JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Equilibrium

In a chemical reaction

A + 2 B ⇌ 2 C + D

, the initial concentration of

B

was 1.5 times of A but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant(K) for the aforesaid chemical reaction is

(A)4✓ Correct
(B)16
(C) $1/4$
(D)1

Answer: (A)

Solution

Let the starting amount of

A

a

, so

B

starts at

1.5 a

, and let

x

A

react. At equilibrium

[A] = a - x

[B] = 1.5 a - 2 x

[C] = 2 x

and

[D] = x

. We are told

A

and

B

end up equal, so

a - x = 1.5 a - 2 x

, giving

x = 0.5 a

. Then

[A] = [B] = 0.5 a

[C] = a

and

[D] = 0.5 a

, so

K_{c} = \frac{[ C ] ^{2} [ D ]}{[ A ] [ B ] ^{2}} = \frac{a ^{2} ( 0.5 a )}{( 0.5 a ) ( 0.5 a ) ^{2}} = \frac{0.5 a ^{3}}{0.125 a ^{3}} = 4

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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