CHEM 107 - ESSENTIALS OF
GENERAL AND ORGANIC
CHEMISTRY - Enthalpy
Question Bank - Set 3
Liberty University
Question 1
Question
Given that the standard enthalpy change of formation (∆H◦
f) for gaseous ni-
trogen dioxide (NO2(g)) is 33.2 kJ/mol and for gaseous dinitrogen tetroxide
(N2O4(g)) is 9.16 kJ/mol, calculate the standard enthalpy change of the reac-
tion:
2NO2(g)→N2O4(g)
Solution
Step 1: Write the thermochemical equation for the reaction in terms of the given
enthalpies of formation.
2NO2(g)→N2O4(g)
Step 2: Determine the enthalpy change for the reaction using Hess’s Law,
which states that the total enthalpy change for a chemical reaction is the same
regardless of the number of steps taken to reach the reaction.
∆H◦
r= Σ∆H◦
f(products) −Σ∆H◦
f(reactants)
Step 3: Substitute the given enthalpies of formation into the equation and
calculate the standard enthalpy change of the reaction.
∆H◦
r= [2 ×∆H◦
f(N2O4)] −[2 ×∆H◦
f(NO2)]
∆H◦
r= [2 ×9.16] −[2 ×33.2]
∆H◦
r= 18.32 −66.4
∆H◦
r=−48.08 kJ/mol
Therefore, the standard enthalpy change of the reaction is −48.08 kJ/mol.
Question 2
Question
Given that the enthalpy change for the reaction 2A(g)+ 3B(g) →C(g) is ∆H=
−1200 kJ. If the enthalpy change for the reaction A(g) →B(g) is -300 kJ,
calculate the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g).
Solution
Step 1: Write the given enthalpy changes for the reactions. The enthalpy change
for the reaction 2A(g) + 3B(g) →C(g) is ∆H=−1200 kJ, and for A(g) →B(g)
is -300 kJ.
Step 2: Calculate the enthalpy change for the desired reaction using Hess’s
Law. To calculate the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g),
we need to consider the given reactions and manipulate them to obtain the
desired reaction.
Step 3: Manipulate the given reactions to obtain the desired reaction. 1.
A(g) →B(g) 2. 2A(g) + 3B(g) →C(g)
To get the desired reaction, we can reverse reaction 1 and multiply reaction
2 by 2: 1. B(g) →A(g) 2. 4A(g) + 6B(g) →2C(g)
Step 4: Determine the enthalpy change for the desired reaction. Since we are
using the reversed reaction 1 and the multiplied reaction 2, we need to change
the signs of the enthalpy changes for those reactions to obtain the enthalpy
change for the desired reaction. −(−300 kJ)+2(−1200 kJ) = 600 kJ−2400 kJ =
−1800 kJ
Therefore, the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g) is
−1800 kJ .
Question 3
Question
A reaction at constant pressure releases 150 kJ of heat and does 50 kJ of work
on the surroundings. If the internal energy decreases by 100 kJ, determine the
change in enthalpy for the reaction.
2
Solution
Step 1: Recall the relationship between enthalpy (H), internal energy (U), and
work done (W):
∆H= ∆U+P∆V
Step 2: Given that the heat released is 150 kJ, the work done on the sur-
roundings is 50 kJ, and the internal energy decreases by 100 kJ, we can rewrite
the equation as:
−∆H=−∆U−P∆V
−∆H=−100 kJ + 50 kJ
Step 3: Solving for the change in enthalpy (∆H), we get:
∆H=−(−100 + 50)
∆H= 50 kJ
Thus, the change in enthalpy for the reaction is 50 kJ.
Question 4
Question
Calculate the change in enthalpy (∆H) when 50.0 g of steam at 100◦C is con-
verted to ice at -10◦C. The specific heat capacity of steam is 2.01 J/g◦C, the
specific heat capacity of water is 4.18 J/g◦C, the specific heat capacity of ice is
2.09 J/g◦C, the heat of fusion of water is 333.5 J/g, and the heat of vaporization
of water is 2257 J/g.
Solution
Step 1: Calculate the heat required to cool the steam to 0◦C and condense it
to water.
The heat required to cool the steam to 0◦C is given by:
q=m×Csteam ×∆T
q= 50.0 g ×2.01 J/g◦C×(0 −100)◦C
q=−10050 J
The heat of vaporization of water is then released when the steam condenses
to water:
q=m×heat of vaporization
q= 50.0 g ×2257 J/g
q=−112850 J
3
Therefore, the total heat released to cool the steam to 0◦C and condense it
to water is:
qtotal =−10050 J −112850 J = −123900 J
Step 2: Calculate the heat required to cool the water to -10◦C and freeze it
to ice.
The heat required to cool the water to -10◦C is given by:
q=m×Cwater ×∆T
q= 50.0 g ×4.18 J/g◦C×(−10 −0)◦C
q=−2090 J
The heat of fusion of water is then released when the water freezes to ice:
q=m×heat of fusion
q= 50.0 g ×333.5 J/g
q=−16675 J
Therefore, the total heat released to cool the water to -10◦C and freeze it to
ice is:
qtotal =−2090 J −16675 J = −18765 J
Step 3: Calculate the total change in enthalpy (∆H). The total change in
enthalpy is the sum of the enthalpy changes for each step:
∆H=−123900 J + (−18765 J) = −142665 J
Therefore, the change in enthalpy (∆H) when 50.0 g of steam at 100◦C is
converted to ice at -10◦C is -142665 J.
Question 5
Question
A reaction is carried out in a bomb calorimeter at constant volume, and it
releases 250 kJ of heat. If the bomb calorimeter itself absorbs 20 kJ of heat
during the reaction, what is the enthalpy change (∆H) for the reaction?
Solution
Step 1: The enthalpy change for the reaction can be calculated using the equa-
tion:
∆H=qreaction −qcalorimeter
where qreaction is the heat released by the reaction and qcalorimeter is the heat
absorbed by the calorimeter.
4
Step 2: Given that the reaction releases 250 kJ of heat and the calorimeter
absorbs 20 kJ of heat, we can substitute these values into the equation:
∆H= 250 kJ −20 kJ
Step 3: Simplifying the above expression, we find:
∆H= 230 kJ
Step 4: Therefore, the enthalpy change for the reaction is 230 kJ .
Question 6
Question
A reaction is carried out in a calorimeter at constant pressure. The initial
temperature of the reactants is 25◦C and the final temperature after mixing is
38◦C. The calorimeter contains 80 g of water and has a heat capacity of 10
Step 1: Calculate the heat absorbed by the water in the calorimeter. The heat
absorbed by the water can be calculated using the formula:
q=mc∆T
where: q= heat absorbed (in joules), m= mass of water (in kg), c= specific
heat capacity of water (4.18 J/g◦C), ∆T= temperature change (in ◦C).
Given that the mass of water is 80 g and the temperature change is 38◦C−
25◦C = 13◦C, we can substitute these values into the formula:
q= (0.08 kg)(4.18 J/g◦C)(13◦C)
Calculating q:
q= 4.18(0.08)(13)
q= 4.18 ×0.104
q= 0.43344 kJ
Therefore, the heat absorbed by the water is 0.43344 kJ.
Step 2: Calculate the enthalpy change of the reaction. The enthalpy change
of the reaction can be calculated using the formula:
∆H=−q
Given that the reaction releases 500 kJ of heat, we can substitute this value into
the formula:
∆H=−500 kJ
Therefore, the enthalpy change of the reaction is −500 kJ.
5
Question 7
Question
A sample of water initially at 15◦C is heated, and its temperature is raised to
85◦C. Calculate the change in enthalpy of the water if the mass of the sample
is 500 g. Assume the specific heat capacity of water is 4.18 J/g◦C.
Solution
Step 1: Calculate the heat energy absorbed by the water to raise its temperature
from 15◦C to 85◦C using the formula:
q=mc∆T
where qis the heat energy, mis the mass of the sample, cis the specific heat
capacity, and ∆Tis the change in temperature.
Step 2: Substitute the given values into the formula:
q= (500 g)(4.18 J/g◦C)(85 −15)
Step 3: Calculate the heat energy absorbed by the water:
q= (500)(4.18)(70)
Step 4: q= 14690 J
Step 5: Calculate the change in enthalpy using the formula:
∆H=q
Step 6: Substitute the value of qinto the formula:
∆H= 14690 J
Answer: The change in enthalpy of the water is 14690 J.
Question 8
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas react
with excess oxygen gas to produce carbon dioxide gas and water vapor, according
to the following balanced equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpy changes:
∆H◦
f(CH4) = −74.6 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
6
Solution
Step 1: Calculate the standard enthalpy change of the reaction using standard
enthalpies of formation.
∆H◦
rxn =Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion.
∆H◦
rxn = [1(−393.5 kJ/mol) + 2(−285.8 kJ/mol)] −[(−74.6 kJ/mol)]
Step 3: Calculate the change in enthalpy of the reaction.
∆H◦
rxn = [−393.5 kJ/mol −571.6 kJ/mol] −[−74.6 kJ/mol]
∆H◦
rxn =−965.1 kJ/mol + 74.6 kJ/mol
∆H◦
rxn =−890.5 kJ/mol
Therefore, the change in enthalpy (∆H) for the reaction of 5.00 moles of
methane gas is -890.5 kJ.
Question 9
Question
Calculate the change in enthalpy (∆H) when 5 moles of methane gas (CH4)
react with excess oxygen gas to produce carbon dioxide gas (CO2) and water
vapor (H2O) according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpies of formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(CH4) = −74.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the given reaction using the
standard enthalpies of formation.
The standard enthalpy change, ∆H◦, for the reaction can be calculated using
the formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
7
Substitute the given standard enthalpies of formation into the formula:
∆H◦= [1(−393.5) + 2(−285.8)] −[1(−74.8) + 2(0)]
Simplify the expression:
∆H◦= (−393.5−571.6) −(−74.8)
∆H◦=−965.1 + 74.8
∆H◦=−890.3 kJ/mol
Therefore, the change in enthalpy (∆H) for the given reaction is -890.3
kJ/mol.
Question 10
Question
Given that the enthalpy change (∆H) for the reaction C(s) + O2(g) →CO2(g)
is -393.5 kJ/mol, calculate the enthalpy change when 10.0 g of carbon reacts
with excess oxygen to form carbon dioxide.
Solution
Step 1: Calculate the number of moles of carbon reacting.
Molar mass of carbon = 12.01 g/mol
Moles of carbon = 10.0 g
12.01 g/mol
= 0.833 mol
Step 2: Use the given enthalpy change to calculate the enthalpy change for
the reaction.
∆H= ∆H◦×moles of reaction
=−393.5 kJ/mol ×0.833 mol
=−327.6 kJ
Therefore, the enthalpy change when 10.0 g of carbon reacts with excess
oxygen to form carbon dioxide is -327.6 kJ.
Question 11
Question
The enthalpy change (∆H) for the reaction
2A(g)+3B(l)→C(s)+4D(g)
8
is −1264 kJ. If the standard enthalpies of formation are −17 kJ/mol for A, −66
kJ/mol for B, −394 kJ/mol for C, and 109 kJ/mol for D, calculate the standard
enthalpy change of the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the formation of products.
The standard enthalpy change for the formation of products is given by:
∆Hproducts =Xn∆Hf
where nis the stoichiometric coefficient and ∆Hfis the standard enthalpy of
formation.
Substitute values:
∆Hproducts = (1)(−394 kJ/mol) + (4)(109 kJ/mol)
∆Hproducts =−394 kJ/mol + 436 kJ/mol
∆Hproducts = 42 kJ/mol
Step 2: Calculate the standard enthalpy change for the formation of reac-
tants. The standard enthalpy change for the formation of reactants is given
by:
∆Hreactants =Xn∆Hf
where nis the stoichiometric coefficient and ∆Hfis the standard enthalpy of
formation.
Substitute values:
∆Hreactants = (2)(−17 kJ/mol) + (3)(−66 kJ/mol)
∆Hreactants =−34 kJ/mol −198 kJ/mol
∆Hreactants =−232 kJ/mol
Step 3: Calculate the standard enthalpy change of the reaction. The stan-
dard enthalpy change of the reaction is given by:
∆H= ∆Hproducts −∆Hreactants
Substitute values:
∆H= 42 kJ/mol −(−232 kJ/mol)
∆H= 42 kJ/mol + 232 kJ/mol
∆H= 274 kJ/mol
Therefore, the standard enthalpy change of the reaction is 274 kJ/mol.
9
Question 12
Question
Consider a reaction where 2 moles of liquid water are converted to steam at 1
atm and 100
°
C. The molar enthalpy of vaporization of water is 40.79 kJ/mol.
Calculate the change in enthalpy for this reaction.
Solution
Step 1: Calculate the heat required to vaporize 1 mole of water. Given that the
molar enthalpy of vaporization of water is 40.79 kJ/mol, the heat required to
vaporize 1 mole of water is 40.79 kJ.
Step 2: Calculate the heat required to vaporize 2 moles of water. Since we
have 2 moles of water, the total heat required to vaporize 2 moles is:
40.79 kJ/mol ×2 mol = 81.58 kJ
Step 3: Calculate the change in enthalpy. The change in enthalpy for this
reaction is the same as the heat required to vaporize 2 moles of water, which is
81.58 kJ. Therefore, the change in enthalpy is 81.58 kJ.
Question 13
Question
Calculate the change in enthalpy for a reaction where 2 moles of nitrogen gas
react with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. The
enthalpies of formation for nitrogen gas, hydrogen gas, and ammonia gas are 0
kJ/mol, 0 kJ/mol, and -46 kJ/mol, respectively.
Solution
Step 1: Write the balanced chemical equation for the reaction.
N2(g)+3H2(g)→2NH3(g)
Step 2: Calculate the total change in enthalpy using the enthalpies of for-
mation. The change in enthalpy (∆H) can be calculated using the formula:
∆H=Xν∆Hf(products) −Xν∆Hf(reactants)
where νrepresents the stoichiometric coefficients and ∆Hfrepresents the en-
thalpy of formation.
