One mole of PCl_{5(g)} was heated in a 1 L c | vedclass

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(B) The dissociation reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Initial moles: $1, 0, 0$Equilibrium moles: $(1-0.1), 0.1, 0.1$

Vedclass · Accepted Answer

$0.455$

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(B) The dissociation reaction is: $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Initial moles: $1, 0, 0$Equilibrium moles: $(1-0.1), 0.1, 0.1$Since the volume is $1 \ L$,the concentrations are $[PCl_{5}] = 0.9 \ M$,$[PCl_{3}] = 0.1 \ M$,and $[Cl_{2}] = 0.1 \ M$.$K_{c} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]} = \frac{0.1 	imes 0.1}{0.9} = \frac{0.01}{0.9} = 0.0111$For the reaction,$\Delta n_{g} = (1+1) - 1 = 1$.Using the relation $K_{p} = K_{c}(RT)^{\Delta n_{g}}$:$K_{p} = 0.0111 	imes (0.082 	imes 500)^{1}$$K_{p} = 0.0111 	imes 41 = 0.4551 \approx 0.455 \ atm$.

One mole of $PCl_{5(g)}$ was heated in a $1 \ L$ closed flask at $500 \ K$. At equilibrium,$0.1 \ mole$ of $Cl_{2(g)}$ was formed. What is its $K_{p}$ (in $atm$)? (Given $R=0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

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