(2 ×0+3×0)] kJ/mol
=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 25
°
C is
-84.68 kJ/mol.
Question 15
Question
A 2.50 g sample of ethylene glycol (C2H6O2) is burned in a bomb calorimeter.
The temperature of the calorimeter increases by 6.50
°
C. If the heat capacity
of the calorimeter is 12.59 kJ/
°
C, calculate the molar heat of combustion of
ethylene glycol. The molar mass of ethylene glycol is 62.07 g/mol.
Solution
Step 1: Calculate the heat released by the calorimeter.
Heat released by the calorimeter = C×∆T
= 12.59 kJ/
°
C×6.50C
= 81.935 kJ
Step 2: Calculate the moles of ethylene glycol burned.
Moles of ethylene glycol = 2.50 g
62.07 g/mol
= 0.04032 mol
Step 3: Calculate the molar heat of combustion of ethylene glycol.
Molar heat of combustion = Heat released by the calorimeter
Moles of ethylene glycol
=81.935 kJ
0.04032 mol
= 2033.12 kJ/mol
Therefore, the molar heat of combustion of ethylene glycol is 2033.12 kJ/mol.
11
Question 16
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite) + 3H2(g)→C2H4(g)
given the following bond dissociation energies:
C-C : 347 kJ/mol
C=C : 614 kJ/mol
H-H : 436 kJ/mol
C-H : 413 kJ/mol
Solution
Step 1: Calculate the total bond dissociation energy for the reactants.
2(C-C) + 3(H-H) + 6(C-H) = 2(347 kJ/mol) + 3(436 kJ/mol) + 6(413 kJ/mol)
= 694 kJ/mol + 1308 kJ/mol + 2478 kJ/mol
= 4480 kJ/mol
Step 2: Calculate the total bond dissociation energy for the products.
C=C + 4(C-H) = 614 kJ/mol + 4(413 kJ/mol)
= 614 kJ/mol + 1652 kJ/mol
= 2266 kJ/mol
Step 3: Calculate the change in bond dissociation energy.
∆H= Energy of bonds broken −Energy of bonds formed
∆H= (4480 kJ/mol) −(2266 kJ/mol)
∆H= 2214 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H= 2214 kJ/mol.
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2C(s)+3H2(g)→C2H6(g)
Given the following standard enthalpy of formation values: C(s) : 0 kJ/mol,
H2(g) : 0 kJ/mol, C2H6(g) : −84.68 kJ/mol.
12
Solution
Step 1: Write the given chemical equation and the formation reaction for
C2H6(g):
Given equation: 2C(s)+3H2(g)→C2H6(g)
Formation reaction for C2H6(g) : 2C(s)+3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change for the formation of C2H6(g):
∆H◦
formation =X∆H◦
products −X∆H◦
reactants
∆H◦
formation =−84.68 kJ/mol −(2(0 kJ/mol) + 3(0 kJ/mol))
∆H◦
formation =−84.68 kJ/mol
Step 3: Remember that the enthalpy change for the reaction is the same as
the enthalpy change for the formation of C2H6(g), but with opposite sign.
∆H◦
reaction =−∆H◦
formation
∆H◦
reaction =−(−84.68 kJ/mol)
∆H◦
reaction = 84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction at 25◦C is
84.68 kJ/mol.
Question 18
Question
Given the standard enthalpy of formation for water gas (H2O(g)) is ∆H◦
f=
−241.8 kJ/mol, calculate the standard enthalpy change (∆H◦) for the reaction
below:
2H2(g) + O2(g) →2H2O(g)
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g) →2H2O(g)
Step 2: Determine the standard enthalpy change (∆H◦) for the reaction
using the standard enthalpy of formation values:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
13
Step 3: Calculate the change in enthalpy based on the given standard en-
thalpy of formation values:
∆H◦= 2 ×∆H◦
f(H2O(g)) −[2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
Step 4: Substitute the known values and calculate:
∆H◦= 2 ×(−241.8 kJ/mol) −[2 ×0 + 0] = −483.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−483.6
kJ/mol.
