Midterm Exam IV
CHEM 181: Introduction to Chemical Principles
November 29, 2012
1. Use some or all of the following data:
∆H◦f (kJ mol−1 )
C(graphite)
C(diamond)
CO2 (g)
H2 O(g)
CH4 (g)
∆H◦bond (kJ mol−1 )
C–C
C–H
H–H
0.
1.9
−393.5
−241.8
−74.6
347
414
435
(a) Assume ∆H◦f is ≈ 0 for C(s, coal). Calculate the ∆H◦rxn for the combustion
of coal and of CH4 .
For the combustion of coal,
C(s, coal) + O2 (g) −→ CO2 (g)
Using heats of formation (and remembering ∆H◦f is zero for oxygen):
◦
∆Hrxn
=
=
=
=
∆Hf◦ (products) − ∆Hf◦ (reactants)
∆Hf◦ (CO2 (g)) − ∆Hf◦ (C(s,coal))
−393.5 − 0
−393.5 kJ mol−1
For CH4 ,
CH4 (g) + 2O2 (g) −→ CO2 (g) + 2H2 O(g)
and
◦
∆Hrxn
= ∆Hf◦ (products) − ∆Hf◦ (reactants)
= ∆Hf◦ (CO2 (g)) + 2∆Hf◦ (H2 O(g)) − ∆Hf◦ (CH4 (g))
= [−393.5 + 2(−241.8)] − [−74.6]
= −802.5 kJ mol−1
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(b) Use the data on the previous page to estimate ∆Hrxn for
2CH4 (g) −→ C2 H6 (g) + H2 (g)
Using your answer, calculate
i. ∆H◦f for C2 H6 .
ii. ∆H◦rxn for the combustion of C2 H6 .
The reaction breaks two C–H bonds and forms one C–C and one H–H
bond. So
◦
◦
◦
∆Hrxn
= ∆Hbond
(reactants) − ∆Hbond
(products)
= [2 · 414] − [347 + 435]
= 46 kJ mol−1
Working the previous part backwards (with ∆H◦f = 0 for H2 ):
◦
∆Hrxn
= ∆Hf◦ (products) − ∆Hf◦ (reactants)
46 kJ mol−1 = ∆Hf◦ (C2 H6 (g)) − 2(−74.6)
∆Hf◦ (C2 H6 (g)) = 46 + 2(−74.6)
= −103.2 kJ mol−1
(This value is a little more negative than the actual value, because C–H
bonds in CH4 are a little stronger than average.) For combustion:
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C2 H6 (g) + O2 (g) −→ 2CO2 (g) + 3H2 O(g)
2
and
◦
∆Hrxn
= ∆Hf◦ (products) − ∆Hf◦ (reactants)
= 2∆Hf◦ (CO2 (g)) + 3∆Hf◦ (H2 O(g)) − ∆Hf◦ (C2 H6 (g))
= [2(−393.5) + 3(−241.8)] − [−103.2]
= −1409 kJ mol−1
(c) Make a quantitative argument using enthalpies of formation and/or bond
enthalpies that supports the statement:
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Combustion of coal produces more CO2 (g) per unit of energy than
any other fossil fuel.
Your argument should go beyond the two examples here (CH4 and C2 H6 ).
The previous part’s approach can be extended to any hydrocarbon:
RH + CH4 (g) −→ RCH3 + H2 (g)
Given the approximations involved with using bond enthalpies, this has
the same +46 kJ mol−1 reaction enthalpy. Heat of formation is then:
◦
∆Hrxn
= ∆Hf◦ (products) − ∆Hf◦ (reactants)
46 kJ mol−1 = ∆Hf◦ (RCH3 ) − (∆Hf◦ (RH) − 74.6)
∆Hf◦ (RCH3 ) = ∆Hf◦ (RH) − 28.6 kJ mol−1
So comparing the two reactions,
RH+?O2 (g) −→?CO2 (g)+?H2 O(g)
and
RCH3 +?O2 (g) −→?CO2 (g)+?H2 O(g)
the second one will produce one extra CO2 and one extra H2 O—that is, the
enthalpy of formation of the combustion products will be −393.5−241.8 =
635.3 kJ mol more negative, and the reactants only 28.6 kJ mol−1 more
negative. That extra CO2 , then, is accompanied by 606.7 kJ mol−1 extra
released heat; that’s much more than the 393.5 kJ mol−1 per CO2 released
burning coal.