Substitute the values into the formula:
∆H= 2(−46 kJ/mol) −[1(0 kJ/mol) + 3(0 kJ/mol)]
10
Step 3: Simplify the expression to find the change in enthalpy.
∆H=−92 kJ/mol −0 kJ/mol
∆H=−92 kJ/mol
Therefore, the change in enthalpy for the reaction is -92 kJ/mol.
Question 14
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with ∆H=−571.6 kJ, calculate the enthalpy change when 4.00 g of H2is
burned in excess O2to form liquid water.
Solution
Step 1: Calculate the moles of hydrogen gas. Given: - Molar mass of hydrogen,
M(H2)=2.016 g/mol - Mass of hydrogen gas, m(H2)=4.00 g Using the formula
n=m
M, where nis moles, we have:
n(H2) = 4.00 g
2.016 g/mol = 1.984 mol
Step 2: Use the stoichiometry of the reaction to find the enthalpy change for
1.00 mol of hydrogen gas. From the balanced chemical equation, the enthalpy
change for the reaction is −571.6 kJ for 2 moles of H2. Therefore, for 1 mole of
H2, the enthalpy change is:
−571.6 kJ
2=−285.8 kJ/mol
Step 3: Calculate the enthalpy change for the given mass of hydrogen gas.
Since 1.984 moles of H2are involved, the enthalpy change for 1.984 moles is:
1.984 mol ×(−285.8 kJ/mol) = −566.7 kJ
Answer: The enthalpy change when 4.00 g of H2is burned in excess O2to
form liquid water is −566.7 kJ.
Question 15
Question
Calculate the change in enthalpy (∆H) for the combustion of propane gas
(C3H8) at constant pressure. The balanced chemical equation for the com-
bustion of propane is:
C3H8(g) + 5
2O2(g)→3CO2(g)+4H2O(g)
11
Given the standard enthalpies of formation (∆H◦
f) for C3H8(g), CO2(g), and
H2O(g) are -103.9 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Calculate ∆H◦for the combustion of propane using the standard en-
thalpies of formation of the reactants and products.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the coefficients of the products and reactants, respectively
in the balanced chemical equation.
Step 2: Substitute the values of the standard enthalpies of formation pro-
vided into the equation.
∆H◦= [3(−393.5) + 4(−241.8)] −[−103.9] kJ/mol
∆H◦= (−1180.5−967.2) + 103.9 kJ/mol = −2043.8 kJ/mol
Therefore, the change in enthalpy for the combustion of propane gas at
constant pressure is ∆H=−2043.8 kJ/mol.
Question 16
Question
A certain reaction has an enthalpy change of -92 kJ/mol. If 2.50 moles of the
reactant are consumed in the reaction, what is the total enthalpy change for the
reaction?
Solution
Step 1: Determine the total enthalpy change for the reaction. Step 2: Use the
given enthalpy change per mole to find the total enthalpy change.
Step 1: Enthalpy change per mole = −92 kJ/molMoles of reactant consumed = 2.50 mol
Step 2: Total enthalpy change = Enthalpy change per mole×Moles of reactant consumed
Total enthalpy change = (−92 kJ/mol) ×(2.50 mol)
Total enthalpy change = −230 kJ
Therefore, the total enthalpy change for the reaction when 2.50 moles of the
reactant are consumed is -230 kJ.
12
Question 17
Question
A reaction has a standard enthalpy change of -394 kJ/mol. If the reaction is
exothermic, what can be said about the enthalpy of the products compared to
the enthalpy of the reactants?
Solution
Step 1: Recall that for an exothermic reaction, the standard enthalpy change is
negative.
Step 2: In this case, the standard enthalpy change is -394 kJ/mol, which
means that the products have lower enthalpy than the reactants.
Step 3: Therefore, the enthalpy of the products is lower than the enthalpy
of the reactants for this exothermic reaction.
Question 18
Question
A sample of nitrogen gas undergoes an isobaric process at a pressure of 1 atm.
During the process, the nitrogen gas absorbs 500 J of heat and performs 300 J
of work. If the initial enthalpy of the nitrogen gas is 2000 J, what is the final
enthalpy of the gas?
Solution
Step 1: Recall the definition of enthalpy: Enthalpy (H) is defined as the sum of
the internal energy (U) of a system and the product of pressure (P) and volume
(V) of the system:
H=U+P V
Step 2: Since the process is isobaric, the change in enthalpy can be expressed
as:
∆H= ∆U+P∆V
Step 3: We are given that the nitrogen gas absorbs 500 J of heat and performs
300 J of work. Therefore, the change in internal energy can be calculated as:
∆U=Q−W= 500 J −300 J = 200 J
Step 4: Substituting the values into the formula for the change in enthalpy:
∆H= 200 J + 1 atm ×∆V
Step 5: But we also know that:
∆H=Hfinal −Hinitial =Hfinal −2000 J
13
Step 6: Equating the two expressions for ∆H:
200 J + 1 atm ×∆V=Hfinal −2000 J
Step 7: Given that the pressure is 1 atm, we can simplify the equation to:
200 J + 1 atm ×∆V=Hfinal −2000 J
Step 8: Since ∆V=W
P, we can substitute in the values to find the change
in volume:
∆V=300 J
1 atm = 300 L ·atm
Step 9: Substituting ∆Vback into the equation and solving for Hfinal:
200 J + 1 atm ×300 L ·atm = Hfinal −2000 J
Step 10: Calculating the final enthalpy:
Hfinal = 200 J + 300 J = 2500 J
Therefore, the final enthalpy of the nitrogen gas is 2500 J.
Question 19
Question
A certain chemical reaction has an enthalpy change of ∆H=−289 kcal/mol. If
1.50 moles of the reactant is consumed in the reaction, what is the total change
in enthalpy in kilocalories?
Solution
Step 1: Determine the total change in enthalpy for 1 mole of the reactant.
Given: ∆H=−289 kcal/mol
Step 2: Calculate the total change in enthalpy for 1 mole of the reactant.
∆Htotal = ∆H×moles of reactant
∆Htotal =−289 kcal/mol ×1.50 moles
∆Htotal =−433.5 kcal
Therefore, the total change in enthalpy for 1.50 moles of the reactant is
−433.5 kcal.
14
Question 20
Question
A reaction is carried out at constant pressure where 100.0 g of a substance
X is converted to substance Y. The reaction releases 1500 J of heat and the
temperature of the surroundings increases by 2.5
°
C. The specific heat capacity of
the surroundings is 4.18 J/(g
°
C). Calculate the enthalpy change for the reaction.
Solution
Step 1: Calculate the heat gained by the surroundings. Given that the temper-
ature change (∆T) is 2.5
°
C and the specific heat capacity of the surroundings
(C) is 4.18 J/(g
°
C), we can use the formula:
q=C×m×∆T
where: q= heat gained by the surroundings, C= specific heat capacity of the
surroundings, m= mass of the surroundings, ∆T= temperature change of the
surroundings.
Substitute the values:
q= 4.18 J/(g
°
C) ×m×2.5
°
C
Step 2: Calculate the mass of the surroundings. Since the reaction releases
heat, the surroundings gain this heat. We can find the mass of the surroundings
using the formula:
q=qreaction =msurroundings ×C×∆T
Rearranging the formula:
msurroundings =qreaction
C×∆T
Step 3: Calculate the enthalpy change for the reaction. The enthalpy change
of the reaction (∆Hreaction) is equal in magnitude but opposite in sign to the
heat gained by the surroundings:
∆Hreaction =−q
Substitute the value obtained for qin Step 1 and calculate ∆Hreaction.
Question 21
Question
A certain reaction has an enthalpy change of ∆H=−486 kJ. If 2.00 moles of
this reaction take place, what is the heat associated with this reaction?
15
Solution
Step 1: Determine the heat associated with the reaction per mole. Given that
the enthalpy change for the reaction is ∆H=−486 kJ, we can find the heat per
mole by dividing by the number of moles:
Heat per mole = ∆H
moles =−486 kJ
2.00 mol
Heat per mole = −243 kJ/mol
Step 2: Calculate the total heat associated with the reaction. To find the
total heat for 2.00 moles of the reaction, we multiply the heat per mole by the
number of moles:
Total heat = Heat per mole ×moles
Total heat = −243 kJ/mol ×2.00 mol
Total heat = −486 kJ
Therefore, the total heat associated with the reaction when 2.00
moles take place is -486 kJ.
Question 22
Question
Calculate the enthalpy change for the reaction below at 298 K:
2H2(g)+O2(g)→2H2O(l)
Given the following standard enthalpy changes of formation: ∆H◦
f[H2(g)] = 0
kJ/mol, ∆H◦
f[O2(g)] = 0 kJ/mol, and ∆H◦
f[H2O(l)] = −285.8 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and apply Hess’s Law. The
enthalpy change for the reaction can be calculated using the standard enthalpy
changes of formation for the reactants and products. We can use Hess’s Law to
obtain the enthalpy change for the desired reaction.
Balanced reaction:
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate the enthalpy change using the standard enthalpy changes
of formation. The enthalpy change for the reaction can be calculated as follows:
∆H◦
rxn =X∆H◦
products −X∆H◦
reactants
16
Substitute the given values:
∆H◦
rxn = 2(∆H◦
f[H2O(l)]) −[2(∆H◦
f[H2(g)]) + ∆H◦
f[O2(g)]]
∆H◦
rxn = 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦
rxn =−571.6 kJ/mol
Therefore, the enthalpy change for the reaction is -571.6 kJ/mol at 298 K.
Question 23
Question
A reaction involving the formation of ammonia gas is represented by the follow-
ing equation:
N2(g)+3H2(g)→2NH3(g)
Given the standard enthalpies of formation (∆H◦
f) for N2(g), H2(g), and N H3(g)
are 0 kJ/mol, 0 kJ/mol, and −46 kJ/mol, respectively, calculate the standard
enthalpy change of the reaction.
Solution
Step 1: Write the balanced chemical equation and standard enthalpy change
formula. The balanced chemical equation is:
N2(g)+3H2(g)→2NH3(g)
The standard enthalpy change of the reaction can be calculated using the for-
mula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Find the standard enthalpy change of the reaction. We know the
formation enthalpies of the products and reactants:
∆H◦
f(NH3) = −46 kJ/mol
∆H◦
f(N2) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Substitute these values into the formula:
∆H◦= [2 ×∆H◦
f(NH3)] −[1 ×∆H◦
f(N2)+3×∆H◦
f(H2)]
∆H◦= [2 ×(−46 kJ/mol)] −[1 ×0 kJ/mol + 3 ×0 kJ/mol]
∆H◦=−92 kJ/mol
Therefore, the standard enthalpy change of the reaction is ∆H◦=−92 kJ/mol.
17
Question 24
Question
Calculate the change in enthalpy (∆H) when 2 moles of nitrogen gas react
completely with 1 mole of oxygen gas to form nitrogen dioxide gas according to
the following balanced equation:
2N2(g)+O2(g)→2NO2(g)
Given that the standard enthalpies of formation (∆H◦
f) for N2(g), O2(g), and
NO2(g) are 0 kJ/mol, 0 kJ/mol, and 33.2 kJ/mol, respectively.
Solution
Step 1: Write the equation for the change in enthalpy using the standard en-
thalpies of formation:
∆H=Xν∆H◦
f
Where νrepresents the stoichiometric coefficients in the balanced equation and
∆H◦
fis the standard enthalpy of formation.
Step 2: Determine the change in enthalpy for the given reaction:
∆H= (2 ×∆H◦
f(NO2)) −(2 ×∆H◦
f(N2)+∆H◦
f(O2))
Step 3: Substitute the given values for the standard enthalpies of formation:
∆H= (2 ×33.2 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
Step 4: Calculate the change in enthalpy:
∆H= 66.4 kJ/mol
Therefore, the change in enthalpy for the given reaction is 66.4 kJ/mol.
Question 25
Question
A reaction involving the conversion of graphite to diamond at standard condi-
tions has an enthalpy change of 2.9 kJ/mol. Calculate the minimum pressure
required to convert 1 mole of graphite to diamond at 25
°
C.
Solution
Step 1: Write down the enthalpy change of the reaction. Given: ∆H= 2.9
kJ/mol
18
Step 2: Recall the definition of standard enthalpy change. The standard
enthalpy change, ∆H◦, is the enthalpy change when reactants in their standard
states are converted to products in their standard states at 1 atm pressure.
Step 3: Determine the pressure needed for the phase change. Since we want
to convert 1 mole of graphite to diamond under standard conditions, we need
to consider the phase change from graphite to diamond and account for the
enthalpy change.
Step 4: Calculate the minimum pressure required using the enthalpy change
and temperature. The enthalpy change at standard conditions is related to the
pressure change via the equation:
∆H= ∆U+P∆V
At constant pressure, the change in enthalpy, ∆H, is approximately equal to
the heat change, ∆U. We can rearrange and solve for pressure:
P=∆H
∆V
Since the reaction involves a solid to solid phase change, the volume change is
negligible. Hence, ∆V≈0 and the pressure required is:
P=∆H
∆V≈2.9 kJ/mol
0=∞
Therefore, the minimum pressure required to convert 1 mole of graphite to
diamond at 25
°
C is theoretically infinite.
Question 26
Question
An ideal gas undergoes a process in which its enthalpy is given by the expression
H= 3.5T2−0.01T3, where His the enthalpy in kJ/mol and Tis the temperature
in K. Determine the heat capacity at constant pressure, Cp, for this gas in terms
of temperature.