Question 19
Question
Calculate the enthalpy change (∆H) for the following reaction:
2 C(graphite) + 3 H2(g) →CH4(g)
Given the following enthalpy changes:
∆H◦
ffor CH4(g) = −74.81 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
∆H◦
ffor H2(g) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the relevant enthalpy
changes.
2 C(graphite) + 3 H2(g) →CH4(g)
Given:
∆H◦
ffor CH4(g) = −74.81 kJ/mol
∆H◦
ffor C(graphite) = 0 kJ/mol
∆H◦
ffor H2(g) = 0 kJ/mol
Step 2: Calculate the enthalpy change for the reaction using standard en-
thalpies of formation.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H= ∆H◦
f(CH4)−(∆H◦
f(C) + 3 ×∆H◦
f(H2))
∆H=−74.81 kJ/mol −(0 + 3 ×0) kJ/mol
∆H=−74.81 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is -74.81 kJ/mol.
14
Question 20
Question
Calculate the enthalpy change (∆H) for the following reaction at 298 K:
2H2(g) + O2(g) →2H2O(l)
Given the following standard enthalpies of formation at 298 K: ∆H◦
f(H2O) =
−285.8 kJ/mol, ∆H◦
f(H2) = 0 kJ/mol, ∆H◦
f(O2) = 0 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and identify the given values.
The balanced chemical equation is:
2H2(g) + O2(g) →2H2O(l)
Given: ∆H◦
f(H2O) = −285.8 kJ/mol ∆H◦
f(H2) = 0 kJ/mol ∆H◦
f(O2) = 0
kJ/mol
Step 2: Calculate the change in enthalpy (∆H) using the standard enthalpies
of formation.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Substitute the given values:
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0]
∆H◦=−571.6 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction is −571.6 kJ/mol .
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
given the following standard enthalpy of formation values: ∆H◦
f(C2H6) =
−84.7 kJ/mol, ∆H◦
f(CO2) = −393.5 kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol,
∆H◦
f(O2) = 0 kJ/mol.
15
Solution
Step 1: First, write the balanced chemical equation for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the given standard enthalpy of formation values into the
equation:
∆H◦= 4 ×∆H◦
f(CO2)+6×∆H◦
f(H2O)−2×∆H◦
f(C2H6)−7×∆H◦
f(O2)
Step 4: Calculate the standard enthalpy change:
∆H◦= 4(−393.5) + 6(−285.8) −2(−84.7) −7(0)
∆H◦=−1574 −1714.8 + 169.4−0
∆H◦=−3119.4 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−3119.4 kJ/mol.
Question 22
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at 298
K:
2 C(graphite) + 3 H2(g) →C2H6(g)
Given the following standard enthalpies of formation at 298 K:
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Solution
Step 1: Write the given balanced chemical equation with the standard enthalpies
of formation:
2 C(graphite) + 3 H2(g) →C2H6(g)
16
Given standard enthalpies of formation:
∆H◦
f[C(graphite)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦=∆H◦
f(C2H6)−2·∆H◦
f(C(graphite)) + 3 ·∆H◦
f(H2)
∆H◦= [−84.68 kJ/mol] −[2 ·0+3·0]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol .
Question 23
Question
The combustion of ethanol (C2H5OH) releases 1367 kJ/mol. Calculate the
enthalpy change for the reaction C2H5OH(g)→C2H4(g) + H2O(g).
Solution
Step 1: Write the balanced chemical equation for the reaction and identify the
given enthalpy change. The balanced chemical equation for the reaction is:
C2H5OH(g)→C2H4(g) + H2O(g)
Given: ∆H=−1367 kJ/mol for the reaction C2H5OH(g)+3O2(g)→2CO2(g)+
3H2O(g)
Step 2: Apply Hess’s Law to find the enthalpy change for the target reaction.
