The percentage yield of as a function of time in the reaction at is given below.

Question

The percentage yield of $S O_{3} (g)$ as a function of time in the reaction $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g), Δ H_{r} < 0$ at $(P, T_{1})$ is given below.

If this reaction is conducted at $(P, T_{2})$ with $T_{2} > T_{1}$ , the percentage yield of $S O_{3}$ as a function of time is represented by

Smart Achievers · Accepted Answer

(4)

As the $T ↑$ rate of reaction increases in the beginning after same time

Being exothermic reaction equilibrium shift backward and yield decreases

Solution

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Ans.
(4)

As the $T ↑$ rate of reaction increases in the beginning after same time

Being exothermic reaction equilibrium shift backward and yield decreases

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Solution

Q. The percentage yield of SO3(g) as a function of time in the reaction SO2(g)+12O2(g)⇌SO3(g), ΔHr<0 at (P,T1) is given below. If this reaction is conducted at (P,T2) with T2>T1, the percentage yield of SO3 as a function of time is represented by

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4

Ans. (4) As the T↑ rate of reaction increases in the beginning after same time Being exothermic reaction equilibrium shift backward and yield decreases

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Ans.
(4)

As the $T ↑$ rate of reaction increases in the beginning after same time

Being exothermic reaction equilibrium shift backward and yield decreases