Solution
Step 1: Recall that the heat capacity at constant pressure, Cp, is defined as the
partial derivative of the enthalpy with respect to temperature:
Cp=∂H
∂T P
Step 2: We are given the enthalpy expression H= 3.5T2−0.01T3, so we
differentiate this expression with respect to temperature T:
∂H
∂T =∂
∂T (3.5T2−0.01T3)
19
Step 3: Calculating the derivative term by term, we get:
∂H
∂T = 7T−0.03T2
Step 4: Therefore, the heat capacity at constant pressure is:
Cp= 7 −0.03TJ/mol
·
K
Question 27
Question
A reaction takes place at constant pressure with a change in enthalpy (∆H) of
-276 kJ. If the reaction releases 123 kJ of heat, what is the change in internal
energy (∆U) for the reaction? Assume no other work is done during the reaction.
Solution
Step 1: Recall the relationship between enthalpy (H), internal energy (U), and
pressure (P):
∆H= ∆U+P∆V
Step 2: Since the reaction takes place at constant pressure (∆V= 0), the
equation simplifies to:
∆H= ∆U
Step 3: Given that ∆H=−276 kJ, we can substitute this into the equation:
−276 kJ = ∆U
Step 4: Since ∆U=−276 kJ, this represents the change in internal energy
for the reaction when 276 kJ of heat is released.
Step 5: However, the reaction releases 123 kJ of heat instead of 276 kJ. To
find the change in internal energy when 123 kJ of heat is released, we need to
adjust ∆Uaccordingly.
Step 6: If the reaction releases 276 kJ of heat but only 123 kJ is actually
released, then the change in internal energy when 123 kJ of heat is released can
be calculated as follows:
∆U=123
276 ×(−276 kJ)
Step 7: Simplifying the calculation:
∆U=123
276 ×(−276) = −123
2
Step 8: Therefore, the change in internal energy (∆U) for the reaction when
123 kJ of heat is released is −123
2kJ.
20
Question 28
Question
Calculate the change in enthalpy when 1 mole of methane gas is burned in excess
oxygen at constant pressure according to the following reaction:
CH4(g)+ 2O2(g)→CO2(g)+ 2H2O(g)
Given the standard enthalpies of formation are: ∆H◦
f(CH4(g)) = −74.8 kJ/mol,
∆H◦
f(CO2(g)) = −393.5 kJ/mol, ∆H◦
f(H2O(g)) = −285.8 kJ/mol.
Solution
Step 1: Write the thermochemical equation for the reaction.
CH4(g)+ 2O2(g)→CO2(g)+ 2H2O(g)
Step 2: Calculate the standard enthalpy of reaction using the standard en-
thalpies of formation.
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = [1×∆H◦
f(CO2(g))+2×∆H◦
f(H2O(g))]−[∆H◦
f(CH4(g))+2×∆H◦
f(O2(g))]
∆H◦
rxn = [1 ×(−393.5) + 2 ×(−285.8)] −[(−74.8) + 2 ×0]
∆H◦
rxn = (−393.5−571.6) −(−74.8) = −965.1 kJ/mol
Therefore, the change in enthalpy when 1 mole of methane gas is burned in
excess oxygen is −965.1 kJ/mol.
Question 29
Question
Calculate the change in enthalpy (∆H) for the following reaction at 25
°
C:
2A(g)+3B(g)→C(g)
Given the following enthalpy changes:
∆H◦
ffor A(g) = −150 kJ/mol
∆H◦
ffor B(g) = −200 kJ/mol
∆H◦
ffor C(g) = −100 kJ/mol
21
Solution
Step 1: Write the balanced chemical equation and calculate the overall change
in enthalpy. The given reaction is:
2A(g)+3B(g)→C(g)
The overall change in enthalpy, ∆H, can be calculated using the formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the enthalpy change for the reaction. Given:
∆H◦
ffor A(g) = −150 kJ/mol
∆H◦
ffor B(g) = −200 kJ/mol
∆H◦
ffor C(g) = −100 kJ/mol
Substitute the values into the formula: ∆H= [1(−100 kJ/mol)]−[2(−150 kJ/mol)+
3(−200 kJ/mol)]
Step 3: Calculate the change in enthalpy. ∆H=−100 kJ/mol+300 kJ/mol+
600 kJ/mol = 800 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H= 800 kJ/mol.
Question 30
Question
A reaction is taking place in a closed system at constant pressure. The reaction
releases 250 kJ of heat and does 100 kJ of work on the surroundings. Calculate
the change in enthalpy (∆H) for the system.
Solution
Step 1: The change in enthalpy is given by the formula:
∆H=q+w
where qis the heat exchanged with the surroundings and wis the work done on
the surroundings.
Step 2: Substitute the given values into the formula:
∆H=−250 kJ + 100 kJ
Step 3: Calculate the change in enthalpy:
∆H=−150 kJ
Step 4: Therefore, the change in enthalpy for the system is ∆H=−150 kJ.
22
Question 31
Question
Calculate the change in enthalpy when 5.0 moles of ammonia gas react with
excess oxygen gas according to the following balanced equation:
4NH3(g)+ 5O2(g)→4NO(g)+ 6H2O(g)
Given that the standard enthalpy of formation for NH3(g),NO(g), and H2O(g)
are -45.9 kJ/mol, 90.3 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using standard
enthalpies of formation. The standard enthalpy change for the reaction can be
calculated by subtracting the sum of the standard enthalpies of formation of the
reactants from the sum of the standard enthalpies of formation of the products.
Given: Standard enthalpy of formation of NH3(g)= -45.9 kJ/mol Standard
enthalpy of formation of NO(g)= 90.3 kJ/mol Standard enthalpy of formation
of H2O(g)= -241.8 kJ/mol
The standard enthalpy change for the reaction is:
∆H◦= (4 ×90.3+6×(−241.8)) −(4 ×(−45.9) + 5 ×0)
∆H◦= (361.2−1452.8) −(−183.6)
∆H◦=−1091.6 + 183.6
∆H◦=−908.0 kJ/mol
Step 2: Calculate the change in enthalpy when 5.0 moles of ammonia gas
react. Given that 4 moles of NH3(g)are involved in the reaction, the change in
enthalpy for 5.0 moles can be calculated using the molar enthalpy change:
∆H=−908.0 kJ/mol
4 mol ×5 mol = −2270.0 kJ
Therefore, the change in enthalpy when 5.0 moles of NH3(g)react is -2270.0
kJ.
Question 32
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas (CH4)
are combusted in excess oxygen gas (O2) to produce carbon dioxide gas (CO2)
23
and water vapor (H2O) according to the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the standard enthalpies of formation (∆H◦
f) for CH4(g) = -74.8 kJ/mol,
CO2(g) = -393.5 kJ/mol, and H2O(g) = -241.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy of formation of the reaction. The
standard enthalpy change of the reaction can be calculated using the standard
enthalpies of formation of the products and reactants:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Given: ∆H◦
f(CH4(g)) = −74.8 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H= [1 ·∆H◦
f(CO2)+2·∆H◦
f(H2O)] −[1 ·∆H◦
f(CH4)+2·∆H◦
f(O2)]
∆H= [1 · −393.5+2· −241.8] −[1 · −74.8+2·0]
∆H= [−393.5−483.6] −[−74.8]
∆H=−877.1 + 74.8
∆H=−802.3 kJ/mol
Step 2: Calculate the change in enthalpy for 5.00 moles of CH4. Given that
5.00 moles of CH4are combusted, the change in enthalpy can be calculated
using the following equation:
∆H=n·∆Hrxn
where nis the number of moles of CH4and ∆Hrxn is the standard enthalpy
change of the reaction.
∆H= 5.00 moles × −802.3 kJ/mol = −4011.5 kJ
Therefore, the change in enthalpy when 5.00 moles of CH4are combusted
is -4011.5 kJ.
Question 33
Question
A certain chemical reaction has an enthalpy change of -287 kJ/mol. If the
reaction produces 3.00 moles of a product, what is the total amount of heat
absorbed or released by the reaction?
24
Solution
Step 1: Identify the given information The enthalpy change (∆H) of the reaction
is -287 kJ/mol. The number of moles of product formed is 3.00 mol.
Step 2: Calculate the total amount of heat absorbed or released To find
the total amount of heat absorbed or released by the reaction, we multiply the
enthalpy change by the number of moles of product formed. Total heat absorbed
or released = ∆H×moles of product
Substitute the given values: Total heat absorbed or released = -287 kJ/mol
×3.00 mol
Step 3: Perform the calculation Total heat = -287 kJ/mol ×3.00 mol =
-861 kJ
Therefore, the total amount of heat absorbed or released by the reaction is
861 kJ, with a negative sign indicating that the reaction releases heat.
Question 34
Question
A chemical reaction is carried out at constant pressure, resulting in a change in
enthalpy of -250 kJ. If 150 kJ of work is done on the system during the reaction,
what is the heat transfer for the reaction?
Solution
Step 1: Recall the definition of enthalpy change (∆H) in terms of heat transfer
(q) and work done on the system (w):
∆H=q+w
Step 2: Plug in the given values into the equation:
−250 kJ = q+ 150 kJ
Step 3: Solve for qby isolating it:
q=−250 kJ −150 kJ
Step 4: Perform the calculation:
q=−400 kJ
Therefore, the heat transfer for the reaction is −400 kJ.
25
∆H◦
r= 18.32 −66.4
∆H◦
r=−48.08 kJ/mol
Therefore, the standard enthalpy change of the reaction is −48.08 kJ/mol.
Question 2
Question
Given that the enthalpy change for the reaction 2A(g)+ 3B(g) →C(g) is ∆H=
−1200 kJ. If the enthalpy change for the reaction A(g) →B(g) is -300 kJ,
calculate the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g).
Solution
Step 1: Write the given enthalpy changes for the reactions. The enthalpy change
for the reaction 2A(g) + 3B(g) →C(g) is ∆H=−1200 kJ, and for A(g) →B(g)
is -300 kJ.
Step 2: Calculate the enthalpy change for the desired reaction using Hess’s
Law. To calculate the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g),
we need to consider the given reactions and manipulate them to obtain the
desired reaction.
Step 3: Manipulate the given reactions to obtain the desired reaction. 1.
A(g) →B(g) 2. 2A(g) + 3B(g) →C(g)
To get the desired reaction, we can reverse reaction 1 and multiply reaction
2 by 2: 1. B(g) →A(g) 2. 4A(g) + 6B(g) →2C(g)
Step 4: Determine the enthalpy change for the desired reaction. Since we are
using the reversed reaction 1 and the multiplied reaction 2, we need to change
the signs of the enthalpy changes for those reactions to obtain the enthalpy
change for the desired reaction. −(−300 kJ)+2(−1200 kJ) = 600 kJ−2400 kJ =
−1800 kJ
Therefore, the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g) is
−1800 kJ .
Question 3
Question
A reaction at constant pressure releases 150 kJ of heat and does 50 kJ of work
on the surroundings. If the internal energy decreases by 100 kJ, determine the
change in enthalpy for the reaction.
2
Solution
Step 1: Recall the relationship between enthalpy (H), internal energy (U), and
work done (W):
∆H= ∆U+P∆V
Step 2: Given that the heat released is 150 kJ, the work done on the sur-
roundings is 50 kJ, and the internal energy decreases by 100 kJ, we can rewrite
the equation as:
−∆H=−∆U−P∆V
−∆H=−100 kJ + 50 kJ
Step 3: Solving for the change in enthalpy (∆H), we get:
∆H=−(−100 + 50)
∆H= 50 kJ
Thus, the change in enthalpy for the reaction is 50 kJ.
Question 4
Question
Calculate the change in enthalpy (∆H) when 50.0 g of steam at 100◦C is con-
verted to ice at -10◦C. The specific heat capacity of steam is 2.01 J/g◦C, the
specific heat capacity of water is 4.18 J/g◦C, the specific heat capacity of ice is
2.09 J/g◦C, the heat of fusion of water is 333.5 J/g, and the heat of vaporization
of water is 2257 J/g.
Solution
Step 1: Calculate the heat required to cool the steam to 0◦C and condense it
to water.
The heat required to cool the steam to 0◦C is given by:
q=m×Csteam ×∆T
q= 50.0 g ×2.01 J/g◦C×(0 −100)◦C
q=−10050 J
The heat of vaporization of water is then released when the steam condenses
to water:
q=m×heat of vaporization
q= 50.0 g ×2257 J/g
q=−112850 J
3
Therefore, the total heat released to cool the steam to 0◦C and condense it
to water is:
qtotal =−10050 J −112850 J = −123900 J
Step 2: Calculate the heat required to cool the water to -10◦C and freeze it
to ice.
The heat required to cool the water to -10◦C is given by:
q=m×Cwater ×∆T
q= 50.0 g ×4.18 J/g◦C×(−10 −0)◦C
q=−2090 J
The heat of fusion of water is then released when the water freezes to ice:
q=m×heat of fusion
q= 50.0 g ×333.5 J/g
q=−16675 J
Therefore, the total heat released to cool the water to -10◦C and freeze it to
ice is:
qtotal =−2090 J −16675 J = −18765 J
Step 3: Calculate the total change in enthalpy (∆H). The total change in
enthalpy is the sum of the enthalpy changes for each step:
∆H=−123900 J + (−18765 J) = −142665 J
Therefore, the change in enthalpy (∆H) when 50.0 g of steam at 100◦C is
converted to ice at -10◦C is -142665 J.
Question 5
Question
A reaction is carried out in a bomb calorimeter at constant volume, and it
releases 250 kJ of heat. If the bomb calorimeter itself absorbs 20 kJ of heat
during the reaction, what is the enthalpy change (∆H) for the reaction?
Solution
Step 1: The enthalpy change for the reaction can be calculated using the equa-
tion:
∆H=qreaction −qcalorimeter
where qreaction is the heat released by the reaction and qcalorimeter is the heat
absorbed by the calorimeter.
4
Step 2: Given that the reaction releases 250 kJ of heat and the calorimeter
absorbs 20 kJ of heat, we can substitute these values into the equation:
∆H= 250 kJ −20 kJ
Step 3: Simplifying the above expression, we find:
∆H= 230 kJ
Step 4: Therefore, the enthalpy change for the reaction is 230 kJ .
