CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Thermochemistry
Question Bank - Set 5
Liberty University
Question 1
Question
Calculate the enthalpy change (∆H) for the following reaction:
2C(s) + 3H2(g)→C2H6(g)
Given the following bond energies: C-C = 347 kJ/mol, C-H = 413 kJ/mol, and
H-H = 436 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy required = 2(C-C) + 3(3C-H) + 3(H-H)
= 2(347 kJ/mol) + 3(3 ×413 kJ/mol) + 3(436 kJ/mol)
= 694 kJ + 3(1239 kJ) + 1308 kJ
= 3710 kJ
Step 2: Calculate the total energy released when the new bonds are formed
in the product.
Energy released = 2(6C-H)
= 2(6 ×413 kJ/mol)
= 4956 kJ
Step 3: Calculate the change in enthalpy (∆H) for the reaction.
∆H= Energy absorbed −Energy released
= 3710 kJ −4956 kJ
=−1246 kJ
Answer: The enthalpy change for the reaction is −1246 kJ.
Question 2
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(l)
Given the following standard enthalpies of formation (∆H◦
f):
∆H◦
f(C2H2(g)) = 226.7 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
= (4 ×∆H◦
f(CO2(g)) + 2 ×∆H◦
f(H2O(l))) −(2 ×∆H◦
f(C2H2(g))) −(5 ×0)
= (4 × −393.5 kJ/mol + 2 × −285.8 kJ/mol) −(2 ×226.7 kJ/mol)
= (−1574 kJ/mol −571.6 kJ/mol) −(453.4 kJ/mol)
=−2145 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is -2145
kJ/mol.
Question 3
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction using
the given bond dissociation energies:
CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g)
Given bond dissociation energies: D(CH) = 413 kJ/mol D(Cl−Cl) = 242 kJ/mol
D(C−Cl) = 339 kJ/mol D(H−Cl) = 431 kJ/mol
2
Solution
Step 1: Calculate the total bond dissociation energy of the reactants.
Dreactants = 1 ×D(CH)+4×D(Cl −Cl)+1×D(C−Cl)+4×D(H−Cl)
= 1 ×413 kJ/mol + 4 ×242 kJ/mol + 1 ×339 kJ/mol + 4 ×431 kJ/mol
= 413 + 968 + 339 + 1724
= 3444 kJ/mol
Step 2: Calculate the total bond dissociation energy of the products.
Dproducts = 1 ×D(C−Cl)+4×D(H−Cl)
= 1 ×339 kJ/mol + 4 ×431 kJ/mol
= 339 + 1724
= 2063 kJ/mol
Step 3: Calculate the change in bond dissociation energy (∆D).
∆D=Dproducts −Dreactants = 2063 kJ/mol −3444 kJ/mol =−1381 kJ/mol
Step 4: Calculate the standard enthalpy change using the equation:
∆H◦= ∆D+ ∆H◦
rxn
Since the reaction is exothermic, ∆H◦
rxn =−1381 kJ/mol Therefore, the stan-
dard enthalpy change for the reaction is ∆H◦=−1381 kJ/mol.
Question 4
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l)
and the standard enthalpies of formation (∆H◦
f) for H2(g), O2(g), and H2O(l)
as 0 kJ/mol, 0 kJ/mol, and -286 kJ/mol, respectively, calculate the standard
enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change using the formula:
∆H◦=Xn∆H◦
f
where nis the stoichiometric coefficient and ∆H◦
fis the standard enthalpy of
formation.
3
Step 2: Substitute the values into the formula and calculate the standard
enthalpy change:
∆H◦= 2 ×∆H◦
f(H2O) −2×∆H◦
f(H2)−∆H◦
f(O2)
∆H◦= 2 ×(−286 kJ/mol) −2×(0 kJ/mol) −0 kJ/mol
∆H◦=−572 kJ/mol
Therefore, the standard enthalpy change for the reaction is -572 kJ/mol.
Question 5
Question
Consider the reaction:
2C(s) + 3H2(g)→C2H6(g)
Given the following information:
Standard enthalpy of formation of C(s): −394 kJ/mol
Standard enthalpy of formation of H2(g) :-286 kJ/molStandardenthalpyofformationofC2H6(g) :-
84 kJ/mol Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Write the balanced equation for the reaction:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values. The standard enthalpy change for the reaction can be
calculated using the following formula:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
where ∆H◦is the standard enthalpy change for the reaction, nand mare
the stoichiometric coefficients of the products and reactants, respectively, and
∆H◦
products and ∆H◦
reactants are the standard enthalpies of formation of the prod-
ucts and reactants, respectively.
Step 3: Substitute the given values into the formula:
∆H◦= (1)(−84) −(2)(−394) −(3)(−286) kJ/mol
∆H◦=−84 + 788 + 858 kJ/mol
∆H◦= 1562 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is 1562 kJ/mol.
4
Question 6
Question
Consider the following reaction:
2H2(g) + O2(g) →2H2O(l)
Given that ∆H◦
ffor H2(g) = 0 kJ/mol, ∆H◦
ffor O2(g) = 0 kJ/mol, and ∆H◦
f
for H2O(l) = −286 kJ/mol, calculate the standard enthalpy change, ∆H◦, for
the above reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2(g) + O2(g) →2H2O(l)
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction. The
standard enthalpy change for a reaction can be calculated using the standard
enthalpies of formation of the reactants and products. The standard enthalpy
change (∆H◦) for a reaction is given by the difference in the sum of the standard
enthalpies of formation of the products and the sum of the standard enthalpies
of formation of the reactants.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Given:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(l)) = −286 kJ/mol
Substitute the values into the equation:
∆H◦= 2(∆H◦
f(H2O(l))) −[2(∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
∆H◦= 2(−286 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦=−572 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−572 kJ/mol.
Question 7
Question
For the reaction:
2H2(g) + O2(g)→2H2O(l)
the standard enthalpy change (∆H◦) is -571.6 kJ/mol. Calculate the standard
enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
5
Solution
Step 1: Write the given balanced chemical equation and the desired equation.
Given balanced chemical equation:
2H2(g) + O2(g)→2H2O(l)
Desired equation:
H2O(l)→H2(g) + 1
2O2(g)
Step 2: Determine the standard enthalpy change for the desired equation.
We can manipulate the given equation to obtain the desired equation as follows:
2H2O(l)→2H2(g) + O2(g)
Step 3: Reverse the given equation.
−2H2O(l)→ −2H2(g)−O2(g)
H2O(l)→H2(g) + 1
2O2(g)
Step 4: Determine the standard enthalpy change for the desired equation.
Since the given enthalpy change is for the forward reaction, we must change the
sign when reversing the equation:
−∆H◦=−(−571.6 kJ/mol) = 571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
is 571.6 kJ/mol.
Question 8
Question
Calculate the enthalpy change (∆H) for the reaction below using the given bond
dissociation energies:
2H2O(g)→2H2(g)+O2(g)
Given bond dissociation energies:
H-H: 436 kJ/mol
O=O: 495 kJ/mol
O-H: 463 kJ/mol
6
Solution
Step 1: Write out the balanced chemical equation for the reaction.
2H2O(g)→2H2(g)+O2(g)
Step 2: Calculate the total bond energy of the reactants. Total bond energy
of reactants = 2(2 mol ×O-H bond energy) + H-H bond energy
Total bond energy of reactants = 2 ×2×463 kJ/mol + 436 kJ/mol
Total bond energy of reactants = 1852 kJ/mol + 436 kJ/mol
Total bond energy of reactants = 2288 kJ
Step 3: Calculate the total bond energy of the products. Total bond energy
of products = 2(2 mol ×H-H bond energy) + O=O bond energy
Total bond energy of products = 2 ×2×436 kJ/mol + 495 kJ/mol
Total bond energy of products = 1744 kJ/mol + 495 kJ/mol
Total bond energy of products = 2239 kJ
Step 4: Calculate the change in bond energy. ∆H= Total bond energy of products−
Total bond energy of reactants
∆H= 2239 kJ - 2288 kJ
∆H=−49 kJ
Therefore, the enthalpy change (∆H) for the reaction is -49 kJ.
Question 9
Question
Calculate the enthalpy change (∆H) for the following reaction:
2C(s) + 3H2(g)→C2H6(g)
given the following bond dissociation energies:
C-C bond: 348 kJ/mol
C-H bond: 412 kJ/mol
H-H bond: 436 kJ/mol
C=C bond: 614 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy to break C-C bonds = 2 ×348 kJ/mol = 696 kJ/mol
Energy to break H-H bonds = 3 ×436 kJ/mol = 1308 kJ/mol
Energy to break C-H bonds = 2 ×412 kJ/mol = 824 kJ/mol
Total energy to break bonds = 696 + 1308 + 824 = 2828 kJ/mol
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Step 2: Calculate the total energy released in the formation of bonds in the
products.
Energy released in forming C-C bonds = 1 ×614 kJ/mol = 614 kJ/mol
Energy released in forming C-H bonds = 6 ×412 kJ/mol = 2472 kJ/mol
Total energy released in forming bonds = 614 + 2472 = 3086 kJ/mol
Step 3: Calculate the change in enthalpy (∆H) for the reaction using the
bond dissociation energies.
∆H= Energy to break bonds−Energy released in forming bonds = 2828 kJ/mol−3086 kJ/mol = −258 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H=−258 kJ/mol.
Question 10
Question
Calculate the enthalpy change (∆H) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
given the following bond dissociation energies:
H−H: 432 kJ/mol
O=O: 497 kJ/mol
O−H: 464 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy in bonds broken = 2 ×Energy of H−Hbond + Energy of O=Obond
= 2 ×432 kJ/mol + 1 ×497 kJ/mol
= 1361 kJ
Step 2: Calculate the total energy released when the new bonds are formed
in the products.
Energy in bonds formed = 2 ×Energy of O−Hbond
= 2 ×464 kJ/mol
= 928 kJ
Step 3: Calculate the change in enthalpy (∆H) for the reaction.
∆H= Energy in bonds broken −Energy in bonds formed
= 1361 kJ −928 kJ
= 433 kJ
Therefore, the enthalpy change (∆H) for the reaction is 433 kJ.
