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

Answer:

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

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,

$-300=-0.008314\times 300 \ln K$

$\ln \left(K\right)=\frac{300}{2.4942}$

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

Also,

K₁ = 10*K₂

K₂ = 0.1 K₁

So,

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

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

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

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