the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29

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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29

8 K when [dihydroxyacetone phosphate]=0.100 M[dihydroxyacetone phosphate]=0.100 M and [glyceraldehyde-3-phosphate]=0.00300 M[glyceraldehyde-3-phosphate]=0.00300 M .

Chemistry

1 answer:

Vera_Pavlovna [14] · Answer 1 · 2021-12-24T13:23:36+03:00

Answer:

ΔG = 16.218 KJ/mol

Explanation:

dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate

ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴ R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )