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

Step-by-step 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 * 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 =
`(-(\Delta G))/(RT)`

=
`(31500)/(8.314 * 310)`

ln K = 12.22

K = antilog (12.22)

=
`2.1 * 10^(5)`

Therefore, we can conclude that value of equilibrium constant for the given reaction is
`2.1 * 10^(5)`.