2 mol PCl_5 is heated in a 4 L closed vesse | vedclass

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(A) Initial moles of $PCl_5 = 2 \ mol$. Volume $= 4 \ L$. At equilibrium,$55\%$ of $PCl_5$ remains undissociated. Moles of $PCl_5$ at equilibrium $= 2

Vedclass · Accepted Answer

$0.184 \ mol \ L^{-1}$

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(A) Initial moles of $PCl_5 = 2 \ mol$. Volume $= 4 \ L$. At equilibrium,$55\%$ of $PCl_5$ remains undissociated. Moles of $PCl_5$ at equilibrium $= 2 	imes 0.55 = 1.1 \ mol$. Moles of $PCl_5$ dissociated $= 2 - 1.1 = 0.9 \ mol$. According to the reaction $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,moles of $PCl_3$ and $Cl_2$ formed are $0.9 \ mol$ each. Concentrations at equilibrium: $[PCl_5] = 1.1 / 4 = 0.275 \ M$,$[PCl_3] = 0.9 / 4 = 0.225 \ M$,$[Cl_2] = 0.9 / 4 = 0.225 \ M$. $K_c = ([PCl_3][Cl_2]) / [PCl_5] = (0.225 	imes 0.225) / 0.275 = 0.050625 / 0.275 \approx 0.184 \ mol \ L^{-1}$.

$2 \ mol$ $PCl_5$ is heated in a $4 \ L$ closed vessel at a definite temperature. At equilibrium,$55\%$ of $PCl_5$ remains undissociated. Find $K_c$. Reaction: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$

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