CHEM 131 - ADVANCED GENERAL
CHEMISTRY I - Thermochemistry
Question Bank - Set 1
Liberty University
Question 1
Question
Given the following reaction at 25
°
C:
2H2(g)+O2(g)→2H2O(l)
The standard enthalpy change of the reaction is -483.6 kJ/mol. Calculate
the standard enthalpy of formation of liquid water (∆H0
f) using the standard
enthalpies of formation given below:
∆H0
f(H2(g)) = 0 kJ/mol
∆H0
f(O2(g)) = 0 kJ/mol
∆H0
f(H2O(l)) = −285.8 kJ/mol
Solution
Step 1: Write the given standard enthalpy change as a combination of the
standard enthalpies of formation of the reactants and products.
∆H0=Xν∆H0
f(products) −Xν∆H0
f(reactants)
Where ∆H0is the standard enthalpy change of the reaction, νis the stoi-
chiometric coefficient, and ∆H0
fis the standard enthalpy of formation.
For the given reaction:
∆H0= (2 ×∆H0
f(H2O(l))) −(2 ×∆H0
f(H2(g)) + ∆H0
f(O2(g)))
Step 2: Substitute the values of the standard enthalpies of formation into
the equation and solve for ∆H0
f(H2O(l)).
∆H0= (2 × −285.8) −(2 ×0 + 0)
∆H0=−571.6 kJ/mol
Therefore, the standard enthalpy of formation of liquid water is −571.6 kJ/mol .
Question 2
Question
Calculate the enthalpy change (∆H) for the combustion of 1 mole of propane
gas (C3H8) at constant pressure, given the following standard enthalpies of
formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(C3H8) = −103.9 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the combustion of 1 mole of
propane:
C3H8+ 5O2→3CO2+ 4H2O
Step 2: Calculate the enthalpy change using the standard enthalpies of for-
mation:
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
= [3 ×∆H◦
f(CO2)+4×∆H◦
f(H2O)] −[1 ×∆H◦
f(C3H8)+5×∆H◦
f(O2)]
= [3(−393.5) + 4(−285.8)] −[−103.9 + 5(0)]
= [−1180.5−1143.2] −[−103.9]
=−2323.7 kJ/mol
Therefore, the enthalpy change for the combustion of 1 mole of propane gas
is ∆H=−2323.7 kJ/mol.
Question 3
Question
Calculate the standard enthalpy change for the reaction below using the given
standard enthalpies of formation:
2C(graphite)+3H2(g)→C2H6(g)
2
Given:
∆H◦
f(C(graphite)) = 0 kcal/mol
∆H◦
f(H2(g)) = 0 kcal/mol
∆H◦
f(C2H6(g)) = −20 kcal/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 equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the standard enthalpies of formation into the equation.
∆H◦=∆H◦
f(C2H6(g))−2∆H◦
f(C(graphite)) + 3∆H◦
f(H2(g))
Step 4: Plug in the given values and calculate.
∆H◦= (−20 kcal/mol) −(2(0 kcal/mol) + 3(0 kcal/mol))
∆H◦=−20 kcal/mol
Therefore, the standard enthalpy change for the reaction is −20 kcal/mol.
Question 4
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
given the following bond dissociation energies: C−Cbond = 348 kJ/mol,
C−Hbond = 413 kJ/mol, O=Obond = 498 kJ/mol, O−Hbond = 463
kJ/mol, C=Obond = 743 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy required to break the bonds in 2C2H6(g) :2(8 ×C−Cbonds) + 12(6 ×C−Hbonds)
= 2(8 ×348 kJ/mol) + 12(6 ×413 kJ/mol)
= 2(2784 kJ) + 12(2478 kJ)
= 5568 kJ + 29736 kJ
= 35304 kJ
3
Energy required to break the bonds in 7O2(g) :7(4 ×O=Obonds)
= 7(4 ×498 kJ/mol)
= 7(1992 kJ)
= 13944 kJ
Step 2: Calculate the total energy released by forming the bonds in the
products.
Energy released by forming the bonds in 4CO2(g) :4(4 ×C=Obonds)
= 4(4 ×743 kJ/mol)
= 4(2972 kJ)
= 11888 kJ
Energy released by forming the bonds in 6H2O(l) :6(2 ×O−Hbonds)
= 6(2 ×463 kJ/mol)
= 6(926 kJ)
= 5556 kJ
Step 3: Calculate the overall energy change for the reaction.
∆H= Energy required to break bonds in reactants −Energy released by forming bonds in products
= (35304 kJ + 13944 kJ) −(11888 kJ + 5556 kJ)
= 49248 kJ −17444 kJ
= 31804 kJ
Therefore, the enthalpy change (∆H) for the reaction is 31,804 kJ.
Question 5
Question
Given the following reaction and enthalpy changes:
2H2(g)+O2(g)→2H2O(g)∆H=−483.6 kJ/mol
Calculate the enthalpy change when 5.00 g of water is formed. (Molar mass
of water = 18.02 g/mol)
4
Solution
Step 1: Calculate the number of moles of water formed. Step 2: Use the given
enthalpy change to calculate the enthalpy change when 5.00 g of water is formed.
Step 1: Calculate the number of moles of water formed. Given: Mass of
water formed = 5.00 g Molar mass of water = 18.02 g/mol
Number of moles of water = Mass
Molar mass Number of moles of water = 5.00 g
18.02 g/mol
Number of moles of water = 0.2778 mol
Step 2: Use the given enthalpy change to calculate the enthalpy change
when 5.00 g of water is formed. Given: ∆H=−483.6 kJ/mol Number of moles
of water = 0.2778 mol
Enthalpy change = Number of moles ×∆HEnthalpy change = 0.2778 mol
×-483.6 kJ/mol
Enthalpy change = -134.5 kJ
Therefore, the enthalpy change when 5.00 g of water is formed is -134.5 kJ.
Question 6
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(s) + 3O2(g)→2CO2(g)
Given the following standard enthalpy of formations:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[O2(g)] = 0 kJ/mol
∆H◦
f[CO2(g)] = −393.5 kJ/mol
Solution
Step 1: Write the balanced chemical equation and calculate the ∆H◦using the
standard enthalpies of formation.
The standard enthalpy change (∆H◦) for the reaction is the sum of the
standard enthalpies of formation of the products minus the sum of the standard
enthalpies of formation of the reactants.
Given reaction: 2C(s) + 3O2(g)→2CO2(g)
The standard enthalpy change is calculated as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the standard enthalpies of formation into the equation:
∆H◦= 2∆H◦
f(CO2(g)) −(2∆H◦
f(C(s)) + 3∆H◦
f(O2(g)))
Now, substitute the given values:
∆H◦= 2(−393.5) −(2(0) + 3(0))
5
∆H◦=−787 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−787 kJ/mol.
Question 7
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of 1 mole of
benzene (C6H6) using the following data:
C6H6(l) + 15
2O2(g)→6CO2(g)+3H2O(l)
∆H◦=−3267.8 kJ
Solution
Step 1: Write the balanced chemical equation for the combustion of 1 mole
benzene.
C6H6(l) + 15
2O2(g) →6CO2(g) + 3H2O(l)
Step 2: Calculate the ∆H◦for the combustion of 1 mole benzene using the
given data.
∆H◦= 6 ×∆H◦
(CO2)+ 3 ×∆H◦
(H2O) −∆H◦
(C6H6)−15
2×∆H◦
(O2)
∆H◦= 6 ×(−393.5) + 3 ×(−285.8) −(−3267.8) −15
2×0
∆H◦=−2443.6−857.4 + 3267.8
∆H◦=−33.2 kJ/mol
Therefore, the standard enthalpy change for the combustion of 1 mole of
benzene is −33.2 kJ/mol.
Question 8
Question
The standard enthalpy of formation of ammonia, ∆H◦
f, is equal to -45.9 kJ/mol.
