Consider the following reaction: $$\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}^{\circ}=+58 \mathrm{~kJ}$$ For each of the following cases ( $\mathrm{a}, \mathrm{b}$ ) the direction in which the equilibrium shifts is: (a) Temperature is decreased (b) Pressure is increased by adding $\mathrm{N}_{2}$ at constant T

(B) (a) towards reactant, (b) no change

(A) (a) towards reactant, (b) towards product

(C) (a) towards product, (b) no change

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Equilibrium

Consider the following reaction:

N_{2} O_{4} ⇌ 2 NO_{2} (g); Δ H^{\circ} = + 58 kJ

For each of the following cases (

a, b

) the direction in which the equilibrium shifts is:
(a) Temperature is decreased
(b) Pressure is increased by adding

N_{2}

at constant T

Answer: (B)

The forward reaction is endothermic (

Δ H^{\circ} = + 58 kJ

). (a) Lowering

T

favours the exothermic (reverse) direction, shifting the equilibrium towards the reactant

N_{2} O_{4}

. (b) Adding

N_{2}

(an inert gas) at constant

T

and volume leaves the partial pressures of

N_{2} O_{4}

and

N O_{2}

unchanged, so

Q = K

— no shift. Hence (a) towards reactant, (b) no change.

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option