Does the number of moles of reaction products increase, decrease or remain same when each of the following equ

Exercises · 6.25

Q. Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a) PCl5

(g) ⇌ PCl3

(g) + Cl2

(g)

(b) CaO (s) + CO2

(g) ⇌ CaCO3 (s)

(g) ⇌ Fe3O4 (s) + 4H2 (g)

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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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For a gaseous equilibrium, decreasing pressure (by increasing volume) shifts the reaction toward the side with more gas moles. The number of moles of products increases in (a), decreases in (b), and remains the same in (c).

The Core Idea: Le Chatelier’s Principle and Gas Moles

When you increase the volume of a container, the pressure of all gases drops. The system responds by shifting in the direction that produces more gas molecules — because more gas molecules at the same temperature would generate a higher pressure, partially counteracting the change. This is a direct application of Le Chatelier’s principle.

The key quantity to compare is $\Delta n_g$ , the change in the number of moles of gas when the reaction proceeds forward:

\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})

If $\Delta n_g > 0$ : forward direction produces more gas → shift forward on pressure decrease → product moles increase.
If $\Delta n_g < 0$ : forward direction consumes gas → shift backward on pressure decrease → product moles decrease.
If $\Delta n_g = 0$ : no net change in gas moles → pressure change has no effect on equilibrium position → product moles remain same.

Watch out

A common mistake is to count solids and liquids when calculating $\Delta n_g$ . Only gaseous species matter — their concentrations change with volume; solids and liquids have fixed concentrations (activity = 1) and do not appear in the equilibrium expression for pressure effects.

Step-by-Step Analysis

1. Reaction (a): $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$

Count gas moles on each side:

Reactants: 1 mole of $\text{PCl}_5$ (gas)
Products: 1 mole of $\text{PCl}_3$ + 1 mole of $\text{Cl}_2$ = 2 moles of gas

\Delta n_g = 2 - 1 = +1

Since $\Delta n_g > 0$ , decreasing pressure (increasing volume) shifts the equilibrium to the right, toward more product. The number of moles of reaction products increases.

Tip

This is the classic dissociation reaction. If you imagine pulling the piston out, the system "wants" to fill the extra space with more gas particles — so PCl₅ breaks apart.

2. Reaction (b): $\text{CaO}(s) + \text{CO}_2(g) \rightleftharpoons \text{CaCO}_3(s)$

Identify gaseous species only:

Reactants: 1 mole of $\text{CO}_2$ (gas) — CaO is solid, ignore
Products: 0 moles of gas — CaCO₃ is solid

$\Delta n_g = 0 - 1 = -1$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set D-41 markMCQ
$\ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)}$
(a) Increasing the concentration of CO
(b) Increasing the concentration of $\ce{H2O}$
(c) Decreasing the concentration of $\ce{CH4}$
(d) Decreasing the concentration of $\ce{H2}$ (A) a and c (B) b and d (C) a and d (D) a and b
›Reveal solutionSolution
Le Chatelier: adding a reactant or removing a product shifts the equilibrium forward — so (a) increasing $[\ce{CO}]$ and (c) decreasing $[\ce{CH4}]$ favour methane.

The equilibrium.
$\ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)}, \qquad K_c = \frac{[\ce{CH4}][\ce{H2O}]}{[\ce{CO}][\ce{H2}]^3}$
Reactants: $\ce{CO}$ , $\ce{H2}$ . Products: $\ce{CH4}$ , $\ce{H2O}$ .

Step 1 — The rule (reaction-quotient form).

A disturbance changes $Q = \dfrac{[\ce{CH4}][\ce{H2O}]}{[\ce{CO}][\ce{H2}]^3}$ . If $Q < K_c$ the system moves forward (more $\ce{CH4}$ ); if $Q > K_c$ it moves backward.

Step 2 — Test each change.

(a) Increase $[\ce{CO}]$ — a reactant term in the denominator grows, so $Q$ decreases, $Q<K_c$ ⇒ shift forward. Methane formation is favoured. ✓

(b) Increase $[\ce{H2O}]$ — a product term in the numerator grows, so $Q$ increases, $Q>K_c$ ⇒ shift backward. Methane formation is opposed. ✗ …
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KCET 2018Set A-11 markMCQ
$K_p$
(g) + HCl\ (g)} $$ (A) 1 (B) 0.5 (C) 1.5 (D) 2
›Reveal solutionSolution
$\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$ — solids and pure liquids are excluded because they do not appear in $K_p$ .

Step 1 — The definition that matters.

In
$K_{p} = K_{c}(RT)^{\Delta n}$
$\Delta n$ is not the change in total moles — it is
$\Delta n = \sum n_{\text{gaseous products}} - \sum n_{\text{gaseous reactants}}$
The reason: $K_p$ is written in terms of partial pressures, and only gases exert a partial pressure. Pure solids and pure liquids have constant activity (taken as 1) and are omitted from both $K_p$ and $K_c$ .

Step 2 — Identify the phases in the given reaction.
$\mathrm{NH_4Cl\,(s)} \rightleftharpoons \mathrm{NH_3\,(g)} + \mathrm{HCl\,(g)}$

Reactant: NH $_4$ Cl is (s) — a solid ⇒ contributes 0 gaseous moles.

Products: NH $_3$ (g) = 1 mol, HCl (g) = 1 mol ⇒ total 2 gaseous moles.

Step 3 — Compute $\Delta n$ .

$\Delta n = 2 - 0 = 2$ …
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KCET 2018Set A-11 markMCQ
$2\mathrm{SO}_2 + \mathrm{O}_2 \rightleftharpoons 2\mathrm{SO}_3$
›Reveal solutionSolution

Divide each species' rate of change by its stoichiometric coefficient to get one common rate of reaction, then read off the rate for $\mathrm{SO_3}$ .

Step 1 — Write the unique rate of reaction.

For a reaction $aA + bB \rightarrow cC$ , all species change at rates proportional to their coefficients, so a single rate is defined as
$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt}$
The minus signs are for reactants (their concentrations fall), and dividing by the coefficient makes the rate independent of which species you monitor.

For $2\mathrm{SO_2} + \mathrm{O_2} \rightleftharpoons 2\mathrm{SO_3}$ :
$\text{Rate} = -\frac{1}{2}\frac{d[\mathrm{SO_2}]}{dt} = -\frac{d[\mathrm{O_2}]}{dt} = +\frac{1}{2}\frac{d[\mathrm{SO_3}]}{dt}$
Step 2 — Relate $\mathrm{SO_3}$ to $\mathrm{O_2}$ .

Equating the last two expressions:
$\frac{1}{2}\frac{d[\mathrm{SO_3}]}{dt} = -\frac{d[\mathrm{O_2}]}{dt}$
$\Rightarrow\quad \frac{d[\mathrm{SO_3}]}{dt} = 2 \times \left(\text{rate of disappearance of } \mathrm{O_2}\right)$ …
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Q.Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

The Core Idea: Le Chatelier’s Principle and Gas Moles

Step-by-Step Analysis

1. Reaction (a): PCl5(g)⇌PCl3(g)+Cl2(g)\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)PCl5​(g)⇌PCl3​(g)+Cl2​(g)

2. Reaction (b): CaO(s)+CO2(g)⇌CaCO3(s)\text{CaO}(s) + \text{CO}_2(g) \rightleftharpoons \text{CaCO}_3(s)CaO(s)+CO2​(g)⇌CaCO3​(s)

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1. Reaction (a): $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$

2. Reaction (b): $\text{CaO}(s) + \text{CO}_2(g) \rightleftharpoons \text{CaCO}_3(s)$

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