Factors Affecting Equilibria

ChemistryEquilibriumFor JEE aspirants

Le Chatelier's principle states that when a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that partially counteracts the change. Adding a reactant pushes the reaction forward; adding a product pushes it back; raising pressure shifts toward fewer gas moles; and raising temperature favours the endothermic direction. A catalyst does not shift the equilibrium; it only speeds up the approach to it. These rules power the industrial synthesis of ammonia, sulphur trioxide, and nitric oxide, and predict phase behaviour of ice, water, and diamond.

Key Rules - Quick Reference

Add reactant: shift forward. Add product: shift reverse.
Increase pressure (decrease volume) shifts equilibrium toward fewer gas moles ( $Δ n_{g}$ side).
Decrease pressure (increase volume) shifts toward more gas moles.
If $Δ n_{g} = 0$ , pressure/volume change has no effect on position of equilibrium.
Increase temperature shifts endothermic direction forward; decrease favours exothermic direction.
Van't Hoff quantitative form: $lo g \frac{K _{2}}{K _{1}} = \frac{Δ H ^{\circ}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$
Catalyst: rates rise equally, equilibrium reached faster, no shift in position.
Inert gas at constant $V$ : no shift. At constant $P$ : shifts toward more gas moles.

1. Le Chatelier's Principle - Statement

Formulated by Henry Louis Le Chatelier in 1884, this principle governs the response of an equilibrium system to any external disturbance:

Statement: If a change is applied to a system at equilibrium, the equilibrium shifts in the direction that tends to relieve the effect of the change, and a new equilibrium is established under the new conditions.

The principle is qualitative but universally applicable. Whenever we perturb temperature, pressure, volume, or concentration, we can predict the direction of shift by asking: which way makes the change smaller?

Figure 1: Le Chatelier's principle - a stress applied to an equilibrium system produces a shift that partially counteracts the stress.

2. Effect of Concentration

Adding or removing a reactant or product changes $Q$ instantly while $K$ stays the same, so the system responds by moving to restore $Q = K$ :

Add reactant: $Q$ decreases, so the reaction shifts forward to consume the added reactant.
Add product: $Q$ increases, so the reaction shifts reverse.
Remove product: $Q$ decreases, reaction shifts forward (a way to drive incomplete reactions to completion).
Add solid or pure liquid: no effect - their activities are constant.

For $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , adding $N_{2}$ or $H_{2}$ shifts the equilibrium forward and increases ammonia yield; removing ammonia (by cooling and liquefying it) also pulls the equilibrium forward.

3. Effect of Pressure and Volume

Changing volume (and hence total pressure) matters only when at least one gaseous species is present, and only when the total number of gas moles changes on going from reactants to products.

For a gas-phase reaction with

Δ n_{g} = n_{gas, products} - n_{gas, reactants}

:
Increase $P$ (decrease $V$ ): equilibrium shifts toward fewer gas moles.
Decrease $P$ (increase $V$ ): equilibrium shifts toward more gas moles.
$Δ n_{g} = 0$ : no effect.

Examples:

$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ , $Δ n_{g} = + 1$ . Higher $P$ shifts reverse (fewer gas moles on reactant side).
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , $Δ n_{g} = - 2$ . Higher $P$ shifts forward.
$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ , $Δ n_{g} = 0$ . No shift on changing $P$ or $V$ .

4. Effect of Temperature

Temperature is unique because it is the only factor that actually changes the value of $K$ itself. From the Van't Hoff equation:

lo g \frac{K _{2}}{K _{1}} = \frac{Δ H ^{\circ}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]

Le Chatelier's principle applied qualitatively:

Endothermic reaction ( $Δ H > 0$ ): heat acts as a reactant. Raising $T$ shifts equilibrium forward; $K$ increases.
Exothermic reaction ( $Δ H < 0$ ): heat acts as a product. Raising $T$ shifts equilibrium reverse; $K$ decreases.

