Question
Phosphorus pentachloride dissociates as follows, in a closed reaction vessel
$$PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)$$
If total pressure at equilibrium of the reaction mixture is $$P$$ and degree of dissociation of $$PC{l_5}$$ is $$x,$$ the partial pressure of $$PC{l_3}$$ will be
A.
$$\left( {\frac{x}{{x - 1}}} \right)P$$
B.
$$\left( {\frac{x}{{1 - x}}} \right)P$$
C.
$$\left( {\frac{x}{{x + 1}}} \right)P$$
D.
$$\left( {\frac{{2x}}{{1 - x}}} \right)P$$
Answer :
$$\left( {\frac{x}{{x + 1}}} \right)P$$
Solution :
$$\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \rightleftharpoons \mathop {PC{l_3}\left( g \right)}\limits_x + \mathop {C{l_2}\left( g \right)}\limits_x $$
Total moles after dissociation $$1 - x + x + x = 1 + x$$
$${p_{PC{l_3}}} = $$ mole fraction of $$PC{l_3}$$ × Total pressure
$$\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\left( {\frac{x}{{1 + x}}} \right)P$$