Question 6
Question
A reaction is carried out in a calorimeter at constant pressure. The initial
temperature of the reactants is 25◦C and the final temperature after mixing is
38◦C. The calorimeter contains 80 g of water and has a heat capacity of 10
Step 1: Calculate the heat absorbed by the water in the calorimeter. The heat
absorbed by the water can be calculated using the formula:
q=mc∆T
where: q= heat absorbed (in joules), m= mass of water (in kg), c= specific
heat capacity of water (4.18 J/g◦C), ∆T= temperature change (in ◦C).
Given that the mass of water is 80 g and the temperature change is 38◦C−
25◦C = 13◦C, we can substitute these values into the formula:
q= (0.08 kg)(4.18 J/g◦C)(13◦C)
Calculating q:
q= 4.18(0.08)(13)
q= 4.18 ×0.104
q= 0.43344 kJ
Therefore, the heat absorbed by the water is 0.43344 kJ.
Step 2: Calculate the enthalpy change of the reaction. The enthalpy change
of the reaction can be calculated using the formula:
∆H=−q
Given that the reaction releases 500 kJ of heat, we can substitute this value into
the formula:
∆H=−500 kJ
Therefore, the enthalpy change of the reaction is −500 kJ.
5
Question 7
Question
A sample of water initially at 15◦C is heated, and its temperature is raised to
85◦C. Calculate the change in enthalpy of the water if the mass of the sample
is 500 g. Assume the specific heat capacity of water is 4.18 J/g◦C.
Solution
Step 1: Calculate the heat energy absorbed by the water to raise its temperature
from 15◦C to 85◦C using the formula:
q=mc∆T
where qis the heat energy, mis the mass of the sample, cis the specific heat
capacity, and ∆Tis the change in temperature.
Step 2: Substitute the given values into the formula:
q= (500 g)(4.18 J/g◦C)(85 −15)
Step 3: Calculate the heat energy absorbed by the water:
q= (500)(4.18)(70)
Step 4: q= 14690 J
Step 5: Calculate the change in enthalpy using the formula:
∆H=q
Step 6: Substitute the value of qinto the formula:
∆H= 14690 J
Answer: The change in enthalpy of the water is 14690 J.
Question 8
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas react
with excess oxygen gas to produce carbon dioxide gas and water vapor, according
to the following balanced equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpy changes:
∆H◦
f(CH4) = −74.6 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
6
Solution
Step 1: Calculate the standard enthalpy change of the reaction using standard
enthalpies of formation.
∆H◦
rxn =Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion.
∆H◦
rxn = [1(−393.5 kJ/mol) + 2(−285.8 kJ/mol)] −[(−74.6 kJ/mol)]
Step 3: Calculate the change in enthalpy of the reaction.
∆H◦
rxn = [−393.5 kJ/mol −571.6 kJ/mol] −[−74.6 kJ/mol]
∆H◦
rxn =−965.1 kJ/mol + 74.6 kJ/mol
∆H◦
rxn =−890.5 kJ/mol
Therefore, the change in enthalpy (∆H) for the reaction of 5.00 moles of
methane gas is -890.5 kJ.
Question 9
Question
Calculate the change in enthalpy (∆H) when 5 moles of methane gas (CH4)
react with excess oxygen gas to produce carbon dioxide gas (CO2) and water
vapor (H2O) according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpies of formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(CH4) = −74.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the given reaction using the
standard enthalpies of formation.
The standard enthalpy change, ∆H◦, for the reaction can be calculated using
the formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
7
Substitute the given standard enthalpies of formation into the formula:
∆H◦= [1(−393.5) + 2(−285.8)] −[1(−74.8) + 2(0)]
Simplify the expression:
∆H◦= (−393.5−571.6) −(−74.8)
∆H◦=−965.1 + 74.8
∆H◦=−890.3 kJ/mol
Therefore, the change in enthalpy (∆H) for the given reaction is -890.3
kJ/mol.
Question 10
Question
Given that the enthalpy change (∆H) for the reaction C(s) + O2(g) →CO2(g)
is -393.5 kJ/mol, calculate the enthalpy change when 10.0 g of carbon reacts
with excess oxygen to form carbon dioxide.
Solution
Step 1: Calculate the number of moles of carbon reacting.
Molar mass of carbon = 12.01 g/mol
Moles of carbon = 10.0 g
12.01 g/mol
= 0.833 mol
Step 2: Use the given enthalpy change to calculate the enthalpy change for
the reaction.
∆H= ∆H◦×moles of reaction
=−393.5 kJ/mol ×0.833 mol
=−327.6 kJ
Therefore, the enthalpy change when 10.0 g of carbon reacts with excess
oxygen to form carbon dioxide is -327.6 kJ.
Question 11
Question
The enthalpy change (∆H) for the reaction
2A(g)+3B(l)→C(s)+4D(g)
8
is −1264 kJ. If the standard enthalpies of formation are −17 kJ/mol for A, −66
kJ/mol for B, −394 kJ/mol for C, and 109 kJ/mol for D, calculate the standard
enthalpy change of the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the formation of products.
The standard enthalpy change for the formation of products is given by:
∆Hproducts =Xn∆Hf
where nis the stoichiometric coefficient and ∆Hfis the standard enthalpy of
formation.
Substitute values:
∆Hproducts = (1)(−394 kJ/mol) + (4)(109 kJ/mol)
∆Hproducts =−394 kJ/mol + 436 kJ/mol
∆Hproducts = 42 kJ/mol
Step 2: Calculate the standard enthalpy change for the formation of reac-
tants. The standard enthalpy change for the formation of reactants is given
by:
∆Hreactants =Xn∆Hf
where nis the stoichiometric coefficient and ∆Hfis the standard enthalpy of
formation.
Substitute values:
∆Hreactants = (2)(−17 kJ/mol) + (3)(−66 kJ/mol)
∆Hreactants =−34 kJ/mol −198 kJ/mol
∆Hreactants =−232 kJ/mol
Step 3: Calculate the standard enthalpy change of the reaction. The stan-
dard enthalpy change of the reaction is given by:
∆H= ∆Hproducts −∆Hreactants
Substitute values:
∆H= 42 kJ/mol −(−232 kJ/mol)
∆H= 42 kJ/mol + 232 kJ/mol
∆H= 274 kJ/mol
Therefore, the standard enthalpy change of the reaction is 274 kJ/mol.
9
Question 12
Question
Consider a reaction where 2 moles of liquid water are converted to steam at 1
atm and 100
°
C. The molar enthalpy of vaporization of water is 40.79 kJ/mol.
Calculate the change in enthalpy for this reaction.
Solution
Step 1: Calculate the heat required to vaporize 1 mole of water. Given that the
molar enthalpy of vaporization of water is 40.79 kJ/mol, the heat required to
vaporize 1 mole of water is 40.79 kJ.
Step 2: Calculate the heat required to vaporize 2 moles of water. Since we
have 2 moles of water, the total heat required to vaporize 2 moles is:
40.79 kJ/mol ×2 mol = 81.58 kJ
Step 3: Calculate the change in enthalpy. The change in enthalpy for this
reaction is the same as the heat required to vaporize 2 moles of water, which is
81.58 kJ. Therefore, the change in enthalpy is 81.58 kJ.
Question 13
Question
Calculate the change in enthalpy for a reaction where 2 moles of nitrogen gas
react with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. The
enthalpies of formation for nitrogen gas, hydrogen gas, and ammonia gas are 0
kJ/mol, 0 kJ/mol, and -46 kJ/mol, respectively.
Solution
Step 1: Write the balanced chemical equation for the reaction.
N2(g)+3H2(g)→2NH3(g)
Step 2: Calculate the total change in enthalpy using the enthalpies of for-
mation. The change in enthalpy (∆H) can be calculated using the formula:
∆H=Xν∆Hf(products) −Xν∆Hf(reactants)
where νrepresents the stoichiometric coefficients and ∆Hfrepresents the en-
thalpy of formation.
Substitute the values into the formula:
∆H= 2(−46 kJ/mol) −[1(0 kJ/mol) + 3(0 kJ/mol)]
10
Step 3: Simplify the expression to find the change in enthalpy.
∆H=−92 kJ/mol −0 kJ/mol
∆H=−92 kJ/mol
Therefore, the change in enthalpy for the reaction is -92 kJ/mol.
Question 14
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with ∆H=−571.6 kJ, calculate the enthalpy change when 4.00 g of H2is
burned in excess O2to form liquid water.
Solution
Step 1: Calculate the moles of hydrogen gas. Given: - Molar mass of hydrogen,
M(H2)=2.016 g/mol - Mass of hydrogen gas, m(H2)=4.00 g Using the formula
n=m
M, where nis moles, we have:
n(H2) = 4.00 g
2.016 g/mol = 1.984 mol
Step 2: Use the stoichiometry of the reaction to find the enthalpy change for
1.00 mol of hydrogen gas. From the balanced chemical equation, the enthalpy
change for the reaction is −571.6 kJ for 2 moles of H2. Therefore, for 1 mole of
H2, the enthalpy change is:
−571.6 kJ
2=−285.8 kJ/mol
Step 3: Calculate the enthalpy change for the given mass of hydrogen gas.
Since 1.984 moles of H2are involved, the enthalpy change for 1.984 moles is:
1.984 mol ×(−285.8 kJ/mol) = −566.7 kJ
Answer: The enthalpy change when 4.00 g of H2is burned in excess O2to
form liquid water is −566.7 kJ.
Question 15
Question
Calculate the change in enthalpy (∆H) for the combustion of propane gas
(C3H8) at constant pressure. The balanced chemical equation for the com-
bustion of propane is:
C3H8(g) + 5
2O2(g)→3CO2(g)+4H2O(g)
11
Given the standard enthalpies of formation (∆H◦
f) for C3H8(g), CO2(g), and
H2O(g) are -103.9 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Calculate ∆H◦for the combustion of propane using the standard en-
thalpies of formation of the reactants and products.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the coefficients of the products and reactants, respectively
in the balanced chemical equation.
Step 2: Substitute the values of the standard enthalpies of formation pro-
vided into the equation.
∆H◦= [3(−393.5) + 4(−241.8)] −[−103.9] kJ/mol
∆H◦= (−1180.5−967.2) + 103.9 kJ/mol = −2043.8 kJ/mol
Therefore, the change in enthalpy for the combustion of propane gas at
constant pressure is ∆H=−2043.8 kJ/mol.
Question 16
Question
A certain reaction has an enthalpy change of -92 kJ/mol. If 2.50 moles of the
reactant are consumed in the reaction, what is the total enthalpy change for the
reaction?
Solution
Step 1: Determine the total enthalpy change for the reaction. Step 2: Use the
given enthalpy change per mole to find the total enthalpy change.
Step 1: Enthalpy change per mole = −92 kJ/molMoles of reactant consumed = 2.50 mol
Step 2: Total enthalpy change = Enthalpy change per mole×Moles of reactant consumed
Total enthalpy change = (−92 kJ/mol) ×(2.50 mol)
Total enthalpy change = −230 kJ
Therefore, the total enthalpy change for the reaction when 2.50 moles of the
reactant are consumed is -230 kJ.
12
Question 17
Question
A reaction has a standard enthalpy change of -394 kJ/mol. If the reaction is
exothermic, what can be said about the enthalpy of the products compared to
the enthalpy of the reactants?
Solution
Step 1: Recall that for an exothermic reaction, the standard enthalpy change is
negative.
Step 2: In this case, the standard enthalpy change is -394 kJ/mol, which
means that the products have lower enthalpy than the reactants.
Step 3: Therefore, the enthalpy of the products is lower than the enthalpy
of the reactants for this exothermic reaction.
Question 18
Question
A sample of nitrogen gas undergoes an isobaric process at a pressure of 1 atm.
During the process, the nitrogen gas absorbs 500 J of heat and performs 300 J
of work. If the initial enthalpy of the nitrogen gas is 2000 J, what is the final
enthalpy of the gas?
Solution
Step 1: Recall the definition of enthalpy: Enthalpy (H) is defined as the sum of
the internal energy (U) of a system and the product of pressure (P) and volume
(V) of the system:
H=U+P V
Step 2: Since the process is isobaric, the change in enthalpy can be expressed
as:
∆H= ∆U+P∆V
Step 3: We are given that the nitrogen gas absorbs 500 J of heat and performs
300 J of work. Therefore, the change in internal energy can be calculated as:
∆U=Q−W= 500 J −300 J = 200 J
Step 4: Substituting the values into the formula for the change in enthalpy:
∆H= 200 J + 1 atm ×∆V
Step 5: But we also know that:
∆H=Hfinal −Hinitial =Hfinal −2000 J
13
Step 6: Equating the two expressions for ∆H:
200 J + 1 atm ×∆V=Hfinal −2000 J
Step 7: Given that the pressure is 1 atm, we can simplify the equation to:
200 J + 1 atm ×∆V=Hfinal −2000 J
Step 8: Since ∆V=W
P, we can substitute in the values to find the change
in volume:
∆V=300 J
1 atm = 300 L ·atm
Step 9: Substituting ∆Vback into the equation and solving for Hfinal:
200 J + 1 atm ×300 L ·atm = Hfinal −2000 J
Step 10: Calculating the final enthalpy:
Hfinal = 200 J + 300 J = 2500 J
Therefore, the final enthalpy of the nitrogen gas is 2500 J.
Question 19
Question
A certain chemical reaction has an enthalpy change of ∆H=−289 kcal/mol. If
1.50 moles of the reactant is consumed in the reaction, what is the total change
in enthalpy in kilocalories?
Solution
Step 1: Determine the total change in enthalpy for 1 mole of the reactant.
Given: ∆H=−289 kcal/mol
Step 2: Calculate the total change in enthalpy for 1 mole of the reactant.
