This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: G&G Ch. 3 Problem 1 (pg. 74) This is an essential calculation of the inherent rate constant of any reaction, which is just another expression of the free energy of the reaction, i.e. the propensity of the system to be driven one way or another across the euilibrium symbol ( ). Reactions are clasically defined as proceeding from left-to-right, with positive (+) free energies indicating unfavorable reactions, and negative ( – ) free energies indicating favorable reactions. And because pH calculations deal with equilibria, we can steal the “iΔf” paradigm from there and apply it here to simplify matters: Fructose-1-P + H 2 O fructose + P i @ 25°C = 298K (RT) i Δ f 0.2 – x +x +x 6.52x10-5 x x K eq = x 2 6.52x10-5 x = 0.2 – 6.52x10-5 = 0.1999348M 613M and because we know from above that 0.2 – x = 6.52x10-5 , i.e. K eq = (0.1999348) 2 6.52x10-5 = ΔG° = – RTlnK eq = – (8.314J/K•mol)(298K)ln(613) = – 15902.9J/mol = Now calculate the free energy: – 15.9kJ/mol We’re given [Fructose-1-P] for both initial and equilibrium states. G&G Ch. 3 (cont’d)G&G Ch....
View Full Document
This note was uploaded on 07/16/2008 for the course BIS 102 taught by Professor Hilt during the Summer '08 term at UC Davis.
- Summer '08