JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Equilibrium

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2 NOCl (g) ⇌ 2 NO (g) + Cl_{2} (g)

In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4

mol / L

. The equilibrium constant at

3 0^{\circ} C

is ______

\times 1 0^{- 4}

Answer: 125

Solution

Initial concentration of NOCl = 2.0 M. At equilibrium, [NO] = 0.4 M. From stoichiometry, if x M of NOCl dissociates, 2x M of NO forms. So x = 0.4/2 = 0.2 M. At equilibrium: [NOCl] = 2.0 - 0.4 = 1.6 M, [NO] = 0.4 M, [Cl₂] = 0.2 M. The equilibrium constant

K_{c} = \frac{[ NO ] ^{2} [ Cl _{2} ]}{[ NOCl ] ^{2}} = \frac{( 0.4 ) ^{2} \times 0.2}{( 1.6 ) ^{2}} = \frac{0.032}{2.56} = 0.0125 = 12.5 \times 1 0^{- 3} = 125 \times 1 0^{- 4}

. Answer: 125.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer

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