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Chemical Reactions - Le Chatelier’s principle

Grade 11IGCSEChemistry

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 conditions (concentration, temperature, or pressure), the system will shift its equilibrium position to counteract the change.

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Concentration: Increasing the concentration of a reactant (e.g., [A][A]) will shift the equilibrium to the right to produce more products. Decreasing a reactant will shift it to the left.

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Temperature: If the forward reaction is exothermic (ΔH<0\Delta H < 0), increasing the temperature will shift the equilibrium to the left (endothermic direction). If the forward reaction is endothermic (ΔH>0\Delta H > 0), increasing the temperature shifts it to the right.

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Pressure: A change in pressure only affects reactions involving gases. Increasing the total pressure shifts the equilibrium towards the side with the fewer number of moles of gas. Decreasing pressure shifts it towards the side with more moles of gas.

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Catalysts: A catalyst increases the rate of both the forward and backward reactions equally. Therefore, it helps reach equilibrium faster but does not change the position of the equilibrium or the yield of products.

📐Formulae

aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD

ΔH=enthalpy change (kJ/mol)\Delta H = \text{enthalpy change (kJ/mol)}

N2(g)+3H2(g)⇌2NH3(g)ΔH=−92 kJ/molN_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92 \text{ kJ/mol}

2SO2(g)+O2(g)⇌2SO3(g)ΔH=−197 kJ/mol2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197 \text{ kJ/mol}

💡Examples

Problem 1:

Predict the effect on the equilibrium position and the yield of NH3NH_3 in the Haber process: N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) (ΔH=−92 kJ/mol\Delta H = -92 \text{ kJ/mol}) if the pressure is increased.

Solution:

The equilibrium shifts to the right, and the yield of NH3NH_3 increases.

Explanation:

There are 1+3=41 + 3 = 4 moles of gas on the reactant side and 22 moles of gas on the product side. According to Le Chatelier’s Principle, increasing the pressure causes the system to shift towards the side with fewer moles of gas to reduce the pressure, which is the product side (NH3NH_3).

Problem 2:

Consider the reversible reaction: 2NO2(g)⇌N2O4(g)2NO_2(g) \rightleftharpoons N_2O_4(g) where NO2NO_2 is brown and N2O4N_2O_4 is colorless. The forward reaction is exothermic (ΔH<0\Delta H < 0). What happens to the color of the mixture if it is heated?

Solution:

The mixture becomes darker brown.

Explanation:

Since the forward reaction is exothermic, the backward reaction is endothermic. Increasing the temperature favors the endothermic reaction. Therefore, the equilibrium shifts to the left, increasing the concentration of NO2(g)NO_2(g), which results in a deeper brown color.

Le Chatelier’s principle - Revision Notes & Key Formulas | IGCSE Grade 11 Chemistry