CHEMISTRY 156
FIRST MIDTERM EXAM
SOLUTIONS
SPRING, 2004
INSERT: READ THIS FIRST
Chemistry 156, Spring, 2004
First Midterm Examination
Useful constants and data relating to problems 1 and 2 are found on this insert. You may
use the back side of this insert for scratch work but enter all work to be graded in the
space provided with each question. Show your work in all questions involving a
calculation and/or derivation in order to receive credit. This examination is worth 270
points.
I) Useful constants
R = 8.31451 J/K-mole = 0.082054 L-atm/K-mole
NA = 6.0221 x 1023 molecules/mole
kB = 1.38065 x 10-23 J/molecule
ln(10) = 2.303
0.00C = 273.15 K
II) Data relating to problem 1
thermodynamic data for the interconversion of native (N) and denatured lysozyme (D)
Cp(N(aq)) = 3172 J/K-mole
Cp(D(aq)) = 11783 J/K-mole
H for N(aq)D(aq) is 360.00 kJ/mole at 313.65 K.
D and N are in equilibrium under standard conditions at 313.65 K.
III) Data relating to problem 2
values of thermodynamic state functions at 25.00 C
species
Hf(kJ/mole)
S(J/K-mole)
O2(g)
H2(g)
205.138
130.684
H2O(l)
-285.830
188.825
NAME:_________SOLUTIONS KEY______________ READ THE INSERT FIRST!
1) (105 points) Professor Blaber at Florida State University has employed calorimetry to
study the interconversion of the native (N) and denatured (D) forms of the enzyme
lysozyme. Thermodynamic data relating to these forms are found on the insert.
a) Calculate H for the denaturation reaction N(aq)D(aq) at 25.00C.
Set up a thermodynamic cycle: N(298.15 K) to N(313.65 K) [step I]
N(313.64 K) to D(313.65 K) [step II]
D(313.65 K) to D(298.15 K) [step III]
H(I) = Cp(N)(313.65-298.15) = 49.166 kJ/mole
H(II) = H(reaction) = 360.00 kJ/mole
H(III) = Cp(D)(298.15-313.65) = -182.636 kJ/mole
H(net) = 49.166 + 360.00 + -182.636 = 226.5 kJ/mole
a) Calculate S for the denaturation reaction N(aq)D(aq) at 25.00C.
Use the same cycle of 3 steps. Each step of the cycle is reversible!
S(I) = Cp(N)ln(313.65/298.15) = 160.76 J/K-mole
S(II) = H(II)/313.65 = 360,000/313.65 = 1147.78 J/K-mole
S(III) = Cp(D)ln(313.65/298.15) = -436.4 J/K-mole
S(net) = 160.76 + 1147.78 + -436.4 = 711.4 J/K-mole
N.B. S(298.15 K) does not equal H(298.15 K)/298.15 as the two phases are not in
equilibrium at 298.15 K.
b) Show with a calculation that the native form (N) is more stable at 25.00C.
Use G to address the question. G(298 K) = H(298 K) - TS(298 K) 226,530 J (298.15)(711.4) = 14.4 kJ > 0 therefore the denaturation reaction does not proceed
spontaneously and the native form is more stable.
2) (95 points) A fuel cell, which is based on the reaction 2 H2(g) + O2(g)  2 H2O(l), is
coupled to an electric motor constructed to lift a weight of 5.0 x 103 Kg. The fuel cell,
operating at 25.00C under a constant atmospheric pressure of 1.000 atm, is charged with
2.000 mole of gaseous dihydrogen.
a) Calculate the enthalpy change, H, and the energy change, E, associated with
the combustion of the hydrogen.
H = 2Hf(H2O(l)) = 2(-285.830 kJ) = -571.kJ
H = E + pV = E + pV so E = H - pV
pV = 2(0) -1(RT) -2RT = -3RT (liquids occupy very little volume and pV = nRT for
ideal gases)
E = -571.66 - -7.43 = -564.22 kJ
b) Suppose that the fuel cell is run very slowly, calculate the heat and the total
work.
The process is run very slowly and hence can be treated as a reversible process.
q = qrev = TS
S = 2S(H2O(1)) -1S(O2(g)) -2S= 2(188.825) - 1(205.138) - 2(130.684) = -88.85 J/K
q = (298.15)(-88.85) = -26.49 kJ
N.B. q does not equal H because w(other) is not zero.
E = q + w so w = - q = -537.74 kJ
c) Compare the performance of the fuel cell run very slowly with one that runs
very quickly. Given a fixed amount of fuel, in which case will the device raise
the weight higher? Explain.
The maximum amount of useful work, raising the weight in this case, is obtained when
the process is run as reversibly, i.e. slowly. Hence, the fuel cell run slowly will raise the
weight higher. The intent of the problem was to raise the issue of reversibility versus
irreversibility.
3) (70 points) Real gases are well described by the van der Waals equation of state:
(V - nb)(p + an2/V2) = nRT
where the parameter b is a measure of the molecular volume and the parameter a, the
attraction between molecules.
a) In order to assess quantitatively the consequences of deviation of free-gas
behavior, first derive using the 4 great differentials general (i.e. no assumptions
about the nature of the substance) equations that relate the following quantities to
functions of just p, V, T, and n.
i) (S/V)T,n
At fixed n, dA = -SdT -pdV. A is a state function so the differential is exact and the cross
derivatives are equal.
-(S/V)T,n = -(p/T)V,n so (S/V)T,n = (p/T)V,n
ii) (E/V)T,n
At fixed n, dE = TdS = pdV. Divide by dV and attach the conditions of fixed n and V,
one obtains
(E/V)T,n = T(S/V)T,n - p. Howeverever from part (i), (S/V)T,n = (p/T)V,n
Therefore, (E/V)T,n = T(p/T)V,n - p
b) Derive for a real gas an expression for (S/V)T,n and interpret the results.
Starting with the van der Waals equation of state, one solves for p and obtains
p = nRT/(V - nb) - an2/V2. Therefore, (S/V)T,n = (p/T)V,n = nR/(V - nb). The result
is nR/V for an ideal gas so the molecular volume is the only feature of non-ideality that
contributes to the dependence of entropy on volume and hence concentration.
c) Derive for a real gas an expression for (E/V)T,n and interpret the results.
Use the above results. (E/V)T,n = T[nR/(V - nb)] - {nRT/(V - nb) - an2/V2} = an2/V2.
The result is zero for an ideal gas. One obtains the striking result that the dependence of E
on volume (concentration) for a non-ideal gas depends solely on the attraction between
the molecules. This calculation shows rather nicely the orthogonality between entropy
and energy.
exam1_2004.doc, Chemistry 156, spring, 2004, WES