InterviewSolution
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InterviewSolution
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The value of `DeltaG^(ɵ)` for the phosphorylation of glycose in glycolysis is `13.8 kJ mol^(-1)`. Find the value of `K_(c)` at `298 K` |
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Answer» `DeltaG^(ɵ) = 13.8 kJ//"mol" = 13.8 xx 10^(3) J//"mol"` Also, `DeltaG^(ɵ) = - RT"ln"K_(c)` Hence, In `K_(c) = - 13.8 xx 10^(3)J//"mol"` `(8.314 K "mol"^(-1)K^(-1) xx 298 K)` In `K_(c) = -5.569` `K_(c) = e^(-5.569)` `K_(c) = 3.81 xx 10^(-3)` |
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