Question

For the aqueous reaction dihydroxyacetone phosphate is the reactant and glyceraldehyde 3 phosphate is the product. dihydroxyacetone phosphate − ⇀ ↽ − glyceraldehyde − 3 − phosphate the standard change in Gibbs free energy is Δ G ° ' = 7.53 kJ/mol . Calculate Δ G for this reaction at 298 K when [dihydroxyacetone phosphate] = 0.100 M and [glyceraldehyde-3-phosphate] = 0.00400 M . Δ G = kJ / mol

Answer 1

Answer : The Gibbs free energy of the reaction is -0.445 kJ/mol.

Explanation :

The given chemical equation is:

$\text{Dihydroxyacetone phosphate}\rightleftharpoons \text{Glyceraldehyde-3-phosphate}$

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

$K_{eq}=\frac{\text{[Glyceraldehyde-3-phosphate]}}{\text{[Dihydroxyacetone phosphate]}}$

Given:

[Glyceraldehyde-3-phosphate] = 0.00400 M

[Dihydroxyacetone phosphate] = 0.100 M

Now put all the given values in above equation, we get:

$K_{eq}=\frac{0.004}{0.100}=0.04$

The relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G=\Delta G^o+RT\ln K_1$

where,

$\Delta G^o$ = Standard Gibbs free energy = 7.53 kJ/mol = 7530 J/mol

(Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = temperature = 298 K

Now put all the given values in above equation, we get:

$\Delta G=7530J/mol+(8.3145J/Kmol)\times 298K\times \ln (0.04)\\\\\Delta G=-445J/mol=-0.455kJ/mol$

Hence, the Gibbs free energy of the reaction is -0.445 kJ/mol.