Chapter 3, Problem 1P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Calculating $K_{\text{eq}}\text{and }\Delta G$ from Concentrations An enzymatic hydrolysis of fructose-1-P.

$\text{Fructose-1-P}+{\text{H}}_2\text{O}\rightleftharpoons \text{fructose }+{\text{P}}_{\text{i}}$ was allowed to proceed to equilibrium at 25⁰C. The original concentration of fructose-1-P was 0.2 M. but when the system had reached equilibrium, the concentration of fructose-1-P was only 6.52 × 10^-5 M. Calculate the equilibrium constant for this reaction and the free energy of hydrolysis of fructose-1-P.

Interpretation Introduction

To calculate: The equilibrium constant for fructose-1-P hydrolysis reaction and the free energy for the reaction.

Introduction:

The mathematical expression for K_eq and $\Delta G^0$ are:

  ${\text{K}}_{\text{eq}}\text{=}\frac{\left[\text{Products}\right]}{\left[\text{Reactants}\right]}$

  $\Delta G^0=-RT\ln K_{eq}$

Where K_eq is equilibrium constant, $\Delta G^0$ is change in Gibbs free energy at standard conditions, R is Gas constant, T is temperature, square brackets represents concentration.

Explanation of Solution

Initial [Fructose-1-P] = $0.2\text{ M}$

[Fructose-1-P] after reaching equilibrium = $\text{6}\text{.52}\times {\text{10}}^{-5}\text{ M}$

  ${\text{Fructose-1-P + H}}_{\text{2}}\text{O }\rightleftharpoons {\text{ Fructose + P}}_i$

So [Fructose] produced = $0.2\text{ M - 6}\text{.52}\times {\text{10}}^{-5}\text{ M = 0}\text{.199 M}$

  ${\text{K}}_{\text{eq}}\text{=}\frac{\left[\text{Fructose}\right]\left[{\text{P}}_i\right]}{\left[\text{Fructose-1-P}\right]}$

  $\begin{array}{l}\text{ = }\frac{\text{0}\text{.199 M}\times \text{0}\text{.199 M}}{\text{6}\text{.52}\times {\text{10}}^{-5}\text{ M}}\\ {\text{K}}_{\text{eq}}=607.37\text{ M}\end{array}$

  $\begin{array}{l}\Delta G^0=-RT\ln K_{eq}\\ \Delta G^0=-8.314\text{ J/mol}\text{.K }\times \text{ 298 K }\times \text{ ln (607}\text{.37)}\\ \Delta G^0=-15879\text{ J/mol}\\ \Delta G^0=-15.9\text{ kJ/mol}\end{array}$