To find the enthalpy change for the target reaction, we will use the enthalpies
of formation of the reactants and products. The enthalpy change for the target
reaction can be found by manipulating the given reaction to add up to the target
reaction. First, we reverse the given reaction:
2CO2(g)+3H2O(g)→C2H5OH(g)+3O2(g)
Step 3: Find the enthalpy change for the reversed reaction. Since ∆His for
the forward reaction, we will negate the given ∆Hvalue to find the enthalpy
change for the reversed reaction:
∆Hrev =−(−1367) kJ/mol = 1367 kJ/mol
17
Step 4: Manipulate the equations to obtain the target reaction. To use
the reversed reaction to obtain the target reaction, we need to cancel out the
substances that are not part of the target reaction.
2CO2(g)+3H2O(g)→C2H5OH(g)+3O2(g)
2CO2(g)+3H2O(g)→C2H4(g) + H2O(g)
Step 5: Calculate the enthalpy change for the target reaction. The enthalpy
change for the target reaction is equal to the negative of the enthalpy change
for the reversed reaction, as the target reaction is the reverse of the reversed
reaction. Therefore, the enthalpy change for the target reaction is:
∆Htarget =−1367 kJ/mol
So, the enthalpy change for the reaction C2H5OH(g)→C2H4(g) + H2O(g)
is −1367 kJ/mol.
Question 24
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2C6H12O6(s) + 17O2(g)→12CO2(g) + 12H2O(l)
Given the standard enthalpies of formation: C6H12O6(s) = −1273.3 kJ/mol,
CO2(g) = −393.5 kJ/mol, H2O(l) = −285.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy change for the reaction based on the
standard enthalpies of formation.
The standard enthalpy change can be calculated using the equation:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
where nand mare the stoichiometric coefficients of the products and reactants,
respectively.
Given: C6H12O6(s) = −1273.3 kJ/mol, CO2(g) = −393.5 kJ/mol, H2O(l) =
−285.8 kJ/mol.
The standard enthalpy change for the reaction is:
∆H◦= 12(−393.5) + 12(−285.8) −2(−1273.3)
Step 2: Solve for ∆H◦.
∆H◦=−4722 + (−3429.6) + 2546.6
18
∆H◦=−4205.0 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction at 25◦C is
-4205.0 kJ/mol.
Question 25
Question
Given the following reaction:
2H2(g)+ O2(g)→2H2O(l)
the standard enthalpy change of formation for water (∆H◦
f) is -285.8 kJ/mol.
Calculate the standard enthalpy change of the reaction in kJ.
Solution
Step 1: Write down the balanced chemical equation with the given enthalpy
change of formation.
2H2(g) + O2(g) →2H2O(l) ∆H◦
f=−285.8 kJ/mol
Step 2: Calculate the standard enthalpy change of the reaction using the
enthalpy change of formation values.
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = 2 ×∆H◦
f(H2O(l)) −[2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
∆H◦
rxn = 2 ×(−285.8) −[2 ×0 + 0] = −571.6 kJ
Therefore, the standard enthalpy change of the reaction is −571.6 kJ.
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of acety-
lene (C2H2(g)) to form carbon dioxide (CO2(g)) and water (H2O(l)) given the
following standard enthalpies of formation:
∆H◦
f(C2H2(g)) = 226.7 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
19
Solution
Step 1: Write the balanced chemical equation for the combustion of acetylene:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(l)
Step 2: Calculate the standard enthalpy change using the given standard
enthalpies of formation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [4∆H◦
f(CO2(g)) + 2∆H◦
f(H2O(l))] −[2∆H◦
f(C2H2(g))]
∆H◦= [4(−393.5) + 2(−285.8)] −[2(226.7)]
∆H◦= [−1574 −571.6] −453.4
∆H◦=−2145.6−453.4
∆H◦=−2599.0 kJ/mol
Therefore, the standard enthalpy change for the combustion of acetylene to
form carbon dioxide and water is −2599.0 kJ/mol.
