4.0 , mol of argon and 5.0 , mol of PCl_5 are | vedclass

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(A) The reaction is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.Initial moles: $n(PCl_5) = 5.0 \, mol$,$n(Ar) = 4.0 \, mol$. Total initial moles

Vedclass · Accepted Answer

$2.25$

Vedclass · Answer

(A) The reaction is: $PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g)$.Initial moles: $n(PCl_5) = 5.0 \, mol$,$n(Ar) = 4.0 \, mol$. Total initial moles = $9.0 \, mol$.Initial pressure $(P_i)$ using $PV = nRT$: $P_i = \frac{9.0 	imes 0.082 	imes 610}{100} = 4.5 \, atm$.Initial partial pressures: $P_{PCl_5} = \frac{5}{9} 	imes 4.5 = 2.5 \, atm$,$P_{Ar} = \frac{4}{9} 	imes 4.5 = 2.0 \, atm$.At equilibrium:$PCl_5 ightleftharpoons PCl_3 + Cl_2$Initial: $2.5 \, atm \quad 0 \quad 0$Change: $-x \quad +x \quad +x$Equilibrium: $(2.5 - x) \quad x \quad x$Total pressure at equilibrium: $P_{total} = P_{PCl_5} + P_{PCl_3} + P_{Cl_2} + P_{Ar} = 6.0 \, atm$.$(2.5 - x) + x + x + 2.0 = 6.0$$4.5 + x = 6.0 \implies x = 1.5 \, atm$.Equilibrium partial pressures: $P_{PCl_5} = 2.5 - 1.5 = 1.0 \, atm$,$P_{PCl_3} = 1.5 \, atm$,$P_{Cl_2} = 1.5 \, atm$.$K_p = \frac{P_{PCl_3} 	imes P_{Cl_2}}{P_{PCl_5}} = \frac{1.5 	imes 1.5}{1.0} = 2.25$.

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