8
Question 11
Question
Given the following reaction:
2H2(g) + O2(g)→2H2O(l)
The standard enthalpy change for the reaction is -483.6 kJ mol−1. Calculate
the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
at the same temperature.
Solution
Step 1: Write the given reaction and its standard enthalpy change. Given
reaction:
2H2(g) + O2(g)→2H2O(l)
∆H=−483.6 kJ mol−1
Step 2: Write the target reaction. Target reaction:
H2O(l)→H2(g) + 1
2O2(g)
Step 3: Determine the standard enthalpy change for the target reaction.
Since the target reaction is the reverse of the given reaction, the sign of the
enthalpy change will be reversed. Therefore, ∆H= 483.6 kJ mol−1
So, the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
is 483.6 kJ mol−1at the same temperature.
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of ethylene
(C2H4) using the following information:
The standard enthalpy of formation of CO2is -393.5 kJ/mol.
The standard enthalpy of formation of H2O is -285.8 kJ/mol.
The standard enthalpy of formation of C2H4is 52.3 kJ/mol.
The balanced equation for the combustion of ethylene is:
C2H4(g) + 3O2(g)→2CO2(g) + 2H2O(l)
9
Solution
Step 1: Calculate the standard enthalpy change for the combustion of ethy-
lene using the standard enthalpy of formation values provided. The standard
enthalpy change can be calculated using the equation:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Where nand mare the stoichiometric coefficients from the balanced equa-
tion.
Step 2: Substitute the given values into the equation and calculate the stan-
dard enthalpy change.
∆H◦= [2(−393.5) + 2(−285.8)] −[1(52.3) + 3(0)]
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= (−787.0) −(52.3) = −834.3 kJ/mol
Therefore, the standard enthalpy change for the combustion of ethylene is
∆H◦=−834.3 kJ/mol.
Question 13
Question
Calculate the enthalpy change for the combustion of ethylene, C2H4(g), to form
carbon dioxide and water using the following data:
Reaction ∆H◦(kJ/mol)
C2H4(g) + 3O2(g) →2CO2(g) + 2H2O(l) -1411
H2(g) + 1
2O2(g) →H2O(l) -286
C(s) + O2(g) →CO2(g) -394
Solution
Step 1: Writing the balanced equation for the combustion of ethylene:
C2H4(g) + 3O2(g) →2CO2(g) + 2H2O(l)
Step 2: Calculating the enthalpy change for the given reaction:
∆H=X∆Hproducts −X∆Hreactants
∆H= [2(2∆HCO2) + 2∆HH2O(l)]−[∆HC2H4+ 3∆HO2]
∆H= [2(2 × −394) + 2 × −286] −[−1411]
∆H= [2(−788) −572] + 1411
∆H=−1576 −572 + 1411
∆H=−737 kJ/mol
Therefore, the enthalpy change for the combustion of ethylene to form carbon
dioxide and water is -737 kJ/mol.
10
Question 14
Question
A reaction has a standard enthalpy change of -418 kJ/mol. If the reaction is
exothermic, determine if the reaction is exothermic or endothermic when carried
out at 25
°
C if the reaction is carried out under nonstandard conditions with a
pressure of 2 atm. Assume the reaction has an ideal gas as a product.
Solution
Step 1: Write the expression for the standard Gibbs free energy change in terms
of the standard enthalpy change and the standard entropy change:
∆G◦= ∆H◦−T∆S◦
Step 2: Use the relationship between standard Gibbs free energy change and
equilibrium constant to find the standard entropy change (∆S◦).
We know that ∆G◦=−RT ln(K) where Kis the equilibrium constant under
standard conditions. Therefore,
∆G◦=−RT ln(K◦)
where K◦is the equilibrium constant under standard conditions.
Step 3: Given that the reaction is exothermic, we know that ∆H◦=−418
kJ/mol.
Step 4: Calculate the standard entropy change using the standard Gibbs
free energy change equation:
∆S◦=∆H◦−∆G◦
T
Step 5: Substitute the known values into the equation and solve for ∆S◦:
∆S◦=−418 −(−RT ln(K◦))
T
Step 6: Given that the pressure is 2 atm, we need to consider the effect of
pressure on the equilibrium constant in the expression for ∆G. The effect of
changing pressure on the equilibrium constant can be accounted for using the
Van’t Hoff equation:
ln K2
K1=∆nRT
R
where ∆nis the change in the number of moles of gas as the reaction occurs.
Step 7: Determine the change in the number of moles of gas (∆n) for the
reaction based on the stoichiometry of the reaction.
Step 8: Substitute the calculated values into the Van’t Hoff equation to find
the new equilibrium constant (K2) at 2 atm.
11
Step 9: Use the Van’t Hoff equation to adjust the equilibrium constant for
the new pressure to calculate ∆Gunder non-standard conditions.
Step 10: Use the equation ∆G= ∆H−T∆Sto determine if the reaction is
exothermic or endothermic under the given nonstandard conditions.
Question 15
Question
Calculate the enthalpy change for the reaction below using the following data:
2H2(g) + O2(g)→2H2O(l)
Given:
2H2(g) + O2(g) →2H2O(l) ∆H = -571.7 kJ
H2O(l) →H2O(g) ∆H = +44.0 kJ
H2(g) →H2O(g) ∆H = -483.6 kJ
Solution
Step 1: Given the following equations:
2H2(g) + O2(g)→2H2O(l) ∆H=−571.7kJ (1)
H2O(l)→H2O(g) ∆H= +44.0kJ (2)
H2(g)→H2O(g) ∆H=−483.6kJ (3)
Step 2: We can use these three equations to find the enthalpy change for the
desired reaction:
2H2(g) + O2(g)→2H2O(l)
Step 3: First, we need to change the phase of 2H2O(l) to 2H2O(g), then
combine the equations:
(1) + (2) ⇒2H2(g) + O2(g)→2H2O(g) ∆H=−571.7 + 44.0 = −527.7kJ (4)
Step 4: Next, we combine equation (3) with equation (4) to obtain the
desired reaction:
(4) + (3) ⇒2H2(g) + O2(g)→2H2O(g) + H2(g) ∆H=−527.7−483.6 = −1011.3kJ
Step 5: Therefore, the enthalpy change for the reaction 2H2(g) + O2(g)→
2H2O(l) is -1011.3 kJ.
12
Question 16
Question
Calculate the standard enthalpy change for the following reaction at 25
°
C:
2C(s) + 5H2(g) →C2H6(g)
Given the following standard enthalpy of formation values:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.7 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction.
2C(s) + 5H2(g) →C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xν∆H◦
f
where νis the stoichiometric coefficient of each substance in the balanced
chemical equation and ∆H◦
fis the standard enthalpy of formation.
Plugging in the values:
∆H◦= 2(0 kJ/mol) + 5(0 kJ/mol) −84.7 kJ/mol
∆H◦=−84.7 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.7 kJ/mol .
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2H2O2→2H2O+O2
Given the following standard enthalpies of formation: ∆H◦
f(H2O2) = −196.1
kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol, ∆H◦
f(O2) = 0 kJ/mol.
13
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2O2→2H2O+O2
Step 2: Determine the standard enthalpy change (∆H◦) for the reaction
using the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= 2 ×∆H◦
f(H2O) + ∆H◦
f(O2)−2×∆H◦
f(H2O2)
∆H◦= 2 ×(−285.8 kJ/mol) + 0 −2×(−196.1 kJ/mol)
∆H◦=−571.6 kJ/mol + 392.2 kJ/mol
∆H◦=−179.4 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction at
25◦C is -179.4 kJ/mol.
Question 18
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−484 kJ
Calculate the enthalpy change for the reaction:
2H2O(l)→2H2(g)+O2(g)
Solution
Step 1: The enthalpy change of the reverse reaction is the negative of the
enthalpy change of the forward reaction.
−∆H=−(−484 kJ)
−∆H= 484 kJ
Therefore, the enthalpy change for the reaction:
2H2O(l)→2H2(g)+O2(g)
is 484 kJ.
14
Question 19
Question
Calculate the enthalpy change for the reaction below using the given standard
enthalpies of formation:
2C(s) + 3H2(g)→C2H6(g)
Given:
∆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 balanced chemical equation for the reaction.
The balanced chemical equation for the reaction is:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the enthalpy change using the standard enthalpies of for-
mation.
The enthalpy change (∆Hrxn) for the reaction can be calculated using the equa-
tion:
∆Hrxn =X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given standard enthalpy of formation values into the equation:
∆Hrxn = (∆H◦
f[C2H6(g)]) −(∆H◦
f[C(s)] + ∆H◦
f[H2(g)])
∆Hrxn = (−84.68) −(0 + 0)
∆Hrxn =−84.68 kJ/mol
Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.
Question 20
Question
Calculate the enthalpy change (∆H◦) for the reaction below based on the given
bond dissociation energies:
2CH4(g) + 3O2(g)→2CO2(g) + 4H2O(l)
15
Given bond dissociation energies:
C-H = 413 kJ/mol
O=O = 498 kJ/mol
O-H = 463 kJ/mol
C=O = 743 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy to break bonds = 2 ×(C-H) + 3 ×(O=O) + 12 ×(O-H)
= 2 ×413 kJ/mol + 3 ×498 kJ/mol + 12 ×463 kJ/mol
= 826 kJ/mol + 1494 kJ/mol + 5556 kJ/mol
= 7876 kJ/mol
Step 2: Calculate the total energy released when new bonds are formed in
the products.
Energy released by forming new bonds = 2 ×(C=O) + 4 ×(O-H)
= 2 ×743 kJ/mol + 4 ×463 kJ/mol
= 1486 kJ/mol + 1852 kJ/mol
= 3338 kJ/mol
Step 3: Calculate the change in enthalpy (∆H◦) for the reaction.
∆H◦= Energy released by forming new bonds −Energy to break bonds
∆H◦= 3338 kJ/mol −7876 kJ/mol
∆H◦=−4538 kJ/mol
Therefore, the enthalpy change (∆H◦) for the reaction is -4538 kJ/mol,
indicating that the reaction is exothermic.
