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A mixture of 1.57 mol of N_2, 1.92 mol of H_2 and 8.13 mol of NH_3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction

7.12     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 K. At this temperature, the equilibrium constant, Kc for the reaction N_{2}_{(g)}+3H_{2}_{(g)}\rightleftharpoons 2NH_{3}_{(g)}  is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

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M manish

We have,
N_{2}_{(g)}+3H_{2}_{(g)}\rightleftharpoons 2NH_{3}_{(g)}

K_c = 1.7 \times 10^{2}

The concentration of species are-
[N_2] = 1.57/20 mol L^{-1}\\ {[H_2]}=1.92/20mol L^{-1}\\ {[NH_3]}=8.13/20mol L^{-1}

We know the formula of 

Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3}
      \\=\frac{(8.13/20)^2}{(1.57/20)(1.92/20)^3}\\ =2.4\times 10^{-3}
The reaction is not in equilibrium. Since Qc>K_c, the equilibrium proceeds in reverse direction.

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