For the reaction, PCl(3)(g)+Cl(2)(g) hAr...

For the reaction, `PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g)`, the position of equilibrium can be shifted to the right by:

Doubling the volume

Increasing the temperature

Addition of equimolar quantities of `PCl_(3) "and" PCl_(5)`

Addition of `Cl_(2)` at constant volume

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The correct Answer is:

Addition of `Cl_(2)` at constant volume will take the reaction to the right

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