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C:
4NH3(g)+5O2(g)→4NO(g)+6H2O(l)
Given the following standard enthalpies of formation:
∆H◦
f(NH3) = −46.1 kJ/mol
∆H◦
f(NO) = 90.3 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
16
Solution
Step 1: Calculate the standard enthalpy change for the reaction using standard
enthalpies of formation. The standard enthalpy change of a reaction 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.
Given:
∆H◦
f(NH3) = −46.1 kJ/mol
∆H◦
f(NO) = 90.3 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
The standard enthalpy change for the reaction can be calculated as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [4∆H◦
f(NO) + 6∆H◦
f(H2O(l))] −[4∆H◦
f(NH3) + 5∆H◦
f(O2)]
∆H◦= [4(90.3) + 6(−285.8)] −[4(−46.1) + 5(0)]
∆H◦= [361.2−1714.8] −[−184.4]
∆H◦= 1053.6 + 184.4
∆H◦= 1238 kJ
Therefore, the standard enthalpy change for the reaction is ∆H◦= 1238 kJ.
Question 22
Question
A reaction has a standard enthalpy change of -335 kJ/mol. If 2.50 mol of the
reaction takes place, what is the total heat change for the reaction?
Solution
Step 1: Determine the total heat change for the reaction in kJ. -335 kJ/mol x
2.50 mol = -837.5 kJ
Answer: The total heat change for the reaction is -837.5 kJ.
17
Question 23
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(graphite) + 3H2(g) →C2H6(g)
given the following standard enthalpy of formation values:
∆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.
2C(graphite) + 3H2(g) →C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients in the balanced equation.
Substitute the given values:
∆H◦= 1(−84.68 kJ/mol) −[2(0 kJ/mol) + 3(0 kJ/mol)]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−84.68
kJ/mol.
Question 24
Question
A reaction has a standard enthalpy change of -350 kJ. If 2.50 moles of this
reaction release 175 kJ of heat, what is the actual enthalpy change of the reaction
in kJ?
18
Solution
Step 1: Calculate the moles of reaction that correspond to the standard enthalpy
change. Given: Standard enthalpy change (∆H◦) = -350 kJ Moles of reaction
for standard enthalpy change (n◦) = 1 mole
Step 2: Use stoichiometry to calculate the moles of reaction for the amount
of heat released. Given: Heat released = 175 kJ Moles of reaction for heat
released (n) = 2.50 moles
Step 3: Calculate the actual enthalpy change of the reaction. From Step 1:
∆H◦
n◦=−350 kJ
1 mol =−350 kJ/mol
From Step 2:
−350 kJ/mol
n=175 kJ
2.50 mol =−70 kJ/mol
Therefore, the actual enthalpy change of the reaction is -70 kJ.
Question 25
Question
Given the following reaction and enthalpy values, calculate the standard en-
thalpy change (∆H◦) for the reaction at 298 K:
2C(s)+ 2H2O(g)→2CO(g)+ 2H2(g)
∆H◦
f(CO(g)) = −110.5 kJ/mol
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
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)
= [2∆H◦
f(CO(g)) + 2∆H◦
f(H2(g))] −[2∆H◦
f(C(s)) + 2∆H◦
f(H2O(g))]
= [2(−110.5) + 2(0)] −[2(0) + 2(−241.8)]
= [−221.0 + 0] −[0 −483.6]
=−221.0 + 483.6
= 262.6 kJ
Therefore, the standard enthalpy change for the reaction at 298 K is 262.6 kJ .
19
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2SO2(g)+O2(g)→2SO3(g)
given the following standard enthalpies of formation:
∆H◦
f(SO2) = −296.8 kJ/mol
∆H◦
f(SO3) = −395.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard en-
thalpies of formation.
2SO2(g)+O2(g)→2SO3(g)
Given:
∆H◦
f(SO2) = −296.8 kJ/mol
∆H◦
f(SO3) = −395.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= 2(−395.7 kJ/mol) −[2(−296.8 kJ/mol) + 0 kJ/mol]
∆H◦=−791.4 kJ/mol −(−593.6 kJ/mol)
∆H◦=−791.4 kJ/mol + 593.6 kJ/mol
∆H◦=−197.8 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −197.8
kJ/mol.
20
Question 27
Question
Calculate the standard enthalpy change (∆H◦) for the reaction
2A(g) + B(g) →3C(g) + D(g)
given the following data:
∆H◦
f(A) = −250 kJ/mol
∆H◦
f(B) = −150 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −50 kJ/mol
Solution
Step 1: Calculate ∆H◦for the reaction using the standard enthalpy of formation
data provided:
We can use the equation
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the coefficients in the balanced chemical equation.
Plugging in the given values, we have:
∆H◦= [3(−400) + (−50)] −[2(−250) + (−150)]
Step 2: Simplify the equation to find the ∆H◦.
Calculating the values, we get:
∆H◦= [−1200 −50] −[−500 −150]
∆H◦=−1250 −(−650)
∆H◦=−1250 + 650
∆H◦=−600 kJ/mol
Therefore, the standard enthalpy change for the reaction is −600 kJ/mol .
21
Question 28
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2C(s) + 3H2(g)→C2H6(g)
Given the standard enthalpy of formation values:
∆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 balanced chemical equation for the reaction.
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change of the reaction using the
standard enthalpies of formation.
∆H◦=X(∆H◦
f(products)) −X(∆H◦
f(reactants))
Step 3: Substitute the given standard enthalpy of formation values into the
equation.
∆H◦=∆H◦
f(C2H6(g))−2·∆H◦
f(C(s)) + 3 ·∆H◦
f(H2(g))
Step 4: Plug in the values and calculate the standard enthalpy change.
∆H◦= [−84.68 kJ/mol] −[2 ·0 kJ/mol + 3 ·0 kJ/mol]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−84.68 kJ/mol.
Question 29
Question
Given the following reaction and enthalpy changes:
2C(s) + 3H2(g)→C2H6(g) ∆H=−84.7 kJ
4C(s) + 6H2(g)→2C2H6(g) ∆H=−118.4 kJ
Calculate the enthalpy change for the following reaction:
C2H6(g)→2C(s) + 3H2(g)
22
Solution
Step 1: Given the enthalpy changes for the two reactions, we need to manipulate
them to find the enthalpy change for the target reaction. First, reverse the first
reaction and multiply it by 2 to match the coefficients of the target reaction:
2(C2H6(g)→2C(s) + 3H2(g))
= 2(2C(s) + 3H2(g)→C2H6(g))
= 4C(s) + 6H2(g)→2C2H6(g) ∆H= 2(−84.7) = −169.4 kJ
Step 2: Next, add the second reaction as is to the previous equation to
eliminate the intermediates:
4C(s)+6H2(g)→2C2H6(g)+4C(s)+6H2(g) ∆H=−169.4 kJ−118.4 kJ = −287.8 kJ
Step 3: Finally, cancel out the common compounds on both sides to get the
enthalpy change for the target reaction:
C2H6(g)→2C(s) + 3H2(g) ∆H=−287.8 kJ
Question 30
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
given the following standard enthalpy of formations:
C2H2(g) = 226 kJ/mol
CO2(g) = −393.5 kJ/mol
H2O(g) = −241.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the reaction by using the
standard enthalpies of formation:
The standard enthalpy change can be calculated using the formula:
∆H◦=X(products) −X(reactants)
Given:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
23
Question 2
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(l)
Given the following standard enthalpies of formation (∆H◦
f):
∆H◦
f(C2H2(g)) = 226.7 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
= (4 ×∆H◦
f(CO2(g)) + 2 ×∆H◦
f(H2O(l))) −(2 ×∆H◦
f(C2H2(g))) −(5 ×0)
= (4 × −393.5 kJ/mol + 2 × −285.8 kJ/mol) −(2 ×226.7 kJ/mol)
= (−1574 kJ/mol −571.6 kJ/mol) −(453.4 kJ/mol)
=−2145 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is -2145
kJ/mol.
Question 3
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction using
the given bond dissociation energies:
CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g)
Given bond dissociation energies: D(CH) = 413 kJ/mol D(Cl−Cl) = 242 kJ/mol
D(C−Cl) = 339 kJ/mol D(H−Cl) = 431 kJ/mol
2
Solution
Step 1: Calculate the total bond dissociation energy of the reactants.
Dreactants = 1 ×D(CH)+4×D(Cl −Cl)+1×D(C−Cl)+4×D(H−Cl)
= 1 ×413 kJ/mol + 4 ×242 kJ/mol + 1 ×339 kJ/mol + 4 ×431 kJ/mol
= 413 + 968 + 339 + 1724
= 3444 kJ/mol
Step 2: Calculate the total bond dissociation energy of the products.
Dproducts = 1 ×D(C−Cl)+4×D(H−Cl)
= 1 ×339 kJ/mol + 4 ×431 kJ/mol
= 339 + 1724
= 2063 kJ/mol
Step 3: Calculate the change in bond dissociation energy (∆D).
∆D=Dproducts −Dreactants = 2063 kJ/mol −3444 kJ/mol =−1381 kJ/mol
Step 4: Calculate the standard enthalpy change using the equation:
∆H◦= ∆D+ ∆H◦
rxn
Since the reaction is exothermic, ∆H◦
rxn =−1381 kJ/mol Therefore, the stan-
dard enthalpy change for the reaction is ∆H◦=−1381 kJ/mol.
Question 4
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l)
and the standard enthalpies of formation (∆H◦
f) for H2(g), O2(g), and H2O(l)
as 0 kJ/mol, 0 kJ/mol, and -286 kJ/mol, respectively, calculate the standard
enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change using the formula:
∆H◦=Xn∆H◦
f
where nis the stoichiometric coefficient and ∆H◦
fis the standard enthalpy of
formation.
3
Step 2: Substitute the values into the formula and calculate the standard
enthalpy change:
∆H◦= 2 ×∆H◦
f(H2O) −2×∆H◦
f(H2)−∆H◦
f(O2)
∆H◦= 2 ×(−286 kJ/mol) −2×(0 kJ/mol) −0 kJ/mol
∆H◦=−572 kJ/mol
Therefore, the standard enthalpy change for the reaction is -572 kJ/mol.
