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Equilibrium - Le Chatelier’s Principle

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium shifts in a direction that tends to counteract or undo the effect of the change.

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Effect of Concentration: Increasing the concentration of reactants or decreasing the concentration of products shifts the equilibrium to the right (forward direction). Conversely, increasing product concentration shifts it to the left (backward direction).

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Effect of Pressure: An increase in pressure shifts the equilibrium towards the side with fewer moles of gaseous species. If Δng=0\Delta n_g = 0, pressure has no effect on the equilibrium position.

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Effect of Temperature: For an exothermic reaction (ΔH<0\Delta H < 0), increasing the temperature shifts the equilibrium to the left. For an endothermic reaction (ΔH>0\Delta H > 0), increasing the temperature shifts the equilibrium to the right.

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Effect of Catalyst: A catalyst lowers the activation energy for both forward and backward reactions equally. It helps reach equilibrium faster but does not change the position of the equilibrium or the value of KeqK_{eq}.

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Addition of Inert Gas: Adding an inert gas at constant volume has no effect on equilibrium. At constant pressure, adding an inert gas shifts the equilibrium towards the side with a greater number of moles (where Δng>0\Delta n_g > 0).

📐Formulae

Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}

Δng=∑nproducts(g)−∑nreactants(g)\Delta n_g = \sum n_{products(g)} - \sum n_{reactants(g)}

log⁡(K2K1)=ΔH∘2.303R(T2−T1T1T2)\log \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)

💡Examples

Problem 1:

Consider the synthesis of ammonia by Haber's process: N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) where ΔH=−92.4 kJ/mol\Delta H = -92.4 \text{ kJ/mol}. Predict the effect of increasing pressure and decreasing temperature.

Solution:

Increasing pressure favors the side with fewer moles (4 moles reactant→2 moles product4 \text{ moles reactant} \rightarrow 2 \text{ moles product}), so it shifts right. Since the reaction is exothermic (ΔH<0\Delta H < 0), decreasing temperature shifts the equilibrium to the right to produce more heat.

Explanation:

According to Le Chatelier's Principle, the system opposes the increase in pressure by reducing volume/moles and opposes the decrease in temperature by shifting in the exothermic direction.

Problem 2:

What happens to the equilibrium PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) if the volume of the container is increased?

Solution:

Increasing the volume is equivalent to decreasing the pressure. The equilibrium will shift towards the side with more gaseous moles, which is the product side (1 mole→2 moles1 \text{ mole} \rightarrow 2 \text{ moles}).

Explanation:

The shift occurs to the right to increase the number of moles and restore the pressure balance, as Δng=(1+1)−1=1>0\Delta n_g = (1+1) - 1 = 1 > 0.

Problem 3:

In the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g), what is the effect of adding an inert gas at constant volume?

Solution:

There is no effect on the equilibrium position.

Explanation:

At constant volume, the partial pressures and molar concentrations of the reacting species remain unchanged, even though the total pressure of the system increases.

Le Chatelier’s Principle - Revision Notes & Key Formulas | ICSE Class 11 Chemistry