Question 27
Question
For the reaction:
2A(g)+3B(g)→4C(g)+2D(g)
the standard enthalpy change (∆rH◦) is −1820 kJ/mol at 298 K. Calculate
the standard enthalpy of formation of compound A if the standard enthalpy of
formation of compound B is −410 kJ/mol, compound C is −790 kJ/mol, and
compound D is −560 kJ/mol.
Solution
Step 1: Determine the standard enthalpy of formation of compound A using
Hess’s law. Step 2: Apply Hess’s law to the enthalpy of the reaction to solve
for ∆H◦
fof compound A.
Step 1: Write the standard enthalpy change of the reaction in terms of the
standard enthalpies of formation of the compounds involved:
2∆H◦
f(A) + 3∆H◦
f(B) = 4∆H◦
f(C) + 2∆H◦
f(D)
Step 2: Substitute the values provided into the equation:
2∆H◦
f(A) + 3(−410 kJ/mol) = 4(−790 kJ/mol) + 2(−560 kJ/mol)
20
2∆H◦
f(A)−1230 kJ/mol = −3160 kJ/mol
2∆H◦
f(A) = −1930 kJ/mol
∆H◦
f(A) = −965 kJ/mol
Therefore, the standard enthalpy of formation of compound A is −965
kJ/mol.
Question 28
Question
Given the following thermochemical equation:
2A(g)+3B(g)→C(g)
The enthalpy change for this reaction is ∆H=−150 kJ. If 5.00 moles of A
and 7.00 moles of B react to form 2.00 moles of C under constant pressure, what
is the enthalpy change for the reaction when 1.00 mole of C is formed under the
same conditions?
Solution
Step 1: Find the enthalpy change for the reaction when 1.00 mole of C is formed.
Given:
2A(g)+3B(g)→C(g)
∆H=−150 kJ
We want to find the enthalpy change for the reaction when 1.00 mole of C is
formed. Using stoichiometry, we can relate the enthalpy changes for the given
reaction:
Moles of C formed = 2
Moles of C formed = 2 ×∆H
2= 2 × −150 kJ = −300 kJ
Therefore, the enthalpy change for the reaction when 1.00 mole of C is formed
is −300 kJ .
21
Question 29
Question
Calculate the enthalpy change (∆H) for the reaction below at 25◦C given the
following information:
2H2(g) + O2(g)→2H2O(l)
∆H◦
f(kJ/mol): H2(g) = 0; O2(g) = 0; H2O(l) = -286
Solution
Step 1: Write the balanced equation for the reaction and determine the ∆Hof
the reaction using the standard enthalpies of formation (∆H◦
f) of the compounds
involved.
2H2(g) + O2(g)→2H2O(l)
∆H=P∆H◦
f(products) −P∆H◦
f(reactants)
Step 2: Substitute the given ∆H◦
fvalues into the equation and solve.
∆H= [2 ·∆H◦
f(H2O(l))] −[2 ·∆H◦
f(H2(g)) + 1 ·∆H◦
f(O2(g))]
∆H= [2 ·(−286)] −[2 ·0+1·0] = −572 kJ/mol
Therefore, the enthalpy change (∆H) for the reaction at 25◦C is -572 kJ/mol.
Question 30
Question
Given the following reaction, calculate the heat of reaction at 298 K using
standard enthalpies of formation:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
Given:
∆H◦
f(C2H6) = −84.68 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.83 kJ/mol
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Solution
Step 1: Calculate the standard enthalpy of reaction using standard enthalpies
of formation.
Given the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
The standard enthalpy of reaction can be calculated using the equation:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Substitute in the values:
∆H◦= [4(−393.5) + 6(−285.83)] −[2(−84.68) + 7(0)]
∆H◦= [−1574 −1714.98] −[−169.36]
∆H◦=−3288.98 + 169.36
∆H◦=−3119.62 kJ
Therefore, the heat of reaction at 298 K is -3119.62 kJ.
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