Question

Pure phosphine originally present at $2.5atm$ and $300K$ decomposes slowly according to the equation:
$4P{H}_3\left(g\right)\rightleftharpoons P_4\left(g\right)+6H_2\left(g\right)$
What is the vapour density of phosphine if it dissociates to the extent of 40%?

• A. 13.07
• B. 22.17
• C. 18.07
• D. 19.2

Answer 1

The correct option is A 13.07

For the reaction:
$4P{H}_3\left(g\right)\rightleftharpoons P_4\left(g\right)+6H_2\left(g\right)$

$\alpha =\frac{D-d}{d(n-1)}$
$\alpha$= dissociation constant
D = theortical vapour Density $=\frac{GMW}{2}$
$d$ = Experimetal Vapour Density
n = No. of moles of product formed.
now
for $P{H}_3⟶GMW=31+3\left(1\right)=34$
$D=\frac{GMH}{2}=\frac{34}{2}=17$
$n=\frac{7}{4},as6H_2+one{P}_4$ isformed for four moles of reactants.
Given $\alpha$= 0.4.

putting all the values,
$0.4=\frac{17-d}{d(\frac{7}{4}-1)}$
$0.4\left(3\right)\left(d\right)=(17-d)4$

$\therefore d=13.07$