Question 5
Question
Consider the reaction:
2C(s) + 3H2(g)→C2H6(g)
Given the following information:
Standard enthalpy of formation of C(s): −394 kJ/mol
Standard enthalpy of formation of H2(g) :-286 kJ/molStandardenthalpyofformationofC2H6(g) :-
84 kJ/mol Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Write the balanced equation for the reaction:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values. The standard enthalpy change for the reaction can be
calculated using the following formula:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
where ∆H◦is the standard enthalpy change for the reaction, nand mare
the stoichiometric coefficients of the products and reactants, respectively, and
∆H◦
products and ∆H◦
reactants are the standard enthalpies of formation of the prod-
ucts and reactants, respectively.
Step 3: Substitute the given values into the formula:
∆H◦= (1)(−84) −(2)(−394) −(3)(−286) kJ/mol
∆H◦=−84 + 788 + 858 kJ/mol
∆H◦= 1562 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is 1562 kJ/mol.
4
Question 6
Question
Consider the following reaction:
2H2(g) + O2(g) →2H2O(l)
Given that ∆H◦
ffor H2(g) = 0 kJ/mol, ∆H◦
ffor O2(g) = 0 kJ/mol, and ∆H◦
f
for H2O(l) = −286 kJ/mol, calculate the standard enthalpy change, ∆H◦, for
the above reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2(g) + O2(g) →2H2O(l)
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction. The
standard enthalpy change for a reaction can be calculated using the standard
enthalpies of formation of the reactants and products. The standard enthalpy
change (∆H◦) for a reaction is given by the difference in the sum of the standard
enthalpies of formation of the products and the sum of the standard enthalpies
of formation of the reactants.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Given:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(l)) = −286 kJ/mol
Substitute the values into the equation:
∆H◦= 2(∆H◦
f(H2O(l))) −[2(∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
∆H◦= 2(−286 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦=−572 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−572 kJ/mol.
Question 7
Question
For the reaction:
2H2(g) + O2(g)→2H2O(l)
the standard enthalpy change (∆H◦) is -571.6 kJ/mol. Calculate the standard
enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
5
Solution
Step 1: Write the given balanced chemical equation and the desired equation.
Given balanced chemical equation:
2H2(g) + O2(g)→2H2O(l)
Desired equation:
H2O(l)→H2(g) + 1
2O2(g)
Step 2: Determine the standard enthalpy change for the desired equation.
We can manipulate the given equation to obtain the desired equation as follows:
2H2O(l)→2H2(g) + O2(g)
Step 3: Reverse the given equation.
−2H2O(l)→ −2H2(g)−O2(g)
H2O(l)→H2(g) + 1
2O2(g)
Step 4: Determine the standard enthalpy change for the desired equation.
Since the given enthalpy change is for the forward reaction, we must change the
sign when reversing the equation:
−∆H◦=−(−571.6 kJ/mol) = 571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
is 571.6 kJ/mol.
Question 8
Question
Calculate the enthalpy change (∆H) for the reaction below using the given bond
dissociation energies:
2H2O(g)→2H2(g)+O2(g)
Given bond dissociation energies:
H-H: 436 kJ/mol
O=O: 495 kJ/mol
O-H: 463 kJ/mol
6
Solution
Step 1: Write out the balanced chemical equation for the reaction.
2H2O(g)→2H2(g)+O2(g)
Step 2: Calculate the total bond energy of the reactants. Total bond energy
of reactants = 2(2 mol ×O-H bond energy) + H-H bond energy
Total bond energy of reactants = 2 ×2×463 kJ/mol + 436 kJ/mol
Total bond energy of reactants = 1852 kJ/mol + 436 kJ/mol
Total bond energy of reactants = 2288 kJ
Step 3: Calculate the total bond energy of the products. Total bond energy
of products = 2(2 mol ×H-H bond energy) + O=O bond energy
Total bond energy of products = 2 ×2×436 kJ/mol + 495 kJ/mol
Total bond energy of products = 1744 kJ/mol + 495 kJ/mol
Total bond energy of products = 2239 kJ
Step 4: Calculate the change in bond energy. ∆H= Total bond energy of products−
Total bond energy of reactants
∆H= 2239 kJ - 2288 kJ
∆H=−49 kJ
Therefore, the enthalpy change (∆H) for the reaction is -49 kJ.
Question 9
Question
Calculate the enthalpy change (∆H) for the following reaction:
2C(s) + 3H2(g)→C2H6(g)
given the following bond dissociation energies:
C-C bond: 348 kJ/mol
C-H bond: 412 kJ/mol
H-H bond: 436 kJ/mol
C=C bond: 614 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy to break C-C bonds = 2 ×348 kJ/mol = 696 kJ/mol
Energy to break H-H bonds = 3 ×436 kJ/mol = 1308 kJ/mol
Energy to break C-H bonds = 2 ×412 kJ/mol = 824 kJ/mol
Total energy to break bonds = 696 + 1308 + 824 = 2828 kJ/mol
7
Step 2: Calculate the total energy released in the formation of bonds in the
products.
Energy released in forming C-C bonds = 1 ×614 kJ/mol = 614 kJ/mol
Energy released in forming C-H bonds = 6 ×412 kJ/mol = 2472 kJ/mol
Total energy released in forming bonds = 614 + 2472 = 3086 kJ/mol
Step 3: Calculate the change in enthalpy (∆H) for the reaction using the
bond dissociation energies.
∆H= Energy to break bonds−Energy released in forming bonds = 2828 kJ/mol−3086 kJ/mol = −258 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H=−258 kJ/mol.
Question 10
Question
Calculate the enthalpy change (∆H) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
given the following bond dissociation energies:
H−H: 432 kJ/mol
O=O: 497 kJ/mol
O−H: 464 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy in bonds broken = 2 ×Energy of H−Hbond + Energy of O=Obond
= 2 ×432 kJ/mol + 1 ×497 kJ/mol
= 1361 kJ
Step 2: Calculate the total energy released when the new bonds are formed
in the products.
Energy in bonds formed = 2 ×Energy of O−Hbond
= 2 ×464 kJ/mol
= 928 kJ
Step 3: Calculate the change in enthalpy (∆H) for the reaction.
∆H= Energy in bonds broken −Energy in bonds formed
= 1361 kJ −928 kJ
= 433 kJ
Therefore, the enthalpy change (∆H) for the reaction is 433 kJ.
8
Question 11
Question
Given the following reaction:
2H2(g) + O2(g)→2H2O(l)
The standard enthalpy change for the reaction is -483.6 kJ mol−1. Calculate
the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
at the same temperature.
Solution
Step 1: Write the given reaction and its standard enthalpy change. Given
reaction:
2H2(g) + O2(g)→2H2O(l)
∆H=−483.6 kJ mol−1
Step 2: Write the target reaction. Target reaction:
H2O(l)→H2(g) + 1
2O2(g)
Step 3: Determine the standard enthalpy change for the target reaction.
Since the target reaction is the reverse of the given reaction, the sign of the
enthalpy change will be reversed. Therefore, ∆H= 483.6 kJ mol−1
So, the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
is 483.6 kJ mol−1at the same temperature.
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of ethylene
(C2H4) using the following information:
The standard enthalpy of formation of CO2is -393.5 kJ/mol.
The standard enthalpy of formation of H2O is -285.8 kJ/mol.
The standard enthalpy of formation of C2H4is 52.3 kJ/mol.
The balanced equation for the combustion of ethylene is:
C2H4(g) + 3O2(g)→2CO2(g) + 2H2O(l)
9
Solution
Step 1: Calculate the standard enthalpy change for the combustion of ethy-
lene using the standard enthalpy of formation values provided. The standard
enthalpy change can be calculated using the equation:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Where nand mare the stoichiometric coefficients from the balanced equa-
tion.
Step 2: Substitute the given values into the equation and calculate the stan-
dard enthalpy change.
∆H◦= [2(−393.5) + 2(−285.8)] −[1(52.3) + 3(0)]
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= (−787.0) −(52.3) = −834.3 kJ/mol
Therefore, the standard enthalpy change for the combustion of ethylene is
∆H◦=−834.3 kJ/mol.
Question 13
Question
Calculate the enthalpy change for the combustion of ethylene, C2H4(g), to form
carbon dioxide and water using the following data:
Reaction ∆H◦(kJ/mol)
C2H4(g) + 3O2(g) →2CO2(g) + 2H2O(l) -1411
H2(g) + 1
2O2(g) →H2O(l) -286
C(s) + O2(g) →CO2(g) -394
Solution
Step 1: Writing the balanced equation for the combustion of ethylene:
C2H4(g) + 3O2(g) →2CO2(g) + 2H2O(l)
Step 2: Calculating the enthalpy change for the given reaction:
∆H=X∆Hproducts −X∆Hreactants
∆H= [2(2∆HCO2) + 2∆HH2O(l)]−[∆HC2H4+ 3∆HO2]
∆H= [2(2 × −394) + 2 × −286] −[−1411]
∆H= [2(−788) −572] + 1411
∆H=−1576 −572 + 1411
∆H=−737 kJ/mol
Therefore, the enthalpy change for the combustion of ethylene to form carbon
dioxide and water is -737 kJ/mol.
10
Question 14
Question
A reaction has a standard enthalpy change of -418 kJ/mol. If the reaction is
exothermic, determine if the reaction is exothermic or endothermic when carried
out at 25
°
C if the reaction is carried out under nonstandard conditions with a
pressure of 2 atm. Assume the reaction has an ideal gas as a product.
Solution
Step 1: Write the expression for the standard Gibbs free energy change in terms
of the standard enthalpy change and the standard entropy change:
∆G◦= ∆H◦−T∆S◦
Step 2: Use the relationship between standard Gibbs free energy change and
equilibrium constant to find the standard entropy change (∆S◦).
We know that ∆G◦=−RT ln(K) where Kis the equilibrium constant under
standard conditions. Therefore,
∆G◦=−RT ln(K◦)
where K◦is the equilibrium constant under standard conditions.
Step 3: Given that the reaction is exothermic, we know that ∆H◦=−418
kJ/mol.