Calculate the standard enthalpy change, ∆H◦, for the following reaction:
2NH3(g)+ 5O2(g)→4NO2(g)+ 6H2O(l)
6
Solution
Step 1: Write the given standard enthalpy of formation for each compound.
∆H◦
f(NH3(g)) = −45.9 kJ/mol
∆H◦
f(NO2(g)) = Unknown, denoted by xkJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
∆H◦=X(Products) −X(Reactants)
=4∆H◦
f(NO2(g)) + 6∆H◦
f(H2O(l))−2∆H◦
f(NH3(g)) + 5(0)
= (4x+ 6(−285.8)) −(2(−45.9) + 0)
= 4x−1714.8 + 91.8
= 4x−1623 kJ
Step 3: Substitute the values and solve for ∆H◦.
∆H◦= 4x−1623
Step 4: Use Hess’s Law, which states that the sum of the enthalpy changes
for a series of reactions is equal to the enthalpy change if the reactions are added
together.
∆H◦= 4∆H◦
f(NO2(g)) + 6∆H◦
f(H2O(l))−2∆H◦
f(NH3(g)) + 5(0)
Step 5: Equate the expressions for ∆H◦to find the value of x.
4x−1623 = 4∆H◦
f(NO2(g)) + 6∆H◦
f(H2O(l))−2∆H◦
f(NH3(g))
Step 6: Solve for x.
4x−1623 = 4(0) + 6(−285.8) −2(−45.9)
4x−1623 = −1714.8 + 571.6
4x=−1143.2 + 1623
4x= 479.8
x= 119.95 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
479.8 kJ .
Question 9
Question
Given the following reaction:
2A+ 3B→C+ 4D
7
If the standard enthalpies of formation are ∆H◦
f:−100 kJ/mol for A, −150
kJ/mol for B, −200 kJ/mol for C, and −50 kJ/mol for D, calculate the standard
enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change using the standard enthalpies
of formation. The standard enthalpy change for the reaction can be calculated
using the equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
where ∆H◦
fis the standard enthalpy of formation.
Step 2: Identify the enthalpies of formation for the products and reactants.
For the given reaction: Products: C and 4D Reactants: 2A and 3B
Step 3: Substitute the values into the equation and solve for ∆H◦.
∆H◦= [(−200 kJ/mol)+4(−50 kJ/mol)]−[2(−100 kJ/mol)+3(−150 kJ/mol)]
∆H◦= (−200 kJ/mol −200 kJ/mol) −(−200 kJ/mol −450 kJ/mol)
∆H◦=−400 kJ/mol + 650 kJ/mol
∆H◦= 250 kJ/mol
Therefore, the standard enthalpy change for the reaction is 250 kJ/mol.
Question 10
Question
Given the following reaction:
2H2(g) + O2(g) →2H2O(l)
The standard enthalpy change for the reaction is -483.7 kJ/mol. Calculate
the standard enthalpy change when 4 moles of H2(g) and 2 moles of O2(g) react
to form water.
Solution
Step 1: Calculate the standard enthalpy change for the given reaction when 1
mole of H2(g) reacts. Given: Standard enthalpy change = -483.7 kJ/mol
Since the reaction involves 2 moles of H2(g), we can calculate the standard
enthalpy change for 1 mole of H2(g) reacting using stoichiometry:
∆H=−483.7 kJ
2mol =−241.85 kJ/mol
8
Step 2: Calculate the standard enthalpy change for the reaction involving 4
moles of H2(g). Since the reaction is multiplied by 2 in the given equation, we
need to double the standard enthalpy change calculated above:
−241.85 kJ/mol ×2 = −483.7 kJ
Thus, the standard enthalpy change for the reaction involving 4 moles of
H2(g) is −483.7 kJ .
Question 11
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−571.6 kJ
Calculate the enthalpy change (∆H) for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
Solution
Step 1: Write the reverse reaction of the given reaction. Step 2: Write the
enthalpy change for the reverse reaction. Step 3: Calculate the enthalpy change
for the reverse reaction.
Step 1: Reverse the given reaction:
H2O(l)→2H2(g)+O2(g)
Step 2: Write the enthalpy change for the reverse reaction:
Since the reverse reaction is the reverse of the given reaction, the enthalpy
change for the reverse reaction is the negative of the enthalpy change for the
given reaction:
∆H=−(−571.6 kJ)
Step 3: Calculate the enthalpy change for the reverse reaction:
∆H= 571.6 kJ
Therefore, the enthalpy change for the reaction H2O(l)→H2(g) + 1
2O2(g)
is 571.6 kJ .
9
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2A(g)+3B(g)→C(g)+4D(g)
Given the following information:
∆H◦for the reaction A(g)→B(g) is -300 kJ/mol
∆H◦for the reaction C(g) + D(g)→A(g)+2B(g) is 400 kJ/mol
∆H◦for the reaction A(g) + D(g)→C(g)+2B(g) is -600 kJ/mol
Solution
To find the ∆H◦for the given reaction, we can use the Hess’s Law: the over-
all enthalpy change for a reaction is the sum of the enthalpy changes of the
individual steps into which the reaction can be divided.
Step 1: First, we need to identify a series of reactions that sum to the
desired overall reaction. By manipulating the given reactions, we can rearrange
and combine them to match the desired overall reaction. Let’s list out the given
reactions and the desired overall reaction:
1. A(g)→B(g) ∆H◦=−300 kJ/mol
2. C(g) + D(g)→A(g)+2B(g) ∆H◦= 400 kJ/mol
3. A(g) + D(g)→C(g)+2B(g) ∆H◦=−600 kJ/mol
4.2A(g)+3B(g)→C(g)+4D(g) (desired overall reaction)
Step 2: Rearrange and combine the given reactions to achieve the desired
overall reaction:
Multiply reaction 1 by 2: 2A(g)→2B(g) ∆H◦=−600 kJ/mol
Add reaction 2 to the above result: 2A(g)+3B(g)→C(g)+2B(g) + D(g) ∆H◦=−200 kJ/mol
Multiply reaction 3 by 2: 2A(g)+2D(g)→2C(g)+4B(g) ∆H◦=−1200 kJ/mol
Add reaction 2 to the above result: 2A(g)+3B(g)→C(g) + D(g)+4B(g) ∆H◦=−800 kJ/mol
Step 3: Therefore, the ∆H◦for the overall reaction is -800 kJ/mol.
Question 13
Question
Given the following reaction:
2H2(g) + O2(g)→2H2O(g)
10
The enthalpy change (∆H) for this reaction is -484 kJ. Calculate the standard
enthalpy of formation (∆H◦
f) of water vapor, given the standard enthalpies of
formation are: ∆H◦
f(H2O) = −286 kJ/mol and ∆H◦
f(O2) = 0 kJ/mol.
Solution
Step 1: Write the balanced equation for the formation of water vapor using the
given reaction and enthalpies of formation.
The formation of water vapor from its elements is:
2H2(g) + O2(g)→2H2O(g)
Step 2: Write the expression for the standard enthalpy change of the reac-
tion.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Given that ∆H=−484 kJ and ∆H◦
f(H2O) = −286 kJ/mol, ∆H◦
f(O2)=0
kJ/mol, and we need to find ∆H◦
f(H2O).
Step 3: Substitute the values into the equation from step 2 and solve for
∆H◦
f(H2O).
−484 kJ = 2(−286) kJ/mol −2(0) kJ/mol
−484 kJ = −572 kJ/mol
Step 4: Calculate the standard enthalpy of formation of water vapor.
−484 kJ = −572 kJ/mol
88 kJ = ∆H◦
f(H2O)
Therefore, the standard enthalpy of formation of water vapor is 88 kJ/mol.
Question 14
Question
Consider the following reaction at 298 K:
2H2O(l)→2H2(g)+O2(g)
Given the standard enthalpies of formation at 298 K: ∆H◦
f(H2O(l)) =
−285.8 kJ/mol, ∆H◦
f(H2(g)) = 0 kJ/mol, and ∆H◦
f(O2(g)) = 0 kJ/mol.
Calculate the standard enthalpy change for this reaction.
