CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Thermochemistry
Question Bank - Set 4
Liberty University
Question 1
Question
A reaction has a standard enthalpy change of -394 kJ/mol and a standard
entropy change of 110 J/(mol
·
K). Calculate the standard free energy change at
298 K.
Solution
Step 1: Calculate the standard free energy change using the equation ∆G=
∆H−T∆Swhere ∆Gis the standard free energy change, ∆His the standard
enthalpy change, Tis the temperature in Kelvin, and ∆Sis the standard entropy
change.
Step 2: Substitute the given values into the equation:
∆G=−394 kJ/mol −(298 K)(0.110 kJ/mol
·
K)
Step 3: Convert the entropy change to kJ/mol
·
K:
0.110 kJ/mol
·
K = 110 ×10−3kJ/mol
·
K
Step 4: Perform the calculation:
∆G=−394 kJ/mol −(298 ×110 ×10−3) kJ/mol
Step 5: Simplify the expression to find the standard free energy change:
∆G=−394 kJ/mol −32.78 kJ/mol
∆G=−426.78 kJ/mol
Therefore, the standard free energy change at 298 K is -426.78 kJ/mol.
Question 2
Question
Consider the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following enthalpy changes:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction.
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpy of formation values given.
∆H◦=Xn∆H◦
f products −Xm∆H◦
f reactants
Step 2: Substitute the given values of standard enthalpies of formation into
the formula:
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
Step 3: Perform the calculation to find the standard enthalpy change of the
reaction:
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction is
-571.6 kJ/mol.
Question 3
Question
A student is investigating the combustion of ethene gas (C2H4) and measures the
enthalpy change for the reaction to be −1411 kJ/mol. Write the balanced chem-
ical equation for the combustion of ethene and calculate the enthalpy change
when 3.00 moles of ethene react.
2
Solution
Step 1: Write the balanced chemical equation for the combustion of ethene:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Step 2: Calculate the enthalpy change for the combustion of 1 mole of
ethene: From the balanced equation, 1 mole of ethene produces −1411 kJ of
heat. Therefore, the enthalpy change for the combustion of 1 mole of ethene is
−1411 kJ/mol.
Step 3: Calculate the enthalpy change when 3.00 moles of ethene react:
Since the enthalpy change for 1 mole of ethene is −1411 kJ/mol, the enthalpy
change when 3.00 moles of ethene react is:
−1411 kJ/mol ×3.00 mol = −4233 kJ
Therefore, when 3.00 moles of ethene react, the enthalpy change is −4233
kJ.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2NOCl(g)→2NO(g) + Cl2(g)
given the following standard enthalpy change values:
∆H◦
f(NOCl(g)) = 51.0 kJ/mol
∆H◦
f(NO(g)) = 90.3 kJ/mol
∆H◦
f(Cl2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation is:
2NOCl(g)→2NO(g) + Cl2(g)
Step 2: Calculate the standard enthalpy change for the reaction using stan-
dard enthalpies of formation.
The standard enthalpy change for a reaction can be calculated using the
equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
3
Given:
∆H◦
f(NOCl(g)) = 51.0 kJ/mol
∆H◦
f(NO(g)) = 90.3 kJ/mol
∆H◦
f(Cl2(g)) = 0 kJ/mol
Plugging in the values:
∆H◦= 2 ×∆H◦
f(NO(g)) + ∆H◦
f(Cl2(g)) −2×∆H◦
f(NOCl(g))
= 2 ×90.3 kJ/mol + 0 kJ/mol −2×51.0 kJ/mol
= 180.6 kJ/mol −102.0 kJ/mol
= 78.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is 78.6 kJ/mol.
Question 5
Question
The combustion of ethylene (C2H4) releases 1411.03 kJ/mol of heat. Calculate
the standard enthalpy change for the combustion of ethylene to form CO2and
H2O.
Solution
Step 1: Write the balanced chemical equation for the combustion of ethylene:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Step 2: Determine the moles of ethylene that react to release 1411.03 kJ of
heat. Since the reaction releases 1411.03 kJ/mol, the heat released by 1 mole
of ethylene is 1411.03 kJ. Therefore, the heat released by x moles of ethylene is
1411.03 kJ/mol ×xmol.
Step 3: Calculate the heat of combustion of ethylene. The heat of combus-
tion of ethylene is the same as the heat released by the reaction. Given that
1411.03 kJ is released by 1 mole of ethylene, the heat of combustion is 1411.03
kJ/mol.
Therefore, the standard enthalpy change for the combustion of ethylene to
form CO2and H2Ois 1411.03 kJ/mol.
4
Question 6
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2C(graphite)+3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
C(graphite) = 0 kcal/mol, H2(g) = 0 kcal/mol, C2H6(g) = −20.2 kcal/mol
Solution
Step 1: Write the balanced chemical equation and determine the standard en-
thalpy change. The standard enthalpy change for a reaction can be calculated
by taking the sum of the standard enthalpies of formation of the products and
subtracting the sum of the standard enthalpies of formation of the reactants.
The balanced chemical equation for the reaction is:
2C(graphite)+3H2(g)→C2H6(g)
The standard enthalpy change (∆H◦) can be calculated as:
∆H◦=XProducts −XReactants
∆H◦= [1×Standard enthalpy of formation of C2H6(g)]−[2×Standard enthalpy of formation of C(graphite)+3×Standard enthalpy of formation of H2(g)]
Step 2: Substitute the given values into the equation.
∆H◦= [−20.2] −[2(0) + 3(0)]
∆H◦=−20.2 kcal/mol
Therefore, the standard enthalpy change for the given reaction is −20.2 kcal/mol .
Question 7
Question
Consider the following reaction:
2A(g) + B(g)→C(g)+3D(g)
Given the following data:
∆H◦
f(kJ/mol)
A(g) : 50, B(g) : 100, C(g) : −200, D(g) : −75
Calculate the standard enthalpy change, ∆H◦, for the reaction.
5
Solution
Step 1: Calculate the standard enthalpy change, ∆H◦, for the reaction using
the standard enthalpies of formation for each substance.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion and calculate.
∆H◦= [(∆H◦
fof C(g))+3(∆H◦
fof D(g))]−[2(∆H◦
fof A(g))+(∆H◦
fof B(g))]
∆H◦= [(−200) + 3(−75)] −[2(50) + 100]
∆H◦= (−200 −225) −(100 + 100)
∆H◦=−425 −200
∆H◦=−625 kJ/mol
Therefore, the standard enthalpy change, ∆H◦, for the reaction is -625
kJ/mol.
Question 8
Question
Given the following reaction:
2A(g)+3B(g)→C(g)+2D(g)
If the standard enthalpy change of formation for compounds A,B,C, and D
are 50 kJ/mol, 30 kJ/mol, -80 kJ/mol, and -100 kJ/mol, respectively, calculate
the standard enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy change of formation for each compound. The standard enthalpy
change for the reaction can be calculated using the formula:
∆H◦=Xν∆H◦
f, products −Xν∆H◦
f, reactants
where ∆H◦is the standard enthalpy change for the reaction, νis the stoichio-
metric coefficient, and ∆H◦
fis the standard enthalpy change of formation.
For the given reaction:
2A(g)+3B(g)→C(g)+2D(g)
6
Plugging in the values of the standard enthalpy change of formation:
∆H◦= (1)(−80 kJ/mol) + (2)(−100 kJ/mol) −(2)(50 kJ/mol) −(3)(30 kJ/mol)
∆H◦=−80 kJ/mol −200 kJ/mol −100 kJ/mol −90 kJ/mol
∆H◦=−470 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -470 kJ/mol.
Question 9
Question
Given the standard enthalpy of combustion of benzene, C6H6(l), is −3267 kJ/mol,
calculate the standard enthalpy of formation of benzene.
Solution
Step 1: Write the balanced chemical equation for the combustion of one mole
of benzene.
The balanced chemical equation for the combustion of benzene is:
C6H6(l) + 15/2O2(g)→6CO2(g)+3H2O(l)
Step 2: Use Hess’s law to relate the standard enthalpy of combustion with
the standard enthalpies of formation.
The standard enthalpy change of combustion is given by:
∆Hcomb =Xνproducts∆H◦
f−Xνreactants∆H◦
f
where νrepresents the stoichiometric coefficients in the balanced equation.
For the combustion of benzene:
−3267 kJ/mol = 6 ×(−393.5) + 3 ×(−285.8) −∆H◦
f
Step 3: Solve for the standard enthalpy of formation of benzene.
−3267 kJ/mol = −2361 −857.4−∆H◦
f
−3267 kJ/mol = −3218.4−∆H◦
f
∆H◦
f= 49.4 kJ/mol
Therefore, the standard enthalpy of formation of benzene is 49.4 kJ/mol.
7
Question 10
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(g)
If the standard enthalpy change for the reaction is −483.6 kJ/mol, calculate
the standard enthalpy change for the reaction:
H2O(g)→2H(g)+O2(g)
Solution
Step 1: Write the given reaction as a sum of desired reactions.
The given reaction is:
−483.6 kJ/mol = 2H2(g)+O2(g)→2H2O(g)
To find the standard enthalpy change for the desired reaction, the given
reaction can be written as a sum of two reactions as follows:
H2O(g)→H2(g) + 1
2O2(g)
H2(g) + 1
2O2(g)→2H(g)+O2(g)
Step 2: Use Hess’s Law to calculate the standard enthalpy change for the
desired reaction.
Since enthalpy is a state function, the overall enthalpy change for a reaction
is the sum of the enthalpy changes of the individual reactions it can be broken
down into:
∆Hdesired = ∆H1+ ∆H2
Therefore, we need to add the enthalpy changes of the two desired reactions
to find the enthalpy change of the overall desired reaction.
Step 3: Calculate the standard enthalpy change for the desired reaction.
The enthalpy change for the first desired reaction (H2O(g)→H2(g) +
1
2O2(g)) can be calculated as follows:
∆H1=−483.6 kJ/mol
The enthalpy change for the second desired reaction (H2(g) + 1
2O2(g)→
2H(g)+O2(g)) is zero because it is the reverse of the given reaction.
Therefore, the standard enthalpy change for the overall desired reaction is:
∆Hdesired =−483.6 kJ/mol
8
Question 11
Question
Consider the reaction:
2A(g) + B(g)→3C(g)+4D(g)
where ∆H◦
ffor A,B,C, and Dare −200 kJ/mol, 100 kJ/mol, 150 kJ/mol,
and 50 kJ/mol, respectively. If the enthalpy change for the reaction is −500 kJ,
calculate the standard enthalpy of formation for compound B.
Solution
Step 1: Calculate the total change in enthalpy using the standard enthalpy of
formation values provided.
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
−500 kJ = (3)(150 kJ/mol + 4)(50 kJ/mol) −(2)(−200 kJ/mol) −(1)(100 kJ/mol)
−500 kJ = 450 kJ + 200 kJ + 200 kJ + 100 kJ
−500 kJ = 950 kJ −200 kJ
−500 kJ = 750 kJ
Step 2: Solve for the standard enthalpy of formation of compound B.