Step 4: Calculate the standard entropy change using the standard Gibbs
free energy change equation:
∆S◦=∆H◦−∆G◦
T
Step 5: Substitute the known values into the equation and solve for ∆S◦:
∆S◦=−418 −(−RT ln(K◦))
T
Step 6: Given that the pressure is 2 atm, we need to consider the effect of
pressure on the equilibrium constant in the expression for ∆G. The effect of
changing pressure on the equilibrium constant can be accounted for using the
Van’t Hoff equation:
ln K2
K1=∆nRT
R
where ∆nis the change in the number of moles of gas as the reaction occurs.
Step 7: Determine the change in the number of moles of gas (∆n) for the
reaction based on the stoichiometry of the reaction.
Step 8: Substitute the calculated values into the Van’t Hoff equation to find
the new equilibrium constant (K2) at 2 atm.
11
Step 9: Use the Van’t Hoff equation to adjust the equilibrium constant for
the new pressure to calculate ∆Gunder non-standard conditions.
Step 10: Use the equation ∆G= ∆H−T∆Sto determine if the reaction is
exothermic or endothermic under the given nonstandard conditions.
Question 15
Question
Calculate the enthalpy change for the reaction below using the following data:
2H2(g) + O2(g)→2H2O(l)
Given:
2H2(g) + O2(g) →2H2O(l) ∆H = -571.7 kJ
H2O(l) →H2O(g) ∆H = +44.0 kJ
H2(g) →H2O(g) ∆H = -483.6 kJ
Solution
Step 1: Given the following equations:
2H2(g) + O2(g)→2H2O(l) ∆H=−571.7kJ (1)
H2O(l)→H2O(g) ∆H= +44.0kJ (2)
H2(g)→H2O(g) ∆H=−483.6kJ (3)
Step 2: We can use these three equations to find the enthalpy change for the
desired reaction:
2H2(g) + O2(g)→2H2O(l)
Step 3: First, we need to change the phase of 2H2O(l) to 2H2O(g), then
combine the equations:
(1) + (2) ⇒2H2(g) + O2(g)→2H2O(g) ∆H=−571.7 + 44.0 = −527.7kJ (4)
Step 4: Next, we combine equation (3) with equation (4) to obtain the
desired reaction:
(4) + (3) ⇒2H2(g) + O2(g)→2H2O(g) + H2(g) ∆H=−527.7−483.6 = −1011.3kJ
Step 5: Therefore, the enthalpy change for the reaction 2H2(g) + O2(g)→
2H2O(l) is -1011.3 kJ.
12
Question 16
Question
Calculate the standard enthalpy change for the following reaction at 25
°
C:
2C(s) + 5H2(g) →C2H6(g)
Given the following standard enthalpy of formation values:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.7 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction.
2C(s) + 5H2(g) →C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xν∆H◦
f
where νis the stoichiometric coefficient of each substance in the balanced
chemical equation and ∆H◦
fis the standard enthalpy of formation.
Plugging in the values:
∆H◦= 2(0 kJ/mol) + 5(0 kJ/mol) −84.7 kJ/mol
∆H◦=−84.7 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.7 kJ/mol .
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2H2O2→2H2O+O2
Given the following standard enthalpies of formation: ∆H◦
f(H2O2) = −196.1
kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol, ∆H◦
f(O2) = 0 kJ/mol.
13
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2O2→2H2O+O2
Step 2: Determine the standard enthalpy change (∆H◦) for the reaction
using the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= 2 ×∆H◦
f(H2O) + ∆H◦
f(O2)−2×∆H◦
f(H2O2)
∆H◦= 2 ×(−285.8 kJ/mol) + 0 −2×(−196.1 kJ/mol)
∆H◦=−571.6 kJ/mol + 392.2 kJ/mol
∆H◦=−179.4 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction at
25◦C is -179.4 kJ/mol.
Question 18
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−484 kJ
Calculate the enthalpy change for the reaction:
2H2O(l)→2H2(g)+O2(g)
Solution
Step 1: The enthalpy change of the reverse reaction is the negative of the
enthalpy change of the forward reaction.
−∆H=−(−484 kJ)
−∆H= 484 kJ
Therefore, the enthalpy change for the reaction:
2H2O(l)→2H2(g)+O2(g)
is 484 kJ.
14
Question 19
Question
Calculate the enthalpy change for the reaction below using the given standard
enthalpies of formation:
2C(s) + 3H2(g)→C2H6(g)
Given:
∆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 balanced chemical equation for the reaction.
The balanced chemical equation for the reaction is:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the enthalpy change using the standard enthalpies of for-
mation.
The enthalpy change (∆Hrxn) for the reaction can be calculated using the equa-
tion:
∆Hrxn =X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given standard enthalpy of formation values into the equation:
∆Hrxn = (∆H◦
f[C2H6(g)]) −(∆H◦
f[C(s)] + ∆H◦
f[H2(g)])
∆Hrxn = (−84.68) −(0 + 0)
∆Hrxn =−84.68 kJ/mol
Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.
Question 20
Question
Calculate the enthalpy change (∆H◦) for the reaction below based on the given
bond dissociation energies:
2CH4(g) + 3O2(g)→2CO2(g) + 4H2O(l)
15
Given bond dissociation energies:
C-H = 413 kJ/mol
O=O = 498 kJ/mol
O-H = 463 kJ/mol
C=O = 743 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy to break bonds = 2 ×(C-H) + 3 ×(O=O) + 12 ×(O-H)
= 2 ×413 kJ/mol + 3 ×498 kJ/mol + 12 ×463 kJ/mol
= 826 kJ/mol + 1494 kJ/mol + 5556 kJ/mol
= 7876 kJ/mol
Step 2: Calculate the total energy released when new bonds are formed in
the products.
Energy released by forming new bonds = 2 ×(C=O) + 4 ×(O-H)
= 2 ×743 kJ/mol + 4 ×463 kJ/mol
= 1486 kJ/mol + 1852 kJ/mol
= 3338 kJ/mol
Step 3: Calculate the change in enthalpy (∆H◦) for the reaction.
∆H◦= Energy released by forming new bonds −Energy to break bonds
∆H◦= 3338 kJ/mol −7876 kJ/mol
∆H◦=−4538 kJ/mol
Therefore, the enthalpy change (∆H◦) for the reaction is -4538 kJ/mol,
indicating that the reaction is exothermic.
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C:
4NH3(g)+5O2(g)→4NO(g)+6H2O(l)
Given the following standard enthalpies of formation:
∆H◦
f(NH3) = −46.1 kJ/mol
∆H◦
f(NO) = 90.3 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
16
Solution
Step 1: Calculate the standard enthalpy change for the reaction using standard
enthalpies of formation. The standard enthalpy change of a reaction 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.
Given:
∆H◦
f(NH3) = −46.1 kJ/mol
∆H◦
f(NO) = 90.3 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
The standard enthalpy change for the reaction can be calculated as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [4∆H◦
f(NO) + 6∆H◦
f(H2O(l))] −[4∆H◦
f(NH3) + 5∆H◦
f(O2)]
∆H◦= [4(90.3) + 6(−285.8)] −[4(−46.1) + 5(0)]
∆H◦= [361.2−1714.8] −[−184.4]
∆H◦= 1053.6 + 184.4
∆H◦= 1238 kJ
Therefore, the standard enthalpy change for the reaction is ∆H◦= 1238 kJ.
Question 22
Question
A reaction has a standard enthalpy change of -335 kJ/mol. If 2.50 mol of the
reaction takes place, what is the total heat change for the reaction?
Solution
Step 1: Determine the total heat change for the reaction in kJ. -335 kJ/mol x
2.50 mol = -837.5 kJ
Answer: The total heat change for the reaction is -837.5 kJ.
17
Question 23
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(graphite) + 3H2(g) →C2H6(g)
given the following standard enthalpy of formation values:
∆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.
2C(graphite) + 3H2(g) →C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients in the balanced equation.
Substitute the given values:
∆H◦= 1(−84.68 kJ/mol) −[2(0 kJ/mol) + 3(0 kJ/mol)]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−84.68
kJ/mol.
Question 24
Question
A reaction has a standard enthalpy change of -350 kJ. If 2.50 moles of this
reaction release 175 kJ of heat, what is the actual enthalpy change of the reaction
in kJ?
18
Solution
Step 1: Calculate the moles of reaction that correspond to the standard enthalpy
change. Given: Standard enthalpy change (∆H◦) = -350 kJ Moles of reaction
for standard enthalpy change (n◦) = 1 mole
Step 2: Use stoichiometry to calculate the moles of reaction for the amount
of heat released. Given: Heat released = 175 kJ Moles of reaction for heat
released (n) = 2.50 moles
Step 3: Calculate the actual enthalpy change of the reaction. From Step 1:
∆H◦
n◦=−350 kJ
1 mol =−350 kJ/mol
From Step 2:
−350 kJ/mol
n=175 kJ
2.50 mol =−70 kJ/mol
Therefore, the actual enthalpy change of the reaction is -70 kJ.
Question 25
Question
Given the following reaction and enthalpy values, calculate the standard en-
thalpy change (∆H◦) for the reaction at 298 K:
2C(s)+ 2H2O(g)→2CO(g)+ 2H2(g)
∆H◦
f(CO(g)) = −110.5 kJ/mol
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
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)
= [2∆H◦
f(CO(g)) + 2∆H◦
f(H2(g))] −[2∆H◦
f(C(s)) + 2∆H◦
f(H2O(g))]
= [2(−110.5) + 2(0)] −[2(0) + 2(−241.8)]
= [−221.0 + 0] −[0 −483.6]
=−221.0 + 483.6
= 262.6 kJ
Therefore, the standard enthalpy change for the reaction at 298 K is 262.6 kJ .
19
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2SO2(g)+O2(g)→2SO3(g)
given the following standard enthalpies of formation:
∆H◦
f(SO2) = −296.8 kJ/mol
∆H◦
f(SO3) = −395.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard en-
thalpies of formation.
2SO2(g)+O2(g)→2SO3(g)
Given:
∆H◦
f(SO2) = −296.8 kJ/mol
∆H◦
f(SO3) = −395.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= 2(−395.7 kJ/mol) −[2(−296.8 kJ/mol) + 0 kJ/mol]
∆H◦=−791.4 kJ/mol −(−593.6 kJ/mol)
∆H◦=−791.4 kJ/mol + 593.6 kJ/mol
∆H◦=−197.8 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −197.8
kJ/mol.
