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Chemical Reactions - Reversible reactions and dynamic equilibrium

Grade 11A LevelChemistry

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

🔑Concepts

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A reversible reaction is one where the products can react together to reform the original reactants. It is represented by the symbol ⇌\rightleftharpoons.

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Dynamic equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

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Le Chatelier’s Principle states that if a change (concentration, temperature, or pressure) is applied to a system at equilibrium, the system will shift its position to counteract the change.

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Increasing the temperature favors the endothermic reaction (ΔH>0\Delta H > 0) to absorb the extra heat.

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Decreasing the temperature favors the exothermic reaction (ΔH<0\Delta H < 0) to release more heat.

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Increasing the pressure favors the side with the fewer number of gaseous molecules to reduce the pressure.

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Increasing the concentration of a reactant will shift the equilibrium to the right to produce more products.

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A catalyst increases the rate of both forward and reverse reactions equally; it helps reach equilibrium faster but does not change the position of the equilibrium or the yield.

📐Formulae

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}

CuSO4⋅5H2O(s)⇌CuSO4(s)+5H2O(l)CuSO_4 \cdot 5H_2O(s) \rightleftharpoons CuSO_4(s) + 5H_2O(l)

H2O(l)⇌H+(aq)+OH−(aq)H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)

💡Examples

Problem 1:

Predict the effect on the yield of NH3NH_3 in the Haber process reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) if the pressure is increased.

Solution:

The yield of NH3NH_3 will increase.

Explanation:

According to Le Chatelier’s Principle, increasing pressure shifts the equilibrium to the side with fewer gas molecules. The reactant side has 1+3=41 + 3 = 4 moles of gas, while the product side has 22 moles of gas. Therefore, the equilibrium shifts to the right (the product side).

Problem 2:

In the exothermic reaction 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), what happens to the equilibrium position if the temperature is increased?

Solution:

The equilibrium position shifts to the left (towards the reactants).

Explanation:

Since the forward reaction is exothermic (releases heat), increasing the temperature causes the system to favor the endothermic reverse reaction to absorb the added thermal energy. This results in a lower yield of SO3SO_3.

Problem 3:

Explain why a catalyst is used in the Contact Process if it does not increase the yield of SO3SO_3.

Solution:

A catalyst is used to increase the rate of reaction.

Explanation:

While a catalyst does not shift the position of equilibrium, it provides an alternative pathway with a lower activation energy (EaE_a). This allows the system to reach the state of dynamic equilibrium much faster, making the industrial process more time-efficient.

Reversible reactions and dynamic equilibrium Grade 11 Notes & Examples