Question

$100mL$ of $P{H}_3$ on heating forms $P_4$ and $H_2$, volume changes in the reaction is:

• A. an increase of 50 mL
• B. an increase of 100 mL
• C. an increase of 150 mL
• D. a decrease of 50 mL

Answer 1

Answer: (A)

an increase of 50 mL

When 100 mL of $P{H}_3$ on heating forms $P_4$ and $H_2$, volume changes in the reaction the desired balanced equation is as :

$\underset{4mL}{\underset{4mol}{4P{H}_3\left(g\right)}}\to P_4\left(s\right)+\underset{6mL}{\underset{6mol}{6H_2\left(g\right)}}$

$\therefore 100mlP{H}_3$ will produce $=\frac{6}{4}\times 100=150mL$ (by volume)

($\because$ Considering ${P_4}_{\left(s\right)}$ will have negligible volume$

Thus, there is an increase of $50mL.$

Hence, the correct option is $A$