∆Htotal = ∆H×moles of reactant
∆Htotal =−289 kcal/mol ×1.50 moles
∆Htotal =−433.5 kcal
Therefore, the total change in enthalpy for 1.50 moles of the reactant is
−433.5 kcal.
14
Question 20
Question
A reaction is carried out at constant pressure where 100.0 g of a substance
X is converted to substance Y. The reaction releases 1500 J of heat and the
temperature of the surroundings increases by 2.5
°
C. The specific heat capacity of
the surroundings is 4.18 J/(g
°
C). Calculate the enthalpy change for the reaction.
Solution
Step 1: Calculate the heat gained by the surroundings. Given that the temper-
ature change (∆T) is 2.5
°
C and the specific heat capacity of the surroundings
(C) is 4.18 J/(g
°
C), we can use the formula:
q=C×m×∆T
where: q= heat gained by the surroundings, C= specific heat capacity of the
surroundings, m= mass of the surroundings, ∆T= temperature change of the
surroundings.
Substitute the values:
q= 4.18 J/(g
°
C) ×m×2.5
°
C
Step 2: Calculate the mass of the surroundings. Since the reaction releases
heat, the surroundings gain this heat. We can find the mass of the surroundings
using the formula:
q=qreaction =msurroundings ×C×∆T
Rearranging the formula:
msurroundings =qreaction
C×∆T
Step 3: Calculate the enthalpy change for the reaction. The enthalpy change
of the reaction (∆Hreaction) is equal in magnitude but opposite in sign to the
heat gained by the surroundings:
∆Hreaction =−q
Substitute the value obtained for qin Step 1 and calculate ∆Hreaction.
Question 21
Question
A certain reaction has an enthalpy change of ∆H=−486 kJ. If 2.00 moles of
this reaction take place, what is the heat associated with this reaction?
15
Solution
Step 1: Determine the heat associated with the reaction per mole. Given that
the enthalpy change for the reaction is ∆H=−486 kJ, we can find the heat per
mole by dividing by the number of moles:
Heat per mole = ∆H
moles =−486 kJ
2.00 mol
Heat per mole = −243 kJ/mol
Step 2: Calculate the total heat associated with the reaction. To find the
total heat for 2.00 moles of the reaction, we multiply the heat per mole by the
number of moles:
Total heat = Heat per mole ×moles
Total heat = −243 kJ/mol ×2.00 mol
Total heat = −486 kJ
Therefore, the total heat associated with the reaction when 2.00
moles take place is -486 kJ.
Question 22
Question
Calculate the enthalpy change for the reaction below at 298 K:
2H2(g)+O2(g)→2H2O(l)
Given the following standard enthalpy changes of formation: ∆H◦
f[H2(g)] = 0
kJ/mol, ∆H◦
f[O2(g)] = 0 kJ/mol, and ∆H◦
f[H2O(l)] = −285.8 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and apply Hess’s Law. The
enthalpy change for the reaction can be calculated using the standard enthalpy
changes of formation for the reactants and products. We can use Hess’s Law to
obtain the enthalpy change for the desired reaction.
Balanced reaction:
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate the enthalpy change using the standard enthalpy changes
of formation. The enthalpy change for the reaction can be calculated as follows:
∆H◦
rxn =X∆H◦
products −X∆H◦
reactants
16
Substitute the given values:
∆H◦
rxn = 2(∆H◦
f[H2O(l)]) −[2(∆H◦
f[H2(g)]) + ∆H◦
f[O2(g)]]
∆H◦
rxn = 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦
rxn =−571.6 kJ/mol
Therefore, the enthalpy change for the reaction is -571.6 kJ/mol at 298 K.
Question 23
Question
A reaction involving the formation of ammonia gas is represented by the follow-
ing equation:
N2(g)+3H2(g)→2NH3(g)
Given the standard enthalpies of formation (∆H◦
f) for N2(g), H2(g), and N H3(g)
are 0 kJ/mol, 0 kJ/mol, and −46 kJ/mol, respectively, calculate the standard
enthalpy change of the reaction.
Solution
Step 1: Write the balanced chemical equation and standard enthalpy change
formula. The balanced chemical equation is:
N2(g)+3H2(g)→2NH3(g)
The standard enthalpy change of the reaction can be calculated using the for-
mula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Find the standard enthalpy change of the reaction. We know the
formation enthalpies of the products and reactants:
∆H◦
f(NH3) = −46 kJ/mol
∆H◦
f(N2) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Substitute these values into the formula:
∆H◦= [2 ×∆H◦
f(NH3)] −[1 ×∆H◦
f(N2)+3×∆H◦
f(H2)]
∆H◦= [2 ×(−46 kJ/mol)] −[1 ×0 kJ/mol + 3 ×0 kJ/mol]
∆H◦=−92 kJ/mol
Therefore, the standard enthalpy change of the reaction is ∆H◦=−92 kJ/mol.
17
Question 24
Question
Calculate the change in enthalpy (∆H) when 2 moles of nitrogen gas react
completely with 1 mole of oxygen gas to form nitrogen dioxide gas according to
the following balanced equation:
2N2(g)+O2(g)→2NO2(g)
Given that the standard enthalpies of formation (∆H◦
f) for N2(g), O2(g), and
NO2(g) are 0 kJ/mol, 0 kJ/mol, and 33.2 kJ/mol, respectively.
Solution
Step 1: Write the equation for the change in enthalpy using the standard en-
thalpies of formation:
∆H=Xν∆H◦
f
Where νrepresents the stoichiometric coefficients in the balanced equation and
∆H◦
fis the standard enthalpy of formation.
Step 2: Determine the change in enthalpy for the given reaction:
∆H= (2 ×∆H◦
f(NO2)) −(2 ×∆H◦
f(N2)+∆H◦
f(O2))
Step 3: Substitute the given values for the standard enthalpies of formation:
∆H= (2 ×33.2 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
Step 4: Calculate the change in enthalpy:
∆H= 66.4 kJ/mol
Therefore, the change in enthalpy for the given reaction is 66.4 kJ/mol.
Question 25
Question
A reaction involving the conversion of graphite to diamond at standard condi-
tions has an enthalpy change of 2.9 kJ/mol. Calculate the minimum pressure
required to convert 1 mole of graphite to diamond at 25
°
C.
Solution
Step 1: Write down the enthalpy change of the reaction. Given: ∆H= 2.9
kJ/mol
18
Step 2: Recall the definition of standard enthalpy change. The standard
enthalpy change, ∆H◦, is the enthalpy change when reactants in their standard
states are converted to products in their standard states at 1 atm pressure.
Step 3: Determine the pressure needed for the phase change. Since we want
to convert 1 mole of graphite to diamond under standard conditions, we need
to consider the phase change from graphite to diamond and account for the
enthalpy change.
Step 4: Calculate the minimum pressure required using the enthalpy change
and temperature. The enthalpy change at standard conditions is related to the
pressure change via the equation:
∆H= ∆U+P∆V
At constant pressure, the change in enthalpy, ∆H, is approximately equal to
the heat change, ∆U. We can rearrange and solve for pressure:
P=∆H
∆V
Since the reaction involves a solid to solid phase change, the volume change is
negligible. Hence, ∆V≈0 and the pressure required is:
P=∆H
∆V≈2.9 kJ/mol
0=∞
Therefore, the minimum pressure required to convert 1 mole of graphite to
diamond at 25
°
C is theoretically infinite.
Question 26
Question
An ideal gas undergoes a process in which its enthalpy is given by the expression
H= 3.5T2−0.01T3, where His the enthalpy in kJ/mol and Tis the temperature
in K. Determine the heat capacity at constant pressure, Cp, for this gas in terms
of temperature.
Solution
Step 1: Recall that the heat capacity at constant pressure, Cp, is defined as the
partial derivative of the enthalpy with respect to temperature:
Cp=∂H
∂T P
Step 2: We are given the enthalpy expression H= 3.5T2−0.01T3, so we
differentiate this expression with respect to temperature T:
∂H
∂T =∂
∂T (3.5T2−0.01T3)
19
Step 3: Calculating the derivative term by term, we get:
∂H
∂T = 7T−0.03T2
Step 4: Therefore, the heat capacity at constant pressure is:
Cp= 7 −0.03TJ/mol
·
K
Question 27
Question
A reaction takes place at constant pressure with a change in enthalpy (∆H) of
-276 kJ. If the reaction releases 123 kJ of heat, what is the change in internal
energy (∆U) for the reaction? Assume no other work is done during the reaction.
Solution
Step 1: Recall the relationship between enthalpy (H), internal energy (U), and
pressure (P):
∆H= ∆U+P∆V
Step 2: Since the reaction takes place at constant pressure (∆V= 0), the
equation simplifies to:
∆H= ∆U
Step 3: Given that ∆H=−276 kJ, we can substitute this into the equation:
−276 kJ = ∆U
Step 4: Since ∆U=−276 kJ, this represents the change in internal energy
for the reaction when 276 kJ of heat is released.
Step 5: However, the reaction releases 123 kJ of heat instead of 276 kJ. To
find the change in internal energy when 123 kJ of heat is released, we need to
adjust ∆Uaccordingly.
Step 6: If the reaction releases 276 kJ of heat but only 123 kJ is actually
released, then the change in internal energy when 123 kJ of heat is released can
be calculated as follows:
∆U=123
276 ×(−276 kJ)
Step 7: Simplifying the calculation:
∆U=123
276 ×(−276) = −123
2
Step 8: Therefore, the change in internal energy (∆U) for the reaction when
123 kJ of heat is released is −123
2kJ.
20
Question 28
Question
Calculate the change in enthalpy when 1 mole of methane gas is burned in excess
oxygen at constant pressure according to the following reaction:
CH4(g)+ 2O2(g)→CO2(g)+ 2H2O(g)
Given the standard enthalpies of formation are: ∆H◦
f(CH4(g)) = −74.8 kJ/mol,
∆H◦
f(CO2(g)) = −393.5 kJ/mol, ∆H◦
f(H2O(g)) = −285.8 kJ/mol.
Solution
Step 1: Write the thermochemical equation for the reaction.
CH4(g)+ 2O2(g)→CO2(g)+ 2H2O(g)
Step 2: Calculate the standard enthalpy of reaction using the standard en-
thalpies of formation.
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = [1×∆H◦
f(CO2(g))+2×∆H◦
f(H2O(g))]−[∆H◦
f(CH4(g))+2×∆H◦
f(O2(g))]
∆H◦
rxn = [1 ×(−393.5) + 2 ×(−285.8)] −[(−74.8) + 2 ×0]
∆H◦
rxn = (−393.5−571.6) −(−74.8) = −965.1 kJ/mol
Therefore, the change in enthalpy when 1 mole of methane gas is burned in
excess oxygen is −965.1 kJ/mol.
Question 29
Question
Calculate the change in enthalpy (∆H) for the following reaction at 25
°
C:
2A(g)+3B(g)→C(g)
Given the following enthalpy changes:
∆H◦
ffor A(g) = −150 kJ/mol
∆H◦
ffor B(g) = −200 kJ/mol
∆H◦
ffor C(g) = −100 kJ/mol
21
Solution
Step 1: Write the balanced chemical equation and calculate the overall change
in enthalpy. The given reaction is:
2A(g)+3B(g)→C(g)
The overall change in enthalpy, ∆H, can be calculated using the formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the enthalpy change for the reaction. Given:
∆H◦
ffor A(g) = −150 kJ/mol
∆H◦
ffor B(g) = −200 kJ/mol
∆H◦
ffor C(g) = −100 kJ/mol
Substitute the values into the formula: ∆H= [1(−100 kJ/mol)]−[2(−150 kJ/mol)+
3(−200 kJ/mol)]
Step 3: Calculate the change in enthalpy. ∆H=−100 kJ/mol+300 kJ/mol+
600 kJ/mol = 800 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H= 800 kJ/mol.
Question 30
Question
A reaction is taking place in a closed system at constant pressure. The reaction
releases 250 kJ of heat and does 100 kJ of work on the surroundings. Calculate
the change in enthalpy (∆H) for the system.
Solution
Step 1: The change in enthalpy is given by the formula:
∆H=q+w
where qis the heat exchanged with the surroundings and wis the work done on
the surroundings.
Step 2: Substitute the given values into the formula:
∆H=−250 kJ + 100 kJ
Step 3: Calculate the change in enthalpy:
∆H=−150 kJ
Step 4: Therefore, the change in enthalpy for the system is ∆H=−150 kJ.
22
Question 31
Question
Calculate the change in enthalpy when 5.0 moles of ammonia gas react with
excess oxygen gas according to the following balanced equation:
4NH3(g)+ 5O2(g)→4NO(g)+ 6H2O(g)
Given that the standard enthalpy of formation for NH3(g),NO(g), and H2O(g)
are -45.9 kJ/mol, 90.3 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using standard
enthalpies of formation. The standard enthalpy change for the reaction can be
calculated by subtracting the sum of the standard enthalpies of formation of the
reactants from the sum of the standard enthalpies of formation of the products.
Given: Standard enthalpy of formation of NH3(g)= -45.9 kJ/mol Standard
enthalpy of formation of NO(g)= 90.3 kJ/mol Standard enthalpy of formation
of H2O(g)= -241.8 kJ/mol
The standard enthalpy change for the reaction is:
∆H◦= (4 ×90.3+6×(−241.8)) −(4 ×(−45.9) + 5 ×0)
∆H◦= (361.2−1452.8) −(−183.6)
∆H◦=−1091.6 + 183.6
∆H◦=−908.0 kJ/mol
Step 2: Calculate the change in enthalpy when 5.0 moles of ammonia gas
react. Given that 4 moles of NH3(g)are involved in the reaction, the change in
enthalpy for 5.0 moles can be calculated using the molar enthalpy change:
∆H=−908.0 kJ/mol
4 mol ×5 mol = −2270.0 kJ
Therefore, the change in enthalpy when 5.0 moles of NH3(g)react is -2270.0
kJ.
