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

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Answer 1 · 2016-08-27T16:45:16

$K_c = 0.27$

THE GIVEN REVERSIBLE GASEOUS REACTION & ICE TABLE

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

I $" "2" mol"" "" "" "" "0 " mol"" "" "" "" "0" mol"$

C $-2alpha" mol"" "" "" "" "2alpha " mol" " "" "" "2alpha" mol"$

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

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

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

At equilibrium the molar concentrations of the components of the mixture are

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

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

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

Equilibrium constant

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

$" "=(0.4xx0.4)/0.6 " mol·L"^-1$

$" "=0.27 " mol·L"^-1$