The volume of the reaction vessel containing an equilibrium mixture in the reaction is increased, when equilibrium is re-established:
$S O_{2} C l_{2} (g) ⇋ S O_{2} (g) + C l_{2} (g)$

the amount of

S O_{2}

will have decreased

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the amount of

S O_{2} C l_{2}

will have increased

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the amount of

C l_{2}

will have increased

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the amount of

C l_{2}

will have remained unchanged

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Solution

The correct option is C the amount of $C l_{2}$ will have increased
The equilibrium reaction is $S O_{2} C l_{2} (g) ⇋ S O_{2} (g) + C l_{2} (g)$ .
The forward reaction proceeds with an increase in the number of moles from 1 to 2. When the volume of the vessel is increased, the pressure decreases. The equilibrium will shift in the forward direction so as to nullify the effect of decreased pressure.
Hence, the amount of chlorine will have increased.

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Similar questions

S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g)

S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g)

S O_{2} C l_{2} (g) ⇌ S O_{2} (g) + C l_{2} (g)

S O_{2} C l_{2 (g)} ⇌ S O_{2 (g)} + C l_{2 (g)}

S O_{2} (g) + C l_{2} (g) ⇋ S O_{2} C l_{2} (l)

S O_{2}, C l_{2}

S O_{2} C l_{2}

S O_{2}

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