Question

For the reaction $\left[Cu\right(N{H}_3{)}_4{]}^{2+}+H_{2}O\to \left[Cu\right(N{H}_3{)}_3{H}_{2}O{]}^{2+}+N{H}_3$
The net rate of reaction at any time is given by,

net rate $=-2.0\times {10}^{-4}\left[Cu\right(N{H}_3{)}_4{]}^{2+}$$=3.0\times {10}^5$ $\left[Cu\right(N{H}_3{)}_3{H}_{2}O{]}^{2+}.\left[N{H}_3\right]$
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 the above

Answer 1

Answer: (D)

All of the above

Net rate$=k_b\left(Cu\right(N{H}_3{)}_3{H}_2{O}^{2+}.\left[N{H}_3\right]$
$kt=2\times {10}^{-4}$
$kb=3\times {10}^{+5}$
$Const=\frac{kt}{kb}=\frac{2\times {10}^{-4}}{3\times {10}^5}$
$=6.6\times {10}^{-10}$
[D]