The value of ∆G•for phosphorylation of the glucose in the glycolysis is 13.8 Kilo joule per mole find the value of kc at 298 Kelvin

Question

Arun · Answer

ΔG=-2.303RT log Kc
where R =8.314jk^-1mol^-1;T=298k;
Given
ΔG=13.8KJ/Mol but has value of R is in joules,we take it as 13.8×10³j/mol
therefore 13.8×10³=-2.303×8.314×298logKc
log Kc =-2.4185
antilog=0.2661×10^-2

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The value of ∆G•for phosphorylation of the glucose in the glycolysis is 13.8 Kilo joule per mole find the value of kc at 298 Kelvin

The value of ∆G•for phosphorylation of the glucose in the glycolysis is 13.8 Kilo joule per mole find the value of kc at 298 Kelvin

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The value of ∆G•for phosphorylation of the glucose in the glycolysis is 13.8 Kilo joule per mole find the value of kc at 298 Kelvin

The value of ∆G•for phosphorylation of the glucose in the glycolysis is 13.8 Kilo joule per mole find the value of kc at 298 Kelvin

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