−500 kJ = 750 kJ −200 kJ −100 kJ + xkJ
−500 kJ = 450 kJ + xkJ
x=−950 kJ
Therefore, the standard enthalpy of formation for compound Bis −950
kJ/mol.
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the reaction below using
standard enthalpies of formation.
2C(s) + 3H2(g)→C2H6(g)
Given the standard enthalpies of formation:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
9
Solution
Step 1: Write the given chemical equation and balance it if needed:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation: The standard enthalpy change can be calculated using
the standard enthalpies of formation of the products and reactants.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
For the given reaction:
∆H◦= ∆H◦
f[C2H6(g)] −[∆H◦
f[C(s)] + ∆H◦
f[H2(g)]]
Substitute the values:
∆H◦=−84.68 kJ/mol −[0 + 0]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol .
Question 13
Question
Calculate the enthalpy change for the reaction below using standard enthalpies
of formation:
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)) = −84.7 kJ/mol
Solution
Step 1: Write down the balanced chemical equation and identify the standard
enthalpies of formation for each compound involved. The balanced chemical
equation is:
2C(graphite)+3H2(g)→C2H6(g)
10
The standard enthalpies of formation for each compound are:
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.7 kJ/mol
Step 2: Calculate the enthalpy change for the reaction using the standard
enthalpies of formation. The enthalpy change (∆H) for the reaction can be
calculated as follows:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H= ∆H◦
f(C2H6(g)) −[∆H◦
f(2C(graphite)) + ∆H◦
f(3H2(g))]
∆H=−84.7 kJ/mol −[0 + 0]
∆H=−84.7 kJ/mol
Therefore, the enthalpy change for the reaction is −84.7 kJ/mol .
Question 14
Question
Calculate the standard enthalpy change for the reaction:
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
Solution
Step 1: Write the balanced chemical equation for the reaction and calculate the
standard enthalpy change.
C(graphite) + H2(g) →C2H6(g)
Given ∆H◦
fvalues:
∆H◦=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
11
∆H◦= [∆H◦
f(C2H6(g))] −[∆H◦
f(C(graphite)) + 3∆H◦
f(H2(g))]
∆H◦= [−84.68] −[0 + 3(0)] = −84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol.
Question 15
Question
A reaction has a standard enthalpy change of ∆H◦=−395 kJ/mol. Calculate
the amount of heat (in joules) released when 5.00 grams of this reaction occurs.
Solution
Step 1: Calculate the molar mass of the reaction. The molar mass of the reaction
can be calculated as follows:
Molar mass = Mass
Moles
Given that the mass is 5.00 grams, and assuming the reaction is the limiting
reactant, we can calculate the number of moles in 5.00 grams using the molar
mass of the reaction.
Step 2: Calculate the heat released per mole. We know that the standard
enthalpy change is -395 kJ/mol. This means that -395 kJ of heat is released for
every 1 mole of the reaction that occurs.
Step 3: Calculate the heat released for 5.00 grams of the reaction. Now that
we have the amount of heat released per mole, we can calculate the amount of
heat released for 5.00 grams of the reaction using the number of moles calculated
in Step 1.
Step 4: Perform the calculations and convert the units. Finally, we convert
the heat released from kJ to J to get the final answer.
Given: ∆H◦=−395 kJ/mol,Mass = 5.00 g
Step 1:
Molar mass = 5.00 g
moles
The molar mass of the reaction will depend on the specific reaction provided.
Step 2: Heat released per mole = -395 kJ/mol
Step 3: Amount of heat released for 5.00 g = (Amount of heat released per
mole) ×(Moles in 5.00 g)
Step 4: Convert the units to get the final answer.
12
Question 16
Question
Calculate the enthalpy change for the reaction below, given the following bond
dissociation energies:
2C2H4(g)+3O2(g)→4CO2(g)+4H2O(g)
Bond dissociation energies: - C-C: 348 kJ/mol - C=C: 611 kJ/mol - O=O:
498 kJ/mol - C=O: 805 kJ/mol - O-H: 463 kJ/mol - C-H: 413 kJ/mol
Solution
Step 1: Calculate the total bond energy of reactants.
2(4 ×C-C + 4 ×C-H) + 3(2 ×O=O)
= 2(4 ×348 + 4 ×413) + 3(2 ×498)
= 2(1392 + 1652) + 3(996)
= 4088 + 2994
= 7082 kJ
Step 2: Calculate the total bond energy of products.
4(4 ×C=O + 4 ×O-H) = 4(4 ×805 + 4 ×463)
= 4(3220 + 1852)
= 4(5072)
= 20288 kJ
Step 3: Calculate the enthalpy change.
∆H= (Total bond energy of reactants) −(Total bond energy of products)
= 7082 −20288
=−13206 kJ
Therefore, the enthalpy change for the reaction is ∆H=−13206 kJ.
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(s)+3H2(g)→C2H6(g)
13
given the following standard enthalpy of formations:
∆H◦
fof C(s) = 0 kJ/mol
∆H◦
fof H2(g) = 0 kJ/mol
∆H◦
fof C2H6(g) = −84.68 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard enthalpy
of formation values. Step 2: Use the standard enthalpy change of formation
formula to calculate the standard enthalpy change for the given reaction.
Step 1: The balanced chemical equation is:
2C(s)+3H2(g)→C2H6(g)
Given:
∆H◦
fof C(s) = 0 kJ/mol
∆H◦
fof H2(g) = 0 kJ/mol
∆H◦
fof C2H6(g) = −84.68 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦):
∆H◦=Xn∆H◦
fof products −Xm∆H◦
fof reactants
∆H◦= (1 × −84.68) −(2 ×0+3×0)
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −84.68 kJ/mol .
Question 18
Question
Consider the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−483.7 kJ
Calculate the enthalpy change for the reaction
N2(g) + 3H2(g)→2NH3(g)
given the following bond dissociation energies:
N-N = 945 kJ/mol,N-H = 389 kJ/mol,H-H = 432 kJ/mol
14
Solution
Step 1: Calculate the bond dissociation energy of the reactants: N2(g) = 1 ×
(N-N) = 1 ×945 = 945 kJ/mol
H2(g)=3×(H-H) = 3 ×432 = 1296 kJ/mol
Total energy of reactants = 945 kJ/mol + 1296 kJ/mol = 2241 kJ/mol
Step 2: Calculate the bond dissociation energy of the products: NH3(g) =
2×(N-H) + 6 ×(H-H) = 2 ×389 + 6 ×432 = 2574 kJ/mol
Total energy of products = 2574 kJ/mol
Step 3: Calculate the enthalpy change for the reaction: ∆H= Energy of bonds broken−
Energy of bonds formed
∆H= 2241 kJ/mol −2574 kJ/mol = −333 kJ/mol
Therefore, the enthalpy change for the reaction N2(g) + 3H2(g)→2NH3(g)
is -333 kJ/mol.
Question 19
Question
A reaction has a standard enthalpy change of -220 kJ/mol and a standard
entropy change of 120 J/(mol
·
K). Calculate the standard Gibbs free energy
change at 298 K.
Solution
Step 1: Recall the relationship between Gibbs free energy (∆G), enthalpy change
(∆H), and entropy change (∆S) at constant temperature (T):
∆G= ∆H−T·∆S
Step 2: Convert the entropy change to kJ/(mol
·
K) to match the units of
enthalpy change:
∆S= 120 J/(mol
·
K) = 0.120 kJ/(mol
·
K)
Step 3: Substitute the values into the equation:
∆G=−220 kJ/mol −(298 K) ·0.120 kJ/(mol
·
K)
Step 4: Calculate the standard Gibbs free energy change:
∆G=−220 kJ/mol −35.76 kJ/mol
∆G=−255.76 kJ/mol
Therefore, the standard Gibbs free energy change at 298 K is -255.76 kJ/mol.
15
Question 20
Question
Given the following reaction:
2A(g)+3B(g)→C(g)+4D(g)
The enthalpy change for the reaction is -3050 kJ. If the enthalpy change of
formation for compounds C and D are -240 kJ/mol and -126 kJ/mol, respec-
tively, calculate the enthalpy change of formation for compound A.
Solution
Step 1: Write out the enthalpy change of formation for compound C and D in
terms of the given reaction.
∆Hf(C) = −240 kJ/mol
∆Hf(D) = −126 kJ/mol
Step 2: Apply Hess’s Law to find the enthalpy change of formation for com-
pound A. The enthalpy change of formation for a compound is the enthalpy
change when 1 mole of the compound is formed from its elements in their stan-
dard states.
Step 3: Express the enthalpy change of the given reaction in terms of the
enthalpy change of formation for compounds A, B, C, and D.
∆Hrxn =Xn∆Hf(products) −Xm∆Hf(reactants)
∆Hrxn =Xn∆Hf(products) −Xm∆Hf(reactants)
∆Hrxn = ∆Hf(C) + 4∆Hf(D)−(2∆Hf(A) + 3∆Hf(B))
∆Hrxn =−240 + 4(−126) −(2∆Hf(A) + 3(0))
Step 4: Substitute the given enthalpy change of the reaction and solve for
the enthalpy change of formation for compound A.
−3050 = −240 −504 −2∆Hf(A)
−3050 = −744 −2∆Hf(A)
−2306 = −2∆Hf(A)
∆Hf(A) = 1153 kJ/mol
Therefore, the enthalpy change of formation for compound A is 1153 kJ/mol.
16
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2(g)+O2(g)→2H2O(l)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol,∆H◦
f(O2(g)) = 0 kJ/mol,∆H◦
f(H2O(l)) = −285.8 kJ/mol
Solution
Step 1: Write the balanced chemical equation with the given standard enthalpies
of formation.
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the values of the standard enthalpies of formation into
the equation.
∆H◦= [2∆H◦
f(H2O(l))] −[2∆H◦
f(H2(g))+∆H◦
f(O2(g))]
Step 4: Substitute the given values into the equation and calculate.
∆H◦= [2(−285.8)] −[2(0) + 0]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−571.6 kJ/mol.
Question 22
Question
Given the following reaction:
2H2S(g)+3O2(g)→2SO2(g)+2H2O(g)
If the standard enthalpy change for the reaction is -1036 kJ, calculate the
standard enthalpy change for the reaction below:
2H2S(g)+2O2(g)→2SO2(g)+2H2O(g)
17
Solution
Step 1: Write the given standard enthalpy change for the first reaction.
∆H1=−1036 kJ
Step 2: Determine the stoichiometry difference between the given reaction
and the desired reaction. In the given reaction, there are 3 moles of O2reacting.
In the desired reaction, there are only 2 moles of O2reacting. Therefore, to
relate the two reactions, we need to consider the enthalpy change for one mole
of O2reacting.
Step 3: Calculate the enthalpy change for the desired reaction using the
stoichiometry difference. Since the enthalpy change is an extensive property,
we can find the enthalpy change for the desired reaction by multiplying the
enthalpy change of the given reaction by the ratio of moles of O2in the desired
reaction to moles of O2in the given reaction.
∆H2=2
3×∆H1
∆H2=2
3×(−1036)
∆H2=−690.67 kJ
Therefore, the standard enthalpy change for the reaction
2H2S(g)+2O2(g)→2SO2(g)+2H2O(g)
is −690.67 kJ.
