Worksheet Chem 337 spring 2010

Worksheet Chem 337 spring 2010 - CHEMISTRY 337 Worksheet...

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CHEMISTRY 337 Worksheet Spring 2010 Name: __Shengnan Zheng __ Seat: _____ 1. Consider the reaction: A + B K C + D. If the equilibrium constant for this reaction is a large number (say, 10,000), what do we know about the standard free- energy change ( G' °) for the reaction? Describe the relationship between K eq ' and G' °. ∆G'° = –RT ln K eq '. If K eq ' is a large (positive) number, the term –RT ln K eq ' (and therefore ∆G'°) has a relatively large, negative value. 2. If a 0.1 M solution of glucose 1-phosphate is incubated with a catalytic amount of phospho-glucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate until equilibrium is reached. At equilibrium, the concentration of glucose 1-phosphate is 4.5 x 10 –3 M and that of glucose 6-phosphate is 8.6 x 10 –2 M. Set up the expressions for the calculation of K eq ' and 4 G' ° for this reaction (in the direction of glucose 6- phosphate formation). ( R = 8.315 J/mol·K; T = 298 K) Keq = [Glc-6-P] = 8.6 x 10 –2 = 19.11 [Glc-1-P] 4.5 x 10 –3 ΔGo’= -RT ln K’eq R=8.315 J/mol ° K T=298 ° K ΔGo’= -8.315 (298) ln 19.1 = -7.3 kJ/mol 3. Use the following half-cell potentials to calculate (a) the overall cell potential and (b) ΔG o .
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Cu 2+ + e - Cu + E 0 = 0.16 V Fe 3+ + e - Fe 2+ E 0 = 0.77 V a. E cell = 0.77V - 0.16V = 0.61V (Note: E cell = Δ V = E right E left ) b. Δ = –nFE° = - (1)( 96485 C/mol)(0.61V) = -58 .85 kJ/mol 4. In glycolysis, the enzyme pyruvate kinase catalyzes this reaction: Phosphoenolpyruvate + ADP pyruvate + ATP Given the information below, show how you would calculate the equilibrium constant for this reaction. ( R = 8.315 J/mol·K; T = 298 K) Reaction 1) ATP ADP + P i o G' ° = –30.5 kJ/mol Reaction 2) phosphoenolpyruvate pyruvate + P i o G' ° = –61.9 kJ/mol Calculate the number of ADP and P i molecules for every one ATP molecule present at equilibrium for the reaction ATP ADP + P i ΔG o = -30.5 kJ/mol The reaction is the sum of reaction 2 and the reverse of reaction 1. Therefore, ∆G'° = –31.4 kJ/mol. ∆G'° = –RT ln K eq ' ln K eq ' = –∆G'°/RT = 31.4 kJ/mol / [(8.315 J/mol·K)(298 K)]
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ln K eq ' = 12.672 K eq ' = 3.19 x 10 5 For every one molecule of ATP, there is one molecule of ADP and one molecule of P i .
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5. Glycerol 3-phosphate dehydrogenase catalyzes the following reversible reaction: Glycerol 3-phosphate + NAD + NADH + H + + dihydroxyacetone phosphate Given the standard reduction potentials below, calculate G' ° for the glycerol 3- phosphate dehydrogenase reaction, proceeding from left to right as shown. Show your work. (The Faraday constant, , is 96.48 kJ/V·mol.) Dihydroxyacetone phosphate + 2e + 2H + -> glycerol 3-phosphate E' ° = –0.29 V NAD + + H + + 2e -> NADH E' ° = –0.32 V Therefore ΔG o' for glycerol-3-phosphate reaction is -5.778kJ/mol. 6.
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This note was uploaded on 04/29/2010 for the course CHEM 337 taught by Professor Birchbichler during the Spring '10 term at Slippery Rock.

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Worksheet Chem 337 spring 2010 - CHEMISTRY 337 Worksheet...

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