Question

The ionisation constant of $N{H}_4^{+}$ in water is $5.6\times 10^{-10}$ at $25^{\circ }C$ . The rate constant for the reaction of $N{H}_4^{+}$ and $O{H}^{-}$ to form $N{H}_3$ and $H_{2}O$ at $25^{\circ }C$ is $3.4\times 10^{10}Lmo{l}^{-1}{s}^{-1}$ . The rate constant for proton transfer from water to $N{H}_3$ is

• A. $6.07\times 10^5$
• B. $0.607\times 10^5$
• C. $60.7\times 10^5$
• D. $6.07\times 10^{10}$

Answer 1

Answer: (A)

$N{H}_4^{+}\rightleftharpoons N{H}_3+H^{+};k_1=5.6\times 10^{10}$ …(i)
$H_{2}O\rightleftharpoons H^{+}+OH;k_2=1\times 10^{14}$ …(ii)
$K=\frac{k_1}{k_2}=\frac{5.6\times 10^{-10}}{1\times 10^{-14}}$
$=5.6\times 10^4$
By subtracting equation (ii) from equation (i) we get
$N{H}_4^{+}+OH\rightleftharpoons N{H}_3+H_{2}O$
From equation
$k_f=3.4\times 10^{10}/mi{l}^{-1}{S}^{-1}$
$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
$k_r=?$
$K=\frac{k_f}{k_r}$
So $k_r=\frac{k_f}{K}=\frac{3.4\times 10^{10}}{5.6\times 10^4}$
$=6.07\times 10^5$