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ProbSet5_Solutions(2)

# ProbSet5_Solutions(2) - UC Berkeley MCB 100A/Chem 130A...

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Hand in answers to the 4 problems marked with a * on Thursday, October 8, 2009. ANSWER KEY Q1. *Two different conformations of a protein are separated in energy by 15 kJ mol -1 . (i) At what temperature will the ratio of their populations be equal to 1:2? (ii) What will be their relative populations at room temperature (300K)? (iii) What energy difference between the two levels would yield a relative population of 1:2 at room temperature? Answer: Δ U = 15 kJ mol -1 .Note that the units are kJ mol -1 , so use R instead of k in the Boltzmann Equation. i. N 2 N 1 = exp( −Δ U RT ) = 1 2 (condition given by the question) Take the logarithm of both sides to give: ln 1 2 = 0.693 = −Δ U RT = 15000 RT (not that the expression for R involves J not kJ, so energy is expressed in J mol -1 here) Therefore: RT = 15000 0.693 = 21645 At this point you can calculate T: T = 21645 8.314 = 2603 K ii. Recall that at room temperature, RT 2.5 kJmol -1 . For this temperature, N 2 N 1 = e 15 2.5 = e 6 = 0.0025 iii. N 2 N 1 = 0.5 = exp( −Δ U 2.5 ) Take the logarithm of both sides to get: -0.693 = −Δ U 2.5 Therefore: Δ U = 1.73 kJ mol -1 Q2. A calorimeter and a heat bath are shown in the diagram below. The system is defined in two different ways, as indicated in (i) and (ii). The calorimeter contains a solution of protein molecules. A solution containing drug molecules is added and energy is released when the drug binds to the protein. For (i) and (ii) below circle whether A, B or C represents the correct situation with respect to heat UC Berkeley. MCB 100A/Chem 130A Problem Set 5 Fall 2010 1 of 9

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transfer. If none of the three situations is applicable, write down the correct description of heat transfer. Answer: (i) A: dqsys < 0 dqsurr > 0 (ii) B: dqsys > 0 dqsurr = 0 Q3. Which of the following properties are extensive: a. Pressureb. Temperature c. Volume d. Density e. Molarity f. Energy g. Number of moles h. Mass i. Amount of heat released Answer: Extensive properties: c, f, g, h, i Q4. Consider 5 L of nitrogen gas, held in a heat bath at 298 K. Assume ideal gas behavior. The gas expands under a constant external pressure of 1 atm until its volume is 24.5 L. Calculate the work done by the gas. How many minutes would a 4 watt light bulb (a refrigerator light) stay lit if the work done by the gas was used with 100% efficiency to light the light bulb? Hint: Watt is the standard unit UC Berkeley. MCB 100A/Chem 130A Problem Set 5 Fall 2010 2 of 9
of power, i.e., energy / time. Answer: volume = 5 L temp = 298 K P ext = 1 atm w = P ext Δ V = (1 atm) x (24.5 – 5 L) w = 19.5 L atm convert to Joules 1 L atm = 101.325 J 19.5 L atm = 1975.84 J power (watt)= energy(joules) time(seconds) time= energy power Therefore: time= 1975.84J 4Jsec -1 = 493.96 sec = 8.23 minutes. Q5. *Glucose combines with oxygen to yield carbon dioxide and water. When this reaction is carried out in an open test tube at 1 atm pressure and 25 degrees Celsius, 2800 kJ of heat is given off per mole of glucose oxidized.

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ProbSet5_Solutions(2) - UC Berkeley MCB 100A/Chem 130A...

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