Q.Explain why adding an inert gas such as argon to the equilibrium mixture $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ has no effect on the position of equilibrium if added at constant volume, but shifts the equilibrium backward if added at constant total pressure.

Both perturbations — adding a reactant and removing a product — push the equilibrium $2H_2(g) + CO(g) \rightleftharpoons CH_3OH(g)$ towards the product side, per Le Chatelier's principle.

The equilibrium is:

$2H_2(g) + CO(g) \rightleftharpoons CH_3OH(g)$

Le Chatelier's principle: if a system at equilibrium is subjected to a change (in concentration, pressure, temperature, etc.), the equilibrium shifts in the direction that tends to counteract (undo) that change.

(i) Addition of H$_2$: Increasing the concentration of a reactant (H$_2$) disturbs the equilibrium. To counteract (reduce) this added H$_2$, the equilibrium shifts in the forward direction (towards the product, CH$_3$OH), consuming the extra H$_2$ (along with CO) and producing more methanol. Net effect: more CH$_3$OH is formed.

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Le Chatelier's principle governs how a system at equilibrium responds to a disturbance: it shifts to partially counteract the change.

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that tends to counteract (partly undo) that change. The main factors are:

1. Concentration: Adding more of a reactant shifts equilibrium towards the products (to consume the extra reactant); removing a product (or adding more product) similarly shifts the equilibrium to compensate.

2. Pressure/volume (relevant for gas-phase equilibria involving unequal moles of gas on the two sides): Increasing pressure (decreasing volume) shifts equilibrium towards the side with fewer moles of gas, since that side occupies less volume; decreasing pressure shifts it towards the side with more moles of gas. If the moles of gas are equal on both sides, pressure/volume changes have no effect on the position of equilibrium.

3. Temperature: Increasing temperature shifts equilibrium in the endothermic direction (the direction that absorbs the added heat); decreasing temperature shifts it in the exothermic direction.

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Le Chatelier's principle: the system counteracts any imposed change; both actions here push the equilibrium forward (toward more CH3OH).

For the equilibrium $2H_2(g) + CO(g) \rightleftharpoons CH_3OH(g)$:

(i) Addition of H2: Increasing the concentration of a reactant (H2) disturbs the equilibrium. By Le Chatelier's principle, the system shifts in the direction that consumes the added reactant — i.e., the forward direction — to partially offset the increase. This produces more CH3OH and consumes more CO.

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By Le Chatelier's principle, higher temperature reduces NH3 yield (since the forward reaction is exothermic) while higher pressure increases NH3 yield (since the forward reaction decreases the total moles of gas).

The reaction N2(g) + 3H2(g) ⇌ 2NH3(g) (the Haber process) is exothermic in the forward direction (ΔH < 0), and the forward reaction reduces the total number of gas moles from 4 (1 + 3) to 2.

Effect of temperature: Since the forward reaction releases heat, by Le Chatelier's principle, raising the temperature shifts the equilibrium in the direction that absorbs the added heat — i.e., backward, toward N2 and H2 — decreasing the yield of NH3. Lowering the temperature favours the forward (exothermic) reaction and increases NH3 yield, although at very low temperature the rate of reaction becomes too slow, so industrially a moderate temperature (around 700 K) is used as a compromise.

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