20
Question 27
Question
Calculate the standard enthalpy change (∆H◦) for the reaction
2A(g) + B(g) →3C(g) + D(g)
given the following data:
∆H◦
f(A) = −250 kJ/mol
∆H◦
f(B) = −150 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −50 kJ/mol
Solution
Step 1: Calculate ∆H◦for the reaction using the standard enthalpy of formation
data provided:
We can use the equation
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the coefficients in the balanced chemical equation.
Plugging in the given values, we have:
∆H◦= [3(−400) + (−50)] −[2(−250) + (−150)]
Step 2: Simplify the equation to find the ∆H◦.
Calculating the values, we get:
∆H◦= [−1200 −50] −[−500 −150]
∆H◦=−1250 −(−650)
∆H◦=−1250 + 650
∆H◦=−600 kJ/mol
Therefore, the standard enthalpy change for the reaction is −600 kJ/mol .
21
Question 28
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2C(s) + 3H2(g)→C2H6(g)
Given the standard enthalpy of formation values:
∆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 balanced chemical equation for the reaction.
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change of the reaction using the
standard enthalpies of formation.
∆H◦=X(∆H◦
f(products)) −X(∆H◦
f(reactants))
Step 3: Substitute the given standard enthalpy of formation values into the
equation.
∆H◦=∆H◦
f(C2H6(g))−2·∆H◦
f(C(s)) + 3 ·∆H◦
f(H2(g))
Step 4: Plug in the values and calculate the standard enthalpy change.
∆H◦= [−84.68 kJ/mol] −[2 ·0 kJ/mol + 3 ·0 kJ/mol]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−84.68 kJ/mol.
Question 29
Question
Given the following reaction and enthalpy changes:
2C(s) + 3H2(g)→C2H6(g) ∆H=−84.7 kJ
4C(s) + 6H2(g)→2C2H6(g) ∆H=−118.4 kJ
Calculate the enthalpy change for the following reaction:
C2H6(g)→2C(s) + 3H2(g)
22
Solution
Step 1: Given the enthalpy changes for the two reactions, we need to manipulate
them to find the enthalpy change for the target reaction. First, reverse the first
reaction and multiply it by 2 to match the coefficients of the target reaction:
2(C2H6(g)→2C(s) + 3H2(g))
= 2(2C(s) + 3H2(g)→C2H6(g))
= 4C(s) + 6H2(g)→2C2H6(g) ∆H= 2(−84.7) = −169.4 kJ
Step 2: Next, add the second reaction as is to the previous equation to
eliminate the intermediates:
4C(s)+6H2(g)→2C2H6(g)+4C(s)+6H2(g) ∆H=−169.4 kJ−118.4 kJ = −287.8 kJ
Step 3: Finally, cancel out the common compounds on both sides to get the
enthalpy change for the target reaction:
C2H6(g)→2C(s) + 3H2(g) ∆H=−287.8 kJ
Question 30
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
given the following standard enthalpy of formations:
C2H2(g) = 226 kJ/mol
CO2(g) = −393.5 kJ/mol
H2O(g) = −241.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the reaction by using the
standard enthalpies of formation:
The standard enthalpy change can be calculated using the formula:
∆H◦=X(products) −X(reactants)
Given:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
23
Question 2
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(l)
Given the following standard enthalpies of formation (∆H◦
f):
∆H◦
f(C2H2(g)) = 226.7 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
= (4 ×∆H◦
f(CO2(g)) + 2 ×∆H◦
f(H2O(l))) −(2 ×∆H◦
f(C2H2(g))) −(5 ×0)
= (4 × −393.5 kJ/mol + 2 × −285.8 kJ/mol) −(2 ×226.7 kJ/mol)
= (−1574 kJ/mol −571.6 kJ/mol) −(453.4 kJ/mol)
=−2145 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is -2145
kJ/mol.
Question 3
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction using
the given bond dissociation energies:
CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g)
Given bond dissociation energies: D(CH) = 413 kJ/mol D(Cl−Cl) = 242 kJ/mol
D(C−Cl) = 339 kJ/mol D(H−Cl) = 431 kJ/mol
2
Solution
Step 1: Calculate the total bond dissociation energy of the reactants.
Dreactants = 1 ×D(CH)+4×D(Cl −Cl)+1×D(C−Cl)+4×D(H−Cl)
= 1 ×413 kJ/mol + 4 ×242 kJ/mol + 1 ×339 kJ/mol + 4 ×431 kJ/mol
= 413 + 968 + 339 + 1724
= 3444 kJ/mol
Step 2: Calculate the total bond dissociation energy of the products.
Dproducts = 1 ×D(C−Cl)+4×D(H−Cl)
= 1 ×339 kJ/mol + 4 ×431 kJ/mol
= 339 + 1724
= 2063 kJ/mol
Step 3: Calculate the change in bond dissociation energy (∆D).
∆D=Dproducts −Dreactants = 2063 kJ/mol −3444 kJ/mol =−1381 kJ/mol
Step 4: Calculate the standard enthalpy change using the equation:
∆H◦= ∆D+ ∆H◦
rxn
Since the reaction is exothermic, ∆H◦
rxn =−1381 kJ/mol Therefore, the stan-
dard enthalpy change for the reaction is ∆H◦=−1381 kJ/mol.
Question 4
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l)
and the standard enthalpies of formation (∆H◦
f) for H2(g), O2(g), and H2O(l)
as 0 kJ/mol, 0 kJ/mol, and -286 kJ/mol, respectively, calculate the standard
enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change using the formula:
∆H◦=Xn∆H◦
f
where nis the stoichiometric coefficient and ∆H◦
fis the standard enthalpy of
formation.
3
Step 2: Substitute the values into the formula and calculate the standard
enthalpy change:
∆H◦= 2 ×∆H◦
f(H2O) −2×∆H◦
f(H2)−∆H◦
f(O2)
∆H◦= 2 ×(−286 kJ/mol) −2×(0 kJ/mol) −0 kJ/mol
∆H◦=−572 kJ/mol
Therefore, the standard enthalpy change for the reaction is -572 kJ/mol.
Question 5
Question
Consider the reaction:
2C(s) + 3H2(g)→C2H6(g)
Given the following information:
Standard enthalpy of formation of C(s): −394 kJ/mol
Standard enthalpy of formation of H2(g) :-286 kJ/molStandardenthalpyofformationofC2H6(g) :-
84 kJ/mol Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Write the balanced equation for the reaction:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values. The standard enthalpy change for the reaction can be
calculated using the following formula:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
where ∆H◦is the standard enthalpy change for the reaction, nand mare
the stoichiometric coefficients of the products and reactants, respectively, and
∆H◦
products and ∆H◦
reactants are the standard enthalpies of formation of the prod-
ucts and reactants, respectively.
Step 3: Substitute the given values into the formula:
∆H◦= (1)(−84) −(2)(−394) −(3)(−286) kJ/mol
∆H◦=−84 + 788 + 858 kJ/mol
∆H◦= 1562 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is 1562 kJ/mol.
4
Question 6
Question
Consider the following reaction:
2H2(g) + O2(g) →2H2O(l)
Given that ∆H◦
ffor H2(g) = 0 kJ/mol, ∆H◦
ffor O2(g) = 0 kJ/mol, and ∆H◦
f
for H2O(l) = −286 kJ/mol, calculate the standard enthalpy change, ∆H◦, for
the above reaction.
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2(g) + O2(g) →2H2O(l)
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction. The
standard enthalpy change for a reaction can be calculated using the standard
enthalpies of formation of the reactants and products. The standard enthalpy
change (∆H◦) for a reaction is given by the difference in the sum of the standard
enthalpies of formation of the products and the sum of the standard enthalpies
of formation of the reactants.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Given:
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2O(l)) = −286 kJ/mol
Substitute the values into the equation:
∆H◦= 2(∆H◦
f(H2O(l))) −[2(∆H◦
f(H2(g)) + ∆H◦
f(O2(g))]
∆H◦= 2(−286 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦=−572 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−572 kJ/mol.
Question 7
Question
For the reaction:
2H2(g) + O2(g)→2H2O(l)
the standard enthalpy change (∆H◦) is -571.6 kJ/mol. Calculate the standard
enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
5
Solution
Step 1: Write the given balanced chemical equation and the desired equation.
Given balanced chemical equation:
2H2(g) + O2(g)→2H2O(l)
Desired equation:
H2O(l)→H2(g) + 1
2O2(g)
Step 2: Determine the standard enthalpy change for the desired equation.
We can manipulate the given equation to obtain the desired equation as follows:
2H2O(l)→2H2(g) + O2(g)
Step 3: Reverse the given equation.
−2H2O(l)→ −2H2(g)−O2(g)
H2O(l)→H2(g) + 1
2O2(g)
Step 4: Determine the standard enthalpy change for the desired equation.
Since the given enthalpy change is for the forward reaction, we must change the
sign when reversing the equation:
−∆H◦=−(−571.6 kJ/mol) = 571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
is 571.6 kJ/mol.
Question 8
Question
Calculate the enthalpy change (∆H) for the reaction below using the given bond
dissociation energies:
2H2O(g)→2H2(g)+O2(g)
Given bond dissociation energies:
H-H: 436 kJ/mol
O=O: 495 kJ/mol
O-H: 463 kJ/mol
6
Solution
Step 1: Write out the balanced chemical equation for the reaction.
2H2O(g)→2H2(g)+O2(g)
Step 2: Calculate the total bond energy of the reactants. Total bond energy
of reactants = 2(2 mol ×O-H bond energy) + H-H bond energy
Total bond energy of reactants = 2 ×2×463 kJ/mol + 436 kJ/mol
Total bond energy of reactants = 1852 kJ/mol + 436 kJ/mol
Total bond energy of reactants = 2288 kJ
Step 3: Calculate the total bond energy of the products. Total bond energy
of products = 2(2 mol ×H-H bond energy) + O=O bond energy
Total bond energy of products = 2 ×2×436 kJ/mol + 495 kJ/mol
Total bond energy of products = 1744 kJ/mol + 495 kJ/mol
Total bond energy of products = 2239 kJ
Step 4: Calculate the change in bond energy. ∆H= Total bond energy of products−
Total bond energy of reactants
∆H= 2239 kJ - 2288 kJ
∆H=−49 kJ
Therefore, the enthalpy change (∆H) for the reaction is -49 kJ.
