Question

The dissociation constants for acetic acid and HCN
at 25°C are 1.5 * 10-5 and 4.5 x 10-0,
respectively. The equilibrium constant for the
equilibrium
CN + CH3COOH + HCN + CH2COOP
would be :-
(1) 3.0 x 10^4
(2) 3.0 x 10^5
(3) 3.0 x10^-5
(4) 3.0 x 10^-4

explanation plz!​

Answer 1

Answer:

3.0 x 10^-4.

Explanation:

Since, in the question we have the first chemical equation as CH3COOH which gets dissociated into CH3COO- + H+ which is having the value of equilibrium constant as 1.5*10^-5. Again, for the second chemical equation we have that HCN dissociates into H+ and CN- which is having the k as 4.5*10^-10.

So, now on adding both the equations we will get the required equation which is CN + CH3COOH which will give HCN + CH3COO-. So, the equilibrium constant for this reaction will be keq= 1.5*10^-5/4.5*10^-10 which on solving we will get 3*10^-4.

Answer 2

Answer:

the correct answer is 3×10^4