Question 23
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H4(g)+ 3O2(g)→4CO2(g)+ 4H2O(l)
given the following standard enthalpy of formation values:
∆H◦
f(C2H4(g)) = 52.5 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
18
Solution
Step 1: Write the given reaction in terms of the standard enthalpies of formation.
2∆H◦
f(C2H4(g)) + 3∆H◦
f(O2(g))→4∆H◦
f(CO2(g)) + 4∆H◦
f(H2O(l))
Step 2: Substitute the given values:
2(52.5) + 3(0) →4(−393.5) + 4(−285.8)
Step 3: Calculate the total change in enthalpy:
105 + 0 → −1574 −1143.2
105 → −2717.2
Therefore, the enthalpy change for the reaction is ∆H=−2712.2 kJ.
Question 24
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−484 kJ
What is the standard enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
Solution
Step 1: Write out the target reaction in terms of the given reaction:
H2O(l)→2H2(g)+O2(g)
Step 2: Use the given reaction to find the enthalpy change:
∆H=−484 kJ
Since we want H2O(l)→2H2(g) + O2(g), we need to flip the given reaction:
−2H2O(l)→2H2(g)+O2(g) ∆H= 484 kJ
Step 3: Divide the enthalpy change by 2 to get the desired reaction:
−484
2=−242 kJ
Therefore, the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
is −242 kJ .
19
Question 25
Question
Given the following information:
The standard enthalpy of formation of methane gas (CH4) is −74.8 kJ/mol.
The standard enthalpy of combustion of methane gas is −890 kJ/mol.
Calculate the standard enthalpy of the following reaction:
2CH4(g)+3O2(g)→2CO2(g)+4H2O(l)
Solution
Step 1: Write the given and required equations
CH4(g)→CO2(g)+2H2O(l) (Given)
2CH4(g)+3O2(g)→2CO2(g)+4H2O(l) (Required)
Step 2: Determine the enthalpy change for the given equation The enthalpy
change for the given equation is the sum of the standard enthalpies of formation
of the products minus the sum of the standard enthalpies of formation of the
reactants:
∆H1= [2 ×∆Hf(CO2)+2×∆Hf(H2O)] −[∆Hf(CH4)]
Substitute the given values:
∆H1= [2×∆Hf(CO2)+2×∆Hf(H2O)]−[∆Hf(CH4)] = [2×(−393.5 kJ/mol)+2×(−285.8 kJ/mol)]−(−74.8 kJ/mol)
∆H1= [−787 kJ/mol −571.6 kJ/mol] + 74.8 kJ/mol = −1383.6 kJ/mol
Step 3: Modify the given equation as needed To obtain the required equation,
we need to double the given equation.
2×(CH4(g)→CO2(g)+2H2O(l)) (Given)
Step 4: Determine the enthalpy change for the required equation The en-
thalpy change for the required equation is twice the enthalpy change for the
given equation:
∆H= 2 ×∆H1= 2 ×(−1383.6 kJ/mol) = −2767.2 kJ/mol
Therefore, the standard enthalpy of the reaction 2CH4(g)+3O2(g)→
2CO2(g)+4H2O(l) is −2767.2 kJ/mol .
20
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2(g) + O2(g)→2H2O(l)
given the following standard enthalpies of formation: H2(g) = 0 kJ/mol, O2(g) =
0 kJ/mol, H2O(l) = −285.8 kJ/mol.
Solution
Step 1: Write out the balanced chemical equation and standard enthalpy change
formula:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
Where nand mare the stoichiometric coefficients in the balanced chemical
equation.
Step 2: Substitute the standard enthalpies of formation into the formula:
∆H◦= 2(−285.8) −[2(0) + 0] = −571.6 kJ/mol
Step 3: Calculate the standard enthalpy change for the reaction:
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction 2H2(g) + O2(g)→
2H2O(l) is ∆H◦=−571.6 kJ/mol.
Question 27
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite)+2H2(g)→C2H4(g)
given the following bond dissociation energies: C−Cbond = 347 kJ/mol, C−H
bond = 413 kJ/mol, and H−Hbond = 436 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Bonds broken = 2(C−C) + 4(C−H) + 2(H−H)
= 2(347) + 4(413) + 2(436)
= 694 + 1652 + 872
= 3218 kJ
21
Step 2: Calculate the total energy released when the bonds in the products
are formed.
Bonds formed = 1(C=C) + 4(C−H)
= 1(612) + 4(413)
= 612 + 1652
= 2264 kJ
Step 3: Calculate the ∆Hfor the reaction.
∆H= Energy required to break bonds −Energy released from forming bonds
= 3218 kJ −2264 kJ
= 954 kJ
Answer: The enthalpy change for the reaction is 954 kJ (endothermic).
Question 28
Question
Calculate the enthalpy change (∆H) when 8.00 g of methane (CH4) is burned
in excess oxygen according to the reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the following molar enthalpies of formation:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the moles of methane burned. Given: Mass of methane,
m= 8.00 g Molar mass of methane, MCH4= 16.05 g/mol (from periodic table)
Using the formula:
moles of CH4=m
MCH4
moles of CH4=8.00 g
16.05 g/mol = 0.498 mol
Step 2: Determine the enthalpy change when 0.498 mol of methane is burned.
Given:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
22
∆H◦
f(H2O) = −285.8 kJ/mol
From the balanced chemical equation, the enthalpy change can be calculated
as:
∆H=n·∆Hproducts −m·∆Hreactants
Where nand mare the stoichiometric coefficients for the products and reactants,
respectively.
Substitute the values:
∆H= (1)(−393.5) + 2(−285.8) −(1)(−74.8)
∆H=−393.5−571.6 + 74.8
∆H=−890.3 kJ
Therefore, the enthalpy change when 8.00 g of methane is burned is −890.3 kJ .
Question 29
Question
Consider the following reaction:
2H2O2(l)→2H2O(l)+O2(g)
Given the standard enthalpies of formation:
∆H◦
f(H2O2(l)) = −196.2 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
is:
2H2O2(l)→2H2O(l)+O2(g)
Step 2: Determine the standard enthalpy change for the reaction using stan-
dard enthalpies of formation. The standard enthalpy change for the reaction
can be calculated using the formula:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of the products and reactants,
respectively, in the balanced chemical equation.
23
Substitute the given values into the formula:
∆H◦= [2(−285.8) + 0] −[2(−196.2)]
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= [−571.6] −[−392.4]
∆H◦=−571.6 + 392.4
∆H◦=−179.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−179.2 kJ/mol.
Question 30
Question
Calculate the enthalpy change for the reaction:
2C(graphite)+3H2(g)→C2H6(g)
Given the following bond dissociation energies:
C−C= 347 kcal/mol, C −H= 99 kcal/mol, H −H= 104 kcal/mol
Solution
Step 1: Calculate the total bond energy of the reactants. The total bond energy
of the reactants can be calculated as follows:
2(C−C) + 3(C−H) + 3(H−H)
Step 2: Substitute the given bond dissociation energies to find the total bond
energy of the reactants.
2(347) + 3(99) + 3(104) = 694 + 297 + 312 = 1303 kcal/mol
Step 3: Calculate the total bond energy of the products. The total bond
energy of the products is:
C−C= 0 kcal/mol (1 mol of C-C bonds broken), C−H= 0 kcal/mol (6 mol of C-H bonds formed)
Step 4: Calculate the enthalpy change. The enthalpy change can be calcu-
lated as the difference between the total bond energy of the reactants and the
total bond energy of the products:
∆H= Total bond energy of reactants −Total bond energy of products
∆H= 1303 kcal/mol −(0 kcal/mol + 0 kcal/mol) = 1303 kcal/mol
Therefore, the enthalpy change for the given reaction is 1303 kcal/mol .
24
Question 2
Question
Consider the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following enthalpy changes:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction.
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpy of formation values given.
∆H◦=Xn∆H◦
f products −Xm∆H◦
f reactants
Step 2: Substitute the given values of standard enthalpies of formation into
the formula:
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
Step 3: Perform the calculation to find the standard enthalpy change of the
reaction:
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction is
-571.6 kJ/mol.
Question 3
Question
A student is investigating the combustion of ethene gas (C2H4) and measures the
enthalpy change for the reaction to be −1411 kJ/mol. Write the balanced chem-
ical equation for the combustion of ethene and calculate the enthalpy change
when 3.00 moles of ethene react.
2
Solution
Step 1: Write the balanced chemical equation for the combustion of ethene:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Step 2: Calculate the enthalpy change for the combustion of 1 mole of
ethene: From the balanced equation, 1 mole of ethene produces −1411 kJ of
heat. Therefore, the enthalpy change for the combustion of 1 mole of ethene is
−1411 kJ/mol.
Step 3: Calculate the enthalpy change when 3.00 moles of ethene react:
Since the enthalpy change for 1 mole of ethene is −1411 kJ/mol, the enthalpy
change when 3.00 moles of ethene react is:
−1411 kJ/mol ×3.00 mol = −4233 kJ
Therefore, when 3.00 moles of ethene react, the enthalpy change is −4233
kJ.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2NOCl(g)→2NO(g) + Cl2(g)
given the following standard enthalpy change values:
∆H◦
f(NOCl(g)) = 51.0 kJ/mol
∆H◦
f(NO(g)) = 90.3 kJ/mol
∆H◦
f(Cl2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation is:
2NOCl(g)→2NO(g) + Cl2(g)
Step 2: Calculate the standard enthalpy change for the reaction using stan-
dard enthalpies of formation.
The standard enthalpy change for a reaction can be calculated using the
equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
3
Given:
∆H◦
f(NOCl(g)) = 51.0 kJ/mol
∆H◦
f(NO(g)) = 90.3 kJ/mol
∆H◦
f(Cl2(g)) = 0 kJ/mol
Plugging in the values:
∆H◦= 2 ×∆H◦
f(NO(g)) + ∆H◦
f(Cl2(g)) −2×∆H◦
f(NOCl(g))
= 2 ×90.3 kJ/mol + 0 kJ/mol −2×51.0 kJ/mol
= 180.6 kJ/mol −102.0 kJ/mol
= 78.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is 78.6 kJ/mol.
Question 5
Question
The combustion of ethylene (C2H4) releases 1411.03 kJ/mol of heat. Calculate
the standard enthalpy change for the combustion of ethylene to form CO2and
H2O.
Solution
Step 1: Write the balanced chemical equation for the combustion of ethylene:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Step 2: Determine the moles of ethylene that react to release 1411.03 kJ of
heat. Since the reaction releases 1411.03 kJ/mol, the heat released by 1 mole
of ethylene is 1411.03 kJ. Therefore, the heat released by x moles of ethylene is
1411.03 kJ/mol ×xmol.
Step 3: Calculate the heat of combustion of ethylene. The heat of combus-
tion of ethylene is the same as the heat released by the reaction. Given that
1411.03 kJ is released by 1 mole of ethylene, the heat of combustion is 1411.03
kJ/mol.
Therefore, the standard enthalpy change for the combustion of ethylene to
form CO2and H2Ois 1411.03 kJ/mol.
