Consider the following equilibrium systems: A⇌2B ΔH o = 20.0 kJ/mol Reaction 1 A+B⇌C ΔH o = -5.4 kJ/mol Reaction 2 A⇌B ΔH o = 0.0 kJ/mol Reaction 3 Predict the change in the equilibrium constant Kc that would occur in each case if the temperature of the reacting system were raised. Reaction 1: There is no change. Kc decreases. Kc increases. Reaction 2: Kc decreases. There is no change. Kc increases.

Question

Aryan Mayer

Chemistry

13 February, 16:54

Consider the following equilibrium systems: A⇌2B ΔH o = 20.0 kJ/mol Reaction 1 A+B⇌C ΔH o = -5.4 kJ/mol Reaction 2 A⇌B ΔH o = 0.0 kJ/mol Reaction 3 Predict the change in the equilibrium constant Kc that would occur in each case if the temperature of the reacting system were raised. Reaction 1: There is no change. Kc decreases. Kc increases. Reaction 2: Kc decreases. There is no change. Kc increases.

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Erickson · Answer 1

Reaction 1: Kc increases

Reaction 2: Kc decreases

Reaction 3: The is no change

Explanation:

Let us consider the following reactions:

Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol

Reaction 2: A + B ⇌ C ΔH° = - 5.4 kJ/mol

Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol

To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that if the temperature is raised (perturbation), the system will react to lower the temperature.

Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.

Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.

Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.