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Thermodynamics Practice Exam Key
Note: All problems included in this practice exam are drawn from problems used in previous semesters. Exams typically include 7 or 8 problems that are a mixture of qualitative problems calling for written explanations and quantitative problems that involve calculations and, in some cases, written explanations.
On the following pages are problems that consider the thermodynamics of chemical or biochemical systems. Read each question carefully and consider how you will approach it before you put pen or pencil to paper. If you are unsure how to answer a question, then move on to another question; working on a new question may suggest an approach to a question that is more troublesome. If a question requires a written response, be sure that you answer in complete sentences and that you directly and clearly address the question. No brain dumps allowed! Generous partial credit is available, but only if you include sufficient work for evaluation and that work is relevant to the question.
Problem Points Maximum Problem Points Maximum 1 11 5 16 2 11 6 19 3 11 7 16 4 16 Total 100
A few constants and thermodynamics values are given here; other information is included within individual problems.
- density ( d ) of water is 1.00 g/mL
- specific heat ( S ) of water is 4.184 J/g•°C
- the gas constant ( R ) is 8.314 J/molrxn•K
- Faraday’s constant ( F ) is 96,485 J/V•mol e–
substance ∆ Hof (kJ/molrxn) ∆ So^ (J/K•molrxn) ∆ Gof (kJ/molrxn) CO 2 (g) –393.5 213.6 –394. CO (g) –110.5 197.9 –137. CH 4 (g) –74.85 186.2 –50. C 6 H 12 O 6 (s) –1274.5 212.1 –910. H+ (aq) 0 0 0 H 2 (g) 0 131.0 0 H 2 O (g) –241.8 188.7 –228. H 2 O (l) –285.8 69.9 –237. O 2 (g) 0 205.0 0 OH– (aq) –229.94 –10.5 –157.
Problem 1. In a blast furnace for producing iron from iron ore, the following sequence of reactions takes place: Fe 2 O 3 (s) → Fe 3 O 4 (s) → FeO (s) → Fe (s). Shown below is a plot of ∆ G as a function of temperature for the last step in this sequence of reactions
FeO( s ) + CO( g ) → Fe( s ) + CO 2 ( g )
Based on this plot, predict the sign for ∆ H and for ∆ S , and explain the reason for your predictions in 1– sentences.
temperature
G
From the plot, we see that the reaction is favorable ( ∆ G < 0 ) at temperatures below ≈ 640 K and unfavorable ( ∆ G > 0 ) at temperatures above ≈ 640 K; given the relationship ∆ G = ∆ H − T ∆ S, this requires that ∆ H is negative and that ∆ S is negative.
Problem 2. The reaction A → B has a standard state free energy, ∆ Go , of 10 kJ/molrxn and a free energy, ∆ G , of 0 kJ/molrxn. Which of the following statements about the system is correct:
- Some A will react with B until equilibrium is reached.
- Some B will react with A until equilibrium is reached.
- The system is at equilibrium and contains more A than B.
- The system is at equilibrium and contains more B than A.
- There is insufficient information to reach a conclusion. Circle the statement that is correct and, in 2–3 sentences, convince me that you are right. Should you choose the last option, then explain what additional information you need to arrive at a conclusion.
Because ∆ G is zero, we know that the system is at equilibrium, eliminating the first two options as possible answers. Because ∆ Go^ is positive, we know that the equilibrium constant is less than one—recall that ∆ Go^ = − RT lnK and that the natural log of a number less than one is negative, which makes ∆ Go^ positive— and at equilibrium there are more reactants than there are products; thus, the third option is correct.
