Question

An equilibrium mixture of $N_2,H_2$ and $N{H}_3$ at $700$K contains $0.036$M $N_2$ and $0.15$M $H_2$. At this temperature, $K_c$ for the reaction $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ is $0.29$. What is the concentration of $N{H}_3$?

Answer 1

${N_2}_{\left(g\right)}+{3H_2}_{\left(g\right)}\rightleftharpoons {2N{H}_3}_{\left(g\right)}$

At eqm $0.036M$ $0.15M$ $x$

$K_C=\frac{[N{H}_3{]}^2}{\left[N_2\right]\times [H_2{]}^3}\Rightarrow 0.29=\frac{x^2}{0.036\times (0.15{)}^3}$

$\Rightarrow x=5.9\times {10}^{-3}M$

$\therefore$ Concentration of $N{H}_3$ is $5.9\times {10}^{-3}M$ .