Which of the following reactions will get affected by increasing the pressure? Also, mention whether change wi

Exercises · 6.26

Q. Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

(i) COCl2

(g) ⇌ CO

(g) + Cl2

(g)

(ii) CH4

(g) + 2S2

(g) ⇌ CS2

(g) + 2H2S

(g)

(iii) CO2

(g) + C (s) ⇌ 2CO

(g)

(iv) 2H2

(g) + CO

(g) ⇌ CH3OH

(g)

(v) CaCO3 (s) ⇌ CaO (s) + CO2

(g)

(vi) 4 NH3

(g) + 5O2

(g) ⇌ 4NO

(g) + 6H2O(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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Concept: Phase Equilibrium — Le Chatelier’s Principle (effect of pressure on gaseous equilibria).

Only reactions where the number of moles of gaseous products differs from the number of moles of gaseous reactants are affected by pressure change. Increasing pressure shifts equilibrium toward the side with fewer gas moles.

Step 1: Count gaseous moles on each side for every reaction.

Step 2: Compare — if $\Delta n_g = 0$ , pressure has no effect.

Step 3: If $\Delta n_g \neq 0$ , increasing pressure favours the side with fewer gas moles.

Reaction	Gaseous moles (LHS)	Gaseous moles (RHS)	$\Delta n_g$	Effect of ↑ pressure
(i) COCl₂ ⇌ CO + Cl₂	1	2	+1	Backward
(ii) CH₄ + 2S₂ ⇌ CS₂ + 2H₂S	3	3	0	No effect

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Pressure affects only reactions where the number of moles of gaseous reactants differs from the number of moles of gaseous products. By Le Chatelier’s principle, increasing pressure shifts equilibrium toward the side with fewer gas moles. Here, reactions (i), (iii), (iv), (v), and (vi) are affected; (ii) is unaffected.

The Core Idea: Why Pressure Matters

Think of pressure as a way to “squeeze” the system. If you increase the pressure on a gas-phase equilibrium, the system will try to reduce that pressure by shifting to the side that occupies less volume — that is, the side with fewer total moles of gas. This is Le Chatelier’s principle in action.

The key step is always the same: count only the gaseous moles on each side. Solids and pure liquids don’t contribute because their volume doesn’t change appreciably with pressure.

aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)

Now let’s apply this to each reaction.

1. Reaction (i): $\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)$

Reactant side: 1 mole of gas (COCl₂)
Product side: 1 + 1 = 2 moles of gas (CO + Cl₂)
$\Delta n_g = 2 - 1 = +1$ (positive)

Since the product side has more gas moles, increasing pressure will shift the equilibrium backward (toward the reactant, COCl₂), where there are fewer gas molecules.

Affected? Yes. Direction: Backward.

2. Reaction (ii): $\text{CH}_4(g) + 2\text{S}_2(g) \rightleftharpoons \text{CS}_2(g) + 2\text{H}_2\text{S}(g)$

Reactant side: 1 + 2 = 3 moles of gas
Product side: 1 + 2 = 3 moles of gas
$\Delta n_g = 3 - 3 = 0$

No change in the total number of gas molecules. Pressure increase has no effect on the equilibrium position.

Affected? No.

3. Reaction (iii): $\text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g)$

Reactant side: 1 mole of gas (CO₂) — solid carbon doesn’t count
Product side: 2 moles of gas (CO)
$\Delta n_g = 2 - 1 = +1$

More gas moles on the product side. Increasing pressure shifts the equilibrium backward (toward CO₂ and solid C).

Affected? Yes. Direction: Backward.

Watch out

A common mistake is to count solid carbon as a gas mole. Don’t — only gaseous species matter for pressure effects.

4. Reaction (iv): $2\text{H}_2(g) + \text{CO}(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$

Reactant side: 2 + 1 = 3 moles of gas
Product side: 1 mole of gas (methanol)
$\Delta n_g = 1 - 3 = -2$ (negative)

Fewer gas moles on the product side. Increasing pressure shifts the equilibrium forward (toward methanol).

Affected? Yes. Direction: Forward.

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

Reactant side: 0 moles of gas (both are solids)
Product side: 1 mole of gas (CO₂)
$\Delta n_g = 1 - 0 = +1$ …

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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.Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

The Core Idea: Why Pressure Matters

1. Reaction (i): COCl2(g)⇌CO(g)+Cl2(g)\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)COCl2​(g)⇌CO(g)+Cl2​(g)

2. Reaction (ii): CH4(g)+2S2(g)⇌CS2(g)+2H2S(g)\text{CH}_4(g) + 2\text{S}_2(g) \rightleftharpoons \text{CS}_2(g) + 2\text{H}_2\text{S}(g)CH4​(g)+2S2​(g)⇌CS2​(g)+2H2​S(g)

3. Reaction (iii): CO2(g)+C(s)⇌2CO(g)\text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g)CO2​(g)+C(s)⇌2CO(g)

4. Reaction (iv): 2H2(g)+CO(g)⇌CH3OH(g)2\text{H}_2(g) + \text{CO}(g) \rightleftharpoons \text{CH}_3\text{OH}(g)2H2​(g)+CO(g)⇌CH3​OH(g)

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

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1. Reaction (i): $\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)$

2. Reaction (ii): $\text{CH}_4(g) + 2\text{S}_2(g) \rightleftharpoons \text{CS}_2(g) + 2\text{H}_2\text{S}(g)$

3. Reaction (iii): $\text{CO}_2(g) + \text{C}(s) \rightleftharpoons 2\text{CO}(g)$

4. Reaction (iv): $2\text{H}_2(g) + \text{CO}(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$

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

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