11
Solution
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation is:
2H2O(l)→2H2(g)+O2(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
The standard enthalpy change for the reaction is given by:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))] −[2∆H◦
f(H2O(l))]
∆H◦= [2(0 kJ/mol) + 0 kJ/mol] −[2(−285.8 kJ/mol)]
∆H◦= 571.6 kJ/mol + 571.6 kJ/mol = 1143.2kJ/mol
Therefore, the standard enthalpy change for the reaction is 1143.2kJ/mol.
Question 15
Question
A reaction has a standard enthalpy change of -328 kJ/mol. If 0.50 mol of the
reaction takes place, how much heat is either absorbed or released?
Solution
Step 1: Determine the heat absorbed or released for the given amount of the
reaction. Step 2: Use the standard enthalpy change to calculate the heat.
Step 1: Given: Standard enthalpy change = -328 kJ/mol Number of moles
of reaction, n = 0.50 mol
Step 2: To find the total heat involved in the reaction, we multiply the
standard enthalpy change by the number of moles of the reaction. Total Heat
= Standard enthalpy change ×Number of moles
Substitute the given values into the equation: Total Heat = -328 kJ/mol ×
0.50 mol
Calculate the total heat released or absorbed: Total Heat = -164 kJ
Therefore, when 0.50 mol of the reaction takes place, 164 kJ of heat is
released.
12
Question 16
Question
Given the reaction:
2H2(g) + O2(g)→2H2O(g)
with the following data:
∆H◦
ffor H2O(g) is -241.8 kJ/mol
∆H◦
ffor H2(g) is 0 kJ/mol
∆H◦
ffor O2(g) is 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 provided.
Given reaction: 2H2(g) + O2(g)→2H2O(g)
The standard enthalpy change for the reaction is the sum of the standard
enthalpies of formation of the products minus the sum of the standard enthalpies
of formation of the reactants:
∆H◦= Σn∆H◦
f(products) −Σm∆H◦
f(reactants)
Where nand mare the stoichiometric coefficients of the products and reac-
tants, respectively.
By substituting the given values: ∆H◦= 2∆H◦
f(H2O)−(2∆H◦
f(H2) +
∆H◦
f(O2))
∆H◦= 2(−241.8 kJ/mol) −(2(0 kJ/mol) + 0 kJ/mol)
∆H◦=−483.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
−483.6 kJ/mol.
Question 17
Question
A sample of 2.50 moles of propane (C3H8) is burned in excess oxygen gas.
The combustion of propane is represented by the following balanced chemical
equation:
C3H8(g) + 5 O2(g) →3 CO2(g) + 4 H2O (g)
Given that the standard enthalpy of formation of CO2is -393.5 kJ/mol
and the standard enthalpy of formation of H2O is -285.8 kJ/mol, calculate the
standard enthalpy change (∆H◦) for the combustion of propane.
13
Solution
Step 1: Calculate the standard enthalpy change for the combustion of propane
by using the enthalpy of formation values provided for CO2and H2O.
Given: Standard enthalpy of formation of CO2: -393.5 kJ/mol Standard
enthalpy of formation of H2O: -285.8 kJ/mol Coefficient of CO2in the balanced
equation: 3 Coefficient of H2O in the balanced equation: 4
The standard enthalpy change for the combustion of propane can be calcu-
lated using the equation:
∆H◦=Xν∆H◦
f
Where: νrepresents the stoichiometric coefficient ∆H◦
frepresents the stan-
dard enthalpy of formation
Plugging the values into the equation:
∆H◦= (3 × −393.5 kJ/mol) + (4 × −285.8 kJ/mol)
Step 2: Calculate the standard enthalpy change.
∆H◦= (−1180.5 kJ/mol) + (−1143.2 kJ/mol)
∆H◦=−2323.7 kJ/mol
Therefore, the standard enthalpy change for the combustion of propane is
−2323.7 kJ/mol .
Question 18
Question
For the reaction:
2H2(g)+O2(g)→2H2O(l)
the enthalpy change (∆H) is −483.6 kJ/mol. Calculate the amount of heat (in
kJ) released when 1.00 g of hydrogen gas is burned in excess oxygen.
Solution
Step 1: Calculate the number of moles of hydrogen gas in 1.00 g.
Molar mass of H2= 2.02 g/mol
Number of moles of H2=1.00 g
2.02 g/mol
= 0.495 mol
14
Step 2: Use the stoichiometry of the reaction to calculate the moles of water
produced. Since 2 moles of hydrogen gas produce 2 moles of water, the 0.495
mol of hydrogen gas will produce 0.495 mol of water.
Step 3: Calculate the amount of heat released using the given enthalpy
change.
Heat released = moles of reaction ×∆H
= 0.495 mol × −483.6 kJ/mol
=−239.322 kJ
≈ −239 kJ
Therefore, when 1.00 g of hydrogen gas is burned in excess oxygen, approx-
imately 239 kJ of heat is released.
Question 19
Question
Calculate the enthalpy change (∆H) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
given the following bond dissociation energies:
H−H= 436 kJ/mol
O=O= 498 kJ/mol
H−O= 463 kJ/mol
Solution
Step 1: Calculate the bond energy for the reactants and products. We have:
Bond energy of 2H-H + Bond energy of 1O=O = 2(436) + 498 = 1370 kJ
Bond energy of 4H-O = 4(463) = 1852 kJ
Step 2: Calculate the change in enthalpy (∆H) for the reaction using the
bond energies calculated above.
∆H=XBonds broken −XBonds formed
= (1370) −(1852) = −482 kJ
Therefore, the enthalpy change for the given reaction is ∆H=−482 kJ.
15
Question 20
Question
Calculate the enthalpy change, ∆H, for the combustion of 1 mole of ethylene
(C2H4) gas according to the following reaction:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Given the following standard enthalpies of formation:
∆H◦
f(C2H4) = 52.3 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of reaction using the standard en-
thalpies of formation.
∆H◦
r=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
r= [2 ×∆H◦
f(CO2)+2×∆H◦
f(H2O)] −[∆H◦
f(C2H4)+3×∆H◦
f(O2)]
∆H◦
r= [2 ×(−393.5) + 2 ×(−285.8)] −[52.3+3×0]
∆H◦
r= [−787.0−571.6] −52.3
∆H◦
r=−1358.6 kJ/mol
Hence, the enthalpy change for the combustion of 1 mole of ethylene gas is
∆H=−1358.6 kJ/mol.
Question 21
Question
Calculate the enthalpy change for the combustion of n-hexane (C6H14) given
that the standard enthalpy of formation for n-hexane is -199.3 kJ/mol and the
standard enthalpy of formation for carbon dioxide and water are -393.5 kJ/mol
and -285.8 kJ/mol, respectively.
16
Solution
Step 1: Write the balanced chemical equation for the combustion of n-hexane:
C6H14 + 9O2→6CO2+ 7H2O
Step 2: Calculate the standard enthalpy change for the combustion of n-
hexane using the standard enthalpies of formation for the reactants and prod-
ucts:
∆H=X∆Hf(products) −X∆Hf(reactants)
Step 3: Substitute the given standard enthalpies of formation into the equa-
tion and calculate:
∆H= [6(−393.5) + 7(−285.8)] −(−199.3)
∆H= [−2361 + (−2000.6)] −(−199.3)
∆H=−4361.6 + 199.3
∆H=−4162.3 kJ/mol
Therefore, the enthalpy change for the combustion of n-hexane is -4162.3
kJ/mol.
Question 22
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2 Fe2O3(s) + 3 C(s) →4 Fe(s) + 3 CO2(g)
given the following standard enthalpies of formation:
∆H◦
f(Fe2O3(s)) = −826.2 kJ/mol
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(Fe(s)) = 0 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
17
Solution
Step 1: Calculate the standard enthalpy change for the reaction based on the
standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion.
