Le Chatelier's principle — predict effect of pressure, temperature, concentration change

Question

For the reaction:

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92 \text{ kJ/mol}

Predict the direction of shift when: (a) Pressure is increased (b) Temperature is increased (c) More $N_2$ is added at constant volume

Solution — Step by Step

When a system at equilibrium is disturbed, it shifts in the direction that partially opposes the disturbance. The key word is partially — the system never fully cancels the change, it only reduces it.

Increasing pressure on a gas-phase reaction favours the side with fewer moles of gas. We count: left side has $1 + 3 = 4$ moles of gas, right side has $2$ moles.

The equilibrium shifts forward (towards products) — towards $NH_3$ . This is exactly why the Haber process runs at high pressure (150–300 atm).

This is where we use $\Delta H$ . The reaction is exothermic ( $\Delta H = -92$ kJ/mol), meaning heat is a product of the forward reaction. Think of it as:

N_2 + 3H_2 \rightleftharpoons 2NH_3 + \text{heat}

Adding temperature is like adding a product. By Le Chatelier’s principle, the system shifts backward (towards reactants) to absorb the extra heat. Yield of $NH_3$ decreases.

Adding a reactant at constant volume increases its concentration. The system responds by shifting forward to consume the added $N_2$ , producing more $NH_3$ .

Note: the equilibrium constant $K_c$ does not change (temperature is constant). Only the position of equilibrium shifts.

Disturbance	Direction of Shift	Effect on $NH_3$ yield
Increase pressure	Forward	Increases
Increase temperature	Backward	Decreases
Add more $N_2$	Forward	Increases

Summary: Pressure increase → forward; temperature increase → backward; adding $N_2$ → forward.

Why This Works

Le Chatelier’s principle is the system’s way of maintaining balance. Think of equilibrium as a tug-of-war — when you push one side, the rope moves, but never as far as you pushed.

For pressure and moles of gas: pressure depends on the number of gas molecules in the container (PV = nRT). Shifting towards fewer moles of gas reduces n, which reduces pressure — exactly what the system needs to oppose the increase.

For temperature and $\Delta H$ : an exothermic reaction releases heat to the surroundings. If we raise the temperature, we’re forcing extra heat into the system. The backward endothermic reaction absorbs this heat, so the system shifts backward. For endothermic reactions ( $\Delta H > 0$ ), the logic flips — increasing temperature favours products.

Alternative Method

Instead of memorising rules, write heat as an explicit “species”:

Exothermic: Reactants $\rightleftharpoons$ Products + Heat
Endothermic: Reactants + Heat $\rightleftharpoons$ Products

Now treat heat exactly like any other substance. Increase temperature = add heat = add a product (exothermic case) → shift backward. This method works every time and you never need to remember a separate rule for temperature.

This “heat as a species” trick also works for writing the effect of temperature on $K_c$ : for exothermic reactions, $K_c$ decreases with increasing temperature (product heat is increased, so equilibrium moves left, so ratio of products/reactants falls).

Common Mistake

Confusing pressure effect with adding an inert gas. If you add an inert gas (like argon) at constant volume, the partial pressures of the reacting gases don’t change. So there is no shift in equilibrium. Many students assume any pressure increase shifts the equilibrium — but only if it changes the partial pressures of the reactants/products does it matter.

If instead you add an inert gas at constant pressure (volume increases), the partial pressures of reacting gases actually decrease, and the equilibrium shifts toward more moles of gas. This is the opposite of compressing the system. This distinction appeared in NEET 2022 and JEE Main 2023.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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