Question

A mixture of $1.57$ moles of $N_2,1.92$ moles of $H_2$ and $8.13$ moles of $N{H}_3$ is introduced into a $20$L reaction vessel at $500$K. At this temperature, the equilibrium constant, $K_c$ for the reaction, $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}is1.7\times {10}^2$. What is the direction of the net reaction?

• A. Forward
• B. Backward
• C. At equilibrium
• D. Data is insufficient

Answer 1

The correct option is B Backward

The given reaction is :-

$N_2+{3H}_2\rightleftharpoons 2{NH}_3$

Now, we have

$Q_C=\frac{{\left[{NH}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$

$=\frac{{\left(\frac{8.13}{20}\right)}^2}{\left(\frac{1.57}{20}\right)\times {\left(\frac{1.92}{20}\right)}^3}$ $(\because concentration=\frac{no.ofmoles}{volume})$

$=\frac{20\times 8.13\times 8.13\times 20}{1.57\times 1.92\times 1.92\times 1.92}$

$=2.38\times {10}^3$

Given, $K_C=1.7\times {10}^2$

Now, $Q_C\ne K_C$, so the reaction is not at equilibrium.

Now, $Q_C>K_C$, so the reaction will go towards backward direction.