∆H◦=4×∆H◦
f(Fe(s)) + 3 ×∆H◦
f(CO2(g))−2×∆H◦
f(Fe2O3(s)) + 3 ×∆H◦
f(C(s))
Step 3: Substitute the values and simplify.
∆H◦= (4 ×0 kJ/mol + 3 × −393.5 kJ/mol)−(2 × −826.2 kJ/mol + 3 ×0 kJ/mol)
∆H◦= (0 kJ/mol −1180.5 kJ/mol) −(−1652.4 kJ/mol)
∆H◦=−1180.5 kJ/mol + 1652.4 kJ/mol
∆H◦= 471.9 kJ/mol
Therefore, the standard enthalpy change for the reaction is 471.9 kJ/mol .
Question 23
Question
For a certain reaction at constant pressure, the enthalpy change (∆H) is found
to be −487 kJ. If the reaction is exothermic, predict whether the entropy change
(∆S) is positive, negative, or zero. Justify your answer.
Solution
Step 1: Recall the relationship between ∆Hand ∆S. The relationship between
∆Hand ∆Sis given by the equation:
∆G= ∆H−T∆S
where ∆Gis the Gibbs free energy change, ∆His the enthalpy change, ∆Sis
the entropy change, and Tis the temperature in Kelvin.
Step 2: Understand the properties of an exothermic reaction. In an exother-
mic reaction, heat is released to the surroundings (∆H < 0). This means that
the system loses heat and the surroundings gain heat.
Step 3: Determine the sign of ∆Sfor an exothermic reaction. Since the
reaction is exothermic (∆H < 0) and Tis a positive value (since temperature
is always positive in Kelvin), for the equation ∆G= ∆H−T∆Sto be nega-
tive (indicating a spontaneous reaction), ∆Smust be positive. Therefore, the
entropy change (∆S) for an exothermic reaction is positive.
Therefore, for the given exothermic reaction with ∆H=−487 kJ, the en-
tropy change (∆S) is positive.
18
Question 24
Question
Calculate the enthalpy change (∆H) for the combustion of 1 mole of propane
gas (C3H8) at constant pressure using the following information:
The enthalpy of formation of C3H8is −103.8 kJ/mol.
The balanced chemical equation for the combustion of propane is:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
The enthalpy of formation of CO2is −393.5 kJ/mol and H2Ois −285.8
kJ/mol.
Solution
Step 1: Find the enthalpy change for the combustion of propane from the given
enthalpies of formation.
∆H=X[products] −X[reactants]
= [3 ×Hf(CO2)] + [4 ×Hf(H2O)] −[Hf(C3H8)+5×Hf(O2)]
= [3 ×(−393.5 kJ/mol)] + [4 ×(−285.8 kJ/mol)] −(−103.8 kJ/mol) −(5 ×0 kJ/mol)
=−1180.5 kJ/mol −1143.2 kJ/mol −(−103.8 kJ/mol)
=−2176.9 kJ/mol
Therefore, the enthalpy change for the combustion of 1 mole of propane gas
is −2176.9 kJ/mol.
Question 25
Question
Calculate the standard enthalpy change for the reaction
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpy of formation values:
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
19
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g)→2H2O(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
∆H◦= (2 ×∆H◦
f(H2O(g))) −(2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g)))
∆H◦= (2 × −241.8 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
∆H◦=−483.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −483.6 kJ/mol.
Question 26
Question
Given the following reaction:
2A(g) + B(g)→3C(g)+4D(g)
with the following enthalpies of formation:
∆H◦
f(kJ/mol)
A(g): 100, B(g): 50, C(g): 200, D(g): 150.
Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Calculate the change in enthalpy (∆H◦) using the enthalpies of forma-
tion:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients for the products and reactants
respectively.
Step 2: Substitute the given values into the equation and solve:
∆H◦= (3 ×200 + 4 ×150) −(2 ×100 + 50)
∆H◦= (600 + 600) −(200 + 50)
∆H◦= 1200 −250
∆H◦= 950 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is 950
kJ/mol.
20
Question 27
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C given the standard enthalpies of formation provided:
2C(s) + 3H2(g)→C2H6(g)
∆H◦
f(C2H6(g)) = −84.68 kJ/mol
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 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 (∆H◦) using the standard
enthalpies of formation.
The standard enthalpy change of a reaction is given by:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nrepresents the stoichiometric coefficient of the products, mrepresents
the stoichiometric coefficient of the reactants, ∆H◦
fis the standard enthalpy of
formation.
For the reaction:
∆H◦=1×∆H◦
f(C2H6(g))−2×∆H◦
f(C(s)) + 3 ×∆H◦
f(H2(g))
Substitute the given values:
∆H◦= (1 × −84.68) −(2 ×0+3×0)
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction at
25
°
C is -84.68 kJ/mol.
Question 28
Question
Calculate the enthalpy change (∆H) for the following reaction at 25
°
C:
2A(g)+3B(g)→4C(g)+2D(g)
21
given the following bond dissociation energies:
A-A : 170 kcal/mol
B-B : 200 kcal/mol
C-C : 250 kcal/mol
D-D : 300 kcal/mol
A-B : 250 kcal/mol
C-D : 400 kcal/mol
Solution
Step 1: Calculate the total bond dissociation energy for the reactants and prod-
ucts.
Reactants: 2(A−A) + 3(B−B) + 2(A−B)
= 2(170) + 3(200) + 2(250)
= 340 + 600 + 500
= 1440 kcal/mol
Products: 4(C−C) + 2(D−D) + 2(C−D)
= 4(250) + 2(300) + 2(400)
= 1000 + 600 + 800
= 2400 kcal/mol
Step 2: Calculate the enthalpy change (∆H) for the reaction using the bond
dissociation energies.
∆H= Total energy of bonds broken −Total energy of bonds formed
= Reactants −Products
= 1440 kcal/mol −2400 kcal/mol
=−960 kcal/mol
Therefore, the enthalpy change for the reaction is ∆H=−960 kcal/mol.
Question 29
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2CH4(g)+3O2(g)−→ 2CO2(g)+4H2O(l)
22
given the following standard enthalpies of formation:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Write the given equation in terms of standard enthalpies of formation.
The standard enthalpy change can be calculated using the equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Plugging in the values from the given standard enthalpies of formation, we get:
∆H◦= [2 ×∆H◦
f(CO2)+4×∆H◦
f(H2O)] −[2 ×∆H◦
f(CH4)+3×∆H◦
f(O2)]
Step 2: Substitute the values and calculate. Plugging in the values:
∆H◦= [2(−393.5) + 4(−285.8)] −[2(−74.8) + 3(0)]
= [−787 −1143.2] −[−149.6]
=−1930.2 + 149.6
=−1780.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−1780.6 kJ/mol.
Question 30
Question
For the reaction
2H2O(g)→2H2(g) + O2(g)
at 25
°
C and 1 atm, the change in enthalpy (∆H) is 483.6 kJ. Calculate the
change in internal energy (∆U) for the reaction.
Given: ∆H=−483.6 kJ, R = 8.314 J/(mol·K)
Solution
Step 1: To find ∆U, we can use the equation:
∆U= ∆H−∆ngRT
where ∆ngis the change in the number of moles of gas and R is the ideal gas
constant.
23
Step 2: Substitute the values of the standard enthalpies of formation into
the equation and solve for ∆H0
f(H2O(l)).
∆H0= (2 × −285.8) −(2 ×0 + 0)
∆H0=−571.6 kJ/mol
Therefore, the standard enthalpy of formation of liquid water is −571.6 kJ/mol .
Question 2
Question
Calculate the enthalpy change (∆H) for the combustion of 1 mole of propane
gas (C3H8) at constant pressure, given the following standard enthalpies of
formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(C3H8) = −103.9 kJ/mol
Solution
Step 1: Write the balanced chemical equation for the combustion of 1 mole of
propane:
C3H8+ 5O2→3CO2+ 4H2O
Step 2: Calculate the enthalpy change using the standard enthalpies of for-
mation:
∆H=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
= [3 ×∆H◦
f(CO2)+4×∆H◦
f(H2O)] −[1 ×∆H◦
f(C3H8)+5×∆H◦
f(O2)]
= [3(−393.5) + 4(−285.8)] −[−103.9 + 5(0)]
= [−1180.5−1143.2] −[−103.9]
=−2323.7 kJ/mol
Therefore, the enthalpy change for the combustion of 1 mole of propane gas
is ∆H=−2323.7 kJ/mol.