4
Question 6
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2C(graphite)+3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
C(graphite) = 0 kcal/mol, H2(g) = 0 kcal/mol, C2H6(g) = −20.2 kcal/mol
Solution
Step 1: Write the balanced chemical equation and determine the standard en-
thalpy change. The standard enthalpy change for a reaction can be calculated
by taking the sum of the standard enthalpies of formation of the products and
subtracting the sum of the standard enthalpies of formation of the reactants.
The balanced chemical equation for the reaction is:
2C(graphite)+3H2(g)→C2H6(g)
The standard enthalpy change (∆H◦) can be calculated as:
∆H◦=XProducts −XReactants
∆H◦= [1×Standard enthalpy of formation of C2H6(g)]−[2×Standard enthalpy of formation of C(graphite)+3×Standard enthalpy of formation of H2(g)]
Step 2: Substitute the given values into the equation.
∆H◦= [−20.2] −[2(0) + 3(0)]
∆H◦=−20.2 kcal/mol
Therefore, the standard enthalpy change for the given reaction is −20.2 kcal/mol .
Question 7
Question
Consider the following reaction:
2A(g) + B(g)→C(g)+3D(g)
Given the following data:
∆H◦
f(kJ/mol)
A(g) : 50, B(g) : 100, C(g) : −200, D(g) : −75
Calculate the standard enthalpy change, ∆H◦, for the reaction.
5
Solution
Step 1: Calculate the standard enthalpy change, ∆H◦, for the reaction using
the standard enthalpies of formation for each substance.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion and calculate.
∆H◦= [(∆H◦
fof C(g))+3(∆H◦
fof D(g))]−[2(∆H◦
fof A(g))+(∆H◦
fof B(g))]
∆H◦= [(−200) + 3(−75)] −[2(50) + 100]
∆H◦= (−200 −225) −(100 + 100)
∆H◦=−425 −200
∆H◦=−625 kJ/mol
Therefore, the standard enthalpy change, ∆H◦, for the reaction is -625
kJ/mol.
Question 8
Question
Given the following reaction:
2A(g)+3B(g)→C(g)+2D(g)
If the standard enthalpy change of formation for compounds A,B,C, and D
are 50 kJ/mol, 30 kJ/mol, -80 kJ/mol, and -100 kJ/mol, respectively, calculate
the standard enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy change of formation for each compound. The standard enthalpy
change for the reaction can be calculated using the formula:
∆H◦=Xν∆H◦
f, products −Xν∆H◦
f, reactants
where ∆H◦is the standard enthalpy change for the reaction, νis the stoichio-
metric coefficient, and ∆H◦
fis the standard enthalpy change of formation.
For the given reaction:
2A(g)+3B(g)→C(g)+2D(g)
6
Plugging in the values of the standard enthalpy change of formation:
∆H◦= (1)(−80 kJ/mol) + (2)(−100 kJ/mol) −(2)(50 kJ/mol) −(3)(30 kJ/mol)
∆H◦=−80 kJ/mol −200 kJ/mol −100 kJ/mol −90 kJ/mol
∆H◦=−470 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -470 kJ/mol.
Question 9
Question
Given the standard enthalpy of combustion of benzene, C6H6(l), is −3267 kJ/mol,
calculate the standard enthalpy of formation of benzene.
Solution
Step 1: Write the balanced chemical equation for the combustion of one mole
of benzene.
The balanced chemical equation for the combustion of benzene is:
C6H6(l) + 15/2O2(g)→6CO2(g)+3H2O(l)
Step 2: Use Hess’s law to relate the standard enthalpy of combustion with
the standard enthalpies of formation.
The standard enthalpy change of combustion is given by:
∆Hcomb =Xνproducts∆H◦
f−Xνreactants∆H◦
f
where νrepresents the stoichiometric coefficients in the balanced equation.
For the combustion of benzene:
−3267 kJ/mol = 6 ×(−393.5) + 3 ×(−285.8) −∆H◦
f
Step 3: Solve for the standard enthalpy of formation of benzene.
−3267 kJ/mol = −2361 −857.4−∆H◦
f
−3267 kJ/mol = −3218.4−∆H◦
f
∆H◦
f= 49.4 kJ/mol
Therefore, the standard enthalpy of formation of benzene is 49.4 kJ/mol.
7
Question 10
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(g)
If the standard enthalpy change for the reaction is −483.6 kJ/mol, calculate
the standard enthalpy change for the reaction:
H2O(g)→2H(g)+O2(g)
Solution
Step 1: Write the given reaction as a sum of desired reactions.
The given reaction is:
−483.6 kJ/mol = 2H2(g)+O2(g)→2H2O(g)
To find the standard enthalpy change for the desired reaction, the given
reaction can be written as a sum of two reactions as follows:
H2O(g)→H2(g) + 1
2O2(g)
H2(g) + 1
2O2(g)→2H(g)+O2(g)
Step 2: Use Hess’s Law to calculate the standard enthalpy change for the
desired reaction.
Since enthalpy is a state function, the overall enthalpy change for a reaction
is the sum of the enthalpy changes of the individual reactions it can be broken
down into:
∆Hdesired = ∆H1+ ∆H2
Therefore, we need to add the enthalpy changes of the two desired reactions
to find the enthalpy change of the overall desired reaction.
Step 3: Calculate the standard enthalpy change for the desired reaction.
The enthalpy change for the first desired reaction (H2O(g)→H2(g) +
1
2O2(g)) can be calculated as follows:
∆H1=−483.6 kJ/mol
The enthalpy change for the second desired reaction (H2(g) + 1
2O2(g)→
2H(g)+O2(g)) is zero because it is the reverse of the given reaction.
Therefore, the standard enthalpy change for the overall desired reaction is:
∆Hdesired =−483.6 kJ/mol
8
Question 11
Question
Consider the reaction:
2A(g) + B(g)→3C(g)+4D(g)
where ∆H◦
ffor A,B,C, and Dare −200 kJ/mol, 100 kJ/mol, 150 kJ/mol,
and 50 kJ/mol, respectively. If the enthalpy change for the reaction is −500 kJ,
calculate the standard enthalpy of formation for compound B.
Solution
Step 1: Calculate the total change in enthalpy using the standard enthalpy of
formation values provided.
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
−500 kJ = (3)(150 kJ/mol + 4)(50 kJ/mol) −(2)(−200 kJ/mol) −(1)(100 kJ/mol)
−500 kJ = 450 kJ + 200 kJ + 200 kJ + 100 kJ
−500 kJ = 950 kJ −200 kJ
−500 kJ = 750 kJ
Step 2: Solve for the standard enthalpy of formation of compound B.
−500 kJ = 750 kJ −200 kJ −100 kJ + xkJ
−500 kJ = 450 kJ + xkJ
x=−950 kJ
Therefore, the standard enthalpy of formation for compound Bis −950
kJ/mol.
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the reaction below using
standard enthalpies of formation.
2C(s) + 3H2(g)→C2H6(g)
Given the standard enthalpies of formation:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
9
Solution
Step 1: Write the given chemical equation and balance it if needed:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation: The standard enthalpy change can be calculated using
the standard enthalpies of formation of the products and reactants.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
For the given reaction:
∆H◦= ∆H◦
f[C2H6(g)] −[∆H◦
f[C(s)] + ∆H◦
f[H2(g)]]
Substitute the values:
∆H◦=−84.68 kJ/mol −[0 + 0]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol .
Question 13
Question
Calculate the enthalpy change for the reaction below using standard enthalpies
of formation:
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)) = −84.7 kJ/mol
Solution
Step 1: Write down the balanced chemical equation and identify the standard
enthalpies of formation for each compound involved. The balanced chemical
equation is:
2C(graphite)+3H2(g)→C2H6(g)
10
The standard enthalpies of formation for each compound are:
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.7 kJ/mol
Step 2: Calculate the enthalpy change for the reaction using the standard
enthalpies of formation. The enthalpy change (∆H) for the reaction can be
calculated as follows:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H= ∆H◦
f(C2H6(g)) −[∆H◦
f(2C(graphite)) + ∆H◦
f(3H2(g))]
∆H=−84.7 kJ/mol −[0 + 0]
∆H=−84.7 kJ/mol
Therefore, the enthalpy change for the reaction is −84.7 kJ/mol .
Question 14
Question
Calculate the standard enthalpy change for the reaction:
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
Solution
Step 1: Write the balanced chemical equation for the reaction and calculate the
standard enthalpy change.
C(graphite) + H2(g) →C2H6(g)
Given ∆H◦
fvalues:
∆H◦=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
11
∆H◦= [∆H◦
f(C2H6(g))] −[∆H◦
f(C(graphite)) + 3∆H◦
f(H2(g))]
∆H◦= [−84.68] −[0 + 3(0)] = −84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol.
Question 15
Question
A reaction has a standard enthalpy change of ∆H◦=−395 kJ/mol. Calculate
the amount of heat (in joules) released when 5.00 grams of this reaction occurs.
Solution
Step 1: Calculate the molar mass of the reaction. The molar mass of the reaction
can be calculated as follows:
Molar mass = Mass
Moles
Given that the mass is 5.00 grams, and assuming the reaction is the limiting
reactant, we can calculate the number of moles in 5.00 grams using the molar
mass of the reaction.
Step 2: Calculate the heat released per mole. We know that the standard
enthalpy change is -395 kJ/mol. This means that -395 kJ of heat is released for
every 1 mole of the reaction that occurs.
Step 3: Calculate the heat released for 5.00 grams of the reaction. Now that
we have the amount of heat released per mole, we can calculate the amount of
heat released for 5.00 grams of the reaction using the number of moles calculated
in Step 1.
Step 4: Perform the calculations and convert the units. Finally, we convert
the heat released from kJ to J to get the final answer.
Given: ∆H◦=−395 kJ/mol,Mass = 5.00 g
Step 1:
Molar mass = 5.00 g
moles
The molar mass of the reaction will depend on the specific reaction provided.
Step 2: Heat released per mole = -395 kJ/mol
Step 3: Amount of heat released for 5.00 g = (Amount of heat released per
mole) ×(Moles in 5.00 g)
Step 4: Convert the units to get the final answer.
12
Question 16
Question
Calculate the enthalpy change for the reaction below, given the following bond
dissociation energies:
2C2H4(g)+3O2(g)→4CO2(g)+4H2O(g)
Bond dissociation energies: - C-C: 348 kJ/mol - C=C: 611 kJ/mol - O=O:
498 kJ/mol - C=O: 805 kJ/mol - O-H: 463 kJ/mol - C-H: 413 kJ/mol
Solution
Step 1: Calculate the total bond energy of reactants.
2(4 ×C-C + 4 ×C-H) + 3(2 ×O=O)
= 2(4 ×348 + 4 ×413) + 3(2 ×498)
= 2(1392 + 1652) + 3(996)
= 4088 + 2994
= 7082 kJ
Step 2: Calculate the total bond energy of products.
4(4 ×C=O + 4 ×O-H) = 4(4 ×805 + 4 ×463)
= 4(3220 + 1852)
= 4(5072)
= 20288 kJ
Step 3: Calculate the enthalpy change.
∆H= (Total bond energy of reactants) −(Total bond energy of products)
= 7082 −20288
=−13206 kJ
Therefore, the enthalpy change for the reaction is ∆H=−13206 kJ.
