Consider the following reaction:
N₂O₄(g) ⇌ 2NO₂(g); ΔHᵒ = +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 N₂ at constant T.

Solution

Chemistry ma’am · 4 steps

Endothermic forward plus cooling means backward; an inert gas at fixed volume changes no partial pressure.

1. Case (a), cooling. The forward reaction absorbs heat (Δ H° = +58 kJ), so think of heat as a reactant. Taking heat away makes the system replace it by running the heat-releasing backward reaction. The equilibrium shifts towards the reactant N₂O₄.
2. Case (b), adding N2. N₂ does not react with N₂O₄ or NO₂, so it is an inert gas here. Pressure rises by adding it, which means the volume is fixed. Total pressure goes up, but the moles of N₂O₄ and NO₂ in the same volume do not change.

   Qₚ = p_(NO₂)²/p_(N₂O₄), pᵢ = (nᵢ RT)/V

3. Check the quotient. Each partial pressure depends only on its own moles, T and V, and none of these change. So Qₚ still equals Kₚ, and the equilibrium does not shift.
4. So. Case (a) shifts towards the reactant and case (b) gives no change, which is option (b).

The idea to keep. Le Chatelier works on partial pressures, not total pressure. An inert gas added at constant volume leaves every partial pressure unchanged, so nothing shifts.

Answer from the NTA final answer key. Solution by PrepWiser.