JEE MAIN - Chemistry (2006 - No. 5)
Phosphorus pentachloride dissociates as follows, in a closed reaction vessel
PCl5 (g) $$\leftrightharpoons$$ PCl3 (g) + Cl2 (g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be
PCl5 (g) $$\leftrightharpoons$$ PCl3 (g) + Cl2 (g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl5 is x, the partial pressure of PCl3 will be
$$\left( {{x \over {x + 1}}} \right)P$$
$$\left( {{2x \over {1 - x}}} \right)P$$
$$\left( {{x \over {x - 1}}} \right)P$$
$$\left( {{x \over {1 - x}}} \right)P$$
Explanation
$$\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \mathop {\,\rightleftharpoons\,PC{l_3}\left( g \right)}\limits_x \,\, + \,\,\mathop {C{l_2}\left( g \right)}\limits_x $$
Total moles after dissociation
$$1 - x + x + x = 1 + x$$
$${P_{PC{l_3}}} = $$ mole fraction of $$PCl{}_3 \times $$ Total pressure
$$ = \left( {{x \over {1 + x}}} \right)P$$
Total moles after dissociation
$$1 - x + x + x = 1 + x$$
$${P_{PC{l_3}}} = $$ mole fraction of $$PCl{}_3 \times $$ Total pressure
$$ = \left( {{x \over {1 + x}}} \right)P$$
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