The equilibrium constant at 298 K for a reaction $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$ is 100 . If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in $\mathrm{mol} \mathrm{L}^{-1}$ ) will be:

JEE Main · 2016 · 03 Apr 2016 · Offline · Equilibrium

The equilibrium constant at 298 K for a reaction

A + B ⇌ C + D

is 100 . If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in

mol L^{- 1}

) will be:

Answer: (B)

Let

x

be the amount of A and B that reacts. Starting from 1 M each, at equilibrium

[A] = [B] = 1 - x

and

[C] = [D] = 1 + x

, so

K = \frac{( 1 + x ) ^{2}}{( 1 - x ) ^{2}} = 100

. Taking the square root,

\frac{1 + x}{1 - x} = 10

, hence

1 + x = 10 - 10 x

11 x = 9

and

x = \frac{9}{11} \approx 0.818

. The equilibrium concentration of D is

1 + 0.818 = 1.818 mol L^{- 1}

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option