CHEM 3411, Fall 2010
Solution Set 2
In this solution set, an underline is used to show the last significant digit of numbers. For instance in
x = 2.51693
the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the example, are not
significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in
calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic steps.
Numbers without underlines (including final answers) are shown with the proper number of sig figs.
1
Exercise 2.19a pg 86
Given
The standard enthalpy of combustion of cyclopropane is −2091 kJ mol−1 at 25◦ C. Additionally, the enthalpy of
formation of its isomer propene is +20.42 kJ mol−1 .
In terms of given variables, this is written:
∆r H ⊖ = −2091 kJ mol−1 cyclopropane combustion
∆f H ⊖ = +20.42 kJ mol−1 propene enthalpy of formation
Find
Calculate . . .
• (a) The enthalpy of formation of cyclopropane
• (b) The enthalpy of isomerization from cyclopropane to propene
Strategy
We begin by constructing the balanced chemical equation for combustion of cyclopropane
2C3 H6 + 9O2 (g) → 6H2 O(l) + 6CO2 (g)
∆r H ⊖ = 2 × −2091 kJ mol−1
For any such reaction the reaction enthalpy can related to the enthalpy of products and reactants by (Equation 2.32
on pg 68 in your text book)
∆r H ⊖ =
∑
Products
νp ∆f Hp⊖ −
∑
νr ∆f Hr⊖
Reactants
which for this reaction is
∆r H ⊖ = 6∆f H ⊖ (H2 O(l)) + 6∆f H ⊖ (CO2 (g)) − 2∆f H ⊖ (C3 H6 ) + −9∆f H ⊖ (O2 (g))
{z
} |
{z
}
|
products
reactants
Using the fact that the standard state of oxygen is O2 (g) (∆f H ⊖ (O2 (g)) = 0) the following expression for ∆f H ⊖ (C3 H6 )
is obtained through rearrangement
1
CHEM 3411, Fall 2010
Solution Set 2
∆f H ⊖ (C3 H6 )
=
1
3∆f H ⊖ (H2 O(l)) + 3∆f H ⊖ (CO2 (g)) − ∆r H ⊖
2
= 3 × −285.83 kJ mol−1 + 3 × −393.51 kJ mol−1 −
= 52.98 kJ mol−1
1
× 2 × −2091 kJ mol−1
2
With this information we can now calculate the enthalpy of isomerization of cyclopropane to propene. The balanced
equation for this process is
C3 H6 (cyclopropane) → C3 H6 (propene)
∆r H ⊖ =?
The enthalpy of this reaction is simply given by
∆r H ⊖
=
=
∆f H ⊖ (propene) − ∆f H ⊖ (cyclopropane)
20.42 kJ mol−1 − 52.98
= −32.56
Solution
• The enthalpy of formation of cyclopropane is ∆f H ⊖ (C3 H6 ) = 53 kJ mol−1 .
• The isomerization enthalpy of cyclopropane to propene is ∆r H ⊖ = −33 kJ mol−1 .
2
CHEM 3411, Fall 2010
Solution Set 2
2
Exercise 2.20a pg 86
Given
When a 120 mg sample of naphthalene, C10 H8 (s), was burned in a bomb calorimeter the temperature rose by 3.05 K.
Additionally, we are to predict the temperature rise for the combustion of 10 mg sample of phenol in the same
calorimeter (the enthalpy of combustion of phenol is ∆c H ⊖ (C6 H5 OH(s)) = −7061 kJ mol−1 ).
In terms of given variables, this is written:
mC10 H8 (s) = 120 mg
∆T = 3.05 K
mC6 H5 OH(s) = 10 mg
∆c H ⊖ (C6 H5 OH(s)) = −7061 kJ mol−1
Find
• Calculate the calorimeter constant.
• By how much will the temperature rise after the phenol sample is burned?
