A mixture of $1.57 m o l$ of $N_{2}$ , $1.92 m o l$ of $H_{2}$ and $8.13 m o l$ 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 $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)$ is $1.7 \times 10^{2}$ . Is the reaction mixture at equilibrium? If not , What is the direction of the net reaction?

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Solution

First we will calculate the reaction quotient $Q_{c}$ . Then we will compare $Q_{c}$ with K (the equilibrium constant). From the relative values of $Q_{c}$ and K, we can decide if the reaction equilibrium will shift in the forward direction or in the reverse direction.
$Q_{c} = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}} = \frac{(\frac{8.13}{20})^{2}}{\frac{1.57}{20} \times (\frac{1.92}{20})^{3}} = 2379$

Thus, $Q_{c} (2379) > K_{c} (170)$ . The reaction is not at equilibrium and the amount of products is greater than the equilibrium amount. Hence, the equilibrium will shift in the backward direction.

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