Question

The equilibrium:
$P_4\left(g\right)+6{Cl}_2\left(g\right)\rightleftharpoons 4P{Cl}_3\left(g\right)$
is attained by mixing equal moles of $P_4$ and ${Cl}_2$ in an evacuated vessel. Then at equilibrium:

• A. $\left[{Cl}_2\right]>\left[P{Cl}_3\right]$
• B. $\left[{Cl}_2\right]>\left[P_4\right]$
• C. $\left[P_4\right]>\left[{Cl}_2\right]$
• D. $\left[P{Cl}_3\right]>\left[P_4\right]$

Answer 1

The correct option is C $\left[P_4\right]>\left[{Cl}_2\right]$

The equilibrium:
$P_4\left(g\right)+6{Cl}_2\left(g\right)\rightleftharpoons 4P{Cl}_3\left(g\right)$

t=0 1 1 0
$t=t_{eqm.}$ $1-\alpha$ $1-6\alpha$ $4\alpha$

is attained by mixing equal moles (take 1 moles of each) of $P_4$ and${Cl}_2$ in an evacuated vessel.

Then at equilibrium,

$\left[P_4\right]=\frac{1-\alpha }{V}$ and $\left[C{l}_2\right]=\frac{1-6\alpha }{V}$

$\therefore \left[P_4\right]>\left[{Cl}_2\right]$
From the reaction stoichiometry, we observe that 1 mole of $P_4$ will react with 6 moles of $C{l}_2$. So more number of moles of
$C{l}_2$ are consumed than those of $P_4$ to reach equilibrium.
Hence, the correct option is C.