Strategy
To determine the calorimeter constant C we’ll need to determine how much heat was released in this combustion
reaction and for this we’ll need to know the combustion enthalpy of naphthalene. We’ll start by constructing the
balanced chemical equation for the combustion of naphthalene.
C10 H8 (s) + 12O2 (g) → 4H2 O(l) + 10CO2 (g)
∆r H ⊖ =?
We can calculate the enthalpy of this reaction from
∆r H ⊖
=
∑
∑
νp ∆f Hp⊖ −
Products
νr ∆f Hr⊖
Reactants
= 4∆f H ⊖ (H2 O(l)) + 10∆f H ⊖ (CO2 (g)) − ∆f H ⊖ (C10 H8 (s)) + −12∆f H ⊖ (O2 (g))
= 4 × −285.83 kJ mol−1 + 10 × −393.51 kJ mol−1 − +78.53 kJ mol−1 − 12 × 0
= −5156.950 kJ mol−1
To calculate the heat released in this specific combustion we’ll next need to know how many moles of naphthalene
were burned.
n=
C10 H8 (s)
120
mg
1
g
1000
mg
1 mol
128.17
g
=
9.363 × 10−4 mol C10 H8 (s)
3
CHEM 3411, Fall 2010
Solution Set 2
We can now easily calculate the heat produced qp in the combustion of this quantity of moles.
qp
= n∆r H ⊖
× −5156.950 kJ
−1
= 9.363 × 10−4
mol
mol
= −4.883 kJ
The calorimeter absorbed this heat qa = −qp = 4.883 kJ. As we know the temperature change in the calorimeter
resulting from the absorption of this heat, we can calculate the calorimeter constant C from q = C∆T .
C
qa
∆T
4.883 kJ
=
3.05 K
= 1.601 kJ K−1
=
As we now know the calorimeter constant we can go ahead and determine the calorimeter temperature change for
burning the phenol sample. The balanced chemical equation for this reaction is
C6 H5 OH(s) + 7O2 (g) → 3H2 O(l) + 6CO2 (g)
∆r H ⊖ =?
and the enthalpy of this reaction is
∆r H ⊖
=
∑
νp ∆f Hp⊖ −
∑
νr ∆f Hr⊖
Reactants
Products
= 3∆f H ⊖ (H2 O(l)) + 6∆f H ⊖ (CO2 (g)) − ∆f H ⊖ (C6 H5 OH(s)) + −7∆f H ⊖ (O2 (g))
= 3 × −285.83 kJ mol−1 + 6 × −393.51 kJ mol−1 − −165.0 kJ mol−1 − 7 × 0
= −3053.55 kJ mol−1
To determine the heat, we’ll need the moles in this combustion.
n=
C6 H5 OH(s)
mg
10
1
g
mg
1000
1 mol
94.11
g
=
1.06 × 10−4 mol C10 H8 (s)
Now we’ll calculate the heat produced in combustion.
qp
= n∆r H ⊖
= 1.06 × 10−4 mol × −3053.55 kJ mol−1
= −0.324 kJ
And lastly we’ll determine the temperature change induced in the calorimeter by absorbing this heat.
∆T
=
qa
C
kJ
0.324
1.601
kJ K−1
= 0.202 K
=
4
CHEM 3411, Fall 2010
Solution Set 2
Solution
• The calorimeter constant is C = 1.6 kJ K−1 .
• Burning phenol will lead to a temperature increase of ∆T = 0.2 K.
5
CHEM 3411, Fall 2010
Solution Set 2
3
Exercise 2.24a pg 87
Given
For the reaction C2 H5 OH(l) + 3O2 (g) → 2CO2 (g) + 3H2 O(g), ∆r U ⊖ = −1373 kJ mol−1 at T = 298 K.
In terms of given variables, this is written:
∆r U ⊖ = −1373 kJ mol−1
T = 298 K
Find
Calculate ∆r H ⊖ of this reaction.
Strategy
The following equation relates the enthalpy change of a reaction to the internal energy change (Equation 2.21 on pg
58 in your text book).