Problem 5. As you may know, the energy content of foods is reported in Calories (with a capital “C”). What you may not know is a Calorie is just another way of reporting the enthalpy of a combustion reaction using units of Cal/g instead of kJ/molrxn. For example, the combustion reaction for glucose, C 6 H 12 O 6 , is
C 6 H 12 O 6 ( s ) + 6O 2 ( g ) → 6 CO 2 ( g ) + 6H 2 O( g )
Mary Poppins sang “A teaspoon of sugar helps the medicine go down.” Given that one Calorie is equivalent to one kilocalorie, that one calorie (with a little “c”) is equivalent to 4.184 J, and that a teaspoon of sugar has 4.0 g of glucose, how many Calories help the medicine go down? First, we need to calculate ∆ Ho^ for the combustion of glucose, which, using the data from the exam’s cover sheet, is
∆ Ho^ = [(6)(− 393_._ 5) + (6)(− 241_._ 8)] − [(1)(− 1274_._ 5) + (6)(0)] = − 2537 kJ/molrxn
and then use this result to convert the mass of glucose into Calories
4_._ 0 g glu ×
− 2537 kJ molrxn
×
molrxn mol glu
×
mol glu 180_._ 2 g glu
×
1000 J
kJ
×
1 cal 4_._ 184 J/goC
×
1 Cal 1000 cal
= − 13 Cal
Thus, 13 Calories help the medicine go down.
Problem 6. One of the most common batteries uses the following redox reaction
2 Zn( s ) + 3MnO 2 ( s ) → Mn 3 O 4 ( s ) + 2ZnO( s )
Given that a fresh battery has a potential, E , of +1.54 V, how much free energy is available for useful work if 0.500 g of Zn reacts completely and if the efficiency of converting energy to useful work is 75%?
∆ Go^ = − nF ∆ Eo^ = − 4 mol e − molrxn
×
96 , 485 J
V • mole −^
× 1_._ 54 V = − 594 , 300 J/molrxn
0_._ 500 g Zn ×
1 mol Zn 65_._ 3 g Zn
×
molrxn 2 mol Zn
×
− 594 , 300 J
molrxn
×
1 kJ 1000 J
× 0_._ 75 = − 1_._ 7 kJ
Problem 7. Most of the nickel in the world comes from a single mine in Canada where the impact of a comet many years ago brought a deeply buried deposit of NiS to the earth’s surface. To obtain pure Ni, the ore is reduced to an impure metallic Ni and purified by reacting with carbon monoxide, CO, to form Ni(CO) 4 , which is isolated and converted back to pure Ni. Depending on the temperature, the reaction is one or the other of these two reactions
reaction ∆ H (kJ/molrxn) ∆ S (J/K•molrxn) Ni (s) + 4CO (g) → Ni(CO) 4 (g) –160.8 –410. Ni (s) + 4CO (g) → Ni(CO) 4 (l) –190.9 –507.
The desired product is Ni(CO) 4 (g) because it is easier to separate a gas from solid impurities than it is to remove a liquid from these same impurities. Using the thermodynamic values provided above, report the range of possible temperatures for which the formation of Ni(CO) 4 (g) is both favorable and more favorable than the formation of Ni(CO) 4 (l). First, we note that the signs of ∆ H and of ∆ S for Ni(CO) 4 (g) indicate that the reaction is favorable only below a critical temperature, so let’s start by calculating this upper boundary for the temperature; thus
∆ G = 0 = ∆ H − Tcrit ∆ S = (− 160_._ 8 kJ/molrxn ) − Tcrit (− 0_._ 4105 kJ/K molrxn )
Tcrit = 392 K
Then, we note that the ∆ G vs. T curves for Ni(CO) 4 (g) and for Ni(CO) 4 (l) must cross as they do not have the same value for ∆ S. Because ∆ S for Ni(CO) 4 (l) is more negative than ∆ S for Ni(CO) 4 (g), the desired product is more favorable at temperatures greater than Tcross, which is
∆ HN i ( CO ) 4 ( g ) − Tcross ∆ SN i ( CO ) 4 ( g ) = ∆ HN i ( CO ) 4 ( l ) − Tcross ∆ SN i ( CO ) 4 ( l )
− 160_._ 8 kJ/molrxn + 0_._ 4105 kJ/K molrxn × Tcross = − 190_._ 9 kJ/molrxn + 0_._ 5076 kJ/K molrxn × Tcross
30_._ 1 kJ/molrxn = 0_._ 0971 kJ/K molrxn × Tcross
Tcross = 310 K
Thus, we recover Ni(CO) 4 (g) when 310 K ≤ T ≤ 392 K.
