Chemistry 4A, Fall 2015
Name___________________________
Midterm 3
Potentially useful constants:
Standard state: P = 1 bar = 105 Pa;
T (0 ◦C) =273.15 K
1 mol gas occupies 22.7 L at STP; R = 8.314 J mol−1 K−1
Relative atomic mass: H=1, C=12, N=14, O=16
Part I. Multiple choices (4 points each question; only one correct answer)
1. The tallest building in US is currently the new One World Trade Center (541 m). Approximately how
much pressure would be required to pump water from ground to the top of the building? ( C
)
A. 1 atm
B. 10 atm
C. 50 atm
D. 500 atm
2. Which one of the below is NOT a basic assumption based on which the kinetic theory of gases is derived?
(
B
)
A.
B.
C.
D.
E.
Gas consists of identical, well-separated, small particles
All molecules in the system assume the same kinetic energy
Straight-line motion between collisions
Collisions with walls and each other are elastic
No forces between molecules except during collisions
3. While the word “vacuum” literally means space void of matter, in practice any vacuum system still has
residual gases. “High vacuum” refers to a system with residual gas pressure of 10-6−10-10 bar. Recall that
for a gas at STP, the average separation between molecules is a few nanometers (10-9 m). Estimate the
average separation between gas molecules in a “high vacuum” system at P =10-9 bar: ( B
)
A. Tens of nanometers
B. A few micrometers (10-6 m)
[
C. A few meters
D. A few kilometers
]
4. Van der Waals equation is P + a(n / V )2 (V − nb) = nRT . Given that for H2 a = 0.25 bar L2 mol-2 and b =
0.0265 L/mol, which set of parameters below could be reasonable for N2? ( B
)
A. a = 1.20 bar L2 mol-2; b = −0.0425 L/mol
B. a = 1.37 bar L2 mol-2; b = 0.0387 L/mol
C. a = 0.20 bar L2 mol-2; b = 0.0156 L/mol
D. a = −0.20 bar L2 mol-2; b = 0.0356 L/mol
5. For 5 gas molecules in the two-chamber system discussed in class. What is the statistical probability for all
molecules being all in the left chamber? ( D
)
A. 0
B. 1/5
C. 1/25
D. 1/32
6. A car engine has an operating temperature of about 150 ◦C. If the surroundings are at 25 ◦C, what is the
theoretical maximum efficiency of the engine? ( B
)
A. 17%
B. 30%
C. 70%
D. 83%
7. A system went through a series of irreversible and reversible processes and was eventually restored to its
initial state. We thus expect that for the system: ( A
)
A. ΔS = 0
B. ΔS ≥ 0
C. ΔS > 0
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D. ΔS < 0
Chemistry 4A, Fall 2015
Name___________________________
8. Given ∆ f H  (diamond) = 1.9 kJ/mol , S  (diamond) = 2.38 J ⋅ mol-1 ⋅ K -1 , and S  (graphite) = 5.74 J ⋅ mol-1 ⋅ K -1 ,
we expect: ( D )
A.
B.
C.
D.
Diamond is thermodynamically stable at all temperatures
Diamond is thermodynamically stable at low temperatures only
Diamond is thermodynamically stable at high temperatures only
Diamond is thermodynamically unstable at all temperatures
Part II. Questions. Short answers are fine as long as to the point.
Question 1. (6 points per sub-question) Analysis: Are the following statements true or false? For each
statement, first state unequivocally whether you think the statement is true or false (2 points). Afterwards, for
statements you answer “true”, discuss the mechanisms. For statements you answer “false”, either discuss the
mechanisms, or give one specific counter-example to disprove the statement. (4 points for the explanation)
(a) One mole of H2 and N2 are injected into the same container and allowed to reach equilibrium. It is thus
expected that the root-mean-square velocity of the H2 are N2 molecules in the container are equal.
FALSE.
At equilibrium, H2 are N2 molecules should have the same temperature, thus the same averaged kinetic
energy, not the same vRMS. H2 should have larger vRMS.
(b) ∆H > ∆U holds true for all reactions.
FALSE.
ΔH – ΔU = RTΔn(g)
Alternative answer: ∆H = ∆U for reactions not involving gases or when Δn(gas)=0
(c) Number of microstates, Ω, is an extensive function and depends linearly on system size.
FALSE.