5. Effect of Catalyst

A catalyst lowers the activation energy of both the forward and reverse reactions by exactly the same amount. Because the ratio of the two rate constants (which equals $K$ ) is unchanged, a catalyst does not shift the position of equilibrium. It only reduces the time needed to reach it.

Industrial reactors use catalysts for kinetic reasons (to reach equilibrium in a practical time), not to increase yield. The maximum yield is set by thermodynamics; the catalyst just makes that yield achievable in minutes rather than years.

6. Effect of Adding Inert Gas

An inert gas (like He or Ar) does not appear in $Q$ or $K$ , so it can only affect equilibrium indirectly through pressure or volume changes:

Condition	Effect on partial pressures	Effect on equilibrium
Constant volume	Partial pressures unchanged	No shift
Constant pressure ( $Δ n_{g} > 0$ )	Partial pressures fall, effectively increases volume	Shifts forward (toward more moles)
Constant pressure ( $Δ n_{g} < 0$ )	Partial pressures fall, effectively increases volume	Shifts reverse (toward more moles)
Constant pressure ( $Δ n_{g} = 0$ )	Partial pressures fall equally	No shift

7. Application: Vapour Pressure of Liquids

For $H_{2} O (ℓ) ⇌ H_{2} O (g)$ , $K_{p} = p_{H_{2} O}$ . The vapour pressure depends only on temperature (from $K$ 's temperature dependence) and is independent of the amount of liquid present. At 25 °C the vapour pressure of water is about 24 mm Hg. Relative humidity is defined as the ratio of the actual partial pressure of water in air to the saturated vapour pressure at that temperature.

8. Application: Solid-Liquid Phase Equilibria

Formation of diamond from graphite:

C (graphite) ⇌ C (diamond)

;

Δ H > 0

(endothermic); diamond has higher density (lower molar volume).

By Le Chatelier: high pressure favours diamond (lower volume), and high temperature favours diamond (endothermic). Diamonds form deep in the Earth's mantle where both conditions are met.

Melting of ice:

H_{2} O (s) ⇌ H_{2} O (ℓ)

;

Δ H > 0

(endothermic); liquid water is denser than ice.

High pressure favours liquid water (lower volume). High temperature favours liquid water (endothermic). This is why ice under a skate blade melts to reduce friction.

9. Application: Boiling of Water

$H_{2} O (ℓ) ⇌ H_{2} O (g)$ . Vapour has much higher volume than liquid. Increasing external pressure shifts the equilibrium backward toward liquid, so boiling requires a higher temperature at higher altitude pressures. Pressure cookers exploit this by raising the boiling point of water to about 120 °C.

10. Application: Haber Process for Ammonia

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); Δ H = - 92 kJ mol^{- 1}

Analysis:

Pressure: $Δ n_{g} = 2 - 4 = - 2$ . Higher $P$ shifts forward. Industrial plants use ~200 atm.
Temperature: Exothermic, so lower $T$ gives higher $K$ and better yield. But lower $T$ also makes the reaction painfully slow. Industry compromises at ~450 °C.
Catalyst: Finely divided iron (with $K_{2} O$ and $A l_{2} O_{3}$ promoters) speeds the reaction without shifting equilibrium.
Concentration: Removing $N H_{3}$ by liquefaction pulls the equilibrium forward, allowing continuous conversion.

11. Application: Contact Process for SO₃

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g); Δ H = - 190 kJ mol^{- 1}

Favourable conditions:

$Δ n_{g} = - 1$ , so high $P$ shifts forward. Industrial plants operate near 1.5 to 2 atm because equilibrium yield is already very high at moderate pressure.
Exothermic, so lower $T$ gives higher $K$ . Compromise at ~450 °C balances yield and rate.
Excess $O_{2}$ pushes equilibrium forward.
Catalyst: vanadium pentoxide ( $V_{2} O_{5}$ ) at ~450 °C.

