Pure PCl_5 is introduced into an evacuated ch | vedclass

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(C) The reaction is $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.Let the initial pressure of $PCl_5$ be $P_0$.At equilibrium,let the degree

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(C) The reaction is $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.Let the initial pressure of $PCl_5$ be $P_0$.At equilibrium,let the degree of dissociation be $\alpha$.The total pressure is $P_{total} = 2.0\ atm$.The equilibrium mixture contains $40\%$ $Cl_2$ by volume,which means the mole fraction of $Cl_2$ is $0.4$.Partial pressure of $Cl_2$ is $P_{Cl_2} = 0.4 	imes 2.0 = 0.8\ atm$.Since the stoichiometry is $1:1:1$,$P_{PCl_3} = P_{Cl_2} = 0.8\ atm$.The partial pressure of $PCl_5$ is $P_{PCl_5} = P_{total} - (P_{PCl_3} + P_{Cl_2}) = 2.0 - (0.8 + 0.8) = 0.4\ atm$.$K_p = \frac{P_{PCl_3} 	imes P_{Cl_2}}{P_{PCl_5}} = \frac{0.8 	imes 0.8}{0.4} = \frac{0.64}{0.4} = 1.6\ atm$.

Pure $PCl_5$ is introduced into an evacuated chamber and comes to equilibrium at $247\, ^oC$ and $2.0\ atm$. The equilibrium gaseous mixture contains $40\%$ chlorine by volume. Calculate $K_p$ at $247\, ^oC$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ in $atm$.

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