∆H = ∆U + ∆ng RT
where ∆ng RT is the change in quantity of gas molecules of the reaction.
In analyzing this reaction
C2 H5 OH(l) + 3O2 (g) → 2CO2 (g) + 3H2 O(g)
we see that 5 moles of gas are produced (2 moles of CO2 (g) and 3 of H2 O(g)) while 3 are consumed (O2 (g)). Therefore
∆ng = 2 and ∆r H ⊖ is easily calculated as
∆r H ⊖
=
∆r U ⊖ + ∆ng RT
= −1373 × 103 J mol−1 +
= −1.36804 × 106 J mol−1
2
mol
gas
−1
−1
mol
× 298
K
× 8.314 J
K
1 mol reaction
= −1368.04 kJ mol−1
Solution
∆r H ⊖ = 1368 kJ mol−1
6
CHEM 3411, Fall 2010
Solution Set 2
4
Exercise 2.26a pg 87
Given
Consider the reaction
2NO2 (g) → N2 O4 (g)
at T = 100◦ C (T = 373 K) and the information in table 2.8
Find
Calculate the standard reaction enthalpy of this reaction at this temperature.
Strategy
For this problem, we’ll use the fact that enthalpy is a state function to create a series of processes identical to this
reaction at the non-standard temperature. By summing summing the change in enthalpy over these steps we’ll arrive
at the enthalpy of this reaction at the non-standard temperature.
These processes can be summarized as cooling 2 moles of NO2 (g) from T = 373 K to the standard temperature
of T = 298 K. At this standard temperature the 2 moles of NO2 (g) can be reacted to form 1 mole of N2 O4 (g)
and we can calculate the reaction enthalpy at this standard temperature. Lastly, we’ll heat the 1 mole of N2 O4 (g)
from T = 298 K to T = 373 K. Summing the enthalpy changes over these three processes will give us the enthalpy
difference between 2 moles of NO2 (g) and 1 mole of N2 O4 (g) at T = 373 K and this is the enthalpy difference of the
reaction of interest.
This is summarized as
cool ∆H
2 mol NO2 (g) at T = 373 K −−−−−−cool
−→
∆r H ⊖ (at T = 298 K)
2 mol NO2 (g)atT = 298 K −−−−−−−−−−−−−−→
∆H
heat
1 mol N2 O4 (g)atT = 298 K −−−−
−→
1 mol N2 O4 (g)atT = 373 K
∆r H ⊖ (at T = 373 K) = ∆Hcool + ∆r H ⊖ (at T = 298 K) + ∆Hheat
The enthalpy of cooling or heating a substance can be determined from the substance’s constant pressure heat
capacity Cp .
∫
Tf
Cp dT
∆H =
Ti
For this problem we’ll assume heat capacities are temperature independent over the temperature range of interest
and this simplifies the calculations.
7
CHEM 3411, Fall 2010
Solution Set 2
∫
∆H
=
Tf
−
Cp dT
Ti
∫
=
Tf
Cp
dT
Ti
=
Cp (Tf − Ti )
=
Cp ∆T
Using heat capacities from our textbook we can calculate the enthalpy of heating the 2 moles of NO2 (g) and cooling
1 mole of N2 O4 (g).
∆Hcool
= ν × Cp ∆T (Cool NO2 (g))
mol−1 × (298 −1
= 2 × 37.20 J
K
K − 373
K)
−1
= = −5.5800 kJ mol
∆Hheat
=
=
=
ν × Cp ∆T (Heat N2 O4 (g))
mol−1 × (373 −1
1 × 77.28 J
K
K − 298
K)
= 5.7960 kJ mol−1
(Note that in these last two expressions the reaction coefficients are exact quantities and don’t affect number of
significant figures.)
In the original solution sets the temperatures were swapped which changed the signs of these enthalpies.
The enthalpy of the reaction at the standard temperature of T = 298 can be calculated from enthalpies of formation.