Question 9
Question
Calculate the enthalpy change (∆H) for the following reaction:
2C(s) + 3H2(g)→C2H6(g)
given the following bond dissociation energies:
C-C bond: 348 kJ/mol
C-H bond: 412 kJ/mol
H-H bond: 436 kJ/mol
C=C bond: 614 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy to break C-C bonds = 2 ×348 kJ/mol = 696 kJ/mol
Energy to break H-H bonds = 3 ×436 kJ/mol = 1308 kJ/mol
Energy to break C-H bonds = 2 ×412 kJ/mol = 824 kJ/mol
Total energy to break bonds = 696 + 1308 + 824 = 2828 kJ/mol
7
Step 2: Calculate the total energy released in the formation of bonds in the
products.
Energy released in forming C-C bonds = 1 ×614 kJ/mol = 614 kJ/mol
Energy released in forming C-H bonds = 6 ×412 kJ/mol = 2472 kJ/mol
Total energy released in forming bonds = 614 + 2472 = 3086 kJ/mol
Step 3: Calculate the change in enthalpy (∆H) for the reaction using the
bond dissociation energies.
∆H= Energy to break bonds−Energy released in forming bonds = 2828 kJ/mol−3086 kJ/mol = −258 kJ/mol
Therefore, the enthalpy change for the reaction is ∆H=−258 kJ/mol.
Question 10
Question
Calculate the enthalpy change (∆H) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
given the following bond dissociation energies:
H−H: 432 kJ/mol
O=O: 497 kJ/mol
O−H: 464 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy in bonds broken = 2 ×Energy of H−Hbond + Energy of O=Obond
= 2 ×432 kJ/mol + 1 ×497 kJ/mol
= 1361 kJ
Step 2: Calculate the total energy released when the new bonds are formed
in the products.
Energy in bonds formed = 2 ×Energy of O−Hbond
= 2 ×464 kJ/mol
= 928 kJ
Step 3: Calculate the change in enthalpy (∆H) for the reaction.
∆H= Energy in bonds broken −Energy in bonds formed
= 1361 kJ −928 kJ
= 433 kJ
Therefore, the enthalpy change (∆H) for the reaction is 433 kJ.
8
Question 11
Question
Given the following reaction:
2H2(g) + O2(g)→2H2O(l)
The standard enthalpy change for the reaction is -483.6 kJ mol−1. Calculate
the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
at the same temperature.
Solution
Step 1: Write the given reaction and its standard enthalpy change. Given
reaction:
2H2(g) + O2(g)→2H2O(l)
∆H=−483.6 kJ mol−1
Step 2: Write the target reaction. Target reaction:
H2O(l)→H2(g) + 1
2O2(g)
Step 3: Determine the standard enthalpy change for the target reaction.
Since the target reaction is the reverse of the given reaction, the sign of the
enthalpy change will be reversed. Therefore, ∆H= 483.6 kJ mol−1
So, the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
is 483.6 kJ mol−1at the same temperature.
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of ethylene
(C2H4) using the following information:
The standard enthalpy of formation of CO2is -393.5 kJ/mol.
The standard enthalpy of formation of H2O is -285.8 kJ/mol.
The standard enthalpy of formation of C2H4is 52.3 kJ/mol.
The balanced equation for the combustion of ethylene is:
C2H4(g) + 3O2(g)→2CO2(g) + 2H2O(l)
9
Solution
Step 1: Calculate the standard enthalpy change for the combustion of ethy-
lene using the standard enthalpy of formation values provided. The standard
enthalpy change can be calculated using the equation:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
Where nand mare the stoichiometric coefficients from the balanced equa-
tion.
Step 2: Substitute the given values into the equation and calculate the stan-
dard enthalpy change.
∆H◦= [2(−393.5) + 2(−285.8)] −[1(52.3) + 3(0)]
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= (−787.0) −(52.3) = −834.3 kJ/mol
Therefore, the standard enthalpy change for the combustion of ethylene is
∆H◦=−834.3 kJ/mol.
Question 13
Question
Calculate the enthalpy change for the combustion of ethylene, C2H4(g), to form
carbon dioxide and water using the following data:
Reaction ∆H◦(kJ/mol)
C2H4(g) + 3O2(g) →2CO2(g) + 2H2O(l) -1411
H2(g) + 1
2O2(g) →H2O(l) -286
C(s) + O2(g) →CO2(g) -394
Solution
Step 1: Writing the balanced equation for the combustion of ethylene:
C2H4(g) + 3O2(g) →2CO2(g) + 2H2O(l)
Step 2: Calculating the enthalpy change for the given reaction:
∆H=X∆Hproducts −X∆Hreactants
∆H= [2(2∆HCO2) + 2∆HH2O(l)]−[∆HC2H4+ 3∆HO2]
∆H= [2(2 × −394) + 2 × −286] −[−1411]
∆H= [2(−788) −572] + 1411
∆H=−1576 −572 + 1411
∆H=−737 kJ/mol
Therefore, the enthalpy change for the combustion of ethylene to form carbon
dioxide and water is -737 kJ/mol.
10
Question 14
Question
A reaction has a standard enthalpy change of -418 kJ/mol. If the reaction is
exothermic, determine if the reaction is exothermic or endothermic when carried
out at 25
°
C if the reaction is carried out under nonstandard conditions with a
pressure of 2 atm. Assume the reaction has an ideal gas as a product.
Solution
Step 1: Write the expression for the standard Gibbs free energy change in terms
of the standard enthalpy change and the standard entropy change:
∆G◦= ∆H◦−T∆S◦
Step 2: Use the relationship between standard Gibbs free energy change and
equilibrium constant to find the standard entropy change (∆S◦).
We know that ∆G◦=−RT ln(K) where Kis the equilibrium constant under
standard conditions. Therefore,
∆G◦=−RT ln(K◦)
where K◦is the equilibrium constant under standard conditions.
Step 3: Given that the reaction is exothermic, we know that ∆H◦=−418
kJ/mol.
Step 4: Calculate the standard entropy change using the standard Gibbs
free energy change equation:
∆S◦=∆H◦−∆G◦
T
Step 5: Substitute the known values into the equation and solve for ∆S◦:
∆S◦=−418 −(−RT ln(K◦))
T
Step 6: Given that the pressure is 2 atm, we need to consider the effect of
pressure on the equilibrium constant in the expression for ∆G. The effect of
changing pressure on the equilibrium constant can be accounted for using the
Van’t Hoff equation:
ln K2
K1=∆nRT
R
where ∆nis the change in the number of moles of gas as the reaction occurs.
Step 7: Determine the change in the number of moles of gas (∆n) for the
reaction based on the stoichiometry of the reaction.
Step 8: Substitute the calculated values into the Van’t Hoff equation to find
the new equilibrium constant (K2) at 2 atm.
11
Step 9: Use the Van’t Hoff equation to adjust the equilibrium constant for
the new pressure to calculate ∆Gunder non-standard conditions.
Step 10: Use the equation ∆G= ∆H−T∆Sto determine if the reaction is
exothermic or endothermic under the given nonstandard conditions.
Question 15
Question
Calculate the enthalpy change for the reaction below using the following data:
2H2(g) + O2(g)→2H2O(l)
Given:
2H2(g) + O2(g) →2H2O(l) ∆H = -571.7 kJ
H2O(l) →H2O(g) ∆H = +44.0 kJ
H2(g) →H2O(g) ∆H = -483.6 kJ
Solution
Step 1: Given the following equations:
2H2(g) + O2(g)→2H2O(l) ∆H=−571.7kJ (1)
H2O(l)→H2O(g) ∆H= +44.0kJ (2)
H2(g)→H2O(g) ∆H=−483.6kJ (3)
Step 2: We can use these three equations to find the enthalpy change for the
desired reaction:
2H2(g) + O2(g)→2H2O(l)
Step 3: First, we need to change the phase of 2H2O(l) to 2H2O(g), then
combine the equations:
(1) + (2) ⇒2H2(g) + O2(g)→2H2O(g) ∆H=−571.7 + 44.0 = −527.7kJ (4)
Step 4: Next, we combine equation (3) with equation (4) to obtain the
desired reaction:
(4) + (3) ⇒2H2(g) + O2(g)→2H2O(g) + H2(g) ∆H=−527.7−483.6 = −1011.3kJ
Step 5: Therefore, the enthalpy change for the reaction 2H2(g) + O2(g)→
2H2O(l) is -1011.3 kJ.
12
Question 16
Question
Calculate the standard enthalpy change for the following reaction at 25
°
C:
2C(s) + 5H2(g) →C2H6(g)
Given the following standard enthalpy of formation values:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.7 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction.
2C(s) + 5H2(g) →C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xν∆H◦
f
where νis the stoichiometric coefficient of each substance in the balanced
chemical equation and ∆H◦
fis the standard enthalpy of formation.
Plugging in the values:
∆H◦= 2(0 kJ/mol) + 5(0 kJ/mol) −84.7 kJ/mol
∆H◦=−84.7 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.7 kJ/mol .
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25◦C:
2H2O2→2H2O+O2
Given the following standard enthalpies of formation: ∆H◦
f(H2O2) = −196.1
kJ/mol, ∆H◦
f(H2O) = −285.8 kJ/mol, ∆H◦
f(O2) = 0 kJ/mol.
13
Solution
Step 1: Write the balanced chemical equation for the reaction.
2H2O2→2H2O+O2
Step 2: Determine the standard enthalpy change (∆H◦) for the reaction
using the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= 2 ×∆H◦
f(H2O) + ∆H◦
f(O2)−2×∆H◦
f(H2O2)
∆H◦= 2 ×(−285.8 kJ/mol) + 0 −2×(−196.1 kJ/mol)
∆H◦=−571.6 kJ/mol + 392.2 kJ/mol
∆H◦=−179.4 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction at
25◦C is -179.4 kJ/mol.
Question 18
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−484 kJ
Calculate the enthalpy change for the reaction:
2H2O(l)→2H2(g)+O2(g)
Solution
Step 1: The enthalpy change of the reverse reaction is the negative of the
enthalpy change of the forward reaction.
