Question

For the equilibrium $N_2+3H_2\rightleftharpoons 2N{H}_3,K_c$ at 1000 $K$ is $2.37\times {10}^{-3}$. If at equilibrium $\left[N_2\right]=2M,\left[H_2\right]=3M$, the concentration of $N{H}_3$ is:

• A. $0.00358M$
• B. $0.0358M$
• C. $0.358M$
• D. $3.58M$

Answer 1

The correct option is C $0.358M$

$N_2+3H_2\rightleftharpoons 2N{H}_3$

$K_E=2.37\times {10}^{-3}$

$\left[N_2\right]=2m$

$\left[H_2\right]=3m$

$\left[N{H}_3\right]=?$

$K_E=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$

$2.37\times {10}^{-3}=\frac{{\left[N{H}_3\right]}^2}{\left[2\right]{\left[3\right]}^3}$

$2.37\times {10}^{-3}=\frac{{\left[N{H}_3\right]}^2}{2\times 27}$

${\left[N{H}_3\right]}^2=54\times 2.37\times {10}^{-3}$

${\left[N{H}_3\right]}^2=12.79\times {10}^{-2}$

$\left[N{H}_3\right]=\sqrt{12.79\times {10}^{-2}}$

$\left[N{H}_3\right]=0.0035M$

Hence, the correct option is $\text{C}$