∆r H ⊖
=
∑
Products
νp ∆f Hp⊖ −
∑
νr ∆f Hr⊖
Reactants
= 1 × 9.16 kJ mol−1 − 2 × 33.18 kJ mol−1
= −57.2000 kJ mol−1
Summing these enthalpies gives the reaction enthalpy at the non-standard temperature
∆r H ⊖ (at T = 373 K)
= ∆Hcool + ∆r H ⊖ (at T = 298 K) + ∆Hheat
= −5.5800 kJ mol−1 + −57.2000 kJ mol−1 + 5.7960 kJ mol−1
= −56.9840 kJ mol−1
Solution
∆r H ⊖ (at T = 373 K) = −56.98 kJ mol−1
8
CHEM 3411, Fall 2010
Solution Set 2
5
Exercise 2.29a pg 87
Given
We are given the following enthalpies . . .
• Enthalpy of sublimation of Mg(s) ∆sub H ⊖ = +167.2 kJ mol−1
Mg(s) → Mg(g)
∆sub H ⊖ = +167.2 kJ mol−1
• First ionization enthalpy of Mg(g) ∆ion1 H ⊖ = 7.646 eV
∆ion1 H ⊖ = 7.646 eV
Mg(g) → Mg+ (g)
• Second ionization enthalpy of Mg(g) ∆ion2 H ⊖ = 15.035 eV
∆ion2 H ⊖ = 15.035 eV
Mg+ (g) → Mg+2 (g)
• Dissociation enthalpy of Cl2 (g) ∆dis H ⊖ = +241.6 kJ mol−1
Cl2 (g) → 2Cl(g)
∆dis H ⊖ = +241.6 kJ mol−1
• Electron gain enthalpy of Cl(g) ∆elec H ⊖ = −3.78 eV
Cl(g) → Cl− (g)
∆elec H ⊖ = −3.78 eV
• Enthalpy of solution of MgCl2 (s) ∆sol H ⊖ = −150.5 kJ mol−1
MgCl2 (s) → Mg+2 (aq) + 2Cl− (aq)
∆sol H ⊖ = −150.5 kJ mol−1
• Enthalpy of hydration of Cl− (g) ∆hyd H ⊖ = −383.7 kJ mol−1
Cl− (g) → Cl− (aq)
∆hyd H ⊖ = −383.7 kJ mol−1
Find
Calculate the enthalpy of hydration ∆hyd H ⊖ of for Mg+2 (g).
Strategy
For this problem we need to construct a thermodynamic cycle which includes the unknown process; hydration of
Mg+2 (g), Mg+2 (g) ∆hyd H ⊖ . If we know the enthalpies all other processes in this cycle, then we can solve for the
single unknown enthalpy. Such a cycle is shown is Figure 5 where one mole of Mg+2 (g) and two moles of Cl− (g) are
solvated through two different pathways. These sum of enthalpies over each of these two paths are thereby equal.
For this cycle we found it necessary to include a process that was not given to us by the textbook: the enthalpy
of formation of MgCl2 (s) from Mg(s) and Cl2 (g). We can calculate the enthalpy for this process from its balanced
chemical equation
Mg(s) + Cl2 (g) → MgCl2 (s)
The enthalpy of this reaction is calculated as
9
CHEM 3411, Fall 2010
Solution Set 2
Figure 1: Thermodynamic cycle diagram
∆r H ⊖
=
∑
∑
νp ∆f Hp⊖ −
Products
νr ∆f Hr⊖
Reactants
= ∆f H ⊖ (MgCl2 (s)) − ∆f H ⊖ (Mg(s)) − ∆f H ⊖ (Cl2 (g))
= −641.32 kJ mol−1 − 0 − 0
= −641.32 kJ mol−1
This gives us the enthalpy of the additional process we need:
• The enthalpy of formation of MgCl2 (s) ∆f H ⊖ = −641.32 kJ mol−1
Mg(s) + Cl2 (g) → MgCl2 (s)
∆f H ⊖ = −641.32 kJ mol−1
To reiterate, we had to obtain this quantity because it doesn’t appear possible to complete this problem with only
the given information. The themodynamic cycle diagram in the figure also illustrates the fact that we can’t complete
the cycle unless we link the MgCl2 (s) to the corresponding stable elements. The formation reaction is precisely the
missing link, and the problem statement didn’t have to provide it because we are able to obtain it using information
from the Table 2.8.
