Question

What is the value of k for this aqueous reaction at 298 k?

Answer 1

Answer : The complete question is attached in answer.

The value of K will be = 2 X$10^{-5}$

Explanation : We can use the formula as;

∆G = -RT ln K

on rearranging we get, K = $e^({-deltaG/RT})$

Therefore, we get,

∆G/RT = (26.81 kJ/mol ) / (0.008314 kJ/mol-K) X (298K) = 10.82

So, K = $e^{-10.82}$

Therefore, K = 2 X $10^{-5}$