Question 32
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas (CH4)
are combusted in excess oxygen gas (O2) to produce carbon dioxide gas (CO2)
23
and water vapor (H2O) according to the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the standard enthalpies of formation (∆H◦
f) for CH4(g) = -74.8 kJ/mol,
CO2(g) = -393.5 kJ/mol, and H2O(g) = -241.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy of formation of the reaction. The
standard enthalpy change of the reaction can be calculated using the standard
enthalpies of formation of the products and reactants:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Given: ∆H◦
f(CH4(g)) = −74.8 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H= [1 ·∆H◦
f(CO2)+2·∆H◦
f(H2O)] −[1 ·∆H◦
f(CH4)+2·∆H◦
f(O2)]
∆H= [1 · −393.5+2· −241.8] −[1 · −74.8+2·0]
∆H= [−393.5−483.6] −[−74.8]
∆H=−877.1 + 74.8
∆H=−802.3 kJ/mol
Step 2: Calculate the change in enthalpy for 5.00 moles of CH4. Given that
5.00 moles of CH4are combusted, the change in enthalpy can be calculated
using the following equation:
∆H=n·∆Hrxn
where nis the number of moles of CH4and ∆Hrxn is the standard enthalpy
change of the reaction.
∆H= 5.00 moles × −802.3 kJ/mol = −4011.5 kJ
Therefore, the change in enthalpy when 5.00 moles of CH4are combusted
is -4011.5 kJ.
Question 33
Question
A certain chemical reaction has an enthalpy change of -287 kJ/mol. If the
reaction produces 3.00 moles of a product, what is the total amount of heat
absorbed or released by the reaction?
24
Solution
Step 1: Identify the given information The enthalpy change (∆H) of the reaction
is -287 kJ/mol. The number of moles of product formed is 3.00 mol.
Step 2: Calculate the total amount of heat absorbed or released To find
the total amount of heat absorbed or released by the reaction, we multiply the
enthalpy change by the number of moles of product formed. Total heat absorbed
or released = ∆H×moles of product
Substitute the given values: Total heat absorbed or released = -287 kJ/mol
×3.00 mol
Step 3: Perform the calculation Total heat = -287 kJ/mol ×3.00 mol =
-861 kJ
Therefore, the total amount of heat absorbed or released by the reaction is
861 kJ, with a negative sign indicating that the reaction releases heat.
Question 34
Question
A chemical reaction is carried out at constant pressure, resulting in a change in
enthalpy of -250 kJ. If 150 kJ of work is done on the system during the reaction,
what is the heat transfer for the reaction?
Solution
Step 1: Recall the definition of enthalpy change (∆H) in terms of heat transfer
(q) and work done on the system (w):
∆H=q+w
Step 2: Plug in the given values into the equation:
−250 kJ = q+ 150 kJ
Step 3: Solve for qby isolating it:
q=−250 kJ −150 kJ
Step 4: Perform the calculation:
q=−400 kJ
Therefore, the heat transfer for the reaction is −400 kJ.
25
∆H◦
r= 18.32 −66.4
∆H◦
r=−48.08 kJ/mol
Therefore, the standard enthalpy change of the reaction is −48.08 kJ/mol.
Question 2
Question
Given that the enthalpy change for the reaction 2A(g)+ 3B(g) →C(g) is ∆H=
−1200 kJ. If the enthalpy change for the reaction A(g) →B(g) is -300 kJ,
calculate the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g).
Solution
Step 1: Write the given enthalpy changes for the reactions. The enthalpy change
for the reaction 2A(g) + 3B(g) →C(g) is ∆H=−1200 kJ, and for A(g) →B(g)
is -300 kJ.
Step 2: Calculate the enthalpy change for the desired reaction using Hess’s
Law. To calculate the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g),
we need to consider the given reactions and manipulate them to obtain the
desired reaction.
Step 3: Manipulate the given reactions to obtain the desired reaction. 1.
A(g) →B(g) 2. 2A(g) + 3B(g) →C(g)
To get the desired reaction, we can reverse reaction 1 and multiply reaction
2 by 2: 1. B(g) →A(g) 2. 4A(g) + 6B(g) →2C(g)
Step 4: Determine the enthalpy change for the desired reaction. Since we are
using the reversed reaction 1 and the multiplied reaction 2, we need to change
the signs of the enthalpy changes for those reactions to obtain the enthalpy
change for the desired reaction. −(−300 kJ)+2(−1200 kJ) = 600 kJ−2400 kJ =
−1800 kJ
Therefore, the enthalpy change for the reaction 2C(g) →4A(g) + 3B(g) is
−1800 kJ .
Question 3
Question
A reaction at constant pressure releases 150 kJ of heat and does 50 kJ of work
on the surroundings. If the internal energy decreases by 100 kJ, determine the
change in enthalpy for the reaction.
2
Solution
Step 1: Recall the relationship between enthalpy (H), internal energy (U), and
work done (W):
∆H= ∆U+P∆V
Step 2: Given that the heat released is 150 kJ, the work done on the sur-
roundings is 50 kJ, and the internal energy decreases by 100 kJ, we can rewrite
the equation as:
−∆H=−∆U−P∆V
−∆H=−100 kJ + 50 kJ
Step 3: Solving for the change in enthalpy (∆H), we get:
∆H=−(−100 + 50)
∆H= 50 kJ
Thus, the change in enthalpy for the reaction is 50 kJ.
Question 4
Question
Calculate the change in enthalpy (∆H) when 50.0 g of steam at 100◦C is con-
verted to ice at -10◦C. The specific heat capacity of steam is 2.01 J/g◦C, the
specific heat capacity of water is 4.18 J/g◦C, the specific heat capacity of ice is
2.09 J/g◦C, the heat of fusion of water is 333.5 J/g, and the heat of vaporization
of water is 2257 J/g.
Solution
Step 1: Calculate the heat required to cool the steam to 0◦C and condense it
to water.
The heat required to cool the steam to 0◦C is given by:
q=m×Csteam ×∆T
q= 50.0 g ×2.01 J/g◦C×(0 −100)◦C
q=−10050 J
The heat of vaporization of water is then released when the steam condenses
to water:
q=m×heat of vaporization
q= 50.0 g ×2257 J/g
q=−112850 J
3
Therefore, the total heat released to cool the steam to 0◦C and condense it
to water is:
qtotal =−10050 J −112850 J = −123900 J
Step 2: Calculate the heat required to cool the water to -10◦C and freeze it
to ice.
The heat required to cool the water to -10◦C is given by:
q=m×Cwater ×∆T
q= 50.0 g ×4.18 J/g◦C×(−10 −0)◦C
q=−2090 J
The heat of fusion of water is then released when the water freezes to ice:
q=m×heat of fusion
q= 50.0 g ×333.5 J/g
q=−16675 J
Therefore, the total heat released to cool the water to -10◦C and freeze it to
ice is:
qtotal =−2090 J −16675 J = −18765 J
Step 3: Calculate the total change in enthalpy (∆H). The total change in
enthalpy is the sum of the enthalpy changes for each step:
∆H=−123900 J + (−18765 J) = −142665 J
Therefore, the change in enthalpy (∆H) when 50.0 g of steam at 100◦C is
converted to ice at -10◦C is -142665 J.
Question 5
Question
A reaction is carried out in a bomb calorimeter at constant volume, and it
releases 250 kJ of heat. If the bomb calorimeter itself absorbs 20 kJ of heat
during the reaction, what is the enthalpy change (∆H) for the reaction?
Solution
Step 1: The enthalpy change for the reaction can be calculated using the equa-
tion:
∆H=qreaction −qcalorimeter
where qreaction is the heat released by the reaction and qcalorimeter is the heat
absorbed by the calorimeter.
4
Step 2: Given that the reaction releases 250 kJ of heat and the calorimeter
absorbs 20 kJ of heat, we can substitute these values into the equation:
∆H= 250 kJ −20 kJ
Step 3: Simplifying the above expression, we find:
∆H= 230 kJ
Step 4: Therefore, the enthalpy change for the reaction is 230 kJ .
Question 6
Question
A reaction is carried out in a calorimeter at constant pressure. The initial
temperature of the reactants is 25◦C and the final temperature after mixing is
38◦C. The calorimeter contains 80 g of water and has a heat capacity of 10
Step 1: Calculate the heat absorbed by the water in the calorimeter. The heat
absorbed by the water can be calculated using the formula:
q=mc∆T
where: q= heat absorbed (in joules), m= mass of water (in kg), c= specific
heat capacity of water (4.18 J/g◦C), ∆T= temperature change (in ◦C).
Given that the mass of water is 80 g and the temperature change is 38◦C−
25◦C = 13◦C, we can substitute these values into the formula:
q= (0.08 kg)(4.18 J/g◦C)(13◦C)
Calculating q:
q= 4.18(0.08)(13)
q= 4.18 ×0.104
q= 0.43344 kJ
Therefore, the heat absorbed by the water is 0.43344 kJ.
Step 2: Calculate the enthalpy change of the reaction. The enthalpy change
of the reaction can be calculated using the formula:
∆H=−q
Given that the reaction releases 500 kJ of heat, we can substitute this value into
the formula:
∆H=−500 kJ
Therefore, the enthalpy change of the reaction is −500 kJ.
5
Question 7
Question
A sample of water initially at 15◦C is heated, and its temperature is raised to
85◦C. Calculate the change in enthalpy of the water if the mass of the sample
is 500 g. Assume the specific heat capacity of water is 4.18 J/g◦C.
Solution
Step 1: Calculate the heat energy absorbed by the water to raise its temperature
from 15◦C to 85◦C using the formula:
q=mc∆T
where qis the heat energy, mis the mass of the sample, cis the specific heat
capacity, and ∆Tis the change in temperature.
Step 2: Substitute the given values into the formula:
q= (500 g)(4.18 J/g◦C)(85 −15)
Step 3: Calculate the heat energy absorbed by the water:
q= (500)(4.18)(70)
Step 4: q= 14690 J
Step 5: Calculate the change in enthalpy using the formula:
∆H=q
Step 6: Substitute the value of qinto the formula:
∆H= 14690 J
Answer: The change in enthalpy of the water is 14690 J.
Question 8
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas react
with excess oxygen gas to produce carbon dioxide gas and water vapor, according
to the following balanced equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpy changes:
∆H◦
f(CH4) = −74.6 kJ/mol
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
6
Solution
Step 1: Calculate the standard enthalpy change of the reaction using standard
enthalpies of formation.
∆H◦
rxn =Xν∆H◦
f(products) −Xν∆H◦
f(reactants)
Step 2: Substitute the given standard enthalpies of formation into the equa-
tion.
∆H◦
rxn = [1(−393.5 kJ/mol) + 2(−285.8 kJ/mol)] −[(−74.6 kJ/mol)]
Step 3: Calculate the change in enthalpy of the reaction.
∆H◦
rxn = [−393.5 kJ/mol −571.6 kJ/mol] −[−74.6 kJ/mol]
∆H◦
rxn =−965.1 kJ/mol + 74.6 kJ/mol
∆H◦
rxn =−890.5 kJ/mol
Therefore, the change in enthalpy (∆H) for the reaction of 5.00 moles of
methane gas is -890.5 kJ.
Question 9
Question
Calculate the change in enthalpy (∆H) when 5 moles of methane gas (CH4)
react with excess oxygen gas to produce carbon dioxide gas (CO2) and water
vapor (H2O) according to the following balanced chemical equation:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the following standard enthalpies of formation:
∆H◦
f(CO2) = −393.5 kJ/mol
∆H◦
f(H2O) = −285.8 kJ/mol
∆H◦
f(CH4) = −74.8 kJ/mol
Solution
Step 1: Calculate the standard enthalpy change for the given reaction using the
standard enthalpies of formation.
The standard enthalpy change, ∆H◦, for the reaction can be calculated using
the formula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
7
Substitute the given standard enthalpies of formation into the formula:
∆H◦= [1(−393.5) + 2(−285.8)] −[1(−74.8) + 2(0)]
Simplify the expression:
∆H◦= (−393.5−571.6) −(−74.8)
∆H◦=−965.1 + 74.8
∆H◦=−890.3 kJ/mol
Therefore, the change in enthalpy (∆H) for the given reaction is -890.3
kJ/mol.
Question 10
Question
Given that the enthalpy change (∆H) for the reaction C(s) + O2(g) →CO2(g)
is -393.5 kJ/mol, calculate the enthalpy change when 10.0 g of carbon reacts
with excess oxygen to form carbon dioxide.
Solution
Step 1: Calculate the number of moles of carbon reacting.
Molar mass of carbon = 12.01 g/mol
Moles of carbon = 10.0 g
12.01 g/mol
= 0.833 mol
Step 2: Use the given enthalpy change to calculate the enthalpy change for
the reaction.
∆H= ∆H◦×moles of reaction
=−393.5 kJ/mol ×0.833 mol
=−327.6 kJ
Therefore, the enthalpy change when 10.0 g of carbon reacts with excess
oxygen to form carbon dioxide is -327.6 kJ.
Question 11
Question
The enthalpy change (∆H) for the reaction
2A(g)+3B(l)→C(s)+4D(g)
8
is −1264 kJ. If the standard enthalpies of formation are −17 kJ/mol for A, −66
kJ/mol for B, −394 kJ/mol for C, and 109 kJ/mol for D, calculate the standard
enthalpy change of the reaction.
Solution
Step 1: Calculate the standard enthalpy change for the formation of products.
The standard enthalpy change for the formation of products is given by:
∆Hproducts =Xn∆Hf
where nis the stoichiometric coefficient and ∆Hfis the standard enthalpy of
formation.
Substitute values:
∆Hproducts = (1)(−394 kJ/mol) + (4)(109 kJ/mol)
∆Hproducts =−394 kJ/mol + 436 kJ/mol
∆Hproducts = 42 kJ/mol
Step 2: Calculate the standard enthalpy change for the formation of reac-
tants. The standard enthalpy change for the formation of reactants is given
by:
∆Hreactants =Xn∆Hf
where nis the stoichiometric coefficient and ∆Hfis the standard enthalpy of
formation.
