Which of the following will shift the reaction $P C l_{3 (g)} + C l_{2 (g)} ⇌ P C l_{5 (g)}$ to the left side:

addition of

P C l_{3 (g)}

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increase in pressure

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decrease in pressure

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catalyst

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Solution

The correct option is B decrease in pressure
The given equilibrium :
${P C l}_{3 (g)} + {C l}_{2 (g)} ⇌ {P C l}_{5 (g)}$
Since, no. of moles of gas are greater in reactant side. So, decrease of pressure will shift the reaction to left side.

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Similar questions

{P C l}_{3} (g) + {C l}_{2} (g) ⇌ {P C l}_{5} (g), Δ H = (+) v e

P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g); Δ H = - v e

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

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

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

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