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(s)+3H2(g)→C2H6(g)
13
given the following standard enthalpy of formations:
∆H◦
fof C(s) = 0 kJ/mol
∆H◦
fof H2(g) = 0 kJ/mol
∆H◦
fof C2H6(g) = −84.68 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard enthalpy
of formation values. Step 2: Use the standard enthalpy change of formation
formula to calculate the standard enthalpy change for the given reaction.
Step 1: The balanced chemical equation is:
2C(s)+3H2(g)→C2H6(g)
Given:
∆H◦
fof C(s) = 0 kJ/mol
∆H◦
fof H2(g) = 0 kJ/mol
∆H◦
fof C2H6(g) = −84.68 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦):
∆H◦=Xn∆H◦
fof products −Xm∆H◦
fof reactants
∆H◦= (1 × −84.68) −(2 ×0+3×0)
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −84.68 kJ/mol .
Question 18
Question
Consider the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−483.7 kJ
Calculate the enthalpy change for the reaction
N2(g) + 3H2(g)→2NH3(g)
given the following bond dissociation energies:
N-N = 945 kJ/mol,N-H = 389 kJ/mol,H-H = 432 kJ/mol
14
Solution
Step 1: Calculate the bond dissociation energy of the reactants: N2(g) = 1 ×
(N-N) = 1 ×945 = 945 kJ/mol
H2(g)=3×(H-H) = 3 ×432 = 1296 kJ/mol
Total energy of reactants = 945 kJ/mol + 1296 kJ/mol = 2241 kJ/mol
Step 2: Calculate the bond dissociation energy of the products: NH3(g) =
2×(N-H) + 6 ×(H-H) = 2 ×389 + 6 ×432 = 2574 kJ/mol
Total energy of products = 2574 kJ/mol
Step 3: Calculate the enthalpy change for the reaction: ∆H= Energy of bonds broken−
Energy of bonds formed
∆H= 2241 kJ/mol −2574 kJ/mol = −333 kJ/mol
Therefore, the enthalpy change for the reaction N2(g) + 3H2(g)→2NH3(g)
is -333 kJ/mol.
Question 19
Question
A reaction has a standard enthalpy change of -220 kJ/mol and a standard
entropy change of 120 J/(mol
·
K). Calculate the standard Gibbs free energy
change at 298 K.
Solution
Step 1: Recall the relationship between Gibbs free energy (∆G), enthalpy change
(∆H), and entropy change (∆S) at constant temperature (T):
∆G= ∆H−T·∆S
Step 2: Convert the entropy change to kJ/(mol
·
K) to match the units of
enthalpy change:
∆S= 120 J/(mol
·
K) = 0.120 kJ/(mol
·
K)
Step 3: Substitute the values into the equation:
∆G=−220 kJ/mol −(298 K) ·0.120 kJ/(mol
·
K)
Step 4: Calculate the standard Gibbs free energy change:
∆G=−220 kJ/mol −35.76 kJ/mol
∆G=−255.76 kJ/mol
Therefore, the standard Gibbs free energy change at 298 K is -255.76 kJ/mol.
15
Question 20
Question
Given the following reaction:
2A(g)+3B(g)→C(g)+4D(g)
The enthalpy change for the reaction is -3050 kJ. If the enthalpy change of
formation for compounds C and D are -240 kJ/mol and -126 kJ/mol, respec-
tively, calculate the enthalpy change of formation for compound A.
Solution
Step 1: Write out the enthalpy change of formation for compound C and D in
terms of the given reaction.
∆Hf(C) = −240 kJ/mol
∆Hf(D) = −126 kJ/mol
Step 2: Apply Hess’s Law to find the enthalpy change of formation for com-
pound A. The enthalpy change of formation for a compound is the enthalpy
change when 1 mole of the compound is formed from its elements in their stan-
dard states.
Step 3: Express the enthalpy change of the given reaction in terms of the
enthalpy change of formation for compounds A, B, C, and D.
∆Hrxn =Xn∆Hf(products) −Xm∆Hf(reactants)
∆Hrxn =Xn∆Hf(products) −Xm∆Hf(reactants)
∆Hrxn = ∆Hf(C) + 4∆Hf(D)−(2∆Hf(A) + 3∆Hf(B))
∆Hrxn =−240 + 4(−126) −(2∆Hf(A) + 3(0))
Step 4: Substitute the given enthalpy change of the reaction and solve for
the enthalpy change of formation for compound A.
−3050 = −240 −504 −2∆Hf(A)
−3050 = −744 −2∆Hf(A)
−2306 = −2∆Hf(A)
∆Hf(A) = 1153 kJ/mol
Therefore, the enthalpy change of formation for compound A is 1153 kJ/mol.
16
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2(g)+O2(g)→2H2O(l)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol,∆H◦
f(O2(g)) = 0 kJ/mol,∆H◦
f(H2O(l)) = −285.8 kJ/mol
Solution
Step 1: Write the balanced chemical equation with the given standard enthalpies
of formation.
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the values of the standard enthalpies of formation into
the equation.
∆H◦= [2∆H◦
f(H2O(l))] −[2∆H◦
f(H2(g))+∆H◦
f(O2(g))]
Step 4: Substitute the given values into the equation and calculate.
∆H◦= [2(−285.8)] −[2(0) + 0]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−571.6 kJ/mol.
Question 22
Question
Given the following reaction:
2H2S(g)+3O2(g)→2SO2(g)+2H2O(g)
If the standard enthalpy change for the reaction is -1036 kJ, calculate the
standard enthalpy change for the reaction below:
2H2S(g)+2O2(g)→2SO2(g)+2H2O(g)
17
Solution
Step 1: Write the given standard enthalpy change for the first reaction.
∆H1=−1036 kJ
Step 2: Determine the stoichiometry difference between the given reaction
and the desired reaction. In the given reaction, there are 3 moles of O2reacting.
In the desired reaction, there are only 2 moles of O2reacting. Therefore, to
relate the two reactions, we need to consider the enthalpy change for one mole
of O2reacting.
Step 3: Calculate the enthalpy change for the desired reaction using the
stoichiometry difference. Since the enthalpy change is an extensive property,
we can find the enthalpy change for the desired reaction by multiplying the
enthalpy change of the given reaction by the ratio of moles of O2in the desired
reaction to moles of O2in the given reaction.
∆H2=2
3×∆H1
∆H2=2
3×(−1036)
∆H2=−690.67 kJ
Therefore, the standard enthalpy change for the reaction
2H2S(g)+2O2(g)→2SO2(g)+2H2O(g)
is −690.67 kJ.
Question 23
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H4(g)+ 3O2(g)→4CO2(g)+ 4H2O(l)
given the following standard enthalpy of formation values:
∆H◦
f(C2H4(g)) = 52.5 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
18
Solution
Step 1: Write the given reaction in terms of the standard enthalpies of formation.
2∆H◦
f(C2H4(g)) + 3∆H◦
f(O2(g))→4∆H◦
f(CO2(g)) + 4∆H◦
f(H2O(l))
Step 2: Substitute the given values:
2(52.5) + 3(0) →4(−393.5) + 4(−285.8)
Step 3: Calculate the total change in enthalpy:
105 + 0 → −1574 −1143.2
105 → −2717.2
Therefore, the enthalpy change for the reaction is ∆H=−2712.2 kJ.
Question 24
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−484 kJ
What is the standard enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
Solution
Step 1: Write out the target reaction in terms of the given reaction:
H2O(l)→2H2(g)+O2(g)
Step 2: Use the given reaction to find the enthalpy change:
∆H=−484 kJ
Since we want H2O(l)→2H2(g) + O2(g), we need to flip the given reaction:
−2H2O(l)→2H2(g)+O2(g) ∆H= 484 kJ
Step 3: Divide the enthalpy change by 2 to get the desired reaction:
−484
2=−242 kJ
Therefore, the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
is −242 kJ .
19
Question 25
Question
Given the following information:
The standard enthalpy of formation of methane gas (CH4) is −74.8 kJ/mol.
The standard enthalpy of combustion of methane gas is −890 kJ/mol.
Calculate the standard enthalpy of the following reaction:
2CH4(g)+3O2(g)→2CO2(g)+4H2O(l)
Solution
Step 1: Write the given and required equations
CH4(g)→CO2(g)+2H2O(l) (Given)
2CH4(g)+3O2(g)→2CO2(g)+4H2O(l) (Required)
Step 2: Determine the enthalpy change for the given equation The enthalpy
change for the given equation is the sum of the standard enthalpies of formation
of the products minus the sum of the standard enthalpies of formation of the
reactants:
∆H1= [2 ×∆Hf(CO2)+2×∆Hf(H2O)] −[∆Hf(CH4)]
Substitute the given values:
∆H1= [2×∆Hf(CO2)+2×∆Hf(H2O)]−[∆Hf(CH4)] = [2×(−393.5 kJ/mol)+2×(−285.8 kJ/mol)]−(−74.8 kJ/mol)
∆H1= [−787 kJ/mol −571.6 kJ/mol] + 74.8 kJ/mol = −1383.6 kJ/mol
Step 3: Modify the given equation as needed To obtain the required equation,
we need to double the given equation.
2×(CH4(g)→CO2(g)+2H2O(l)) (Given)
Step 4: Determine the enthalpy change for the required equation The en-
thalpy change for the required equation is twice the enthalpy change for the
given equation:
∆H= 2 ×∆H1= 2 ×(−1383.6 kJ/mol) = −2767.2 kJ/mol
Therefore, the standard enthalpy of the reaction 2CH4(g)+3O2(g)→
2CO2(g)+4H2O(l) is −2767.2 kJ/mol .
20
Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2(g) + O2(g)→2H2O(l)
given the following standard enthalpies of formation: H2(g) = 0 kJ/mol, O2(g) =
0 kJ/mol, H2O(l) = −285.8 kJ/mol.
Solution
Step 1: Write out the balanced chemical equation and standard enthalpy change
formula:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
Where nand mare the stoichiometric coefficients in the balanced chemical
equation.
Step 2: Substitute the standard enthalpies of formation into the formula:
∆H◦= 2(−285.8) −[2(0) + 0] = −571.6 kJ/mol
Step 3: Calculate the standard enthalpy change for the reaction:
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction 2H2(g) + O2(g)→
2H2O(l) is ∆H◦=−571.6 kJ/mol.
Question 27
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite)+2H2(g)→C2H4(g)
given the following bond dissociation energies: C−Cbond = 347 kJ/mol, C−H
bond = 413 kJ/mol, and H−Hbond = 436 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Bonds broken = 2(C−C) + 4(C−H) + 2(H−H)
= 2(347) + 4(413) + 2(436)
= 694 + 1652 + 872
= 3218 kJ
21
Step 2: Calculate the total energy released when the bonds in the products
are formed.
Bonds formed = 1(C=C) + 4(C−H)
= 1(612) + 4(413)
= 612 + 1652
= 2264 kJ
Step 3: Calculate the ∆Hfor the reaction.
∆H= Energy required to break bonds −Energy released from forming bonds
= 3218 kJ −2264 kJ
= 954 kJ
Answer: The enthalpy change for the reaction is 954 kJ (endothermic).
