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HW11_sol_2006

# HW11_sol_2006 - 2R03 Practice exercises for week 11 Note...

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2R03 Practice exercises for week 11 Nov 23 2006 Note: All gases are assumed to be ideal and all thermochemical data are for 298.15K unless otherwise specified. Also assume both entropies and enthalpies to be temperature independent unless otherwise specified. 1. When a reaction such as A (g) + 2B (g) C (g) is in an equilibrium state, which change can shift the reaction towards the product? Explain (a) Remove 1 mol A (g) from the system while keeping the total pressure constant. (b) Add an inert gas while keeping the total volume constant. (c) Compress the volume of the container to half the initial volume. (d) Add 1 mol of C (g) to the system while keeping the total pressure constant. Solution : (c). When a gaseous system at equilibrium is compressed, the composition adjusts so as to reduce the total number of gas molecules. So in this case it would shift the equilibrium to the right (less gas molecules). Addition of products or removal of reactants shifts the equilibrium to the left. Addition of an inert gas at constant volume increases the total pressure but also decreases proportionally the mole fractions the partial pressures are not affected the equilibrium composition doesn’t change. 2a. The native (active) form of an enzyme is in equilibrium with its unfolded (inactive) form, and their relative abundances change with temperature. The ratio of concentrations (active over inactive) was found to be 390 at 50 o C and 6.2 at 100 o C, estimate the absolute value of the enthalpy of unfolding (| r H 0 |). (a) 82.8 kJ·mol -1 (b) 35.9 kJ·mol -1 (c) 13.9 kJ·mol -1 Solution. (a) because: lnK = -∆G/RT = - ∆H/R + T∆S/R So: ln(k’/k) = - ∆H/RT’ + ∆H/RT = ∆H/R(1/T – 1/T’) ( Van’t Hoff equation) So: ln (k 50 /K 100 ) = ∆H/R (1/373 – 1/323), solving for ∆H = 82.8 kJ.mol -1 2b. Determine the sign of r H 0 for the unfolding of the enzyme in the previous exercise: (a) Positive (b) Negative Solution. (a) Unfolding requires heat

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