12. Application: Birkeland-Eyde Process for NO

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g); Δ H = + 180 kJ mol^{- 1}

Favourable conditions:

$Δ n_{g} = 0$ , so pressure has no effect on equilibrium position.
Endothermic, so high $T$ shifts forward. Industrial process uses electric arcs at 2000 to 3000 °C.
Excess $N_{2}$ or $O_{2}$ shifts forward.

Historically important as the first industrial route to fix atmospheric nitrogen, but the enormous energy cost made it uneconomical once the Haber-Bosch process became viable.

13. Application: CuSO₄·5H₂O and Relative Humidity

C u S O_{4} \cdot 5 H_{2} O (s) ⇌ C u S O_{4} \cdot 3 H_{2} O (s) + 2 H_{2} O (g); K_{p} = (p_{H_{2} O})^{2}

At 25 °C, $K_{p} = 1.086 \times 1 0^{- 4}$ atm $^{2}$ , so equilibrium water vapour pressure is $1.086 \times 1 0^{- 4} \approx 1.04 \times 1 0^{- 2}$ atm $\approx$ 8 mm Hg. If ambient humidity gives $p_{H_{2} O} > 8$ mm Hg, the equilibrium shifts left and the trihydrate absorbs water to become the pentahydrate. Below 8 mm Hg, the pentahydrate effloresces (loses water) to form the trihydrate. Since saturation vapour pressure at 25 °C is 24 mm Hg, efflorescence occurs when relative humidity drops below $8/24 \approx 33%$ .

14. Summary Table of Factors

Change applied	Direction of shift	Change in $K$ ?
Add reactant	Forward	No
Add product	Reverse	No
Increase $P$ (decrease $V$ )	Toward fewer gas moles	No
Decrease $P$ (increase $V$ )	Toward more gas moles	No
Increase $T$ (exothermic)	Reverse	$K$ decreases
Increase $T$ (endothermic)	Forward	$K$ increases
Add catalyst	No shift	No
Add inert gas at constant $V$	No shift	No
Add inert gas at constant $P$	Toward more gas moles (if any)	No

Solved Example 1

For

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

, the volume of a closed reaction vessel is halved. Predict the effect on the forward and reverse rates and on the equilibrium position.

Solution:

Halving the volume doubles all concentrations. In the forward-rate expression $r_{f} = k_{f} [S O_{2}]^{2} [O_{2}]$ , all three concentrations double, so $r_{f}$ increases by $2^{2} \times 2 = 8$ times. In the reverse expression $r_{r} = k_{r} [S O_{3}]^{2}$ , $r_{r}$ increases by $2^{2} = 4$ times.

$r_{f}$ now exceeds $r_{r}$ , so the equilibrium shifts forward (toward fewer gas moles). This matches Le Chatelier: $Δ n_{g} = 2 - 3 = - 1$ , so higher pressure favours the forward direction.

Solved Example 2

Under what conditions will the following reactions be favoured in the forward direction?
(i)

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) + 23

kcal
(ii)

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g) + 45

kcal
(iii)

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g) - 43.2

kcal
(iv)

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g) - X

kcal

Solution:

(i) Exothermic, $Δ n_{g} = - 2$ : low $T$ , high $P$ , excess $N_{2}$ and $H_{2}$ .
(ii) Exothermic, $Δ n_{g} = - 1$ : low $T$ , high $P$ , excess $S O_{2}$ and $O_{2}$ .
(iii) Endothermic, $Δ n_{g} = 0$ : high $T$ , any $P$ , excess $N_{2}$ and $O_{2}$ .
(iv) Endothermic, $Δ n_{g} = + 1$ : high $T$ , low $P$ , excess $P C l_{5}$ .

Solved Example 3

A system at equilibrium:

P C l_{5} (g) + heat ⇌ P C l_{3} (g) + C l_{2} (g)

. Why does the temperature of the system decrease when

P C l_{3}

is being removed at constant volume?

