Question

The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol . Calculate ΔG for this reaction at 298 K when [dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00100 M .

Answer 1

ΔG = -3879.6 J/mol = -3.88 kJ/mol

Step-by-step explanation:

Step 1: Data given

The standard change in Gibbs free energy is ΔG°′=7.53 kJ/mol

Temperature = 298 K

[dihydroxyacetone phosphate]=0.100 M

[glyceraldehyde-3-phosphate]=0.00100 M .

Step 2: Calculate ΔG for this reaction

ΔG = ΔG° + RT ln ([glyceraldehyde-3-phosphate]/ [dihydroxyacetone phosphate])

⇒with ΔG° = 7.53 kJ/mol = 7

⇒with R = 8.314 J/mol*K

⇒with T = 298 K

⇒ with [glyceraldehyde-3-phosphate]=0.00100 M

⇒ with [dihydroxyacetone phosphate]=0.100 M

ΔG = 7530 J/mol + 8.314 * 298 * ln(0.001/0.1)

ΔG = -3879.6 J/mol = -3.88 kJ/mol