Substitute values:
∆Hreactants = (2)(−17 kJ/mol) + (3)(−66 kJ/mol)
∆Hreactants =−34 kJ/mol −198 kJ/mol
∆Hreactants =−232 kJ/mol
Step 3: Calculate the standard enthalpy change of the reaction. The stan-
dard enthalpy change of the reaction is given by:
∆H= ∆Hproducts −∆Hreactants
Substitute values:
∆H= 42 kJ/mol −(−232 kJ/mol)
∆H= 42 kJ/mol + 232 kJ/mol
∆H= 274 kJ/mol
Therefore, the standard enthalpy change of the reaction is 274 kJ/mol.
9
Question 12
Question
Consider a reaction where 2 moles of liquid water are converted to steam at 1
atm and 100
°
C. The molar enthalpy of vaporization of water is 40.79 kJ/mol.
Calculate the change in enthalpy for this reaction.
Solution
Step 1: Calculate the heat required to vaporize 1 mole of water. Given that the
molar enthalpy of vaporization of water is 40.79 kJ/mol, the heat required to
vaporize 1 mole of water is 40.79 kJ.
Step 2: Calculate the heat required to vaporize 2 moles of water. Since we
have 2 moles of water, the total heat required to vaporize 2 moles is:
40.79 kJ/mol ×2 mol = 81.58 kJ
Step 3: Calculate the change in enthalpy. The change in enthalpy for this
reaction is the same as the heat required to vaporize 2 moles of water, which is
81.58 kJ. Therefore, the change in enthalpy is 81.58 kJ.
Question 13
Question
Calculate the change in enthalpy for a reaction where 2 moles of nitrogen gas
react with 3 moles of hydrogen gas to produce 2 moles of ammonia gas. The
enthalpies of formation for nitrogen gas, hydrogen gas, and ammonia gas are 0
kJ/mol, 0 kJ/mol, and -46 kJ/mol, respectively.
Solution
Step 1: Write the balanced chemical equation for the reaction.
N2(g)+3H2(g)→2NH3(g)
Step 2: Calculate the total change in enthalpy using the enthalpies of for-
mation. The change in enthalpy (∆H) can be calculated using the formula:
∆H=Xν∆Hf(products) −Xν∆Hf(reactants)
where νrepresents the stoichiometric coefficients and ∆Hfrepresents the en-
thalpy of formation.
Substitute the values into the formula:
∆H= 2(−46 kJ/mol) −[1(0 kJ/mol) + 3(0 kJ/mol)]
10
Step 3: Simplify the expression to find the change in enthalpy.
∆H=−92 kJ/mol −0 kJ/mol
∆H=−92 kJ/mol
Therefore, the change in enthalpy for the reaction is -92 kJ/mol.
Question 14
Question
Given the reaction:
2H2(g)+O2(g)→2H2O(l)
with ∆H=−571.6 kJ, calculate the enthalpy change when 4.00 g of H2is
burned in excess O2to form liquid water.
Solution
Step 1: Calculate the moles of hydrogen gas. Given: - Molar mass of hydrogen,
M(H2)=2.016 g/mol - Mass of hydrogen gas, m(H2)=4.00 g Using the formula
n=m
M, where nis moles, we have:
n(H2) = 4.00 g
2.016 g/mol = 1.984 mol
Step 2: Use the stoichiometry of the reaction to find the enthalpy change for
1.00 mol of hydrogen gas. From the balanced chemical equation, the enthalpy
change for the reaction is −571.6 kJ for 2 moles of H2. Therefore, for 1 mole of
H2, the enthalpy change is:
−571.6 kJ
2=−285.8 kJ/mol
Step 3: Calculate the enthalpy change for the given mass of hydrogen gas.
Since 1.984 moles of H2are involved, the enthalpy change for 1.984 moles is:
1.984 mol ×(−285.8 kJ/mol) = −566.7 kJ
Answer: The enthalpy change when 4.00 g of H2is burned in excess O2to
form liquid water is −566.7 kJ.
Question 15
Question
Calculate the change in enthalpy (∆H) for the combustion of propane gas
(C3H8) at constant pressure. The balanced chemical equation for the com-
bustion of propane is:
C3H8(g) + 5
2O2(g)→3CO2(g)+4H2O(g)
11
Given the standard enthalpies of formation (∆H◦
f) for C3H8(g), CO2(g), and
H2O(g) are -103.9 kJ/mol, -393.5 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Calculate ∆H◦for the combustion of propane using the standard en-
thalpies of formation of the reactants and products.
∆H◦=Xn∆H◦
f(products) −Xm∆H◦
f(reactants)
where nand mare the coefficients of the products and reactants, respectively
in the balanced chemical equation.
Step 2: Substitute the values of the standard enthalpies of formation pro-
vided into the equation.
∆H◦= [3(−393.5) + 4(−241.8)] −[−103.9] kJ/mol
∆H◦= (−1180.5−967.2) + 103.9 kJ/mol = −2043.8 kJ/mol
Therefore, the change in enthalpy for the combustion of propane gas at
constant pressure is ∆H=−2043.8 kJ/mol.
Question 16
Question
A certain reaction has an enthalpy change of -92 kJ/mol. If 2.50 moles of the
reactant are consumed in the reaction, what is the total enthalpy change for the
reaction?
Solution
Step 1: Determine the total enthalpy change for the reaction. Step 2: Use the
given enthalpy change per mole to find the total enthalpy change.
Step 1: Enthalpy change per mole = −92 kJ/molMoles of reactant consumed = 2.50 mol
Step 2: Total enthalpy change = Enthalpy change per mole×Moles of reactant consumed
Total enthalpy change = (−92 kJ/mol) ×(2.50 mol)
Total enthalpy change = −230 kJ
Therefore, the total enthalpy change for the reaction when 2.50 moles of the
reactant are consumed is -230 kJ.
12
Question 17
Question
A reaction has a standard enthalpy change of -394 kJ/mol. If the reaction is
exothermic, what can be said about the enthalpy of the products compared to
the enthalpy of the reactants?
Solution
Step 1: Recall that for an exothermic reaction, the standard enthalpy change is
negative.
Step 2: In this case, the standard enthalpy change is -394 kJ/mol, which
means that the products have lower enthalpy than the reactants.
Step 3: Therefore, the enthalpy of the products is lower than the enthalpy
of the reactants for this exothermic reaction.
Question 18
Question
A sample of nitrogen gas undergoes an isobaric process at a pressure of 1 atm.
During the process, the nitrogen gas absorbs 500 J of heat and performs 300 J
of work. If the initial enthalpy of the nitrogen gas is 2000 J, what is the final
enthalpy of the gas?
Solution
Step 1: Recall the definition of enthalpy: Enthalpy (H) is defined as the sum of
the internal energy (U) of a system and the product of pressure (P) and volume
(V) of the system:
H=U+P V
Step 2: Since the process is isobaric, the change in enthalpy can be expressed
as:
∆H= ∆U+P∆V
Step 3: We are given that the nitrogen gas absorbs 500 J of heat and performs
300 J of work. Therefore, the change in internal energy can be calculated as:
∆U=Q−W= 500 J −300 J = 200 J
Step 4: Substituting the values into the formula for the change in enthalpy:
∆H= 200 J + 1 atm ×∆V
Step 5: But we also know that:
∆H=Hfinal −Hinitial =Hfinal −2000 J
13
Step 6: Equating the two expressions for ∆H:
200 J + 1 atm ×∆V=Hfinal −2000 J
Step 7: Given that the pressure is 1 atm, we can simplify the equation to:
200 J + 1 atm ×∆V=Hfinal −2000 J
Step 8: Since ∆V=W
P, we can substitute in the values to find the change
in volume:
∆V=300 J
1 atm = 300 L ·atm
Step 9: Substituting ∆Vback into the equation and solving for Hfinal:
200 J + 1 atm ×300 L ·atm = Hfinal −2000 J
Step 10: Calculating the final enthalpy:
Hfinal = 200 J + 300 J = 2500 J
Therefore, the final enthalpy of the nitrogen gas is 2500 J.
Question 19
Question
A certain chemical reaction has an enthalpy change of ∆H=−289 kcal/mol. If
1.50 moles of the reactant is consumed in the reaction, what is the total change
in enthalpy in kilocalories?
Solution
Step 1: Determine the total change in enthalpy for 1 mole of the reactant.
Given: ∆H=−289 kcal/mol
Step 2: Calculate the total change in enthalpy for 1 mole of the reactant.
∆Htotal = ∆H×moles of reactant
∆Htotal =−289 kcal/mol ×1.50 moles
∆Htotal =−433.5 kcal
Therefore, the total change in enthalpy for 1.50 moles of the reactant is
−433.5 kcal.
14
Question 20
Question
A reaction is carried out at constant pressure where 100.0 g of a substance
X is converted to substance Y. The reaction releases 1500 J of heat and the
temperature of the surroundings increases by 2.5
°
C. The specific heat capacity of
the surroundings is 4.18 J/(g
°
C). Calculate the enthalpy change for the reaction.
Solution
Step 1: Calculate the heat gained by the surroundings. Given that the temper-
ature change (∆T) is 2.5
°
C and the specific heat capacity of the surroundings
(C) is 4.18 J/(g
°
C), we can use the formula:
q=C×m×∆T
where: q= heat gained by the surroundings, C= specific heat capacity of the
surroundings, m= mass of the surroundings, ∆T= temperature change of the
surroundings.
Substitute the values:
q= 4.18 J/(g
°
C) ×m×2.5
°
C
Step 2: Calculate the mass of the surroundings. Since the reaction releases
heat, the surroundings gain this heat. We can find the mass of the surroundings
using the formula:
q=qreaction =msurroundings ×C×∆T
Rearranging the formula:
msurroundings =qreaction
C×∆T
Step 3: Calculate the enthalpy change for the reaction. The enthalpy change
of the reaction (∆Hreaction) is equal in magnitude but opposite in sign to the
heat gained by the surroundings:
∆Hreaction =−q
Substitute the value obtained for qin Step 1 and calculate ∆Hreaction.
Question 21
Question
A certain reaction has an enthalpy change of ∆H=−486 kJ. If 2.00 moles of
this reaction take place, what is the heat associated with this reaction?
15
Solution
Step 1: Determine the heat associated with the reaction per mole. Given that
the enthalpy change for the reaction is ∆H=−486 kJ, we can find the heat per
mole by dividing by the number of moles:
Heat per mole = ∆H
moles =−486 kJ
2.00 mol
Heat per mole = −243 kJ/mol
Step 2: Calculate the total heat associated with the reaction. To find the
total heat for 2.00 moles of the reaction, we multiply the heat per mole by the
number of moles:
Total heat = Heat per mole ×moles
Total heat = −243 kJ/mol ×2.00 mol
Total heat = −486 kJ
Therefore, the total heat associated with the reaction when 2.00
moles take place is -486 kJ.
Question 22
Question
Calculate the enthalpy change for the reaction below at 298 K:
2H2(g)+O2(g)→2H2O(l)
Given the following standard enthalpy changes of formation: ∆H◦
f[H2(g)] = 0
kJ/mol, ∆H◦
f[O2(g)] = 0 kJ/mol, and ∆H◦
f[H2O(l)] = −285.8 kJ/mol.
Solution
Step 1: Write the balanced chemical equation and apply Hess’s Law. The
enthalpy change for the reaction can be calculated using the standard enthalpy
changes of formation for the reactants and products. We can use Hess’s Law to
obtain the enthalpy change for the desired reaction.
Balanced reaction:
2H2(g)+O2(g)→2H2O(l)
Step 2: Calculate the enthalpy change using the standard enthalpy changes
of formation. The enthalpy change for the reaction can be calculated as follows:
∆H◦
rxn =X∆H◦
products −X∆H◦
reactants
16
Substitute the given values:
∆H◦
rxn = 2(∆H◦
f[H2O(l)]) −[2(∆H◦
f[H2(g)]) + ∆H◦
f[O2(g)]]
∆H◦
rxn = 2(−285.8 kJ/mol) −[2(0 kJ/mol) + 0 kJ/mol]
∆H◦
rxn =−571.6 kJ/mol
Therefore, the enthalpy change for the reaction is -571.6 kJ/mol at 298 K.
Question 23
Question
A reaction involving the formation of ammonia gas is represented by the follow-
ing equation:
N2(g)+3H2(g)→2NH3(g)
Given the standard enthalpies of formation (∆H◦
f) for N2(g), H2(g), and N H3(g)
are 0 kJ/mol, 0 kJ/mol, and −46 kJ/mol, respectively, calculate the standard
enthalpy change of the reaction.
Solution
Step 1: Write the balanced chemical equation and standard enthalpy change
formula. The balanced chemical equation is:
N2(g)+3H2(g)→2NH3(g)
The standard enthalpy change of the reaction can be calculated using the for-
mula:
∆H◦=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Find the standard enthalpy change of the reaction. We know the
formation enthalpies of the products and reactants:
∆H◦
f(NH3) = −46 kJ/mol
∆H◦
f(N2) = 0 kJ/mol
∆H◦
f(H2) = 0 kJ/mol
Substitute these values into the formula:
∆H◦= [2 ×∆H◦
f(NH3)] −[1 ×∆H◦
f(N2)+3×∆H◦
f(H2)]
∆H◦= [2 ×(−46 kJ/mol)] −[1 ×0 kJ/mol + 3 ×0 kJ/mol]
∆H◦=−92 kJ/mol
Therefore, the standard enthalpy change of the reaction is ∆H◦=−92 kJ/mol.
