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  4. 100 mL of PH3 on heating forms P and H2. The volume change in the reaction is

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Q. $100 \,mL$ of $PH_3$ on heating forms $P$ and $H_2$. The volume change in the reaction is

Equilibrium

Solution:

$\underset{4 \,vol}{ 4PH_3(g) } \rightleftharpoons 4P(s) + \underset{6 \,vol}{6H_2 (g)}$
$100\, mL \frac{100 \times 6}{4} = 150\, mL$
Change in volume $= (150 - 100) mL = 50\, mL$