# A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500

Question:

A mixture of $1.57 \mathrm{~mol}$ of $\mathrm{N}_{2}, 1.92 \mathrm{~mol}$ of $\mathrm{H}_{2}$ and $8.13 \mathrm{~mol}$ of $\mathrm{NH}_{3}$ is introduced into a $20 \mathrm{~L}$ reaction vessel at $500 \mathrm{~K}$. At this temperature, the equilibrium constant, $K_{c}$ for the reaction $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})$ is $1.7 \times 10^{2}$

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Solution:

The given reaction is:

$\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longleftrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})}$

The given concentration of various species is

$\left[\mathrm{N}_{2}\right]=\frac{1.57}{20} \mathrm{~mol} \mathrm{~L}^{-1} \quad\left[\mathrm{H}_{2}\right]=\frac{1.92}{20} \mathrm{~mol} \mathrm{~L}^{-1}$

$\left[\mathrm{NH}_{3}\right]=\frac{8.13}{20} \mathrm{~mol} \mathrm{~L}^{-1}$

Now, reaction quotient $Q_{c}$ is:

$Q_{\mathrm{C}}=\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}$

$=\frac{\left(\frac{(8.13)}{20}\right)^{2}}{\left(\frac{1.57}{20}\right)\left(\frac{1.92}{20}\right)^{3}}$

$=2.4 \times 10^{3}$

Since $Q_{\mathrm{C}} \neq K_{\mathrm{C}}$, the reaction mixture is not at equilibrium.

Again, $Q_{\mathrm{C}}>K_{\mathrm{C}}$. Hence, the reaction will proceed in the reverse direction.