Favorable condition for product formation in the given reaction. $S O_{2} + \frac{1}{2} O_{2} S O_{3} (g)$

High pressure

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High temperature & low pressure

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Low temperature & high pressure

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Low temperature & low pressure

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Solution

The correct option is A High pressure
$S O_{2} + \frac{1}{2} O_{2} S O_{3} (g)$ At high pressure, forward shift So $S O_{3}$ mole are increased.

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For the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g), Δ H = - v e$ . The formation of $N H_{3} (g)$ is favored a

A_{2} (g) + B_{2} (g) ⇌ X_{2} (g) Δ_{r} H = - X k J

A_{2}

B_{2}

⇌

X_{2}

△_{r}

In the following reversible reaction $2 S O_{2} + O_{2} ⇌ 2 S O_{3} + Q C a l$

Most suitable condition for the higher production of $S O_{3}$ is

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S O_{3}

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