Question 3
Question
Calculate the standard enthalpy change for the reaction below using the given
standard enthalpies of formation:
2C(graphite)+3H2(g)→C2H6(g)
2
Given:
∆H◦
f(C(graphite)) = 0 kcal/mol
∆H◦
f(H2(g)) = 0 kcal/mol
∆H◦
f(C2H6(g)) = −20 kcal/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 equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 3: Substitute the standard enthalpies of formation into the equation.
∆H◦=∆H◦
f(C2H6(g))−2∆H◦
f(C(graphite)) + 3∆H◦
f(H2(g))
Step 4: Plug in the given values and calculate.
∆H◦= (−20 kcal/mol) −(2(0 kcal/mol) + 3(0 kcal/mol))
∆H◦=−20 kcal/mol
Therefore, the standard enthalpy change for the reaction is −20 kcal/mol.
Question 4
Question
Calculate the enthalpy change (∆H) for the reaction:
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(l)
given the following bond dissociation energies: C−Cbond = 348 kJ/mol,
C−Hbond = 413 kJ/mol, O=Obond = 498 kJ/mol, O−Hbond = 463
kJ/mol, C=Obond = 743 kJ/mol.
Solution
Step 1: Calculate the total energy required to break the bonds in the reactants.
Energy required to break the bonds in 2C2H6(g) :2(8 ×C−Cbonds) + 12(6 ×C−Hbonds)
= 2(8 ×348 kJ/mol) + 12(6 ×413 kJ/mol)
= 2(2784 kJ) + 12(2478 kJ)
= 5568 kJ + 29736 kJ
= 35304 kJ
3
Energy required to break the bonds in 7O2(g) :7(4 ×O=Obonds)
= 7(4 ×498 kJ/mol)
= 7(1992 kJ)
= 13944 kJ
Step 2: Calculate the total energy released by forming the bonds in the
products.
Energy released by forming the bonds in 4CO2(g) :4(4 ×C=Obonds)
= 4(4 ×743 kJ/mol)
= 4(2972 kJ)
= 11888 kJ
Energy released by forming the bonds in 6H2O(l) :6(2 ×O−Hbonds)
= 6(2 ×463 kJ/mol)
= 6(926 kJ)
= 5556 kJ
Step 3: Calculate the overall energy change for the reaction.
∆H= Energy required to break bonds in reactants −Energy released by forming bonds in products
= (35304 kJ + 13944 kJ) −(11888 kJ + 5556 kJ)
= 49248 kJ −17444 kJ
= 31804 kJ
Therefore, the enthalpy change (∆H) for the reaction is 31,804 kJ.
Question 5
Question
Given the following reaction and enthalpy changes:
2H2(g)+O2(g)→2H2O(g)∆H=−483.6 kJ/mol
Calculate the enthalpy change when 5.00 g of water is formed. (Molar mass
of water = 18.02 g/mol)
4
Solution
Step 1: Calculate the number of moles of water formed. Step 2: Use the given
enthalpy change to calculate the enthalpy change when 5.00 g of water is formed.
Step 1: Calculate the number of moles of water formed. Given: Mass of
water formed = 5.00 g Molar mass of water = 18.02 g/mol
Number of moles of water = Mass
Molar mass Number of moles of water = 5.00 g
18.02 g/mol
Number of moles of water = 0.2778 mol
Step 2: Use the given enthalpy change to calculate the enthalpy change
when 5.00 g of water is formed. Given: ∆H=−483.6 kJ/mol Number of moles
of water = 0.2778 mol
Enthalpy change = Number of moles ×∆HEnthalpy change = 0.2778 mol
×-483.6 kJ/mol
Enthalpy change = -134.5 kJ
Therefore, the enthalpy change when 5.00 g of water is formed is -134.5 kJ.
Question 6
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2C(s) + 3O2(g)→2CO2(g)
Given the following standard enthalpy of formations:
∆H◦
f[C(s)] = 0 kJ/mol
∆H◦
f[O2(g)] = 0 kJ/mol
∆H◦
f[CO2(g)] = −393.5 kJ/mol
Solution
Step 1: Write the balanced chemical equation and calculate the ∆H◦using the
standard enthalpies of formation.
The standard enthalpy change (∆H◦) for the reaction is the sum of the
standard enthalpies of formation of the products minus the sum of the standard
enthalpies of formation of the reactants.
Given reaction: 2C(s) + 3O2(g)→2CO2(g)
The standard enthalpy change is calculated as:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the standard enthalpies of formation into the equation:
∆H◦= 2∆H◦
f(CO2(g)) −(2∆H◦
f(C(s)) + 3∆H◦
f(O2(g)))
Now, substitute the given values:
∆H◦= 2(−393.5) −(2(0) + 3(0))
5
∆H◦=−787 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−787 kJ/mol.
Question 7
Question
Calculate the standard enthalpy change (∆H◦) for the combustion of 1 mole of
benzene (C6H6) using the following data:
C6H6(l) + 15
2O2(g)→6CO2(g)+3H2O(l)
∆H◦=−3267.8 kJ
Solution
Step 1: Write the balanced chemical equation for the combustion of 1 mole
benzene.
C6H6(l) + 15
2O2(g) →6CO2(g) + 3H2O(l)
Step 2: Calculate the ∆H◦for the combustion of 1 mole benzene using the
given data.
∆H◦= 6 ×∆H◦
(CO2)+ 3 ×∆H◦
(H2O) −∆H◦
(C6H6)−15
2×∆H◦
(O2)
∆H◦= 6 ×(−393.5) + 3 ×(−285.8) −(−3267.8) −15
2×0
∆H◦=−2443.6−857.4 + 3267.8
∆H◦=−33.2 kJ/mol
Therefore, the standard enthalpy change for the combustion of 1 mole of
benzene is −33.2 kJ/mol.
Question 8
Question
The standard enthalpy of formation of ammonia, ∆H◦
f, is equal to -45.9 kJ/mol.
Calculate the standard enthalpy change, ∆H◦, for the following reaction:
2NH3(g)+ 5O2(g)→4NO2(g)+ 6H2O(l)
6
Solution
Step 1: Write the given standard enthalpy of formation for each compound.
∆H◦
f(NH3(g)) = −45.9 kJ/mol
∆H◦
f(NO2(g)) = Unknown, denoted by xkJ/mol
∆H◦
f(H2O(l)) = −285.8 kJ/mol
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
∆H◦=X(Products) −X(Reactants)
=4∆H◦
f(NO2(g)) + 6∆H◦
f(H2O(l))−2∆H◦
f(NH3(g)) + 5(0)
= (4x+ 6(−285.8)) −(2(−45.9) + 0)
= 4x−1714.8 + 91.8
= 4x−1623 kJ
Step 3: Substitute the values and solve for ∆H◦.
∆H◦= 4x−1623
Step 4: Use Hess’s Law, which states that the sum of the enthalpy changes
for a series of reactions is equal to the enthalpy change if the reactions are added
together.
∆H◦= 4∆H◦
f(NO2(g)) + 6∆H◦
f(H2O(l))−2∆H◦
f(NH3(g)) + 5(0)
Step 5: Equate the expressions for ∆H◦to find the value of x.
4x−1623 = 4∆H◦
f(NO2(g)) + 6∆H◦
f(H2O(l))−2∆H◦
f(NH3(g))
Step 6: Solve for x.
4x−1623 = 4(0) + 6(−285.8) −2(−45.9)
4x−1623 = −1714.8 + 571.6
4x=−1143.2 + 1623
4x= 479.8
x= 119.95 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
479.8 kJ .
