Question

Phosphoglucoisomerase interconverts glucose 6‑phosphate to, and from, glucose 1‑phosphate. glucose 6 - phosphate phosphoglucoisomerase ⇌ glucose 1 - phosphate glucose 6-phosphate⇌phosphoglucoisomeraseglucose 1-phosphate After mixing equal amounts of the two molecules, the solution achieved equilibrium at 25 ∘ C . 25 ∘C. The concentration at equilibrium of glucose 1‑phosphate is 0.01 M, 0.01 M, and the concentration at equilibrium of glucose 6‑phosphate is 0.19 M. 0.19 M. Calculate the equilibrium constant, K eq , Keq, and the standard free energy change, Δ G ∘ , ΔG∘, of the reaction mixture.

Answer 1

Answer: The equilibrium constant of the reaction is 0.0526 and standard Gibbs free energy of the reaction is -7.296 kJ/mol

Explanation:

For the given chemical equation:

$\text{Glucose-6-phosphate}\rightleftharpoons \text{Glucose-1-phosphate}$

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[\text{Glucose-1-phosphate}]}{[\text{Glucose-6-phosphate}]}$

We are given:

$[\text{Glucose-1-phosphate}]=0.01M$

$[\text{Glucose-6-phosphate}]=0.19M$

Putting values in above expression, we get:

$K_{eq}=\frac{0.01}{0.19}=0.0526$

The equation used to calculate standard Gibbs free energy of the reaction follows:

$\Delta G^o=+RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy

R = Gas constant = 8.314 J/K mol

T = Temperature = $25^oC=[25+273]K=298K$

$K_{eq}$ = Equilibrium constant of the reaction = 0.0526

Putting values in above equation, we get:

$\Delta G^o=8.314J/K.mol\times 298K\times \ln (0.0526)\\\\\Delta G^o=-7296.5J/mol=7.296kJ/mol$

Hence, the equilibrium constant of the reaction is 0.0526 and standard Gibbs free energy of the reaction is -7.296 kJ/mol