In the reaction PCl_5 rightleftharpoons PCl_3 | vedclass

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(C) The equilibrium constant expression for the reaction $PCl_5 ightleftharpoons PCl_3 + Cl_2$ is $K_p = \frac{p_{PCl_3} 	imes p_{Cl_2}}{p_{PCl_5}}

Vedclass · Accepted Answer

$2.4$

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(C) The equilibrium constant expression for the reaction $PCl_5 ightleftharpoons PCl_3 + Cl_2$ is $K_p = \frac{p_{PCl_3} 	imes p_{Cl_2}}{p_{PCl_5}}$.Initially,$K_p = \frac{0.3 	imes 0.2}{0.6} = \frac{0.06}{0.6} = 0.1$.When the partial pressures of $PCl_3$ and $Cl_2$ are doubled,the new partial pressures are $p'_{PCl_3} = 2 	imes 0.3 = 0.6 \ atm$ and $p'_{Cl_2} = 2 	imes 0.2 = 0.4 \ atm$.Let the new partial pressure of $PCl_5$ be $x$.Since $K_p$ remains constant at $0.1$,we have $0.1 = \frac{0.6 	imes 0.4}{x}$.Solving for $x$: $x = \frac{0.24}{0.1} = 2.4 \ atm$.

In the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,the partial pressures of $PCl_3$,$Cl_2$,and $PCl_5$ are $0.3 \ atm$,$0.2 \ atm$,and $0.6 \ atm$ respectively. If the partial pressures of $PCl_3$ and $Cl_2$ are doubled,what will be the partial pressure of $PCl_5$ in $atm$?

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