Question

Equilibrium constant, K_c for the reaction
\(N_2 (g) + 3H_2 (g) ⇋ 2NH_3 (g)\)
At 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L^–1 N₂, 2.0 mol L^–1 H₂ and 0.5 mol L^–1 NH₃. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Answer 1

The given reaction is : 
                                                     \(N_2(g)  + 3H_2(g) ↔ 2NH_3(g)\)
At a particular time :           3.0 mol L^-1   2.0 mol L^-1 0.5 mol L^-1

Now, we know that,
\(Q_c = \frac {[NH_3]^2}{[N_2][H_2]^3}\) 

\(Q_c= \frac {(0.5)^2}{(3.0)(2.0)^3}\)
\(Q_c= 0.0104\)
It is given that \(K_c = 0.061\).
Since \(Q_c ≠ K_c\) , the reaction is not at equilibrium.
Since \(Q_c < K_c\), the reaction will proceed in the forward direction to reach equilibrium.