Ncert Exercises & Solutions

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 N2 (g) + 3H2 (g) 2NH3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

The given reaction is:

$N_{2 (g)} + 3 H_{2 (g)} \leftrightarrow 2 {NH}_{3 (g)}$

The given concentration of various species is

$[N_{2}] = \frac{1.57}{20} mol L^{- 1} [H_{2}] = \frac{1.92}{20} mol L^{- 1}$
$[{NH}_{3}] = \frac{8.13}{20} mol L^{- 1}$

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

$Q_{c} = \frac{{[{NH}_{3}]}^{2}}{[N_{2}] {[H_{2}]}^{3}}$
$= \frac{{(\frac{(8.13)}{20})}^{2}}{(\frac{1.57}{20}) {(\frac{1.92}{20})}^{3}}$
$= 2.4 \times 10^{3}$

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

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

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