The volume of the reaction vessel containing an equilibrium mixture in the reaction,

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

is increased. When equilibrium is re-established :

the mass of

S O_{2} (g)

will decrease

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the mass of

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

will increase

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the mass of

C l_{2} (g)

will increase

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the mass of

C l_{2} (g)

will remain unchanged

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Solution

The correct option is B the mass of $C l_{2} (g)$ will increase
An increase in volume for the equilibrium having,

$K_{c} = \frac{[S O_{2}] [C l_{2}]}{[S O_{2} C l_{2}]} = \frac{[m o l e o f S O_{2}] [m o l e o f C l_{2}]}{[m o l e o f S O_{2} C l_{2}] \times V o l u m e}$

will increase the mole of $C l_{2}$ or $S O_{2}$ to maintain $K_{c}$ values constant.

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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}

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

C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g)

S O_{2}

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

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