Question 9
Question
Given the following reaction:
2A+ 3B→C+ 4D
7
If the standard enthalpies of formation are ∆H◦
f:−100 kJ/mol for A, −150
kJ/mol for B, −200 kJ/mol for C, and −50 kJ/mol for D, calculate the standard
enthalpy change for the reaction.
Solution
Step 1: Calculate the standard enthalpy change using the standard enthalpies
of formation. The standard enthalpy change for the reaction can be calculated
using the equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
where ∆H◦
fis the standard enthalpy of formation.
Step 2: Identify the enthalpies of formation for the products and reactants.
For the given reaction: Products: C and 4D Reactants: 2A and 3B
Step 3: Substitute the values into the equation and solve for ∆H◦.
∆H◦= [(−200 kJ/mol)+4(−50 kJ/mol)]−[2(−100 kJ/mol)+3(−150 kJ/mol)]
∆H◦= (−200 kJ/mol −200 kJ/mol) −(−200 kJ/mol −450 kJ/mol)
∆H◦=−400 kJ/mol + 650 kJ/mol
∆H◦= 250 kJ/mol
Therefore, the standard enthalpy change for the reaction is 250 kJ/mol.
Question 10
Question
Given the following reaction:
2H2(g) + O2(g) →2H2O(l)
The standard enthalpy change for the reaction is -483.7 kJ/mol. Calculate
the standard enthalpy change when 4 moles of H2(g) and 2 moles of O2(g) react
to form water.
Solution
Step 1: Calculate the standard enthalpy change for the given reaction when 1
mole of H2(g) reacts. Given: Standard enthalpy change = -483.7 kJ/mol
Since the reaction involves 2 moles of H2(g), we can calculate the standard
enthalpy change for 1 mole of H2(g) reacting using stoichiometry:
∆H=−483.7 kJ
2mol =−241.85 kJ/mol
8
Step 2: Calculate the standard enthalpy change for the reaction involving 4
moles of H2(g). Since the reaction is multiplied by 2 in the given equation, we
need to double the standard enthalpy change calculated above:
−241.85 kJ/mol ×2 = −483.7 kJ
Thus, the standard enthalpy change for the reaction involving 4 moles of
H2(g) is −483.7 kJ .
Question 11
Question
Given the following reaction:
2H2(g)+O2(g)→2H2O(l) ∆H=−571.6 kJ
Calculate the enthalpy change (∆H) for the reaction:
H2O(l)→H2(g) + 1
2O2(g)
Solution
Step 1: Write the reverse reaction of the given reaction. Step 2: Write the
enthalpy change for the reverse reaction. Step 3: Calculate the enthalpy change
for the reverse reaction.
Step 1: Reverse the given reaction:
H2O(l)→2H2(g)+O2(g)
Step 2: Write the enthalpy change for the reverse reaction:
Since the reverse reaction is the reverse of the given reaction, the enthalpy
change for the reverse reaction is the negative of the enthalpy change for the
given reaction:
∆H=−(−571.6 kJ)
Step 3: Calculate the enthalpy change for the reverse reaction:
∆H= 571.6 kJ
Therefore, the enthalpy change for the reaction H2O(l)→H2(g) + 1
2O2(g)
is 571.6 kJ .
9
Question 12
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2A(g)+3B(g)→C(g)+4D(g)
Given the following information:
∆H◦for the reaction A(g)→B(g) is -300 kJ/mol
∆H◦for the reaction C(g) + D(g)→A(g)+2B(g) is 400 kJ/mol
∆H◦for the reaction A(g) + D(g)→C(g)+2B(g) is -600 kJ/mol
Solution
To find the ∆H◦for the given reaction, we can use the Hess’s Law: the over-
all enthalpy change for a reaction is the sum of the enthalpy changes of the
individual steps into which the reaction can be divided.
Step 1: First, we need to identify a series of reactions that sum to the
desired overall reaction. By manipulating the given reactions, we can rearrange
and combine them to match the desired overall reaction. Let’s list out the given
reactions and the desired overall reaction:
1. A(g)→B(g) ∆H◦=−300 kJ/mol
2. C(g) + D(g)→A(g)+2B(g) ∆H◦= 400 kJ/mol
3. A(g) + D(g)→C(g)+2B(g) ∆H◦=−600 kJ/mol
4.2A(g)+3B(g)→C(g)+4D(g) (desired overall reaction)
Step 2: Rearrange and combine the given reactions to achieve the desired
overall reaction:
Multiply reaction 1 by 2: 2A(g)→2B(g) ∆H◦=−600 kJ/mol
Add reaction 2 to the above result: 2A(g)+3B(g)→C(g)+2B(g) + D(g) ∆H◦=−200 kJ/mol
Multiply reaction 3 by 2: 2A(g)+2D(g)→2C(g)+4B(g) ∆H◦=−1200 kJ/mol
Add reaction 2 to the above result: 2A(g)+3B(g)→C(g) + D(g)+4B(g) ∆H◦=−800 kJ/mol
Step 3: Therefore, the ∆H◦for the overall reaction is -800 kJ/mol.
Question 13
Question
Given the following reaction:
2H2(g) + O2(g)→2H2O(g)
10
The enthalpy change (∆H) for this reaction is -484 kJ. Calculate the standard
enthalpy of formation (∆H◦
f) of water vapor, given the standard enthalpies of
formation are: ∆H◦
f(H2O) = −286 kJ/mol and ∆H◦
f(O2) = 0 kJ/mol.
Solution
Step 1: Write the balanced equation for the formation of water vapor using the
given reaction and enthalpies of formation.
The formation of water vapor from its elements is:
2H2(g) + O2(g)→2H2O(g)
Step 2: Write the expression for the standard enthalpy change of the reac-
tion.
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Given that ∆H=−484 kJ and ∆H◦
f(H2O) = −286 kJ/mol, ∆H◦
f(O2)=0
kJ/mol, and we need to find ∆H◦
f(H2O).
Step 3: Substitute the values into the equation from step 2 and solve for
∆H◦
f(H2O).
−484 kJ = 2(−286) kJ/mol −2(0) kJ/mol
−484 kJ = −572 kJ/mol
Step 4: Calculate the standard enthalpy of formation of water vapor.
−484 kJ = −572 kJ/mol
88 kJ = ∆H◦
f(H2O)
Therefore, the standard enthalpy of formation of water vapor is 88 kJ/mol.
Question 14
Question
Consider the following reaction at 298 K:
2H2O(l)→2H2(g)+O2(g)
Given the standard enthalpies of formation at 298 K: ∆H◦
f(H2O(l)) =
−285.8 kJ/mol, ∆H◦
f(H2(g)) = 0 kJ/mol, and ∆H◦
f(O2(g)) = 0 kJ/mol.
Calculate the standard enthalpy change for this reaction.
11
Solution
Step 1: Write the balanced chemical equation for the reaction.
The balanced chemical equation is:
2H2O(l)→2H2(g)+O2(g)
Step 2: Calculate the standard enthalpy change using the standard en-
thalpies of formation.
The standard enthalpy change for the reaction is given by:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Substitute the values:
∆H◦= [2∆H◦
f(H2(g)) + ∆H◦
f(O2(g))] −[2∆H◦
f(H2O(l))]
∆H◦= [2(0 kJ/mol) + 0 kJ/mol] −[2(−285.8 kJ/mol)]
∆H◦= 571.6 kJ/mol + 571.6 kJ/mol = 1143.2kJ/mol
Therefore, the standard enthalpy change for the reaction is 1143.2kJ/mol.
Question 15
Question
A reaction has a standard enthalpy change of -328 kJ/mol. If 0.50 mol of the
reaction takes place, how much heat is either absorbed or released?
Solution
Step 1: Determine the heat absorbed or released for the given amount of the
reaction. Step 2: Use the standard enthalpy change to calculate the heat.
