The preparation of $S O_{3}$ (g) by reaction $S O_{2} (g) + \frac{1}{2} O_{2} ⇌ S O_{3} (g)$ is an exothermic reaction. If the preparation follows the shown temperature-pressure relationship for its % yield at temperatures $T_{1}, T_{2}$ and $T_{3}$ , then the correct relation between $T_{1}, T_{2}$ and $T_{3}$ is:

T_{3} > T_{2} > T_{1}

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T_{1} > T_{2} > T_{3}

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T_{1} = T_{2} = T_{1}

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T_{1} > T_{3} > T_{2}

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Solution

The correct option is C $T_{1} > T_{2} > T_{3}$
The equilibrium reaction $S O_{2} (g) + \frac{1}{2} O_{2} ⇌ S O_{3} (g)$ is an exothermic reaction. When temperature is lowered, the forward reaction is favoured. This increases the yield of $S O_{3}$ . As seen from the graph, this will be possible when $T_{1} > T_{2} > T_{3}$ .

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The preparation of SO3 (g) by reaction SO2(g)+12O2⇌SO3(g) is an exothermic reaction. If the preparation follows the shown temperature-pressure relationship for its % yield at temperatures T1,T2 and T3, then the correct relation between T1,T2 and T3 is:

The correct option is C T1>T2>T3The equilibrium reaction SO2(g)+12O2⇌SO3(g) is an exothermic reaction. When temperature is lowered, the forward reaction is favoured. This increases the yield of SO3. As seen from the graph, this will be possible when T1>T2>T3.