140 mm pressure is developed at equilibrium when PCl5 at 100mm is subjected to dissociation. Then KP for PCl3 +Cl2⇌PCl5 is (in atm-1) nearly

StoichiometryInitial pressure 100 0 0Equilibrium pressure (100 -p) P P Given 100 - P + P + P = 140 At equilibrium Kp'for reverse reaction,

140 mm pressure is developed at equilibrium when PCl5 at 100mm is subjected to dissociation. Then KP for PCl3 +Cl2⇌PCl5 is (in atm-1) nearly

StoichiometryInitial pressure 100 0 0Equilibrium pressure (100 -p) P P Given 100 - P + P + P = 140 At equilibrium Kp'for reverse reaction,

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140 mm pressure is developed at equilibrium when PCl₅ at 100mm is subjected to dissociation. Then K_P for PCl₃+Cl₂ $⇌$ PCl₅ is (in atm^-1) nearly

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Detailed Solution

\large \mathop {PC{l_5}\left( g \right)}\limits^{1\,mole}

Given 100 - P + P + P = 140

At equilibrium

\large {P_{PC{l_3}}} = {P_{C{l_2}}} = 40mm

\large {P_{PC{l_5}}} = \left( {100 - 40} \right) = 60mm

\large {K_P} = \frac{{{P_{PC{l_3}}}.{P_{C{l_2}}}}}{{{P_{PC{l_5}}}}}

\large {K_P} = \frac{{40 \times 40}}{{60}}

\large {K_P} = \frac{{40 \times 2}}{{3}}

$K_{p}^{'}$ for reverse reaction,

\large PC{l_3} + C{l_2}\, \rightleftharpoons \,PC{l_5}

\large {K_P} = \frac{3}{{2 \times 40}} = 0.0375

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Stoichiometry	$\large \mathop {PC{l_5}\left( g \right)}\limits^{1\,mole}$	$\large \rightleftharpoons$	$\large \large \mathop {PC{l_3}\left( g \right)}\limits^{1\,mole} {\mkern 1mu} {\kern 1pt} +$	$\large {\mkern 1mu} \mathop {C{l_2}\left( g \right)}\limits^{1\,mole}$
Initial pressure	100		0	0
Equilibrium pressure	(100 -p)		P	P

140 mm pressure is developed at equilibrium when PCl5 at 100mm is subjected to dissociation. Then KP for PCl3 +Cl2⇌PCl5 is (in atm-1) nearly

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140 mm pressure is developed at equilibrium when PCl₅ at 100mm is subjected to dissociation. Then K_P for PCl₃+Cl₂ $⇌$ PCl₅ is (in atm^-1) nearly