The given question is incomplete. The complete question is as follows.
Citrate synthase catalyzes the reaction
Oxaloacetate + acetyl-CoA 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 () of a reaction and equilibrium constant (K) is as follows.
where, T = temperature in Kelvin
The given data is as follows.
T = 310 K, (as 1 kJ = 1000 J)
Now, putting the given values into the above formula as follows.
ln K =
=
ln K = 12.22
K = antilog (12.22)
=
Therefore, we can conclude that value of equilibrium constant for the given reaction is .
Answer: They are called substances and mixtures
Explanation:
When the reaction equation is:
HF ↔ H+ + F-
and when the Ka expression
= concentration of products/concentration of reactions
so, Ka = [H+][F-]/[HF]
when we assume:
[H+] = [F-] = X
and [HF] = 0.35 - X
So, by substitution:
6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X
∴ X = 0.015 M
∴[H+] = X = 0.015
when PH = -㏒[H+]
∴PH = -㏒0.015
= 1.8