In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon a

NCERT Exemplar · Q19

Q. In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?

(i) H2

(g) + I2

(g) ⇌ 2HI

(g)

(ii) PCl5

(g) ⇌ PCl3

(g) + Cl2

(g)

(iii) N2

(g) + 3H2

(g) ⇌ 2NH3

(g)

(iv) The equilibrium will remain unaffected in all the three cases.

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Concept understanding — Phase Equilibrium

Phase Equilibrium: From Intuition to Precision

Imagine a glass of ice water on a warm day. The ice cubes are melting, but the water around them stays cold. If you wait long enough, the ice stops melting — not because the room got colder, but because the system has reached a balance. The ice and water coexist without any net change. That balance is phase equilibrium.

The Intuition: A Tug-of-War Between Two Tendencies

Every substance wants to exist in the phase (solid, liquid, or gas) that has the lowest energy under the given conditions. But there's a catch: nature also loves disorder (entropy). A solid is low-energy but highly ordered; a gas is high-energy but very disordered. At a specific temperature and pressure, these two opposing drives — minimising energy and maximising disorder — exactly cancel each other out. The result? Two phases can coexist peacefully, with molecules constantly jumping between them but no net change in the amount of each phase.

Think of it as a tug-of-war. On one side, molecules in the solid phase are tightly bound and low in energy. On the other side, molecules in the liquid phase have more freedom and higher entropy. At the melting point, the rope doesn't move — the rates of melting and freezing are equal.

The Precise Statement

Phase equilibrium is the condition in which two or more phases of a substance coexist in thermodynamic equilibrium, with no net transfer of mass between phases. For a pure substance, this occurs at a unique temperature for a given pressure (or vice versa), defined by the equality of the chemical potential (or Gibbs free energy per mole) in all phases.

In simpler terms: at equilibrium, the driving force for a molecule to move from phase A to phase B is exactly the same as the driving force to move from B to A. The system is dynamic — molecules are still crossing the boundary — but macroscopically static.

The Key Condition: Equal Chemical Potential

The rigorous condition for phase equilibrium in a pure substance is:

\mu_{\text{solid}} = \mu_{\text{liquid}} = \mu_{\text{gas}}

where $\mu$ (mu) is the chemical potential — the Gibbs free energy per mole. When chemical potentials are equal, no phase has a "preference" to grow at the expense of another.

Important

For a pure substance, phase equilibrium occurs only along specific lines on a phase diagram (the melting curve, boiling curve, and sublimation curve). At the triple point, all three phases coexist.

A Concrete Example: Water at 0°C

At 1 atm pressure and 0°C, ice and liquid water are in equilibrium. Here's what's happening at the molecular level:

Water molecules at the surface of the ice are vibrating. Some gain enough energy to break free and enter the liquid.
Simultaneously, water molecules in the liquid near the ice surface lose energy and attach to the ice crystal.
The rates of these two processes are equal. So the amount of ice and the amount of liquid water remain constant — even though individual molecules are constantly switching sides. …

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Adding an inert gas at constant volume does not change the partial pressures of the reactants or products, so the equilibrium position remains unchanged in all three reactions. The answer is (iv).

Why inert gases matter (or don't) in equilibrium

When we add an inert gas like argon to an equilibrium mixture, the effect depends entirely on whether the partial pressures of the reacting species change. Equilibrium constants are written in terms of partial pressures (or concentrations), not total pressure.

At constant volume, adding argon increases the total pressure in the container, but the number of moles of each reacting gas stays the same, and the volume they occupy stays the same. Since partial pressure is given by

P_i = \frac{n_i RT}{V}

and neither $n_i$ , $T$ , nor $V$ changes for any reactant or product, none of their partial pressures change. The equilibrium constant $K_p$ depends only on these partial pressures, so the equilibrium position is unaffected.

Tip

At constant pressure (not volume), adding an inert gas forces the volume to expand to keep total pressure constant. That does change partial pressures and can shift the equilibrium toward the side with more moles of gas.

Let's verify this reasoning for each reaction:

Step-by-step analysis

Reaction (i): $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$

The equilibrium constant is

K_p = \frac{P_{\mathrm{HI}}^2}{P_{\mathrm{H_2}} \cdot P_{\mathrm{I_2}}}

Adding argon at constant volume leaves $P_{\mathrm{H_2}}$ , $P_{\mathrm{I_2}}$ , and $P_{\mathrm{HI}}$ unchanged. The ratio $K_p$ remains satisfied, so the equilibrium does not shift.

Reaction (ii): $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$

Here $\Delta n = 2 - 1 = 1$ (more moles on the product side). The equilibrium constant is

K_p = \frac{P_{\mathrm{PCl_3}} \cdot P_{\mathrm{Cl_2}}}{P_{\mathrm{PCl_5}}}

Again, at constant volume, the partial pressures of $\mathrm{PCl_5}$ , $\mathrm{PCl_3}$ , and $\mathrm{Cl_2}$ are unchanged by the addition of argon. The equilibrium remains where it was. …

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Q.In which of the following reactions, the equilibrium remains unaffected on addition of small amount of argon at constant volume?

Why inert gases matter (or don't) in equilibrium

Step-by-step analysis

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