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Chemical Reactions - Reversible Reactions and Equilibrium

Grade 9IB

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

🔑Concepts

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A reversible reaction is a chemical reaction where the reactants form products, which can then react together to reform the original reactants. This is represented by the symbol ⇌\rightleftharpoons.

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Dynamic equilibrium is reached in a closed system when the rate of the forward reaction is exactly equal to the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant.

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A closed system is one where no reactants or products can escape, and no external matter can enter, which is a requirement for establishing equilibrium.

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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 position of equilibrium will shift to counteract that change.

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Effect of Concentration: Increasing the concentration of a reactant will shift the equilibrium to the right (towards products) to consume the added substance.

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Effect of Pressure: In reactions involving gases, increasing the total pressure will shift the equilibrium toward the side with the fewer number of moles of gas.

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Effect of Temperature: Increasing the temperature shifts the equilibrium in the direction of the endothermic reaction (absorbs heat), while decreasing it favors the exothermic reaction (releases heat).

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

📐Formulae

A+B⇌C+DA + B \rightleftharpoons C + D

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

H2O(l)⇌H(aq)++OH(aq)−H_{2}O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}

CuSO4⋅5H2O(s) (blue)⇌CuSO4(s) (white)+5H2O(l)CuSO_{4} \cdot 5H_{2}O_{(s)} \text{ (blue)} \rightleftharpoons CuSO_{4(s)} \text{ (white)} + 5H_{2}O_{(l)}

💡Examples

Problem 1:

In the Haber process, the reaction is given by: 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}. Predict the effect of increasing the temperature on the yield of ammonia (NH3NH_{3}).

Solution:

The yield of NH3NH_{3} will decrease.

Explanation:

The forward reaction is exothermic (ΔH\Delta H is negative), meaning it releases heat. According to Le Chatelier's Principle, increasing the temperature will cause the equilibrium to shift in the endothermic direction (the reverse reaction) to absorb the added heat. This results in less NH3NH_{3} being produced.

Problem 2:

For the gaseous reaction PCl5(g)⇌PCl3(g)+Cl2(g)PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}, determine the direction of the equilibrium shift if the volume of the container is decreased (increasing the pressure).

Solution:

The equilibrium shifts to the left (towards PCl5PCl_{5}).

Explanation:

Decreasing the volume increases the pressure. The system counteracts this by shifting to the side with fewer moles of gas. The reactant side has 11 mole of gas (PCl5PCl_{5}), while the product side has 22 moles of gas (1 PCl3+1 Cl21 \text{ } PCl_{3} + 1 \text{ } Cl_{2}). Therefore, the equilibrium shifts to the left.

Problem 3:

If more N2N_{2} is added to the system N2(g)+3H2(g)⇌2NH3(g)N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}, what happens to the concentration of H2H_{2}?

Solution:

The concentration of H2H_{2} will decrease.

Explanation:

Adding more N2N_{2} increases the concentration of a reactant. To counteract this, the equilibrium shifts to the right to consume the added N2N_{2}. As it shifts right, it also consumes H2H_{2} to produce more NH3NH_{3}, thereby decreasing the concentration of H2H_{2}.