17
Question 24
Question
Calculate the change in enthalpy (∆H) when 2 moles of nitrogen gas react
completely with 1 mole of oxygen gas to form nitrogen dioxide gas according to
the following balanced equation:
2N2(g)+O2(g)→2NO2(g)
Given that the standard enthalpies of formation (∆H◦
f) for N2(g), O2(g), and
NO2(g) are 0 kJ/mol, 0 kJ/mol, and 33.2 kJ/mol, respectively.
Solution
Step 1: Write the equation for the change in enthalpy using the standard en-
thalpies of formation:
∆H=Xν∆H◦
f
Where νrepresents the stoichiometric coefficients in the balanced equation and
∆H◦
fis the standard enthalpy of formation.
Step 2: Determine the change in enthalpy for the given reaction:
∆H= (2 ×∆H◦
f(NO2)) −(2 ×∆H◦
f(N2)+∆H◦
f(O2))
Step 3: Substitute the given values for the standard enthalpies of formation:
∆H= (2 ×33.2 kJ/mol) −(2 ×0 kJ/mol + 0 kJ/mol)
Step 4: Calculate the change in enthalpy:
∆H= 66.4 kJ/mol
Therefore, the change in enthalpy for the given reaction is 66.4 kJ/mol.
Question 25
Question
A reaction involving the conversion of graphite to diamond at standard condi-
tions has an enthalpy change of 2.9 kJ/mol. Calculate the minimum pressure
required to convert 1 mole of graphite to diamond at 25
°
C.
Solution
Step 1: Write down the enthalpy change of the reaction. Given: ∆H= 2.9
kJ/mol
18
Step 2: Recall the definition of standard enthalpy change. The standard
enthalpy change, ∆H◦, is the enthalpy change when reactants in their standard
states are converted to products in their standard states at 1 atm pressure.
Step 3: Determine the pressure needed for the phase change. Since we want
to convert 1 mole of graphite to diamond under standard conditions, we need
to consider the phase change from graphite to diamond and account for the
enthalpy change.
Step 4: Calculate the minimum pressure required using the enthalpy change
and temperature. The enthalpy change at standard conditions is related to the
pressure change via the equation:
∆H= ∆U+P∆V
At constant pressure, the change in enthalpy, ∆H, is approximately equal to
the heat change, ∆U. We can rearrange and solve for pressure:
P=∆H
∆V
Since the reaction involves a solid to solid phase change, the volume change is
negligible. Hence, ∆V≈0 and the pressure required is:
P=∆H
∆V≈2.9 kJ/mol
0=∞
Therefore, the minimum pressure required to convert 1 mole of graphite to
diamond at 25
°
C is theoretically infinite.
Question 26
Question
An ideal gas undergoes a process in which its enthalpy is given by the expression
H= 3.5T2−0.01T3, where His the enthalpy in kJ/mol and Tis the temperature
in K. Determine the heat capacity at constant pressure, Cp, for this gas in terms
of temperature.
Solution
Step 1: Recall that the heat capacity at constant pressure, Cp, is defined as the
partial derivative of the enthalpy with respect to temperature:
Cp=∂H
∂T P
Step 2: We are given the enthalpy expression H= 3.5T2−0.01T3, so we
differentiate this expression with respect to temperature T:
∂H
∂T =∂
∂T (3.5T2−0.01T3)
19
Step 3: Calculating the derivative term by term, we get:
∂H
∂T = 7T−0.03T2
Step 4: Therefore, the heat capacity at constant pressure is:
Cp= 7 −0.03TJ/mol
·
K
Question 27
Question
A reaction takes place at constant pressure with a change in enthalpy (∆H) of
-276 kJ. If the reaction releases 123 kJ of heat, what is the change in internal
energy (∆U) for the reaction? Assume no other work is done during the reaction.
Solution
Step 1: Recall the relationship between enthalpy (H), internal energy (U), and
pressure (P):
∆H= ∆U+P∆V
Step 2: Since the reaction takes place at constant pressure (∆V= 0), the
equation simplifies to:
∆H= ∆U
Step 3: Given that ∆H=−276 kJ, we can substitute this into the equation:
−276 kJ = ∆U
Step 4: Since ∆U=−276 kJ, this represents the change in internal energy
for the reaction when 276 kJ of heat is released.
Step 5: However, the reaction releases 123 kJ of heat instead of 276 kJ. To
find the change in internal energy when 123 kJ of heat is released, we need to
adjust ∆Uaccordingly.
Step 6: If the reaction releases 276 kJ of heat but only 123 kJ is actually
released, then the change in internal energy when 123 kJ of heat is released can
be calculated as follows:
∆U=123
276 ×(−276 kJ)
Step 7: Simplifying the calculation:
∆U=123
276 ×(−276) = −123
2
Step 8: Therefore, the change in internal energy (∆U) for the reaction when
123 kJ of heat is released is −123
2kJ.
20
Question 28
Question
Calculate the change in enthalpy when 1 mole of methane gas is burned in excess
oxygen at constant pressure according to the following reaction:
CH4(g)+ 2O2(g)→CO2(g)+ 2H2O(g)
Given the standard enthalpies of formation are: ∆H◦
f(CH4(g)) = −74.8 kJ/mol,
∆H◦
f(CO2(g)) = −393.5 kJ/mol, ∆H◦
f(H2O(g)) = −285.8 kJ/mol.
Solution
Step 1: Write the thermochemical equation for the reaction.
CH4(g)+ 2O2(g)→CO2(g)+ 2H2O(g)
Step 2: Calculate the standard enthalpy of reaction using the standard en-
thalpies of formation.
∆H◦
rxn =X∆H◦
f(products) −X∆H◦
f(reactants)
∆H◦
rxn = [1×∆H◦
f(CO2(g))+2×∆H◦
f(H2O(g))]−[∆H◦
f(CH4(g))+2×∆H◦
f(O2(g))]
∆H◦
rxn = [1 ×(−393.5) + 2 ×(−285.8)] −[(−74.8) + 2 ×0]
∆H◦
rxn = (−393.5−571.6) −(−74.8) = −965.1 kJ/mol
Therefore, the change in enthalpy when 1 mole of methane gas is burned in
excess oxygen is −965.1 kJ/mol.
Question 29
Question
Calculate the change in enthalpy (∆H) for the following reaction at 25
°
C:
2A(g)+3B(g)→C(g)
Given the following enthalpy changes:
∆H◦
ffor A(g) = −150 kJ/mol
∆H◦
ffor B(g) = −200 kJ/mol
∆H◦
ffor C(g) = −100 kJ/mol
21
Solution
Step 1: Write the balanced chemical equation and calculate the overall change
in enthalpy. The given reaction is:
2A(g)+3B(g)→C(g)
The overall change in enthalpy, ∆H, can be calculated using the formula:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Step 2: Calculate the enthalpy change for the reaction. Given:
∆H◦
ffor A(g) = −150 kJ/mol
∆H◦
ffor B(g) = −200 kJ/mol
∆H◦
ffor C(g) = −100 kJ/mol
Substitute the values into the formula: ∆H= [1(−100 kJ/mol)]−[2(−150 kJ/mol)+
3(−200 kJ/mol)]
Step 3: Calculate the change in enthalpy. ∆H=−100 kJ/mol+300 kJ/mol+
600 kJ/mol = 800 kJ/mol
Therefore, the change in enthalpy for the reaction is ∆H= 800 kJ/mol.
Question 30
Question
A reaction is taking place in a closed system at constant pressure. The reaction
releases 250 kJ of heat and does 100 kJ of work on the surroundings. Calculate
the change in enthalpy (∆H) for the system.
Solution
Step 1: The change in enthalpy is given by the formula:
∆H=q+w
where qis the heat exchanged with the surroundings and wis the work done on
the surroundings.
Step 2: Substitute the given values into the formula:
∆H=−250 kJ + 100 kJ
Step 3: Calculate the change in enthalpy:
∆H=−150 kJ
Step 4: Therefore, the change in enthalpy for the system is ∆H=−150 kJ.
22
Question 31
Question
Calculate the change in enthalpy when 5.0 moles of ammonia gas react with
excess oxygen gas according to the following balanced equation:
4NH3(g)+ 5O2(g)→4NO(g)+ 6H2O(g)
Given that the standard enthalpy of formation for NH3(g),NO(g), and H2O(g)
are -45.9 kJ/mol, 90.3 kJ/mol, and -241.8 kJ/mol, respectively.
Solution
Step 1: Calculate the standard enthalpy change for the reaction using standard
enthalpies of formation. The standard enthalpy change for the reaction can be
calculated by subtracting the sum of the standard enthalpies of formation of the
reactants from the sum of the standard enthalpies of formation of the products.
Given: Standard enthalpy of formation of NH3(g)= -45.9 kJ/mol Standard
enthalpy of formation of NO(g)= 90.3 kJ/mol Standard enthalpy of formation
of H2O(g)= -241.8 kJ/mol
The standard enthalpy change for the reaction is:
∆H◦= (4 ×90.3+6×(−241.8)) −(4 ×(−45.9) + 5 ×0)
∆H◦= (361.2−1452.8) −(−183.6)
∆H◦=−1091.6 + 183.6
∆H◦=−908.0 kJ/mol
Step 2: Calculate the change in enthalpy when 5.0 moles of ammonia gas
react. Given that 4 moles of NH3(g)are involved in the reaction, the change in
enthalpy for 5.0 moles can be calculated using the molar enthalpy change:
∆H=−908.0 kJ/mol
4 mol ×5 mol = −2270.0 kJ
Therefore, the change in enthalpy when 5.0 moles of NH3(g)react is -2270.0
kJ.
Question 32
Question
Calculate the change in enthalpy (∆H) when 5.00 moles of methane gas (CH4)
are combusted in excess oxygen gas (O2) to produce carbon dioxide gas (CO2)
23
and water vapor (H2O) according to the following reaction:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)
Given the standard enthalpies of formation (∆H◦
f) for CH4(g) = -74.8 kJ/mol,
CO2(g) = -393.5 kJ/mol, and H2O(g) = -241.8 kJ/mol.
Solution
Step 1: Calculate the standard enthalpy of formation of the reaction. The
standard enthalpy change of the reaction can be calculated using the standard
enthalpies of formation of the products and reactants:
∆H=X∆H◦
f(products) −X∆H◦
f(reactants)
Given: ∆H◦
f(CH4(g)) = −74.8 kJ/mol
∆H◦
f(CO2(g)) = −393.5 kJ/mol
∆H◦
f(H2O(g)) = −241.8 kJ/mol
∆H= [1 ·∆H◦
f(CO2)+2·∆H◦
f(H2O)] −[1 ·∆H◦
f(CH4)+2·∆H◦
f(O2)]
∆H= [1 · −393.5+2· −241.8] −[1 · −74.8+2·0]
∆H= [−393.5−483.6] −[−74.8]
∆H=−877.1 + 74.8
∆H=−802.3 kJ/mol
Step 2: Calculate the change in enthalpy for 5.00 moles of CH4. Given that
5.00 moles of CH4are combusted, the change in enthalpy can be calculated
using the following equation:
∆H=n·∆Hrxn
where nis the number of moles of CH4and ∆Hrxn is the standard enthalpy
change of the reaction.
∆H= 5.00 moles × −802.3 kJ/mol = −4011.5 kJ
Therefore, the change in enthalpy when 5.00 moles of CH4are combusted
is -4011.5 kJ.
Question 33
Question
A certain chemical reaction has an enthalpy change of -287 kJ/mol. If the
reaction produces 3.00 moles of a product, what is the total amount of heat
absorbed or released by the reaction?
24
Solution
Step 1: Identify the given information The enthalpy change (∆H) of the reaction
is -287 kJ/mol. The number of moles of product formed is 3.00 mol.
Step 2: Calculate the total amount of heat absorbed or released To find
the total amount of heat absorbed or released by the reaction, we multiply the
enthalpy change by the number of moles of product formed. Total heat absorbed
or released = ∆H×moles of product
Substitute the given values: Total heat absorbed or released = -287 kJ/mol
×3.00 mol
Step 3: Perform the calculation Total heat = -287 kJ/mol ×3.00 mol =
-861 kJ
Therefore, the total amount of heat absorbed or released by the reaction is
861 kJ, with a negative sign indicating that the reaction releases heat.
Question 34
Question
A chemical reaction is carried out at constant pressure, resulting in a change in
enthalpy of -250 kJ. If 150 kJ of work is done on the system during the reaction,
what is the heat transfer for the reaction?
Solution
Step 1: Recall the definition of enthalpy change (∆H) in terms of heat transfer
(q) and work done on the system (w):
∆H=q+w
Step 2: Plug in the given values into the equation:
−250 kJ = q+ 150 kJ
Step 3: Solve for qby isolating it:
q=−250 kJ −150 kJ
Step 4: Perform the calculation:
q=−400 kJ
Therefore, the heat transfer for the reaction is −400 kJ.
25
Question 35
Question
Calculate the change in enthalpy (∆H) when 10.0 moles of gaseous ammonia
(NH3) react completely with excess oxygen to form nitrogen monoxide (NO)
and water vapor (H2O) according to the following balanced chemical equation:
4 NH3(g) + 5 O2(g)→4 NO(g) + 6 H2O(g)
Given the enthalpies of formation per mole at 298 K are: ∆H◦
f(NH3) =
−45.9 kJ/mol, ∆H◦
f(NO) = 90.3 kJ/mol, and ∆H◦
f(H2O) = −241.8 kJ/mol.
Solution
Step 1: Calculate the enthalpy change of the reaction using the enthalpies of
formation.
∆H=Xn∆Hf(products) −Xm∆Hf(reactants)
= [4 ×∆H◦
f(NO) + 6 ×∆H◦
f(H2O)] −[4 ×∆H◦
f(NH3)+5×∆H◦
f(O2)]
= [4 ×90.3+6×(−241.8)] −[4 ×(−45.9) + 5 ×0]
= [361.2−1450.8] −[−183.6]
=−1089.6 + 183.6
=−906.0 kJ
Therefore, the change in enthalpy for the given reaction is ∆H=−906.0 kJ.
26
