Question

For the formation of $NH _{3}$ from $N _{2}$ and $H _{2}$ at $500\, K$, the concentration of $N _{2}, H _{2}$ and $NH _{3}$ at equilibrium are $1.5 \times 10^{-2} M , 3.0 \times 10^{-2} M$ and $1.2 \times 10^{-2} M$, respectively. The equilibrium constant for the reverse reaction is

• A. $3.56 \times 10^{2}$
• B. $2.81 \times 10^{-3}$
• C. $3.56 \times 10^{-2}$
• D. $2.81 \times 10^{3}$

Answer 1

Answer: (B)

Given: Concentration of

$\left[ N _{2}\right] =1.5 \times 10^{-2} \,M $

$\left[ H _{2}\right] =3.0 \times 10^{-2} \,M $

$\left[ NH _{3}\right] =1.2 \times 10^{-2}\, M$

The reaction for the formation of $NH _{3}$ is

$N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3}$

Thus, the reverse reaction will be

$2 NH _{3} \rightleftharpoons N _{2}+3 H _{2} $ and equilibrium constant $(K_{c})$ will be

$K_{C}=\frac{\left[ N _{2}\right] \cdot\left[ H _{2}\right]^{3}}{\left[ NH _{3}\right]^{2}}=\frac{\left[1.5 \times 10^{-2}\right]\left[3.0 \times 10^{-2}\right]^{3}}{\left[1.2 \times 10^{-2}\right]^{2}}$

$=\frac{\left(1.5 \times 10^{-2}\right)\left(27.0 \times 10^{-6}\right)}{1.44 \times 10^{-4}}=\frac{40.5}{1.44} \times 10^{-4}$

$=28.125 \times 10^{-4}$

or $K_{C}=2.81 \times 10^{-3}$