Consider the following reaction N 2 O 4( g ) leftharpoons 2 NO 2( g ) ; Δ H 0=+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

Question

Consider the following reaction N 2 O 4( g ) leftharpoons 2 NO 2( g ) ; Δ H 0=+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 (A) (a) towards reactant, (b) no change (B) (a) towards product, (b) towards reactant (C) (a) towards product, (b) no change (D) (a) towards reactant, (b) towards product

Tardigrade Admin · Accepted Answer

Δ H °>0 T downarrow equation shifts back ward. N 2 is treated as inert gas in this case hence no effect on equilibrium

Q. Consider the following reaction
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); Δ H^{0} = + 58 k J$
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 $T$

(a) towards reactant, (b) no change

(a) towards product, (b) towards reactant

(a) towards product, (b) no change

(a) towards reactant, (b) towards product

$Δ H^{\circ} > 0 T ↓$ equation shifts back ward.

$N_{2}$ is treated as inert gas in this case hence no effect on equilibrium

Q. Consider the following reaction N2​O4​(g)⇌2NO2​(g);ΔH0=+58kJ 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 N2​ at constant T