JEE Main · 2015 · 04 Apr 2015 · Offline · Equilibrium

The standard Gibbs energy change at 300 K for the reaction

2 A ⇌ B + C

is 2494.2 J . At a given time, the composition of the reaction mixture is

[A] = \frac{1}{2}, [B] = 2

and

[C] = \frac{1}{2}

. The reaction proceeds in the: [

R = 8.314 J / K / mol, e = 2.718

]

Answer: (A)

Solution

First turn the standard free energy into an equilibrium constant through

Δ G^{\circ} = - R T ln K

2494.2 = - 8.314 \times 300 \times ln K

gives

ln K = - 1

, so

K = e^{- 1} \approx 0.37

. Now test the current mixture with the reaction quotient

Q = \frac{[ B ] [ C ]}{[ A ] ^{2}} = \frac{2 \times 0.5}{( 0.5 ) ^{2}} = 4

. Since

Q = 4

far exceeds

K \approx 0.37

, there is too much product, so the reaction runs backwards to reach equilibrium.

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option