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

The amount of

C l_{2}

remains unchanged.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

The amount of

C l_{2}

increases.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

The amount of

S O_{2} C l_{2}

increases.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

The amount of

S O_{2}

decreases.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B The amount of $C l_{2}$ increases.
Checking $Δ n_{g}$ for the given reaction:

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

$Δ n_{g} = 2 - 1 = 1 (i . e . Δ n_{g} > 0)$

For $Δ n_{g} > 0$ when volume is increased, $\frac{n_{g}}{V} ↓$ the reaction goes where $\frac{n_{g}}{V} ↑$ i.e. the reaction goes where $n_{g} ↑$ .
It increases in the forward direction.
So, the reaction goes in forward direction.
Hence, when equilibrium is re-established, the amount of products formed increases .
$\Rightarrow$ Amount of $C l_{2}$ increases.

Suggest Corrections

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

Join BYJU'S Learning Program