The value of $K_{c}$ for the reaction, $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ ; will increase if the volume of reaction vessel is increased.
If true enter 1, if false enter 0.

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Solution

$Δ n = 2 - (2 + 1) = - 1$ .
Thus, the reaction proceeds with decrease in the number of moles of gaseous species.
When the volume of the reaction vessel is increased, the pressure of the system decreases, the equilibrium will shift in the backward direction so that the effect of decreased pressure is nullified. Hence, the value of the equilibrium constant will decrease.

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