Ω depends exponentially on system size. Instead, S=klnΩ is extensive and depends linearly on system size.
(d) Exothermic reactions are always thermodynamically spontaneous at low temperature.
TRUE.
ΔG = ΔH – TΔS. ΔH decides ΔG in the low-T limit.
(e) Endothermic reactions are always thermodynamically spontaneous at high temperature.
FALSE.
ΔG = ΔH – TΔS. ΔS decides ΔG in the high-T limit.
Another possible answer: counter example: when ΔH > 0 and ΔS < 0 no temperature is spontaneous
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Chemistry 4A, Fall 2015
Name___________________________
Question 2.
(a) (5 Points) What is the relationship between the Boltzmann constant kB and the universal gas constant R?
kB = R/NA
Alternative answer:
kB is used to calculate energy when counting molecules. R is used to calculate energy when counting in the
unit of moles.
(b) (6 Points) When calculating ΔG of a reaction, we often use a relative scale ∆ f H  to calculate ∆H  , but
for calculating ∆S  we use the absolute S  . Explain why.
The absolute value of H is not meaningful, whereas ΔH is well defined. In contrast, 3rd law tells us S (0 K)
= 0, so the absolute value of S is well defined. (Also correct: the absolute value of S can also be defined
according to klnΩ).
Question 3. (12 Points) Octane (C8H18) is a major component of gasoline. Its combustion can be written as:
→ 16 CO 2 + 18 H 2 O
2 C8 H18 + 25 O 2 
∆H  = −1.1 × 10 4 kJ/mol
(a) (6 points) At STP, how much volume of air is needed to burn 1 mol octane? Assume the partial pressure
of O2 in the air is 0.2 bar. Keep two significant figures for the final result.
V (O2) = 25/2 mol * 22.7 L mol-1 = 2.8 × 102 L
V (air) = V (O2) / 0.2 = 1.4 × 103 L
(b) (6 points) How many moles of water can be boiled (consider 25 ◦C→100 ◦C) by the heat released in
burning 1 mol octane? Recall that 1 cal = 4.18 J. Keep two significant figures for the final result.
qrxn = (−1.1×104 kJ/mol * ½ mol of reaction) = −5.5×103 kJ
For water: −qrxn = cmΔT
m=
5.5 × 103 kJ
- qrxn
= 18 kg
=
4.18 J K -1 g -1 × 75 K
c∆T
n = m/M = 1.0×103 mol
Alternative answer: the word “boiled” might be understood as water entirely vaporized?
m=
5.5 × 103 kJ
= 120 mol
(4.18 J K -1 g -1 × 75 K × 18 g/mol) + 41 kJ/mol
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Chemistry 4A, Fall 2015
Name___________________________
Question 4. (15 Points) Consider water-vapor equilibrium: H2O(l) ⇌ H2O(g). Given ∆H l→g (H 2 O) = 41 kJ/mol
and the boiling point of water is 100 ◦C at 1 bar, and assume no T-dependence for ∆H l→g (H 2 O) and ∆S l→g (H 2 O) :
(a) (4 points) Calculate ∆S l→g (H 2 O) .
∆S l→g (H 2 O) =
∆H l→g (H 2 O)
Tb
=
41 kJ/mol
= 110 J K -1 mol-1
373.15 K
(b) (6 points) What is ∆Gl→g (H 2 O) at 100 ◦C and at 0 ◦C, respectively?
At 100 ◦C: ∆Gl→g (H 2 O) = 0 (Boiling point: Vapor and liquid are in equilibrium under standard state)
At 0 ◦C: ∆Gl→g (H 2 O) = ∆H l→g (H 2 O) - T∆S l→g (H 2 O) = 41 kJ/mol - 273.15 K × 110 J K -1 mol-1 = 11 kJ/mol
(c) (5 points) Using the results above, estimate the vapor pressure of water at 0 ◦C. OK to use expressions like
e−x as final result.
∆G  = − RT ln K
⇒ ln K = -
∆G 
- 11 kJ/mol
=
= -4.8
8.314 J K -1 mol-1 × 273.15 K
RT
For the H2O(l)⇌H2O(g) equilibrium:
K = PH2O(g) / Pref (Pure liquid does not contribute to the calculation of Q or K)
∴ PH2O(g) = KP ref = e −4.8 bar
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