Youngstown State University
Department of Chemistry
Chemistry 3739
Final Exam - 200 pts.
Name __SAMPLE FINAL EXAM
December 9, 2009
Chap.1-9,16-19, E&R
ANSWER ANY 8 QUESTIONS - (25 pts. Each) – CIRCLE THEM
Equations and Constants appear on last page.
1. Explain the physical source of the difference between Cv,m and Cp,m for a mole of ideal,
monatomic gas at 25 oC, and show algebraically that this difference equals R.
[Cp = Cv + T(∂P/∂T)V(∂V/∂T)P].
2. For the reaction C(gr) + H2O(g) → CO(g) + H2(g), ΔHorx = 131.28 kJ/mol at 298 K.
Given the Cp,m are 8.53, 33.58, 29.12 and 28.28 J K-1 mol-1for graphite, H2O(g), CO(g) and
H2(g), respectively, find the ΔHrx at 125 oC. As a first approximation, assume the heat
capacities are temperature independent.
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Chem 3739 Final Exam
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3. The Second Law of Thermodynamics says that the entropy of the universe increases for
every spontaneous process, or ΔSuniv≥ 0, where 0 is the case for idealized reversible
processes. Show that this is the case for 150 kJ heat flowing from a high temperature
reservoir at 400 oC to a low temperature reservoir at 37 oC. This corresponds to from a hot
poker to a human hand. [Reservoirs do not change temperature when heat is added to or
removed from them.] In this case, you can assume one of the reservoirs is the “system”.
*4. 1.75 moles of an ideal gas with Cv,m = (3/2)R are transformed from an initial state T = 750 K
and P = 1.75 bar to a final state of T = 350 K and P = 5.25 bar. Calculate ΔU, ΔH and ΔS for
this process.
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Chem 3739 Final Exam
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*5. For the reaction NO2(g) ═ NO(g) + (1/2) O2(g) the values of Kp = 0.0649 at 600. K and
2.78 at 900. K. Find the values of ΔHR (assuming it is temperature independent), ΔGoR and
ΔSR (also assuming it is temperature independent).
6. By combining the first and second laws, and the definition G = H – TS, the total
differential of the Helmholtz free energy (A = f(T,V)), dA = –SdT – PdV, we may identify
the coefficients –S as (∂A/∂T)P and –P as (∂A/∂V)T. With Maxwell’s equations using the
Euler condition we see the (∂S/∂V)T = (∂2A/∂V∂T) = (∂2A/∂T∂V) = (∂P/∂T)V.
Find (∂S/∂V)T for an ideal gas, and connect it to β and κ, where these are defined below.
β = (1/V)(∂V/∂T)P
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and
κ = (1/V)(∂V/∂P)T
Chem 3739 Final Exam
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7. A solution is made up of 248.7 g of ethanol (MW 46.04 g/mol) and 145.9 g water (MW
18.02 g/mol). The volume of the solution is 450.9 cm3. At this mole fraction, the partial
molal volume of water is 17.0 cm3. What is the partial molal volume of ethanol under these
circumstances?
8. The Danielle Cell (Zn│Zn2+││Cu2+│Cu) is composed of the two half cells Zn2+ + 2e- → Zn
and Cu2+ + 2e- → Cu. The cell voltage is 1.10 V at 25 oC. Assume that ΔH and ΔS for this
reaction are temperature independent. Find these quantities if the cell voltage is measured as
1.15 V at 35 oC and 1.20 V at 45oC.
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Chem 3739 Final Exam
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9. Find the mean free path (λ) of an Ar atom at a pressure 1.0 atm (1.01 x 105Pa) at 25 oC. The
diameter (d) of an argon atom is 0.34 nm. The mean free path can be calculated from
the average velocity <v> divided by the collision number z, where σ = π(r)2, the collision
cross section. [N/V] is the particle density at these conditions.
z = [N/V](2)1/2σ(<v>)
*10. The reaction rate as a function of the initial reactant pressure was investigated for the
reaction 2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g) , and the following data were
obtained. Find the parameters of the rate law expression (rate constant and orders in each
reactant). For 15 extra points, suggest a plausible mechanism.
Run #
1
2
3
4
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Po(H2)(kPa)
53.3
53.3
38.5
19.6
Po(NO)(kPa)
40.0
20.3
53.3
53.3
Rate(kPa s-1)
0.137
0.033
0.213
0.105
Chem 3739 Final Exam
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Constants and Equations:
[P + a(n2/V2)][V – nb] = nRT
PV = nRT
R = 0.08205 lit.atm.mol-1K-1 = 8.314 x 10-2 bar.dm3mol-1 K-1 = R = 8.314 J.mol-1K-1
1 atm = 1.013 bar = 1 x 105 Pa
CV = (3/2)R (monat/ideal)
dU = dq + dw ,
dU = CvdT,
dH = CpdT,
Smix= -nR{xAlnxA + xBlnxB + ….}
 = CP /CV
C
∆S =   p
 T
dw = -PextdV (PV work)
dH = dU + d(PV), Svap = Hvap/Tb
H = U + PV
A = U – TS
G = H - TS
CP = CV + R (mol-1)
(T2/T1) = (V2/V1)1-γ (adiab.rev.) dS = (dqrev/T)

 dT ∆S = nRln(V2/V1) + nCp ln(T2/T1) (∂P/∂T)V(∂T/∂V)P(∂V/∂P)T = -1

𝑻𝟐
ΔHT2 = ΔHT1 + ∫𝑻𝟏 (𝜟𝑪P) dT
If V = f(T,P) then:
dV = (∂V/∂T)PdT + (∂V/∂P)TdP, and, if exact, {∂/∂P(∂/∂T)} = {∂/∂T(∂/∂P)}
G = H - TS
dU = dq – PdV
dS = dq/T Go = - RTlnKp G = Go + RTlnQ
Svap = Hvap/Tb
ln Pvap = - (∆Hvap/R)(1/T) + C
ln Kp = - (∆Hr/R)(1/T) + C
NA = 6.02 x 1023 mol-1
So = k lnW
k = 1.38 x 10-23 J/K
Smix= -nR{xAlnxA + xBlnxB + ….}
Ptot = xAPA* + xBPB*
∂lnK/∂(1/T) = -∆H/R
Ptotal = (nA + nB)RT/V = PA + PB
aA = γA xA = PA/PA*
Ptot = aAPA* + aBPB*
Φ(r) = [A/r]e-Kr
K = [(4π/kTD)Σi qi2Ni]1/2
ΔGo = -nFEo
Φ(r) = {qα/Dr} – {qαK/D}
I = (1/2)Σi mi zi2
F = 96,485 J/V
d[lnP]/dt = -kt (1st order)
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(Real)
D = Dielectric Constant
ln γ± = -0.509│z+ z-│I1/2 (water at 25oC)
E = Eo - {RT/nF}lnQ(a)
f(vx) = [m/2πkT]1/2exp{-(m/2kT)vx2}
(Ideal)
<v> = [8RT/πM]1/2
d[P-1]/dt = kt (2nd order)
R = 8.314 J.mol-1.K-1
dP/dt = -kt (0 order)
t1/2 = ln2/k (1st order)