2 moles PCl5 are heated in a 2L flask.At equilibrium 40% PCl5 dissociates . what is the equilibrium constant for this reaction?

Question

Luna Chen · Accepted Answer

#K_c = 0.27#

THE ICE TABLE AND REVERSIBLE GASEOUS REACTION GIVEN

#" "PCl_5(g)" "rightleftharpoons" "PCl_3(g)" "+" "Cl_2(g)#

I #" "2" mol"" "" "" "" "0 " mol"" "" "" "" "0" mol"#

C #-2alpha" mol"" "" "" "" "2alpha " mol" " "" "" "2alpha" mol"#

E #" "2(1-alpha)" mol"" "2alpha" mol"" "" "" "2alpha" " "mol"#

#"Where degree of dissociation, "alpha=40%=0.4#

#"Volume of the equilibrium mixture, V = 2 L"#

The mixture's components' molar concentrations at equilibrium are

#[PCl_5(g) ] =(2(1-alpha))/V=(2(1-0.4))/2= "0.6 mol·L"^-1#

#[PCl_3(g) ]=(2(alpha))/V=(2xx0.4)/2= "0.4 mol·L"^-1#

#[Cl_2(g) ]=(2(alpha))/V=(2xx0.4)/2="0.4 mol·L"^-1#

Constant of equilibrium

#" "K_c=([PCl_3(g) ]xx[Cl_2(g) ]) /[(PCl_5(g) ) ]#

#" "=(0.4xx0.4)/0.6 " mol·L"^-1#

#" "=0.27 " mol·L"^-1#

Cooper Anderson · Answer

undefined The equilibrium constant (Kc) for the reaction PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) is equal to the concentration of the products (PCl3 and Cl2) raised to their stoichiometric coefficients divided by the concentration of the reactant (PCl5) raised to its stoichiometric coefficient. Given that 40% of PCl5 dissociates, the equilibrium concentration of PCl5 is 0.6 moles/L, while the equilibrium concentrations of PCl3 and Cl2 are both 0.4 moles/L. Therefore, Kc = [(0.4)^2] / (0.6) = 0.27.