Unformatted text preview: Name___________________________________________ TA_________________________________ 20.110/7.10 and 5.60
x For each problem, please put a square around your final answer.
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The exam is worth 100 points.
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This exam has 9 pages (+ 1 page for constants). Problem 1: / 35 pts Grader: Problem 2: / 20 pts Grader: Problem 3: / 30 pts Grader: Problem 4: / 15 pts Grader: TOTAL: /100 pts 1 Name___________________________________________ TA_________________________________ 20.110/7.10 and 5.60
1. When you are cut, it triggers a set of reactions that result in a scab. One of the first reactions consists of two proteins: antihemophillic factor (AHF) and factor IXa (CF). These bind to each other to form tenase (TEN), which then further reacts to form the clot. The reaction is the following: At body temperature (37 oC), the equilibrium constant (K) for this reaction is 5.08 nM-‐1, and the concentrations of AHF and CF are 0.7 nM and 0.9 nM, respectively. (a) (10 pts) What is [TEN] at equilibrium at body temperature? Answer: ܭൌ ሾ݁ݏܽ݊݁ݐሿ ሾܨܪܣሿሾܨܥሿ ሾ݁ݏܽ݊݁ݐሿ ൌ כ ܭሾܨܪܣሿ כሾܨܥሿ ሾ݁ݏܽ݊݁ݐሿ ൌ ሺͷǤͲͺ݊ିܯଵ ሻ כሺͲǤ݊ܯሻ כሺͲǤͻ݊ܯሻ ൌ Ǥ 2 Name___________________________________________ TA_________________________________ (b) (10 pts) If you measure [TEN] = 7.8 nM, with the same AHF/CF concentrations in part a, what is 'G and which direction is spontaneous? ܳ ൌ ܳ ൌ ݉ݑ݅ݎܾ݈݅݅ݑݍ݁ݐܣǣ οܩ ൌ െܴ݈ܶ݊ ܭ ο ܩൌ െ ൬ͺǤ͵ͳͶ ିͲͳ כଷ Ǥͺ݊ܯ
ൌ ͳʹǤ͵ͺ݊ିܯଵ ሺͲǤ݊ܯሻሺͲǤͻ݊ܯሻ
ο ܩൌ οܩ െ ܴ݈ܶ݊ܳ ሾ݁ݏܽ݊݁ݐሿ ሾܨܪܣሿሾܨܥሿ ݇ܬ
൰ כሺ͵ͳͲǤͳͷܭሻ כሺͷǤͲͺሻ െ ൬ͺǤ͵ͳͶ ିͲͳ כଷ
൰ כሺ͵ͳͲǤͳͷܭሻ כሺͳʹǤ͵ͺሻ ൌ Ǥ Ȁ ݉ܭ כ ݈
݉ܭ כ ݈ The reaction is spontaneous in the reverse direction because ο ܩ is positive. 3 Name___________________________________________ TA_________________________________ (c) (15 pts) Blood clotting is affected by body temperature. Calculate the equilibrium constant at 35 oC (hypothermia) and 39 oC (high fever). Higher concentrations of tenase lead to larger blood clots. Will you form a larger clot when suffering from hypothermia or fever? The enthalpy of the reaction is -‐45.35 kJ/mol and it can be assumed to be independent of temperature. ܵ݊ܽݒ݄݁ݐ݄ݐ݅ݓ݃݊݅ݐݎܽݐᇱ ݊݅ݐܽݑݍ݂݂݁ܪݐሺ݁ݎݑݏݏ݁ݎݐ݊ܽݐݏ݊ܿݐ݈ܽ݀݅ܽݒݏ݄݄݅ܿ݅ݓሻǣ ݀
ሺ ܭሻ ൌ ݀ܶ
ܴܶ ଶ ݂͵݁݃݊ܽݎ݁ݎݑݐܽݎ݁݉݁ݐ݄݁ݐݎ݁ݒݕ݈݄ܽݐ݊݁ݐ݊ܽݐݏ݊ܿݎͷ െ ͵ͻ ܥǣ ܭଶ
െο ܪ ͳ
൬ ൰ ൌ
൬ െ ൰ ܭଵ
ܶଶ ܶଵ ݂͵ݎͷ ܥሺ͵ͲͺǤͳͷܭሻ: ln (K2) = ln (5.08) -‐ ೖ
಼כ ିସହǤଷହ ଵ ቀ ଷ଼Ǥଵହ െ ଵ
ଷଵǤଵହ ቁ ൌ ͳǤ͵ͻ ܭଶሺଷହ ሻ ൌ Ǥ ૢି ݕ݈ݎ݈ܽ݅݉݅ݏǡ ܽ ܶݐൌ ͵ͳʹǤͳͷܭǣ ܭଶሺଷଽ ሻ ൌ Ǥ ି A larger value of K implies a greater equilibrium concentration of tenase versus the product of AHF and CF. Therefore, you will form more blood-‐clotting factors and a larger clot with hypothermia (at 35 oC). 4 Name___________________________________________ TA_________________________________ Vapor pressure (kPa) 2. Global warming is occurring partially due to a rise in carbon dioxide (CO2) concentration in the atmosphere. Increases in temperature cause air to be able to hold more water. One of the potential consequences of this is extreme precipitation, where the intensity of an event has been shown increases with available moisture. The plot below shows data taken of the vapor pressure