Question

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

Answer 1

For non-standard conditions (i.e different concentrations) we have to use modified non-standard Nerns't equation here

ΔG = ΔG^o' + RTlnQ

given ΔG^o' = 7.53 kJ/mol = 7530 J/mol since 1 kJ = 1000 J

T = 298 K; R = 8.3145 J/mol.K

Q = [products] / [reactants] = [0.00400] / [0.100] = 0.04

ΔG = ΔG^o' + RTlnQ

ΔG = 7530 J/mol + [8.3145 J/mol.K x 298 K x ln[0.04]

ΔG = -445.476227668 J/mol = -0.44547622766 kJ/mol

Answer: -0.445 kJ/mol

r the aqueous reaction E - CH2OH -Он Н- CH2-O CH2 glyceraldehyde - 3- phosphate dihydroxyacetone phosphate the standard change in Gibbs free energy is AG" = 7.53 kJ/mol. Calculate AG for this reaction at 298 K when [dihydroxyacetone phosphate] 0.100 M and [glyceraldehyde-3-phosphate] 0.00400 M The constant R = 8.3145 J/(K mol) Number > Please observe negative sign before number 0.445 -0.445 kJ/ mol AG=

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