Question

One enzyme-catalyzed reaction in a biochemical cycle has an equilibrium constant that is 10 times the equilibrium constant of a second reaction. If the standard Gibbs energy of the former reaction is −300 kJ mol−1, what is the standard reaction Gibbs energy of the second reaction?

Answer 1

`\Delta{G^0}_2 =-294.25\ kJ/mol`

Step-by-step explanation:

The relation between standard Gibbs energy and equilibrium reaction is shown below as:

`\Delta{G^0} =-RT \ln K`

R is Gas constant having value = 0.008314 kJ / K mol

Given temperature, T = 300 K (Source Original)

Given,
`\Delta{G^0}_1=-300\ kJ/mol`

So,

`-300=-0.008314* 300 \ln K`

`\ln \left(K\right)=(300)/(2.4942)`

`\ln \left(K\right)=120.27904`

Also,

K₁ = 10*K₂

K₂ = 0.1 K₁

So,

`\Delta{G^0}_2 =-0.008314* 300 \ln (0.1* K_1)`

`\Delta{G^0}_2 =-0.008314* 300 (\ln 0.1+ \ln K_1)`

`\Delta{G^0}_2 =-0.008314* 300 (-2.3026+120.27904)`

`\Delta{G^0}_2 =-294.25\ kJ/mol`