If we equate the sum of the enthalpies of the two solvation paths in this thermodynamic cycle we get the following
equation:
10
CHEM 3411, Fall 2010
Solution Set 2
2 × ∆hyd H ⊖ (Cl− (g)) + ∆hyd H ⊖ (Mg+2 (g)) =
[
]
−∆ion1 H ⊖ (Mg(g)) + −∆ion2 H ⊖ (Mg+ (g)) + −∆sub H ⊖ (Mg(s))
[
]
−2 × ∆elec H ⊖ (Cl− (g)) + −∆dis H ⊖ (Cl2 (g))
+∆f H ⊖ (MgCl2 (s)) + ∆sol H ⊖ (MgCl2 (s))
This equation can be solved for our unknown ∆hyd H ⊖ (Mg+2 (g))
∆hyd H ⊖ (Mg+2 (g))
=
[
]
−∆ion1 H ⊖ (Mg(g)) + −∆ion2 H ⊖ (Mg+ (g)) + −∆sub H ⊖ (Mg(s))
[
]
−2 × ∆elec H ⊖ (Cl− (g)) + −∆dis H ⊖ (Cl2 (g))
+∆f H ⊖ (MgCl2 (s)) + ∆sol H ⊖ (MgCl2 (s)) +
−2 × ∆hyd H ⊖ (Cl− (g))
]
[
= −737.722 kJ mol−1 + −1450.650 kJ mol−1 + −167.2 kJ mol−1
[
]
−2 × −364.71 kJ mol−1 + −241.6 kJ mol−1 + +
−641.32 kJ mol−1 + −150.5 kJ mol−1 +
−2 × −383.7 kJ mol−1
= −3346.012 kJ mol−1
where the energies in electron-volts have been converted by the following expression
∆ion1 H ⊖ =
∆ion2 H ⊖ =
∆elec H ⊖ =
7.646
eV
1.60218 × 10−19
J
1
eV
1 kJ
1000
J
6.0221 × 1023 electrons
= 737.722 kJ mol−1
1 mol
15.035
eV
1.60218 × 10−19
J
1
eV
1 kJ
1000
J
6.0221 × 1023 electrons
= 1450.650 kJ mol−1
1 mol
−3.78
eV
1.60218 × 10−19
J
1
eV
1 kJ
1000
J
6.0221 × 1023 electrons
= −364.71 kJ mol−1
1 mol
Solution
∆hyd H ⊖ (Mg+2 (g)) = −3346.0 kJ mol−1
11
CHEM 3411, Fall 2010
Solution Set 2
6
Exercise 2.30a pg 87
Given
When a certain freon used in refrigeration was expanded adiabatically from an initial pressure of pi = 32 atm and a
Ti = 0◦ C to a final pressure of pf = 1.00 atm, the temperature fell by ∆T = 22 K.
In terms of given variables, this is written:
pi = 32 atm
Ti = 0◦ C
pf = 1.00 atm
∆T = −22 K
Find
Calculate the Joule-Thomson coefficient, µ, at T = 0◦ C assuming it remains constant over this temperature range.
Strategy
The Joule-Thomson coefficient is defined as
(
µ=
∂T
∂p
)
H
If we assume µ is constant over the range of temperature and pressures changes then the derivative can be replaced
with finite differences.
µ=
∆T
∆p
Substituting in our temperature and pressure changes gives
µ
=
−22 K
1.00 atm−32 atm
−1
= 0.7097 K atm
Solution
µ = 0.71 K atm−1
12