Consider the reversible industrial reaction represented by the thermochemical equation bel… — JAMB UTME

Consider the reversible industrial reaction represented by the thermochemical equation below:

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197\text{ kJ mol}^{-1}

If finely divided vanadium(V) oxide ( $V_2O_5$ ) catalyst is introduced into the reaction vessel at constant volume while the temperature of the system is lowered, which of the following best describes the combined effect on the equilibrium yield of $SO_3(g)$ and the rate of reaching equilibrium?

The yield of $SO_3(g)$ increases because the exothermic forward reaction is favored by cooling, and the equilibrium state is reached more rapidly due to the catalyst lowering activation energy.Answer
B
The yield of $SO_3(g)$ increases further because the catalyst preferentially accelerates the forward reaction, while the lower temperature slows down the overall rate.
C
The yield of $SO_3(g)$ decreases because a negative enthalpy change ($\Delta H < 0$) signifies an endothermic process that is disfavored when temperature is reduced.
D
The yield of $SO_3(g)$ remains unchanged because the catalytic effect counteracts the equilibrium shift induced by the temperature reduction.

Answer

The yield of

SO_3(g)

increases because the exothermic forward reaction is favored by cooling, and the equilibrium state is reached more rapidly due to the catalyst lowering activation energy.

Lowering the temperature favors the heat-producing (exothermic) forward reaction because $\Delta H < 0$, resulting in an increased equilibrium yield of

SO_3(g)

. Simultaneously, adding vanadium(V) oxide provides an alternative reaction pathway with lower activation energy, speeding up the attainment of dynamic equilibrium without altering the equilibrium position.

Step-by-Step Solution

Analyze the thermochemical equation for enthalpy change ($\Delta H$).

$\Delta H = -197\text{ kJ mol}^{-1}$, which confirms the forward reaction is exothermic (releases heat).

Determining whether the forward reaction is exothermic or endothermic is necessary to predict the impact of temperature changes according to Le Chatelier's principle.

Apply Le Chatelier's principle to the temperature decrease.

Lowering the temperature shifts the equilibrium in the heat-producing (exothermic forward) direction, increasing the yield of

SO_3(g)

The system counteracts the loss of thermal energy by shifting toward the side that produces heat.

Evaluate the effect of adding a catalyst (

V_2O_5

The catalyst lowers activation energy for both forward and reverse pathways, accelerating the rate at which equilibrium is attained without affecting the position of equilibrium or yield.

Catalysts increase reaction rates equally in both directions and have zero effect on thermodynamic equilibrium position.

Key Concept

Effect of temperature and catalysts on dynamic equilibrium (Le Chatelier's Principle)

Estimated Time:1m 30s

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