Question 28
Question
Calculate the enthalpy change (∆H) when 8.00 g of methane (CH4) is burned
in excess oxygen according to the reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the following molar enthalpies of formation:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the moles of methane burned. Given: Mass of methane,
m= 8.00 g Molar mass of methane, MCH4= 16.05 g/mol (from periodic table)
Using the formula:
moles of CH4=m
MCH4
moles of CH4=8.00 g
16.05 g/mol = 0.498 mol
Step 2: Determine the enthalpy change when 0.498 mol of methane is burned.
Given:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
22
∆H◦
f(H2O) = −285.8 kJ/mol
From the balanced chemical equation, the enthalpy change can be calculated
as:
∆H=n·∆Hproducts −m·∆Hreactants
Where nand mare the stoichiometric coefficients for the products and reactants,
respectively.
Substitute the values:
∆H= (1)(−393.5) + 2(−285.8) −(1)(−74.8)
∆H=−393.5−571.6 + 74.8
∆H=−890.3 kJ
Therefore, the enthalpy change when 8.00 g of methane is burned is −890.3 kJ .
Question 29
Question
Consider the following reaction:
2H2O2(l)→2H2O(l)+O2(g)
Given the standard enthalpies of formation:
∆H◦
f(H2O2(l)) = −196.2 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
is:
2H2O2(l)→2H2O(l)+O2(g)
Step 2: Determine the standard enthalpy change for the reaction using stan-
dard enthalpies of formation. The standard enthalpy change for the reaction
can be calculated using the formula:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of the products and reactants,
respectively, in the balanced chemical equation.
23
Substitute the given values into the formula:
∆H◦= [2(−285.8) + 0] −[2(−196.2)]
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= [−571.6] −[−392.4]
∆H◦=−571.6 + 392.4
∆H◦=−179.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−179.2 kJ/mol.
Question 30
Question
Calculate the enthalpy change for the reaction:
2C(graphite)+3H2(g)→C2H6(g)
Given the following bond dissociation energies:
C−C= 347 kcal/mol, C −H= 99 kcal/mol, H −H= 104 kcal/mol
Solution
Step 1: Calculate the total bond energy of the reactants. The total bond energy
of the reactants can be calculated as follows:
2(C−C) + 3(C−H) + 3(H−H)
Step 2: Substitute the given bond dissociation energies to find the total bond
energy of the reactants.
2(347) + 3(99) + 3(104) = 694 + 297 + 312 = 1303 kcal/mol
Step 3: Calculate the total bond energy of the products. The total bond
energy of the products is:
C−C= 0 kcal/mol (1 mol of C-C bonds broken), C−H= 0 kcal/mol (6 mol of C-H bonds formed)
Step 4: Calculate the enthalpy change. The enthalpy change can be calcu-
lated as the difference between the total bond energy of the reactants and the
total bond energy of the products:
∆H= Total bond energy of reactants −Total bond energy of products
∆H= 1303 kcal/mol −(0 kcal/mol + 0 kcal/mol) = 1303 kcal/mol
Therefore, the enthalpy change for the given reaction is 1303 kcal/mol .
24
Question 2
Question
Consider the reaction:
2H2(g)+O2(g)→2H2O(l)
Given the following enthalpy changes:
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy change, ∆H◦, for the reaction.
Solution
Step 1: Calculate the standard enthalpy change (∆H◦) for the reaction using
the standard enthalpy of formation values given.
∆H◦=Xn∆H◦
f products −Xm∆H◦
f reactants
Step 2: Substitute the given values of standard enthalpies of formation into
the formula:
∆H◦= 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
Step 3: Perform the calculation to find the standard enthalpy change of the
reaction:
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction is
-571.6 kJ/mol.
Question 3
Question
A student is investigating the combustion of ethene gas (C2H4) and measures the
enthalpy change for the reaction to be −1411 kJ/mol. Write the balanced chem-
ical equation for the combustion of ethene and calculate the enthalpy change
when 3.00 moles of ethene react.
2
Solution
Step 1: Write the balanced chemical equation for the combustion of ethene:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Step 2: Calculate the enthalpy change for the combustion of 1 mole of
ethene: From the balanced equation, 1 mole of ethene produces −1411 kJ of
heat. Therefore, the enthalpy change for the combustion of 1 mole of ethene is
−1411 kJ/mol.
Step 3: Calculate the enthalpy change when 3.00 moles of ethene react:
Since the enthalpy change for 1 mole of ethene is −1411 kJ/mol, the enthalpy
change when 3.00 moles of ethene react is:
−1411 kJ/mol ×3.00 mol = −4233 kJ
Therefore, when 3.00 moles of ethene react, the enthalpy change is −4233
kJ.
Question 4
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2NOCl(g)→2NO(g) + Cl2(g)
given the following standard enthalpy change values:
∆H◦
f(NOCl(g)) = 51.0 kJ/mol
∆H◦
f(NO(g)) = 90.3 kJ/mol
∆H◦
f(Cl2(g)) = 0 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation is:
2NOCl(g)→2NO(g) + Cl2(g)
Step 2: Calculate the standard enthalpy change for the reaction using stan-
dard enthalpies of formation.
The standard enthalpy change for a reaction can be calculated using the
equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
3
Given:
∆H◦
f(NOCl(g)) = 51.0 kJ/mol
∆H◦
f(NO(g)) = 90.3 kJ/mol
∆H◦
f(Cl2(g)) = 0 kJ/mol
Plugging in the values:
∆H◦= 2 ×∆H◦
f(NO(g)) + ∆H◦
f(Cl2(g)) −2×∆H◦
f(NOCl(g))
= 2 ×90.3 kJ/mol + 0 kJ/mol −2×51.0 kJ/mol
= 180.6 kJ/mol −102.0 kJ/mol
= 78.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is 78.6 kJ/mol.
Question 5
Question
The combustion of ethylene (C2H4) releases 1411.03 kJ/mol of heat. Calculate
the standard enthalpy change for the combustion of ethylene to form CO2and
H2O.
Solution
Step 1: Write the balanced chemical equation for the combustion of ethylene:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Step 2: Determine the moles of ethylene that react to release 1411.03 kJ of
heat. Since the reaction releases 1411.03 kJ/mol, the heat released by 1 mole
of ethylene is 1411.03 kJ. Therefore, the heat released by x moles of ethylene is
1411.03 kJ/mol ×xmol.
Step 3: Calculate the heat of combustion of ethylene. The heat of combus-
tion of ethylene is the same as the heat released by the reaction. Given that
1411.03 kJ is released by 1 mole of ethylene, the heat of combustion is 1411.03
kJ/mol.
Therefore, the standard enthalpy change for the combustion of ethylene to
form CO2and H2Ois 1411.03 kJ/mol.
4
Question 6
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2C(graphite)+3H2(g)→C2H6(g)
Given the following standard enthalpies of formation:
C(graphite) = 0 kcal/mol, H2(g) = 0 kcal/mol, C2H6(g) = −20.2 kcal/mol
Solution
Step 1: Write the balanced chemical equation and determine the standard en-
thalpy change. The standard enthalpy change for a reaction can be calculated
by taking the sum of the standard enthalpies of formation of the products and
subtracting the sum of the standard enthalpies of formation of the reactants.
The balanced chemical equation for the reaction is:
2C(graphite)+3H2(g)→C2H6(g)
The standard enthalpy change (∆H◦) can be calculated as:
∆H◦=XProducts −XReactants
∆H◦= [1×Standard enthalpy of formation of C2H6(g)]−[2×Standard enthalpy of formation of C(graphite)+3×Standard enthalpy of formation of H2(g)]
Step 2: Substitute the given values into the equation.
∆H◦= [−20.2] −[2(0) + 3(0)]
∆H◦=−20.2 kcal/mol
Therefore, the standard enthalpy change for the given reaction is −20.2 kcal/mol .
Question 7
Question
Consider the following reaction:
2A(g) + B(g)→C(g)+3D(g)
Given the following data:
∆H◦
f(kJ/mol)
A(g) : 50, B(g) : 100, C(g) : −200, D(g) : −75
Calculate the standard enthalpy change, ∆H◦, for the reaction.
5
Solution
Step 1: Calculate the standard enthalpy change, ∆H◦, for the reaction using
the standard enthalpies of formation for each substance.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion and calculate.
∆H◦= [(∆H◦
fof C(g))+3(∆H◦
fof D(g))]−[2(∆H◦
fof A(g))+(∆H◦
fof B(g))]
∆H◦= [(−200) + 3(−75)] −[2(50) + 100]
∆H◦= (−200 −225) −(100 + 100)
∆H◦=−425 −200
∆H◦=−625 kJ/mol
Therefore, the standard enthalpy change, ∆H◦, for the reaction is -625
kJ/mol.
Question 8
Question
Given the following reaction:
2A(g)+3B(g)→C(g)+2D(g)
If the standard enthalpy change of formation for compounds A,B,C, and D
are 50 kJ/mol, 30 kJ/mol, -80 kJ/mol, and -100 kJ/mol, respectively, calculate
the standard enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using the stan-
dard enthalpy change of formation for each compound. The standard enthalpy
change for the reaction can be calculated using the formula:
∆H◦=Xν∆H◦
f, products −Xν∆H◦
f, reactants
where ∆H◦is the standard enthalpy change for the reaction, νis the stoichio-
metric coefficient, and ∆H◦
fis the standard enthalpy change of formation.
For the given reaction:
2A(g)+3B(g)→C(g)+2D(g)
6
Plugging in the values of the standard enthalpy change of formation:
∆H◦= (1)(−80 kJ/mol) + (2)(−100 kJ/mol) −(2)(50 kJ/mol) −(3)(30 kJ/mol)
∆H◦=−80 kJ/mol −200 kJ/mol −100 kJ/mol −90 kJ/mol
∆H◦=−470 kJ/mol
Therefore, the standard enthalpy change for the given reaction is -470 kJ/mol.
Question 9
Question
Given the standard enthalpy of combustion of benzene, C6H6(l), is −3267 kJ/mol,
calculate the standard enthalpy of formation of benzene.
Solution
Step 1: Write the balanced chemical equation for the combustion of one mole
of benzene.
The balanced chemical equation for the combustion of benzene is:
C6H6(l) + 15/2O2(g)→6CO2(g)+3H2O(l)
Step 2: Use Hess’s law to relate the standard enthalpy of combustion with
the standard enthalpies of formation.
The standard enthalpy change of combustion is given by:
∆Hcomb =Xνproducts∆H◦
f−Xνreactants∆H◦
f
where νrepresents the stoichiometric coefficients in the balanced equation.
For the combustion of benzene:
−3267 kJ/mol = 6 ×(−393.5) + 3 ×(−285.8) −∆H◦
f
Step 3: Solve for the standard enthalpy of formation of benzene.
−3267 kJ/mol = −2361 −857.4−∆H◦
f
−3267 kJ/mol = −3218.4−∆H◦
f
∆H◦
f= 49.4 kJ/mol
Therefore, the standard enthalpy of formation of benzene is 49.4 kJ/mol.
