Backward reaction is favoured by increase in the pressure of which of the following equilibrium?

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

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N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

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P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

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Solution

The correct option is D $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$
The increase of pressure will shift the equilibrium position in the direction of a decrease in the number of moles.
For the reaction $P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ the forward reaction occurs with an increase in the number of moles from 1 to 2 Also, the reverse reaction occurs with a decrease in the number of moles from 2 to 1
when the pressure of the system is increased, the backward reaction will be favoured as the reverse reaction occurs with a decrease in the number of moles.

Suggest Corrections

Similar questions

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)

2 H I_{(g)} ⟶ H_{2 (g)} + I_{2 (g)}

2 N H_{3 (g)} ⟶ N_{2 (g)} + 3 H_{2 (g)}

2 N O_{(g)} ⟶ N_{2 (g)} + O_{2 (g)}

P C l_{5 (g)} ⟶ P C l_{3 (g)} + C l_{2 (g)}

C O_{2} (g) + C (s) \to 2 C O (g)

Δ H = + 172.5

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

Δ H = - 91.8

N_{2} (g) + O_{2} (g) \to 2 N O (g)

Δ H = + 181

2 H_{2} O (g) \to 2 H_{2} (g) + O_{2} (g); Δ H = 484.6

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