Step 1: Given: Standard enthalpy change = -328 kJ/mol Number of moles
of reaction, n = 0.50 mol
Step 2: To find the total heat involved in the reaction, we multiply the
standard enthalpy change by the number of moles of the reaction. Total Heat
= Standard enthalpy change ×Number of moles
Substitute the given values into the equation: Total Heat = -328 kJ/mol ×
0.50 mol
Calculate the total heat released or absorbed: Total Heat = -164 kJ
Therefore, when 0.50 mol of the reaction takes place, 164 kJ of heat is
released.
12
Question 16
Question
Given the reaction:
2H2(g) + O2(g)→2H2O(g)
with the following data:
∆H◦
ffor H2O(g) is -241.8 kJ/mol
∆H◦
ffor H2(g) is 0 kJ/mol
∆H◦
ffor O2(g) is 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 provided.
Given reaction: 2H2(g) + O2(g)→2H2O(g)
The standard enthalpy change for the reaction is the sum of the standard
enthalpies of formation of the products minus the sum of the standard enthalpies
of formation of the reactants:
∆H◦= Σn∆H◦
f(products) −Σm∆H◦
f(reactants)
Where nand mare the stoichiometric coefficients of the products and reac-
tants, respectively.
By substituting the given values: ∆H◦= 2∆H◦
f(H2O)−(2∆H◦
f(H2) +
∆H◦
f(O2))
∆H◦= 2(−241.8 kJ/mol) −(2(0 kJ/mol) + 0 kJ/mol)
∆H◦=−483.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction is ∆H◦=
−483.6 kJ/mol.
Question 17
Question
A sample of 2.50 moles of propane (C3H8) is burned in excess oxygen gas.
The combustion of propane is represented by the following balanced chemical
equation:
C3H8(g) + 5 O2(g) →3 CO2(g) + 4 H2O (g)
Given that the standard enthalpy of formation of CO2is -393.5 kJ/mol
and the standard enthalpy of formation of H2O is -285.8 kJ/mol, calculate the
standard enthalpy change (∆H◦) for the combustion of propane.
13
Solution
Step 1: Calculate the standard enthalpy change for the combustion of propane
by using the enthalpy of formation values provided for CO2and H2O.
Given: Standard enthalpy of formation of CO2: -393.5 kJ/mol Standard
enthalpy of formation of H2O: -285.8 kJ/mol Coefficient of CO2in the balanced
equation: 3 Coefficient of H2O in the balanced equation: 4
The standard enthalpy change for the combustion of propane can be calcu-
lated using the equation:
∆H◦=Xν∆H◦
f
Where: νrepresents the stoichiometric coefficient ∆H◦
frepresents the stan-
dard enthalpy of formation
Plugging the values into the equation:
∆H◦= (3 × −393.5 kJ/mol) + (4 × −285.8 kJ/mol)
Step 2: Calculate the standard enthalpy change.
∆H◦= (−1180.5 kJ/mol) + (−1143.2 kJ/mol)
∆H◦=−2323.7 kJ/mol
Therefore, the standard enthalpy change for the combustion of propane is
−2323.7 kJ/mol .
Question 18
Question
For the reaction:
2H2(g)+O2(g)→2H2O(l)
the enthalpy change (∆H) is −483.6 kJ/mol. Calculate the amount of heat (in
kJ) released when 1.00 g of hydrogen gas is burned in excess oxygen.
Solution
Step 1: Calculate the number of moles of hydrogen gas in 1.00 g.
Molar mass of H2= 2.02 g/mol
Number of moles of H2=1.00 g
2.02 g/mol
= 0.495 mol
14
Step 2: Use the stoichiometry of the reaction to calculate the moles of water
produced. Since 2 moles of hydrogen gas produce 2 moles of water, the 0.495
mol of hydrogen gas will produce 0.495 mol of water.
Step 3: Calculate the amount of heat released using the given enthalpy
change.
Heat released = moles of reaction ×∆H
= 0.495 mol × −483.6 kJ/mol
=−239.322 kJ
≈ −239 kJ
Therefore, when 1.00 g of hydrogen gas is burned in excess oxygen, approx-
imately 239 kJ of heat is released.
Question 19
Question
Calculate the enthalpy change (∆H) for the following reaction:
2H2(g) + O2(g)→2H2O(l)
given the following bond dissociation energies:
H−H= 436 kJ/mol
O=O= 498 kJ/mol
H−O= 463 kJ/mol
Solution
Step 1: Calculate the bond energy for the reactants and products. We have:
Bond energy of 2H-H + Bond energy of 1O=O = 2(436) + 498 = 1370 kJ
Bond energy of 4H-O = 4(463) = 1852 kJ
Step 2: Calculate the change in enthalpy (∆H) for the reaction using the
bond energies calculated above.
∆H=XBonds broken −XBonds formed
= (1370) −(1852) = −482 kJ
Therefore, the enthalpy change for the given reaction is ∆H=−482 kJ.
15
Question 20
Question
Calculate the enthalpy change, ∆H, for the combustion of 1 mole of ethylene
(C2H4) gas according to the following reaction:
C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
Given the following standard enthalpies of formation:
∆H◦
f(C2H4) = 52.3 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy of reaction using the standard en-
thalpies of formation.
∆H◦
r=X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
r= [2 ×∆H◦
f(CO2)+2×∆H◦
f(H2O)] −[∆H◦
f(C2H4)+3×∆H◦
f(O2)]
∆H◦
r= [2 ×(−393.5) + 2 ×(−285.8)] −[52.3+3×0]
∆H◦
r= [−787.0−571.6] −52.3
∆H◦
r=−1358.6 kJ/mol
Hence, the enthalpy change for the combustion of 1 mole of ethylene gas is
∆H=−1358.6 kJ/mol.
Question 21
Question
Calculate the enthalpy change for the combustion of n-hexane (C6H14) given
that the standard enthalpy of formation for n-hexane is -199.3 kJ/mol and the
standard enthalpy of formation for carbon dioxide and water are -393.5 kJ/mol
and -285.8 kJ/mol, respectively.
16
Solution
Step 1: Write the balanced chemical equation for the combustion of n-hexane:
C6H14 + 9O2→6CO2+ 7H2O
Step 2: Calculate the standard enthalpy change for the combustion of n-
hexane using the standard enthalpies of formation for the reactants and prod-
ucts:
∆H=X∆Hf(products) −X∆Hf(reactants)
Step 3: Substitute the given standard enthalpies of formation into the equa-
tion and calculate:
∆H= [6(−393.5) + 7(−285.8)] −(−199.3)
∆H= [−2361 + (−2000.6)] −(−199.3)
∆H=−4361.6 + 199.3
∆H=−4162.3 kJ/mol
Therefore, the enthalpy change for the combustion of n-hexane is -4162.3
kJ/mol.
Question 22
Question
Calculate the standard enthalpy change (∆H◦) for the reaction:
2 Fe2O3(s) + 3 C(s) →4 Fe(s) + 3 CO2(g)
given the following standard enthalpies of formation:
∆H◦
f(Fe2O3(s)) = −826.2 kJ/mol
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(Fe(s)) = 0 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
17
Solution
Step 1: Calculate the standard enthalpy change for the reaction based on the
standard enthalpies of formation.
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion.
∆H◦=4×∆H◦
f(Fe(s)) + 3 ×∆H◦
f(CO2(g))−2×∆H◦
f(Fe2O3(s)) + 3 ×∆H◦
f(C(s))
Step 3: Substitute the values and simplify.
∆H◦= (4 ×0 kJ/mol + 3 × −393.5 kJ/mol)−(2 × −826.2 kJ/mol + 3 ×0 kJ/mol)
∆H◦= (0 kJ/mol −1180.5 kJ/mol) −(−1652.4 kJ/mol)
∆H◦=−1180.5 kJ/mol + 1652.4 kJ/mol
∆H◦= 471.9 kJ/mol
Therefore, the standard enthalpy change for the reaction is 471.9 kJ/mol .
Question 23
Question
For a certain reaction at constant pressure, the enthalpy change (∆H) is found
to be −487 kJ. If the reaction is exothermic, predict whether the entropy change
(∆S) is positive, negative, or zero. Justify your answer.