7
Question 10
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(g)
If the standard enthalpy change for the reaction is −483.6 kJ/mol, calculate
the standard enthalpy change for the reaction:
H2O(g)→2H(g)+O2(g)
Solution
Step 1: Write the given reaction as a sum of desired reactions.
The given reaction is:
−483.6 kJ/mol = 2H2(g)+O2(g)→2H2O(g)
To find the standard enthalpy change for the desired reaction, the given
reaction can be written as a sum of two reactions as follows:
H2O(g)→H2(g) + 1
2O2(g)
H2(g) + 1
2O2(g)→2H(g)+O2(g)
Step 2: Use Hess’s Law to calculate the standard enthalpy change for the
desired reaction.
Since enthalpy is a state function, the overall enthalpy change for a reaction
is the sum of the enthalpy changes of the individual reactions it can be broken
down into:
∆Hdesired = ∆H1+ ∆H2
Therefore, we need to add the enthalpy changes of the two desired reactions
to find the enthalpy change of the overall desired reaction.
Step 3: Calculate the standard enthalpy change for the desired reaction.
The enthalpy change for the first desired reaction (H2O(g)→H2(g) +
1
2O2(g)) can be calculated as follows:
∆H1=−483.6 kJ/mol
The enthalpy change for the second desired reaction (H2(g) + 1
2O2(g)→
2H(g)+O2(g)) is zero because it is the reverse of the given reaction.
Therefore, the standard enthalpy change for the overall desired reaction is:
∆Hdesired =−483.6 kJ/mol
8
Question 11
Question
Consider the reaction:
2A(g) + B(g)→3C(g)+4D(g)
where ∆H◦
ffor A,B,C, and Dare −200 kJ/mol, 100 kJ/mol, 150 kJ/mol,
and 50 kJ/mol, respectively. If the enthalpy change for the reaction is −500 kJ,
calculate the standard enthalpy of formation for compound B.
Solution
Step 1: Calculate the total change in enthalpy using the standard enthalpy of
formation values provided.
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
−500 kJ = (3)(150 kJ/mol + 4)(50 kJ/mol) −(2)(−200 kJ/mol) −(1)(100 kJ/mol)
−500 kJ = 450 kJ + 200 kJ + 200 kJ + 100 kJ
−500 kJ = 950 kJ −200 kJ
−500 kJ = 750 kJ
Step 2: Solve for the standard enthalpy of formation of compound B.
−500 kJ = 750 kJ −200 kJ −100 kJ + xkJ
−500 kJ = 450 kJ + xkJ
x=−950 kJ
Therefore, the standard enthalpy of formation for compound Bis −950
kJ/mol.
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the reaction below using
standard enthalpies of formation.
2C(s) + 3H2(g)→C2H6(g)
Given the standard enthalpies of formation:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[H2(g)] = 0 kJ/mol
∆H◦
f[C2H6(g)] = −84.68 kJ/mol
9
Solution
Step 1: Write the given chemical equation and balance it if needed:
2C(s) + 3H2(g)→C2H6(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation: The standard enthalpy change can be calculated using
the standard enthalpies of formation of the products and reactants.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
For the given reaction:
∆H◦= ∆H◦
f[C2H6(g)] −[∆H◦
f[C(s)] + ∆H◦
f[H2(g)]]
Substitute the values:
∆H◦=−84.68 kJ/mol −[0 + 0]
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol .
Question 13
Question
Calculate the enthalpy change for the reaction below using standard enthalpies
of formation:
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)) = −84.7 kJ/mol
Solution
Step 1: Write down the balanced chemical equation and identify the standard
enthalpies of formation for each compound involved. The balanced chemical
equation is:
2C(graphite)+3H2(g)→C2H6(g)
10
The standard enthalpies of formation for each compound are:
∆H◦
f(C(graphite)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
∆H◦
f(C2H6(g)) = −84.7 kJ/mol
Step 2: Calculate the enthalpy change for the reaction using the standard
enthalpies of formation. The enthalpy change (∆H) for the reaction can be
calculated as follows:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H= ∆H◦
f(C2H6(g)) −[∆H◦
f(2C(graphite)) + ∆H◦
f(3H2(g))]
∆H=−84.7 kJ/mol −[0 + 0]
∆H=−84.7 kJ/mol
Therefore, the enthalpy change for the reaction is −84.7 kJ/mol .
Question 14
Question
Calculate the standard enthalpy change for the reaction:
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
Solution
Step 1: Write the balanced chemical equation for the reaction and calculate the
standard enthalpy change.
C(graphite) + H2(g) →C2H6(g)
Given ∆H◦
fvalues:
∆H◦=Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
11
∆H◦= [∆H◦
f(C2H6(g))] −[∆H◦
f(C(graphite)) + 3∆H◦
f(H2(g))]
∆H◦= [−84.68] −[0 + 3(0)] = −84.68 kJ/mol
Therefore, the standard enthalpy change for the reaction is −84.68 kJ/mol.
Question 15
Question
A reaction has a standard enthalpy change of ∆H◦=−395 kJ/mol. Calculate
the amount of heat (in joules) released when 5.00 grams of this reaction occurs.
Solution
Step 1: Calculate the molar mass of the reaction. The molar mass of the reaction
can be calculated as follows:
Molar mass = Mass
Moles
Given that the mass is 5.00 grams, and assuming the reaction is the limiting
reactant, we can calculate the number of moles in 5.00 grams using the molar
mass of the reaction.
Step 2: Calculate the heat released per mole. We know that the standard
enthalpy change is -395 kJ/mol. This means that -395 kJ of heat is released for
every 1 mole of the reaction that occurs.
Step 3: Calculate the heat released for 5.00 grams of the reaction. Now that
we have the amount of heat released per mole, we can calculate the amount of
heat released for 5.00 grams of the reaction using the number of moles calculated
in Step 1.
Step 4: Perform the calculations and convert the units. Finally, we convert
the heat released from kJ to J to get the final answer.
Given: ∆H◦=−395 kJ/mol,Mass = 5.00 g
Step 1:
Molar mass = 5.00 g
moles
The molar mass of the reaction will depend on the specific reaction provided.
Step 2: Heat released per mole = -395 kJ/mol
Step 3: Amount of heat released for 5.00 g = (Amount of heat released per
mole) ×(Moles in 5.00 g)
Step 4: Convert the units to get the final answer.
12
Question 16
Question
Calculate the enthalpy change for the reaction below, given the following bond
dissociation energies:
2C2H4(g)+3O2(g)→4CO2(g)+4H2O(g)
Bond dissociation energies: - C-C: 348 kJ/mol - C=C: 611 kJ/mol - O=O:
498 kJ/mol - C=O: 805 kJ/mol - O-H: 463 kJ/mol - C-H: 413 kJ/mol
Solution
Step 1: Calculate the total bond energy of reactants.
2(4 ×C-C + 4 ×C-H) + 3(2 ×O=O)
= 2(4 ×348 + 4 ×413) + 3(2 ×498)
= 2(1392 + 1652) + 3(996)
= 4088 + 2994
= 7082 kJ
Step 2: Calculate the total bond energy of products.
4(4 ×C=O + 4 ×O-H) = 4(4 ×805 + 4 ×463)
= 4(3220 + 1852)
= 4(5072)
= 20288 kJ
Step 3: Calculate the enthalpy change.
∆H= (Total bond energy of reactants) −(Total bond energy of products)
= 7082 −20288
=−13206 kJ
Therefore, the enthalpy change for the reaction is ∆H=−13206 kJ.
Question 17
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(s)+3H2(g)→C2H6(g)
13
given the following standard enthalpy of formations:
∆H◦
fof C(s) = 0 kJ/mol
∆H◦
fof H2(g) = 0 kJ/mol
∆H◦
fof C2H6(g) = −84.68 kJ/mol
Solution
Step 1: Write the balanced chemical equation and identify the standard enthalpy
of formation values. Step 2: Use the standard enthalpy change of formation
formula to calculate the standard enthalpy change for the given reaction.
Step 1: The balanced chemical equation is:
2C(s)+3H2(g)→C2H6(g)
Given:
∆H◦
fof C(s) = 0 kJ/mol
∆H◦
fof H2(g) = 0 kJ/mol
∆H◦
fof C2H6(g) = −84.68 kJ/mol
Step 2: Calculate the standard enthalpy change (∆H◦):
∆H◦=Xn∆H◦
fof products −Xm∆H◦
fof reactants
∆H◦= (1 × −84.68) −(2 ×0+3×0)
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −84.68 kJ/mol .
Question 18
Question
Consider the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−483.7 kJ
Calculate the enthalpy change for the reaction
N2(g) + 3H2(g)→2NH3(g)
given the following bond dissociation energies:
N-N = 945 kJ/mol,N-H = 389 kJ/mol,H-H = 432 kJ/mol
14
Solution
Step 1: Calculate the bond dissociation energy of the reactants: N2(g) = 1 ×
(N-N) = 1 ×945 = 945 kJ/mol
H2(g)=3×(H-H) = 3 ×432 = 1296 kJ/mol
Total energy of reactants = 945 kJ/mol + 1296 kJ/mol = 2241 kJ/mol
Step 2: Calculate the bond dissociation energy of the products: NH3(g) =
2×(N-H) + 6 ×(H-H) = 2 ×389 + 6 ×432 = 2574 kJ/mol
Total energy of products = 2574 kJ/mol
Step 3: Calculate the enthalpy change for the reaction: ∆H= Energy of bonds broken−
Energy of bonds formed
∆H= 2241 kJ/mol −2574 kJ/mol = −333 kJ/mol
Therefore, the enthalpy change for the reaction N2(g) + 3H2(g)→2NH3(g)
is -333 kJ/mol.
Question 19
Question
A reaction has a standard enthalpy change of -220 kJ/mol and a standard
entropy change of 120 J/(mol
·
K). Calculate the standard Gibbs free energy
change at 298 K.
Solution
Step 1: Recall the relationship between Gibbs free energy (∆G), enthalpy change
(∆H), and entropy change (∆S) at constant temperature (T):
∆G= ∆H−T·∆S
Step 2: Convert the entropy change to kJ/(mol
·
K) to match the units of
enthalpy change:
∆S= 120 J/(mol
·
K) = 0.120 kJ/(mol
·
K)
Step 3: Substitute the values into the equation:
∆G=−220 kJ/mol −(298 K) ·0.120 kJ/(mol
·
K)
Step 4: Calculate the standard Gibbs free energy change:
∆G=−220 kJ/mol −35.76 kJ/mol
∆G=−255.76 kJ/mol
Therefore, the standard Gibbs free energy change at 298 K is -255.76 kJ/mol.
15
Question 20
Question
Given the following reaction:
2A(g)+3B(g)→C(g)+4D(g)
The enthalpy change for the reaction is -3050 kJ. If the enthalpy change of
formation for compounds C and D are -240 kJ/mol and -126 kJ/mol, respec-
tively, calculate the enthalpy change of formation for compound A.
Solution
Step 1: Write out the enthalpy change of formation for compound C and D in
terms of the given reaction.
∆Hf(C) = −240 kJ/mol
∆Hf(D) = −126 kJ/mol
Step 2: Apply Hess’s Law to find the enthalpy change of formation for com-
pound A. The enthalpy change of formation for a compound is the enthalpy
change when 1 mole of the compound is formed from its elements in their stan-
dard states.