Another approach would make fossil-fuel hydrocarbons from coal by first
reacting the coal with H2 . For each C–C and H–H bond broken, there
are two new C–H bonds formed; this is the reverse of the calculation done
in part (b), and so the reaction enthalpy is −46 kJ mol−1 per H2 added.
This does not change the amount of CO2 produced upon combustion, but
because it produces an extra water, it makes the reaction enthalpy more
negative by 241.8 − 46 = 195.8 kJ mol−1 .
2. A 0.0500 L sample of 0.500 M Ba(NO3 )2 is added to 0.0500 L of 0.500 M MgSO4
in a calorimeter with a total heat capacity of 455 J K−1 . The observed increase
in temperature is 1.43 K.
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(a) Calculate the value of ∆Hrxn (ignoring any dependence on concentration)
for the equation
Ba(NO3 )2 (aq) + MgSO4 (aq) −→ BaSO4 (s) + Mg(NO3 )2 (aq)
(You will need this value of ∆Hrxn for the next part of the problem. If you
do not get an answer, you can continue the calculation assuming ∆Hrxn =
−30 kJ mol−1 .)
There are
−2
(5 × 10
mol
L) 0.5
= 2.5 × 10−2 mol
L
of BaSO4 (s) formed. The total heat released is
q = C∆T
= (455 J K−1 )(1.43 K)
= 650.6 J
Since the surroundings are increasing in temperature, enthalpy change for
the system is negative, and
∆Hrxn = −
650.6 J
= −26 kJ mol−1
2.5 × 10−2 mol
(b) At equilibrium at 20 ◦ C, the concentration of a saturated solution of BaSO4
in water is 1.0487 × 10−5 M. What is the concentration of a saturated
solution at 80 ◦ C?
Change these temperatures to Kelvin, right off: 293 K and 353 K. The
important equilibrium is
Ba2+ (aq) + SO2−
4 (aq) BaSO4 (s)
with
K=
1
[Ba2+ ][SO2−
4 ]
since we know the equilibrium concentrations at 20 ◦ C, we know that
K293K =
1
= 9.093 × 109
(1.0487 × 10−5 )2
4
This is very large, in keeping with a sparingly soluble BaSO4 . Since the
reaction as written releases heat, LeChatelier’s principle says that raising
temperature should push equilibrium towards reactants—that is, greater
solubility. Calculating the temperature dependence of equilibrium requires
the van’t Hoff equation:
K353K
∆H
1
1
ln
= −
−
K293K
R
353 K 293 K
−26 × 103 J mol−1
−4
= −
K−1
−1
−1 −5.801 × 10
8.314 J mol K
= −1.814
K353K
= e−1.814
K293K
= 0.163
K353K = 0.163K293K
= 1.482 × 109
Plugging this back into the definition of K gives
[Ba2+ ] = [SO2−
4 ] = √
1
= 2.598 × 10−5 M
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1.482 × 10
(It might be conceptually easier to reverse the reaction—deal with solvation
rather than precipitation—and flip the sign of ∆H.)
3. The equilibrium constant at 25.0 ◦ C for the equation
Co3+ (aq) + 6NH3 (aq) [Co(NH3 )6 ]3+ (aq)
is
Kf = 2.0 × 107 M−6
(a) Calculate the value of ∆G◦rxn at 25.0 ◦ C. In which direction is the reaction
spontaneous when Co3+ (aq), NH3 (aq), and [Co(NH3 )6 ]3+ are at standard
conditions?
∆G◦rxn = −RT ln K
= −(8.314 J K−1 )(298.15 K) ln(2.0 × 107 )
= −41.67 kJ mol−1
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Reaction is spontaneous in the forward direction, as written.
(b) Calculate the value of ∆Grxn when [Co3+ ] = 0.0050 M, [NH3 ] = 0.1 M, and
[[Co(NH3 )6 ]3+ ] = 1.00 M. In what direction is the reaction spontaneous
under these conditions?
[Co(NH3 )3+
6 ]
3+
[Co ][NH3 ]6
1
=
0.005 · 0.16
= 2 × 108
Q
= RT ln
K
Q =
∆Grxn
= (8.314 J K−1 )(298.15 K) ln
2 × 108
2.0 × 107
= 5.71 kJ mol−1
Reaction is spontaneous in the backward direction (products to reactants),
as written.
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