Question

For the reaction,

$\left[{Cu}\left({NH}_3\right)_4\right]^{2+}+{H}_2 {O} \rightleftharpoons\left[{Cu}\left({NH}_3\right)_3 {H}_2 {O}\right]^{2+}+{NH}_3$

the net rate of reaction at any time is given by :

Net rate $=2.0 \times 10^{-4}\left[\left[{Cu}\left({NH}_3\right)_4\right]^{2+}\right]-$

$3.0 \times 10^5\left[\left[{Cu}\left({NH}_3\right)_3 {H}_2 {O}\right]^{2+}\right]\left[{NH}_3\right]$

Then correct statement is (are)

(a) rate constant for forward reaction $=2 \times 10^{-4}$

(b) rate constant for backward reaction $=3 \times 10^5$

(c) equilibrium constant for the reaction $=6.6 \times 10^{-10}$

(d) all of these.

Answer 1

Correct Option : (D)

Explanation :

Net rate of reaction $=$ rate of forward reaction $-$ rate of backward reaction $=K_f[$ reactant $]-K_b$ [product $]$

On comparing this relation with given equation, we get, $K_f=2.0 \times 10^{-4}, K_b=3.0 \times 10^5$

Also, $K_c=\frac{K_f}{K_b}$ at equilibrium

$K_c=\frac{2 \times 10^{-4}}{3 \times 10^5}=6.6 \times 10^{-10}$