The standard change in gibbs free energy is δ°′=7.53 kj/molδg°′=7.53 kj/mol . calculate δδg for this reaction at 298 k298 k when [dihydroxyacetone phosphate]=0.100 m[dihydroxyacetone phosphate]=0.100 m and [glyceraldehyde-3-phosphate]=0.0030 0 m[glyceraldehyde-3-phosphate]=0.003 00 m .

ΔG = 16.218 KJ/mol

Explanation:

∴ 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 )

b) mutualism

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