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

Aug 27, 2016

${K}_{c} = 0.27$

#### Explanation:

THE GIVEN REVERSIBLE GASEOUS REACTION & ICE TABLE

$\text{ "PCl_5(g)" "rightleftharpoons" "PCl_3(g)" "+" } C {l}_{2} \left(g\right)$

I $\text{ "2" mol"" "" "" "" "0 " mol"" "" "" "" "0" mol}$

C $- 2 \alpha \text{ mol"" "" "" "" "2alpha " mol" " "" "" "2alpha" mol}$

E $\text{ "2(1-alpha)" mol"" "2alpha" mol"" "" "" "2alpha" " "mol}$

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

$\text{Volume of the equilibrium mixture, V = 2 L}$

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

$\left[P C {l}_{5} \left(g\right)\right] = \frac{2 \left(1 - \alpha\right)}{V} = \frac{2 \left(1 - 0.4\right)}{2} = {\text{0.6 mol·L}}^{-} 1$

$\left[P C {l}_{3} \left(g\right)\right] = \frac{2 \left(\alpha\right)}{V} = \frac{2 \times 0.4}{2} = {\text{0.4 mol·L}}^{-} 1$

$\left[C {l}_{2} \left(g\right)\right] = \frac{2 \left(\alpha\right)}{V} = \frac{2 \times 0.4}{2} = {\text{0.4 mol·L}}^{-} 1$

Equilibrium constant

$\text{ } {K}_{c} = \frac{\left[P C {l}_{3} \left(g\right)\right] \times \left[C {l}_{2} \left(g\right)\right]}{\left(P C {l}_{5} \left(g\right)\right)}$

${\text{ "=(0.4xx0.4)/0.6 " mol·L}}^{-} 1$

${\text{ "=0.27 " mol·L}}^{-} 1$