JEE MAIN - Chemistry (2022 - 27th June Morning Shift - No. 13)
2NOCl(g) $$\rightleftharpoons$$ 2NO(g) + Cl2(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 30$$^\circ$$C is ______________ $$\times$$ 10$$-$$4.
Answer
125
Explanation
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Given that at equilibrium, concentration of NO = 0.4 mol/L
$$\therefore$$ 2x = 0.4
$$\Rightarrow$$ x = 0.2
$$\therefore$$ Concentration of NOCl at equilibrium,
[NOCl]eq = 2 $$-$$ 2 $$\times$$ 0.2 = 1.6
and [NO]eq = 0.4
and [Cl2]eq = 0.2
We know,
$${K_C} = {{{{[NO]}^2}[C{l_2}]} \over {{{[NOCl]}^2}}}$$
$$ = {{{{[0.4]}^2}[0.2]} \over {{{[1.6]}^2}}}$$
$$ \Rightarrow {K_C} = 12.5 \times {10^{ - 3}}$$
$$ \Rightarrow {K_C} = 125 \times {10^{ - 4}}$$
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