of water as a function of temperature for two geographical locations over the course of several years. Each data point is a day and, while the air is not always fully saturated, there is a clear maximum, above which higher vapor pressures cannot be achieved for a given temperature. For these maximum values, it is ok to assume that the data was gathered when the system is at equilibrium. Temperature (oC) (20 pts) Using this data, calculate the 'Ho of vaporization of water in the atmosphere, assuming it is independent of Temperature. Take the Clausius-Clapeyron- Equation and integrate. You cannot accurately determine the slope at a given T. ߲
ൌ ߲ܶ ܴܶ ଶ
ሺሻ ൌ െ οܪ
ܥ ܴܶ ଶ
൬ ൰ ൌ െ
ሺ െ ሻ ଵ
ܴ ܶଶ ܶଵ
Choosing on any of the points on the coexistence line allows to calculate the vaporization enthalpy of water. To maximize the accuracy, it is advised to maximize the temperature difference between the points. The vaporization enthalpy is approximately ǻH= 43kJ/mol. One could also make a plot of ln p vs 1/T. The slope (with a unit of K) is equal to ǻH/R. 5 Name___________________________________________ TA_________________________________ 4. The shells of marine organisms contain calcium carbonate, CaCO3, largely in a crystalline form known as calcite. A second crystalline form of calcium carbonate is known as aragonite. The molar mass of CaCO3 is 100 g/mol. Some physical and thermodynamic properties of calcite and aragonite are given here: a. (5 pts) Based on the data given, would you expect an isolated sample of calcite at T=298 K and P = 1.00 bar to convert to aragonite, given sufficient time? Explain. ǯǡσGr0 =σGf,Ara0 -‐σGf,Cal0 =-‐1127.7 kJ/mol-‐(-‐1128.8 kJ/mol)=1.1 kJ/mol σGr0>0. So, the conversion cannot happen spontaneously. Or, alternatively, they can use σHr0 =σHf,Ara0 -‐σHf,Cal0 =-‐0.1 kJ/mol σSr0 =σSf,Ara0 -‐σSf,Cal0 =-‐4.2 J K-‐1 mol-‐1 σGr0=σHr0 Ȃ TσSr0 =1.15 kJ/mol 6 Name___________________________________________ TA_________________________________ b. (10 pts) What pressure must be achieved to induce the conversion of calcite to aragonite at T = 298 K? Assume both calcite and aragonite are incompressible at T = 298 K. When pressure is changed for an isothermal process, for incompressible material, σG =VσP. For 1 mole of calcite, if the conversion is under equilibrium σGr =σGr0 +σVσPൌ ͳǤͳ ൬ ଵ
ଶǤଽଷ െ ଵ
ଶǤଵ ൰ ሺܲଶ െ ͳͲହ ܲܽሻ ൌ Ͳ Unit conversions, 1 kJ=1000 Pa m3, 1 ml =10-‐6 m3 σGr =3971.15 bar So, the pressure must be at least 3971.15 bar to induce the conversion of calcite to aragonite at T=298 K. Note: In step 2, if the students want to draw the cycle and calculate the VσP for calcite and aragonite separately instead of using σVσP, they finally get the same equation and the same answer. 7 Name___________________________________________ TA_________________________________ Parameters and Conversions: Gas Constant: R = 8.314 J / mol-‐K = 0.082 L atm/mol-‐K = 1.986 cal/mol-‐K = 8.314 L kPa/mol-‐K Volume: 1 m3 = 1000 L Temperature: [oC] = [K] -‐ 273 ǯǣ 6.02 x 1023 Pressure: 1 Pa = 10-‐5 bar = 1 N/m2 = 9.87x10-‐6 atm Energy: 1 J = 1 N-‐m = 1 kg(m/s)2 1 cal = 4.184 J 8 ...
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- Fall '08
- pts Grader