For the reaction PCl_5 rightleftharpoons PCl_ | vedclass

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(A) The reaction is $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.Given moles at equilibrium: $n(PCl_5) = 2 \ mol$,$n(PCl_3) = 2 \ mol$,$n(Cl_2) =

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(A) The reaction is $PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g)$.Given moles at equilibrium: $n(PCl_5) = 2 \ mol$,$n(PCl_3) = 2 \ mol$,$n(Cl_2) = 2 \ mol$.Total moles $n_{total} = 2 + 2 + 2 = 6 \ mol$.Total pressure $P_{total} = 3 \ atm$.Partial pressure of each component is given by $p_i = (n_i / n_{total}) 	imes P_{total}$.$p(PCl_5) = (2 / 6) 	imes 3 = 1 \ atm$.$p(PCl_3) = (2 / 6) 	imes 3 = 1 \ atm$.$p(Cl_2) = (2 / 6) 	imes 3 = 1 \ atm$.$K_p = \frac{p(PCl_3) 	imes p(Cl_2)}{p(PCl_5)} = \frac{1 	imes 1}{1} = 1 \ atm$.

For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,at equilibrium,the number of moles of $PCl_5$,$PCl_3$,and $Cl_2$ are $2 \ mol$ each,and the total pressure is $3 \ atm$. What is the value of $K_p$ in $atm$?

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