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The value of DeltaG^(ɵ) for the phosphor...

The value of `DeltaG^(ɵ)` for the phosphorylation of glucose in glycolysis is `13.8 kJ//"mol"`. Find the value of `K_(c)` at `298 K`.

Text Solution

Verified by Experts

`DeltaG^(ɵ)=13.8 kJ "mol" =13.8xx10^(3)J "mol"`
Also, `DeltaG^(ɵ)=-RT "ln" K_(c )`
Hence, In `K_(c )=-13.8xx10^(3) J "mol"^(-1)xx(8.314 J "mol"^(-1) K^(-1)xx298 K)`
In `K_(c )=-5.569`
`K_(c )=e^(-5.569)`
`K_(c )=3.81xx10^(-3)`
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