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, what is the standard reaction Gibbs energy of the second reaction?

Answer 1

Solution: Calculate the standard reaction Gibbs energy of the second reaction

By the formula: ΔrG=ΔrG∘+RTln⁡k

⇒ΔrG∘=−RTln⁡k−ΔrG

But ΔrG=0 at equilibrium

⇒ΔrG∘=−RTln⁡k

First reaction

We get ΔrG1∘=−RTln⁡k1⇒ln⁡k1=−ΔrG1∘RT

Since ΔrG1∘=300kJ/mol

T=37+273=310K (temperature in body)

R=8.31J/(mol\cdotK)

⇒ln⁡k1=−3008.31×310=0.116

Second reaction

We have ΔrG2∘=−RTln⁡k2

Since

• R=8.31J/(mol\cdotK)
• T=310K
• K1=10K2⇒K2=0.1K1

\(\Rightarrow \Delta_r G_2^\circ = -8.31 \times 310 \ln (0.1K_1) = 5632.85239 kJ/mol

Therefore, \(\Delta_r G_2^\circ = 5632.85239 kJ/mol

Therefore, \(\Delta_r G_2^\circ = 5632.85239 kJ/mol