Question

Given $K_a$ values for $N{H}_4^{+}$ and $HCN$ are $5.76\times {10}^{-10}$ and $4.8\times {10}^{-10}$ respectively. What is the equilibrium constant for the following reaction?

$N{H}_{4(aq.)}^{+}+C{N}_{(aq.)}^{-}\rightleftharpoons N{H}_{3(aq.)}+HC{N}_{(aq.)}$

• A. $0.83$
• B. $1.2$
• C. $8.0\times {10}^{-11}$
• D. $27.6\times {10}^{-10}$

Answer 1

The correct option is B $1.2$

Given:

$K_a$ value for $N{H}_4^{+}$ = $5.76\times {10}^{-10}$

$K_a$ value for $HCN$ = $4.8\times {10}^{-10}$

$N{H}_4^{+}+C{N}^{-}\leftrightharpoons N{H}_3+HCN$

Solution:

From the given values of $K_a$ we need to find out the equilibrium constant for the given equation.

The reaction can be well understood by following the order of the reaction. First ammonia dissociates by the following basis,

$N{H}_4^{+}\leftrightharpoons N{H}_3+H^{+}$ (1)

The above equation can be considered as the first-order reaction.

Then the reaction of hydrogen cyanide can be shown as:

$HCN\leftrightharpoons H^{+}+C{N}^{-}$ (2)

This reaction is said to follow the second-order reaction.

The final equation on subtracting equation 1 and 2 can be obtained as:

$N{H}_4^{+}+C{N}^{-}\leftrightharpoons N{H}_3+HCN$

$K_{eq}=\frac{K_1}{K_2}$

$K_{eq}=\frac{5.76\times {10}^{-10}}{4.8\times {10}^{-10}}$

$K_{eq}=\frac{576}{480}$

$K_{eq}=\frac{24}{20}$

$K_{eq}=\frac{12}{10}$

$K_{eq}=1.2$

The correct option is B.