Question

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

• A. $3\times {10}^4{s}^{-1}$
• B. $6\times {10}^6{s}^{-1}$
• C. $2.6\times {10}^3{s}^{-1}$
• D. $6\times {10}^5{s}^{-1}$

Answer 1

Answer: (D)

$6\times {10}^5{s}^{-1}$

$N{H}_3+H_{2}O\stackrel{k_f}{\underset{k_b}{\rightleftharpoons }}N{H}_4^{\oplus }+\stackrel{\ominus }{O}H(k_b=3.4\times {10}^{10})$..........(i)

$N{H}_4^{\oplus }+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{\oplus }(k_a=5.6\times {10}^{-10})$

$k_{base}\left(N{H}_3\right)=\frac{k_f}{k_b}=\frac{k_w}{k_{acid}\left(N{H}_4^{\oplus }\right)}$

$(\because k_{acid}\times k_{base}=k_w)$

$\frac{k_f}{3.4\times {10}^{10}}=\frac{{10}^{-14}}{5.6\times {10}^{-10}}\Rightarrow k_f=6.07\times {10}^5$