Solution:

Removing $P C l_{3}$ lowers $Q$ , so the reaction shifts forward. Since the forward reaction is endothermic, it absorbs heat from the surroundings (which is the system itself, if adiabatic), causing the temperature to fall.

Solved Example 4

For

N H_{4} I (g) ⇌ N H_{3} (g) + H I (g)

Δ H > 0

. What is the effect on

K

of increasing temperature?

Solution:

The forward reaction is endothermic. By Le Chatelier (and Van't Hoff quantitatively), increasing $T$ shifts the reaction forward, so $K$ increases.

Solved Example 5

K_{p}

for

2 B a O_{2} (s) ⇌ 2 B a O (s) + O_{2} (g)

1.6 \times 1 0^{- 4}

atm at 400 °C, with

Δ H = - 25.14

kcal mol

^{- 1}

. Find

K_{p}

at 500 °C, and the moles of

O_{2}

produced in a 2 L vessel at that temperature.

Solution:

Van't Hoff: $lo g \frac{K _{2}}{K _{1}} = \frac{Δ H ^{\circ}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$

$= \frac{- 25140}{2.303 \times 2} [\frac{1}{673} - \frac{1}{773}] = \frac{- 25140}{4.606} \times \frac{100}{520229} \approx - 1.05$

$K_{2} = 1.6 \times 1 0^{- 4} \times 1 0^{- 1.05} = 1.46 \times 1 0^{- 5}$ atm $= p_{O_{2}}$ .

$n_{O_{2}} = \frac{p _{O_{2}} V}{R T} = \frac{1.46 \times 1 0 ^{- 5} \times 2}{0.0821 \times 773} \approx 4.6 \times 1 0^{- 7}$ mol.

Solved Example 6

In the Haber synthesis

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, an inert gas (

A r

) is added at (a) constant volume, (b) constant total pressure. Predict the effect on ammonia yield in each case.

Solution:

(a) At constant $V$ , partial pressures of $N_{2}$ , $H_{2}$ , and $N H_{3}$ are unchanged, so $Q$ is unchanged and the equilibrium does not shift. Yield is unaffected.

(b) At constant $P$ , adding $A r$ requires the volume to expand, which lowers the partial pressures of all reacting species. Since $Δ n_{g} = - 2$ (forward reaction reduces gas moles), the shift is toward the side with more moles, i.e., reverse. Ammonia yield decreases.

Solved Example 7

C u S O_{4} \cdot 5 H_{2} O (s) ⇌ C u S O_{4} \cdot 3 H_{2} O (s) + 2 H_{2} O (g)

with

K_{p} = 1.086 \times 1 0^{- 4}

atm

^{2}

at 25 °C. Vapour pressure of water at 25 °C is 24 mm Hg. Below what relative humidity will

C u S O_{4} \cdot 5 H_{2} O

lose water?

Solution:

$K_{p} = (p_{H_{2} O})^{2}$ , so $p_{H_{2} O} = 1.086 \times 1 0^{- 4} = 1.04 \times 1 0^{- 2}$ atm $\approx 8$ mm Hg.

Efflorescence (loss of water) occurs when ambient $p_{H_{2} O} < 8$ mm Hg, i.e., relative humidity $< 8/24 \approx 33%$ .

Common Mistakes to Avoid

Watch out

Believing a catalyst shifts equilibrium. It only speeds up the approach; the position of equilibrium is untouched.
Applying the pressure rule to reactions where $Δ n_{g} = 0$ . Pressure changes have no effect on the equilibrium position in that case.
Treating "adding inert gas" as always doing nothing. It depends on whether $V$ or $P$ is held constant.
Forgetting that only temperature changes the value of $K$ . All other factors change the position of equilibrium but not $K$ .
Reasoning that "the reaction is exothermic, so high $T$ is bad, so use very low $T$ ". Very low $T$ makes the rate too slow to be useful; industrial processes always compromise.
Including pure solids and pure liquids when counting $Δ n_{g}$ . Only gaseous species count.
Confusing effect of pressure with effect of concentration. Adding more of a gaseous reactant is a concentration change, not a pressure change, if $V$ increases proportionally.
Assuming that adding a reactant always improves yield significantly. Le Chatelier gives direction, not magnitude - the size of the shift depends on how far from equilibrium the system was pushed.