Solution
Step 1: Recall the relationship between ∆Hand ∆S. The relationship between
∆Hand ∆Sis given by the equation:
∆G= ∆H−T∆S
where ∆Gis the Gibbs free energy change, ∆His the enthalpy change, ∆Sis
the entropy change, and Tis the temperature in Kelvin.
Step 2: Understand the properties of an exothermic reaction. In an exother-
mic reaction, heat is released to the surroundings (∆H < 0). This means that
the system loses heat and the surroundings gain heat.
Step 3: Determine the sign of ∆Sfor an exothermic reaction. Since the
reaction is exothermic (∆H < 0) and Tis a positive value (since temperature
is always positive in Kelvin), for the equation ∆G= ∆H−T∆Sto be nega-
tive (indicating a spontaneous reaction), ∆Smust be positive. Therefore, the
entropy change (∆S) for an exothermic reaction is positive.
Therefore, for the given exothermic reaction with ∆H=−487 kJ, the en-
tropy change (∆S) is positive.
18
Question 24
Question
Calculate the enthalpy change (∆H) for the combustion of 1 mole of propane
gas (C3H8) at constant pressure using the following information:
The enthalpy of formation of C3H8is −103.8 kJ/mol.
The balanced chemical equation for the combustion of propane is:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
The enthalpy of formation of CO2is −393.5 kJ/mol and H2Ois −285.8
kJ/mol.
Solution
Step 1: Find the enthalpy change for the combustion of propane from the given
enthalpies of formation.
∆H=X[products] −X[reactants]
= [3 ×Hf(CO2)] + [4 ×Hf(H2O)] −[Hf(C3H8)+5×Hf(O2)]
= [3 ×(−393.5 kJ/mol)] + [4 ×(−285.8 kJ/mol)] −(−103.8 kJ/mol) −(5 ×0 kJ/mol)
=−1180.5 kJ/mol −1143.2 kJ/mol −(−103.8 kJ/mol)
=−2176.9 kJ/mol
Therefore, the enthalpy change for the combustion of 1 mole of propane gas
is −2176.9 kJ/mol.
Question 25
Question
Calculate the standard enthalpy change for the reaction
2H2(g) + O2(g)→2H2O(g)
given the following standard enthalpy of formation values:
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H◦
f(O2(g)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 kJ/mol
19
Solution
Step 1: Write the balanced chemical equation for the reaction:
2H2(g) + O2(g)→2H2O(g)
Step 2: Calculate the standard enthalpy change using the standard enthalpy
of formation values.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
∆H◦= (2 ×∆H◦
f(H2O(g))) −(2 ×∆H◦
f(H2(g)) + ∆H◦
f(O2(g)))
∆H◦= (2 × −241.8 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
∆H◦=−483.6 kJ/mol
Therefore, the standard enthalpy change for the given reaction is −483.6 kJ/mol.
Question 26
Question
Given the following reaction:
2A(g) + B(g)→3C(g)+4D(g)
with the following enthalpies of formation:
∆H◦
f(kJ/mol)
A(g): 100, B(g): 50, C(g): 200, D(g): 150.
Calculate the standard enthalpy change (∆H◦) for the reaction.
Solution
Step 1: Calculate the change in enthalpy (∆H◦) using the enthalpies of forma-
tion:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the stoichiometric coefficients for the products and reactants
respectively.
Step 2: Substitute the given values into the equation and solve:
∆H◦= (3 ×200 + 4 ×150) −(2 ×100 + 50)
∆H◦= (600 + 600) −(200 + 50)
∆H◦= 1200 −250
∆H◦= 950 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the reaction is 950
kJ/mol.
20
Question 27
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction at
25
°
C given the standard enthalpies of formation provided:
2C(s) + 3H2(g)→C2H6(g)
∆H◦
f(C2H6(g)) = −84.68 kJ/mol
∆H◦
f(C(s)) = 0 kJ/mol
∆H◦
f(H2(g)) = 0 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 (∆H◦) using the standard
enthalpies of formation.
The standard enthalpy change of a reaction is given by:
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nrepresents the stoichiometric coefficient of the products, mrepresents
the stoichiometric coefficient of the reactants, ∆H◦
fis the standard enthalpy of
formation.
For the reaction:
∆H◦=1×∆H◦
f(C2H6(g))−2×∆H◦
f(C(s)) + 3 ×∆H◦
f(H2(g))
Substitute the given values:
∆H◦= (1 × −84.68) −(2 ×0+3×0)
∆H◦=−84.68 kJ/mol
Therefore, the standard enthalpy change (∆H◦) for the given reaction at
25
°
C is -84.68 kJ/mol.
Question 28
Question
Calculate the enthalpy change (∆H) for the following reaction at 25
°
C:
2A(g)+3B(g)→4C(g)+2D(g)
21
given the following bond dissociation energies:
A-A : 170 kcal/mol
B-B : 200 kcal/mol
C-C : 250 kcal/mol
D-D : 300 kcal/mol
A-B : 250 kcal/mol
C-D : 400 kcal/mol
Solution
Step 1: Calculate the total bond dissociation energy for the reactants and prod-
ucts.
Reactants: 2(A−A) + 3(B−B) + 2(A−B)
= 2(170) + 3(200) + 2(250)
= 340 + 600 + 500
= 1440 kcal/mol
Products: 4(C−C) + 2(D−D) + 2(C−D)
= 4(250) + 2(300) + 2(400)
= 1000 + 600 + 800
= 2400 kcal/mol
Step 2: Calculate the enthalpy change (∆H) for the reaction using the bond
dissociation energies.
∆H= Total energy of bonds broken −Total energy of bonds formed
= Reactants −Products
= 1440 kcal/mol −2400 kcal/mol
=−960 kcal/mol
Therefore, the enthalpy change for the reaction is ∆H=−960 kcal/mol.
Question 29
Question
Calculate the standard enthalpy change (∆H◦) for the following reaction:
2CH4(g)+3O2(g)−→ 2CO2(g)+4H2O(l)
22
given the following standard enthalpies of formation:
∆H◦
f(CH4) = −74.8 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(O2) = 0 kJ/mol
Solution
Step 1: Write the given equation in terms of standard enthalpies of formation.
The standard enthalpy change can be calculated using the equation:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Plugging in the values from the given standard enthalpies of formation, we get:
∆H◦= [2 ×∆H◦
f(CO2)+4×∆H◦
f(H2O)] −[2 ×∆H◦
f(CH4)+3×∆H◦
f(O2)]
Step 2: Substitute the values and calculate. Plugging in the values:
∆H◦= [2(−393.5) + 4(−285.8)] −[2(−74.8) + 3(0)]
= [−787 −1143.2] −[−149.6]
=−1930.2 + 149.6
=−1780.6 kJ/mol
Therefore, the standard enthalpy change for the reaction is ∆H◦=−1780.6 kJ/mol.
Question 30
Question
For the reaction
2H2O(g)→2H2(g) + O2(g)
at 25
°
C and 1 atm, the change in enthalpy (∆H) is 483.6 kJ. Calculate the
change in internal energy (∆U) for the reaction.
Given: ∆H=−483.6 kJ, R = 8.314 J/(mol·K)
Solution
Step 1: To find ∆U, we can use the equation:
∆U= ∆H−∆ngRT
where ∆ngis the change in the number of moles of gas and R is the ideal gas
constant.
23
Step 2: First, we need to find ∆ng, which is the difference in the number
of moles of gas on the product side and the reactant side. In this case, ∆ng=
2−2 = 0 (2 moles of water gas turn into 2 moles of gas products).
Step 3: Now, substitute the given values into the equation:
∆U=−483.6 kJ −0×8.314 J/(mol ·K) ×(25 + 273.15) K
Step 4: Perform the calculation:
∆U=−483.6×103J = −483600 J
Therefore, the change in internal energy for the reaction is −483600 J or
−483.6 kJ.
24