Step 3: Express the enthalpy change of the given reaction in terms of the
enthalpy change of formation for compounds A, B, C, and D.
∆Hrxn =Xn∆Hf(products) −Xm∆Hf(reactants)
∆Hrxn =Xn∆Hf(products) −Xm∆Hf(reactants)
∆Hrxn = ∆Hf(C) + 4∆Hf(D)−(2∆Hf(A) + 3∆Hf(B))
∆Hrxn =−240 + 4(−126) −(2∆Hf(A) + 3(0))
Step 4: Substitute the given enthalpy change of the reaction and solve for
the enthalpy change of formation for compound A.
−3050 = −240 −504 −2∆Hf(A)
−3050 = −744 −2∆Hf(A)
−2306 = −2∆Hf(A)
∆Hf(A) = 1153 kJ/mol
Therefore, the enthalpy change of formation for compound A is 1153 kJ/mol.
16
Question 21
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2(g)+O2(g)→2H2O(l)
given the following standard enthalpies of formation:
∆H◦
f(H2(g)) = 0 kJ/mol,∆H◦
f(O2(g)) = 0 kJ/mol,∆H◦
f(H2O(l)) = −285.8 kJ/mol
Solution
Step 1: Write the balanced chemical equation with the given standard enthalpies
of formation.
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the values of the standard enthalpies of formation into
the equation.
∆H◦= [2∆H◦
f(H2O(l))] −[2∆H◦
f(H2(g))+∆H◦
f(O2(g))]
Step 4: Substitute the given values into the equation and calculate.
∆H◦= [2(−285.8)] −[2(0) + 0]
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−571.6 kJ/mol.
Question 22
Question
Given the following reaction:
2H2S(g)+3O2(g)→2SO2(g)+2H2O(g)
If the standard enthalpy change for the reaction is -1036 kJ, calculate the
standard enthalpy change for the reaction below:
2H2S(g)+2O2(g)→2SO2(g)+2H2O(g)
17
Solution
Step 1: Write the given standard enthalpy change for the first reaction.
∆H1=−1036 kJ
Step 2: Determine the stoichiometry difference between the given reaction
and the desired reaction. In the given reaction, there are 3 moles of O2reacting.
In the desired reaction, there are only 2 moles of O2reacting. Therefore, to
relate the two reactions, we need to consider the enthalpy change for one mole
of O2reacting.
Step 3: Calculate the enthalpy change for the desired reaction using the
stoichiometry difference. Since the enthalpy change is an extensive property,
we can find the enthalpy change for the desired reaction by multiplying the
enthalpy change of the given reaction by the ratio of moles of O2in the desired
reaction to moles of O2in the given reaction.
∆H2=2
3×∆H1
∆H2=2
3×(−1036)
∆H2=−690.67 kJ
Therefore, the standard enthalpy change for the reaction
2H2S(g)+2O2(g)→2SO2(g)+2H2O(g)
is −690.67 kJ.
Question 23
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H4(g)+ 3O2(g)→4CO2(g)+ 4H2O(l)
given the following standard enthalpy of formation values:
∆H◦
f(C2H4(g)) = 52.5 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
18
Solution
Step 1: Write the given reaction in terms of the standard enthalpies of formation.
2∆H◦
f(C2H4(g)) + 3∆H◦
f(O2(g))→4∆H◦
f(CO2(g)) + 4∆H◦
f(H2O(l))
Step 2: Substitute the given values:
2(52.5) + 3(0) →4(−393.5) + 4(−285.8)
Step 3: Calculate the total change in enthalpy:
105 + 0 → −1574 −1143.2
105 → −2717.2
Therefore, the enthalpy change for the reaction is ∆H=−2712.2 kJ.
Question 24
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−484 kJ
What is the standard enthalpy change for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
Solution
Step 1: Write out the target reaction in terms of the given reaction:
H2O(l)→2H2(g)+O2(g)
Step 2: Use the given reaction to find the enthalpy change:
∆H=−484 kJ
Since we want H2O(l)→2H2(g) + O2(g), we need to flip the given reaction:
−2H2O(l)→2H2(g)+O2(g) ∆H= 484 kJ
Step 3: Divide the enthalpy change by 2 to get the desired reaction:
−484
2=−242 kJ
Therefore, the standard enthalpy change for the reaction
H2O(l)→H2(g) + 1
2O2(g)
is −242 kJ .
19
Question 25
Question
Given the following information:
The standard enthalpy of formation of methane gas (CH4) is −74.8 kJ/mol.
The standard enthalpy of combustion of methane gas is −890 kJ/mol.
Calculate the standard enthalpy of the following reaction:
2CH4(g)+3O2(g)→2CO2(g)+4H2O(l)
Solution
Step 1: Write the given and required equations
CH4(g)→CO2(g)+2H2O(l) (Given)
2CH4(g)+3O2(g)→2CO2(g)+4H2O(l) (Required)
Step 2: Determine the enthalpy change for the given equation The enthalpy
change for the given equation is the sum of the standard enthalpies of formation
of the products minus the sum of the standard enthalpies of formation of the
reactants:
∆H1= [2 ×∆Hf(CO2)+2×∆Hf(H2O)] −[∆Hf(CH4)]
Substitute the given values:
∆H1= [2×∆Hf(CO2)+2×∆Hf(H2O)]−[∆Hf(CH4)] = [2×(−393.5 kJ/mol)+2×(−285.8 kJ/mol)]−(−74.8 kJ/mol)
∆H1= [−787 kJ/mol −571.6 kJ/mol] + 74.8 kJ/mol = −1383.6 kJ/mol
Step 3: Modify the given equation as needed To obtain the required equation,
we need to double the given equation.
2×(CH4(g)→CO2(g)+2H2O(l)) (Given)
Step 4: Determine the enthalpy change for the required equation The en-
thalpy change for the required equation is twice the enthalpy change for the
given equation:
∆H= 2 ×∆H1= 2 ×(−1383.6 kJ/mol) = −2767.2 kJ/mol
Therefore, the standard enthalpy of the reaction 2CH4(g)+3O2(g)→
2CO2(g)+4H2O(l) is −2767.2 kJ/mol .
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Question 26
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2H2(g) + O2(g)→2H2O(l)
given the following standard enthalpies of formation: H2(g) = 0 kJ/mol, O2(g) =
0 kJ/mol, H2O(l) = −285.8 kJ/mol.
Solution
Step 1: Write out the balanced chemical equation and standard enthalpy change
formula:
∆H◦=Xn∆H◦
products −Xm∆H◦
reactants
Where nand mare the stoichiometric coefficients in the balanced chemical
equation.
Step 2: Substitute the standard enthalpies of formation into the formula:
∆H◦= 2(−285.8) −[2(0) + 0] = −571.6 kJ/mol
Step 3: Calculate the standard enthalpy change for the reaction:
∆H◦=−571.6 kJ/mol
Therefore, the standard enthalpy change for the reaction 2H2(g) + O2(g)→
2H2O(l) is ∆H◦=−571.6 kJ/mol.
Question 27
Question
Calculate the enthalpy change (∆H) for the reaction:
2C(graphite)+2H2(g)→C2H4(g)
given the following bond dissociation energies: C−Cbond = 347 kJ/mol, C−H
bond = 413 kJ/mol, and H−Hbond = 436 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Bonds broken = 2(C−C) + 4(C−H) + 2(H−H)
= 2(347) + 4(413) + 2(436)
= 694 + 1652 + 872
= 3218 kJ
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Step 2: Calculate the total energy released when the bonds in the products
are formed.
Bonds formed = 1(C=C) + 4(C−H)
= 1(612) + 4(413)
= 612 + 1652
= 2264 kJ
Step 3: Calculate the ∆Hfor the reaction.
∆H= Energy required to break bonds −Energy released from forming bonds
= 3218 kJ −2264 kJ
= 954 kJ
Answer: The enthalpy change for the reaction is 954 kJ (endothermic).
Question 28
Question
Calculate the enthalpy change (∆H) when 8.00 g of methane (CH4) is burned
in excess oxygen according to the reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
Given the following molar enthalpies of formation:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the moles of methane burned. Given: Mass of methane,
m= 8.00 g Molar mass of methane, MCH4= 16.05 g/mol (from periodic table)
Using the formula:
moles of CH4=m
MCH4
moles of CH4=8.00 g
16.05 g/mol = 0.498 mol
Step 2: Determine the enthalpy change when 0.498 mol of methane is burned.
Given:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
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∆H◦
f(H2O) = −285.8 kJ/mol
From the balanced chemical equation, the enthalpy change can be calculated
as:
∆H=n·∆Hproducts −m·∆Hreactants
Where nand mare the stoichiometric coefficients for the products and reactants,
respectively.
Substitute the values:
∆H= (1)(−393.5) + 2(−285.8) −(1)(−74.8)
∆H=−393.5−571.6 + 74.8
∆H=−890.3 kJ
Therefore, the enthalpy change when 8.00 g of methane is burned is −890.3 kJ .
Question 29
Question
Consider the following reaction:
2H2O2(l)→2H2O(l)+O2(g)
Given the standard enthalpies of formation:
∆H◦
f(H2O2(l)) = −196.2 kJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Write the balanced chemical equation. The balanced chemical equation
is:
2H2O2(l)→2H2O(l)+O2(g)
Step 2: Determine the standard enthalpy change for the reaction using stan-
dard enthalpies of formation. The standard enthalpy change for the reaction
can be calculated using the formula:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients of the products and reactants,
respectively, in the balanced chemical equation.
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Substitute the given values into the formula:
∆H◦= [2(−285.8) + 0] −[2(−196.2)]
Step 3: Calculate the standard enthalpy change for the reaction.
∆H◦= [−571.6] −[−392.4]
∆H◦=−571.6 + 392.4
∆H◦=−179.2 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−179.2 kJ/mol.
Question 30
Question
Calculate the enthalpy change for the reaction:
2C(graphite)+3H2(g)→C2H6(g)
Given the following bond dissociation energies:
C−C= 347 kcal/mol, C −H= 99 kcal/mol, H −H= 104 kcal/mol
Solution
Step 1: Calculate the total bond energy of the reactants. The total bond energy
of the reactants can be calculated as follows:
2(C−C) + 3(C−H) + 3(H−H)
Step 2: Substitute the given bond dissociation energies to find the total bond
energy of the reactants.
2(347) + 3(99) + 3(104) = 694 + 297 + 312 = 1303 kcal/mol
Step 3: Calculate the total bond energy of the products. The total bond
energy of the products is:
C−C= 0 kcal/mol (1 mol of C-C bonds broken), C−H= 0 kcal/mol (6 mol of C-H bonds formed)
Step 4: Calculate the enthalpy change. The enthalpy change can be calcu-
lated as the difference between the total bond energy of the reactants and the
total bond energy of the products:
∆H= Total bond energy of reactants −Total bond energy of products
∆H= 1303 kcal/mol −(0 kcal/mol + 0 kcal/mol) = 1303 kcal/mol
Therefore, the enthalpy change for the given reaction is 1303 kcal/mol .
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