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Le Chatelier's Principle for Del N Greater Than Zero

The oxidation...

Question

The oxidation of ${S O}_{2}$ by $O_{2}$ to ${S O}_{3}$ is an exothermic reaction. The yield of ${S O}_{3}$ will be increased for the exothermic reaction if:

A

temperature is increases and pressure is kept constant

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B

temperature is reduced and pressure is increased

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C

both temperature and pressure are increses

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D

both temperature and pressure are decreased

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Solution

The correct option is B temperature is reduced and pressure is increased
Given, $S O_{2} (g) + \frac{1}{2} O_{2} (g) \to S O_{3} (g)$ ; $Δ H = - v e$ (Exothermic)
$△ n_{g} = 1 - \frac{1}{2} - 1 = - \frac{1}{2}$

Since, the reaction is exothermic. So on increasing temperature, reaction goes on backward direction. Hence yield of $S O_{3}$ decreases. Also, on increasing pressure volume is decreased. So reaction goes on forward direction. Hence yield of $S O_{3}$ increases.

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Similar questions

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