−∆H=−(−484 kJ)
−∆H= 484 kJ
Therefore, the enthalpy change for the reaction:
2H2O(l)→2H2(g)+O2(g)
is 484 kJ.
14
Question 19
Question
Calculate the enthalpy change for the reaction below using the given standard
enthalpies of formation:
2C(s) + 3H2(g)→C2H6(g)
Given:
∆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 balanced chemical equation for the reaction.
The balanced chemical equation for the reaction is:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the enthalpy change using the standard enthalpies of for-
mation.
The enthalpy change (∆Hrxn) for the reaction can be calculated using the equa-
tion:
∆Hrxn =X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the given standard enthalpy of formation values into the equation:
∆Hrxn = (∆H◦
f[C2H6(g)]) −(∆H◦
f[C(s)] + ∆H◦
f[H2(g)])
∆Hrxn = (−84.68) −(0 + 0)
∆Hrxn =−84.68 kJ/mol
Therefore, the enthalpy change for the reaction is -84.68 kJ/mol.
Question 20
Question
Calculate the enthalpy change (∆H◦) for the reaction below based on the given
bond dissociation energies:
2CH4(g) + 3O2(g)→2CO2(g) + 4H2O(l)
15
Given bond dissociation energies:
C-H = 413 kJ/mol
O=O = 498 kJ/mol
O-H = 463 kJ/mol
C=O = 743 kJ/mol
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy to break bonds = 2 ×(C-H) + 3 ×(O=O) + 12 ×(O-H)
= 2 ×413 kJ/mol + 3 ×498 kJ/mol + 12 ×463 kJ/mol
= 826 kJ/mol + 1494 kJ/mol + 5556 kJ/mol
= 7876 kJ/mol
Step 2: Calculate the total energy released when new bonds are formed in
the products.
Energy released by forming new bonds = 2 ×(C=O) + 4 ×(O-H)
= 2 ×743 kJ/mol + 4 ×463 kJ/mol
= 1486 kJ/mol + 1852 kJ/mol
= 3338 kJ/mol
Step 3: Calculate the change in enthalpy (∆H◦) for the reaction.
∆H◦= Energy released by forming new bonds −Energy to break bonds
∆H◦= 3338 kJ/mol −7876 kJ/mol
∆H◦=−4538 kJ/mol
Therefore, the enthalpy change (∆H◦) for the reaction is -4538 kJ/mol,
indicating that the reaction is exothermic.
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C:
4NH3(g)+5O2(g)→4NO(g)+6H2O(l)
Given the following standard enthalpies of formation:
∆H◦
f(NH3) = −46.1 kJ/mol
∆H◦
f(NO) = 90.3 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
16
Solution
Step 1: Calculate the standard enthalpy change for the reaction using standard
enthalpies of formation. The standard enthalpy change of a reaction 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.
Given:
∆H◦
f(NH3) = −46.1 kJ/mol
∆H◦
f(NO) = 90.3 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
The standard enthalpy change for the reaction can be calculated as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= [4∆H◦
f(NO) + 6∆H◦
f(H2O(l))] −[4∆H◦
f(NH3) + 5∆H◦
f(O2)]
∆H◦= [4(90.3) + 6(−285.8)] −[4(−46.1) + 5(0)]
∆H◦= [361.2−1714.8] −[−184.4]
∆H◦= 1053.6 + 184.4
∆H◦= 1238 kJ
Therefore, the standard enthalpy change for the reaction is ∆H◦= 1238 kJ.
Question 22
Question
A reaction has a standard enthalpy change of -335 kJ/mol. If 2.50 mol of the
reaction takes place, what is the total heat change for the reaction?
Solution
Step 1: Determine the total heat change for the reaction in kJ. -335 kJ/mol x
2.50 mol = -837.5 kJ
Answer: The total heat change for the reaction is -837.5 kJ.
17
Question 23
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(graphite) + 3H2(g) →C2H6(g)
given the following standard enthalpy of formation values:
∆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.
2C(graphite) + 3H2(g) →C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients in the balanced equation.
Substitute the given values:
∆H◦= 1(−84.68 kJ/mol) −[2(0 kJ/mol) + 3(0 kJ/mol)]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−84.68
kJ/mol.
Question 24
Question
A reaction has a standard enthalpy change of -350 kJ. If 2.50 moles of this
reaction release 175 kJ of heat, what is the actual enthalpy change of the reaction
in kJ?
18
Solution
Step 1: Calculate the moles of reaction that correspond to the standard enthalpy
change. Given: Standard enthalpy change (∆H◦) = -350 kJ Moles of reaction
for standard enthalpy change (n◦) = 1 mole
Step 2: Use stoichiometry to calculate the moles of reaction for the amount
of heat released. Given: Heat released = 175 kJ Moles of reaction for heat
released (n) = 2.50 moles
Step 3: Calculate the actual enthalpy change of the reaction. From Step 1:
∆H◦
n◦=−350 kJ
1 mol =−350 kJ/mol
From Step 2:
−350 kJ/mol
n=175 kJ
2.50 mol =−70 kJ/mol
Therefore, the actual enthalpy change of the reaction is -70 kJ.
Question 25
Question
Given the following reaction and enthalpy values, calculate the standard en-
thalpy change (∆H◦) for the reaction at 298 K:
2C(s)+ 2H2O(g)→2CO(g)+ 2H2(g)
∆H◦
f(CO(g)) = −110.5 kJ/mol
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
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)
= [2∆H◦
f(CO(g)) + 2∆H◦
f(H2(g))] −[2∆H◦
f(C(s)) + 2∆H◦
f(H2O(g))]
= [2(−110.5) + 2(0)] −[2(0) + 2(−241.8)]
= [−221.0 + 0] −[0 −483.6]
=−221.0 + 483.6
= 262.6 kJ
Therefore, the standard enthalpy change for the reaction at 298 K is 262.6 kJ .
19
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2SO2(g)+O2(g)→2SO3(g)
given the following standard enthalpies of formation:
∆H◦
f(SO2) = −296.8 kJ/mol
∆H◦
f(SO3) = −395.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard en-
thalpies of formation.
2SO2(g)+O2(g)→2SO3(g)
Given:
∆H◦
f(SO2) = −296.8 kJ/mol
∆H◦
f(SO3) = −395.7 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦= 2(−395.7 kJ/mol) −[2(−296.8 kJ/mol) + 0 kJ/mol]
∆H◦=−791.4 kJ/mol −(−593.6 kJ/mol)
∆H◦=−791.4 kJ/mol + 593.6 kJ/mol
∆H◦=−197.8 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −197.8
kJ/mol.
20
Question 27
Question
Calculate the standard enthalpy change (∆H◦) for the reaction
2A(g) + B(g) →3C(g) + D(g)
given the following data:
∆H◦
f(A) = −250 kJ/mol
∆H◦
f(B) = −150 kJ/mol
∆H◦
f(C) = −400 kJ/mol
∆H◦
f(D) = −50 kJ/mol
Solution
Step 1: Calculate ∆H◦for the reaction using the standard enthalpy of formation
data provided:
We can use the equation
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the coefficients in the balanced chemical equation.
Plugging in the given values, we have:
∆H◦= [3(−400) + (−50)] −[2(−250) + (−150)]
Step 2: Simplify the equation to find the ∆H◦.
Calculating the values, we get:
∆H◦= [−1200 −50] −[−500 −150]
∆H◦=−1250 −(−650)
∆H◦=−1250 + 650
∆H◦=−600 kJ/mol
Therefore, the standard enthalpy change for the reaction is −600 kJ/mol .
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Question 28
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2C(s) + 3H2(g)→C2H6(g)
Given the standard enthalpy of formation values:
∆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 balanced chemical equation for the reaction.
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change of the reaction using the
standard enthalpies of formation.
∆H◦=X(∆H◦
f(products)) −X(∆H◦
f(reactants))
Step 3: Substitute the given standard enthalpy of formation values into the
equation.
∆H◦=∆H◦
f(C2H6(g))−2·∆H◦
f(C(s)) + 3 ·∆H◦
f(H2(g))
Step 4: Plug in the values and calculate the standard enthalpy change.
∆H◦= [−84.68 kJ/mol] −[2 ·0 kJ/mol + 3 ·0 kJ/mol]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−84.68 kJ/mol.
Question 29
Question
Given the following reaction and enthalpy changes:
2C(s) + 3H2(g)→C2H6(g) ∆H=−84.7 kJ
4C(s) + 6H2(g)→2C2H6(g) ∆H=−118.4 kJ
Calculate the enthalpy change for the following reaction:
C2H6(g)→2C(s) + 3H2(g)
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Solution
Step 1: Given the enthalpy changes for the two reactions, we need to manipulate
them to find the enthalpy change for the target reaction. First, reverse the first
reaction and multiply it by 2 to match the coefficients of the target reaction:
2(C2H6(g)→2C(s) + 3H2(g))
= 2(2C(s) + 3H2(g)→C2H6(g))
= 4C(s) + 6H2(g)→2C2H6(g) ∆H= 2(−84.7) = −169.4 kJ
Step 2: Next, add the second reaction as is to the previous equation to
eliminate the intermediates:
4C(s)+6H2(g)→2C2H6(g)+4C(s)+6H2(g) ∆H=−169.4 kJ−118.4 kJ = −287.8 kJ
Step 3: Finally, cancel out the common compounds on both sides to get the
enthalpy change for the target reaction:
C2H6(g)→2C(s) + 3H2(g) ∆H=−287.8 kJ
Question 30
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
given the following standard enthalpy of formations:
C2H2(g) = 226 kJ/mol
CO2(g) = −393.5 kJ/mol
H2O(g) = −241.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the reaction by using the
standard enthalpies of formation:
The standard enthalpy change can be calculated using the formula:
∆H◦=X(products) −X(reactants)
Given:
2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)
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The standard enthalpy change will be:
4(−393.5 kJ/mol) + 2(−241.8 kJ/mol) −2(226 kJ/mol) −5(0)
∆H◦=−1574 kJ/mol −483.6 kJ/mol −452 kJ/mol
∆H◦=−2509.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
−2509.6 kJ/mol.
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