Question

Citrate synthase catalyzes the reaction: ????x????????o????c???????????????????? + ????c????????y???? − ????o???? → c???????????????????????? + H???? − ????o???? The standard free energy change for the reaction is −31.5 ???????? ∙ mo????−1. Calculate the equilibrium constant for this reaction at 37℃.

Answer 1

The given question is incomplete. The complete question is as follows.

Citrate synthase catalyzes the reaction

Oxaloacetate + acetyl-CoA $\rightarrow$ citrate + HS-CoA

The standard free energy change for the reaction is -31.5 kJ*mol^-1

( a) Calculate the equilibrium constant for this reaction a 37degrees C

Explanation:

(a).  It is known that , relation between change in free energy ($\Delta G$) of a reaction and equilibrium constant (K) is as follows.

             $\Delta G = -RT \times ln K$  

where,  T = temperature in Kelvin

The given data is as follows.

         T = 310 K,       $\Delta G = -31.5 kJ /mol = -31500 J/mol$  (as 1 kJ = 1000 J)

Now, putting the given values into the above formula as follows.

     ln K = $\frac{-(\Delta G)}{RT}$

            = $\frac{31500}{8.314 \times 310}$

      ln K = 12.22

         K = antilog (12.22)

           = $2.1 \times 10^{5}$

Therefore, we can conclude that value of equilibrium constant for the given reaction is $2.1 \times 10^{5}$.