Frequently Asked Questions

Q1. What is Le Chatelier's principle in simple words?

When you disturb a system that is at equilibrium (by changing concentration, pressure, or temperature), the system responds by shifting in whichever direction partially undoes the disturbance. So if you add a reactant, the reaction eats it up by shifting forward; if you heat an exothermic reaction, the equilibrium moves back toward reactants.

Q2. Does a catalyst change the position of equilibrium?

No. A catalyst lowers the activation energy of the forward and reverse reactions by the same amount, so both rates increase equally. The ratio of the rate constants (which equals $K$ ) is unchanged, and so is the position of equilibrium. The only thing a catalyst changes is the time needed to reach equilibrium.

Q3. Why does adding an inert gas sometimes shift equilibrium and sometimes not?

Inert gas does not appear in $Q$ or $K$ , so it can only affect equilibrium indirectly by changing partial pressures. At constant volume, the partial pressures of reacting species are unchanged and there is no shift. At constant pressure, the volume must increase to accommodate the inert gas, which lowers the partial pressures of the reacting species - equivalent to expanding the system, so equilibrium shifts toward more gas moles.

Q4. How does temperature affect equilibrium?

Temperature is the only factor that changes $K$ itself. For endothermic reactions ( $Δ H > 0$ ), increasing $T$ increases $K$ and shifts equilibrium forward. For exothermic reactions ( $Δ H < 0$ ), increasing $T$ decreases $K$ and shifts equilibrium reverse. The Van't Hoff equation gives the quantitative relation.

Q5. Why does the Haber process use only moderate temperature and pressure?

Ammonia synthesis is exothermic, so thermodynamics favours low $T$ ; and $Δ n_{g} = - 2$ , so thermodynamics favours high $P$ . But at low $T$ the reaction is too slow to be practical, and very high $P$ requires expensive reactor construction. Industry compromises at about 450 °C and 200 atm with an iron catalyst - fast enough to be economic while still giving useful yield.

Q6. What happens if we add more product to a system at equilibrium?

Adding more product increases $Q$ above $K$ , so the reaction shifts in reverse to consume the added product and reduce $Q$ back to $K$ . A new equilibrium is reached with product concentration higher than before but reactant concentrations also higher.

Q7. How can we drive an incomplete reaction to near-completion?

By continuously removing a product as it forms (so $Q$ never approaches $K$ ), the reaction keeps shifting forward. Distilling off water in an esterification, or liquefying $N H_{3}$ in the Haber process, or venting $C O_{2}$ from a decomposition all use this principle. Excess of a cheap reactant can also push the reaction forward.

Q8. Does increasing pressure always favour forward reaction?

No. Increasing pressure shifts equilibrium toward the side with fewer gas moles. If the forward reaction reduces gas moles ( $Δ n_{g} < 0$ ), higher pressure favours it. If the forward reaction increases gas moles ( $Δ n_{g} > 0$ ), higher pressure favours the reverse. If $Δ n_{g} = 0$ , pressure has no effect.

Q9. Why does ice melt under pressure?

For $H_{2} O (s) ⇌ H_{2} O (ℓ)$ , liquid water has higher density and lower volume than ice. Increasing pressure shifts equilibrium toward the lower-volume liquid side, so ice melts. This is one of the few substances that behaves this way; most solids are denser than their liquids and are stabilised by pressure.

Previous year questions on Factors Affecting Equilibria

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