Chemical EquilibriumMedium

Question

The amounts of 0.8 mol of PCl5 and 0.2 mole of PCl3 are mixed in a 1 L flask. At equilibrium, 0.4 mole of PCl3 is present. The equilibrium constant for the reaction, PCl5(g) $\rightleftharpoons$ PCl3(g) + Cl2(g) will be

Options

A.0.05 mol L–1
B.0.13 mol L–1
C.0.013 mol L–1
D.0.60 mol L–1

Solution

PCl5(g) $\rightleftharpoons$ PCl3(g) + Cl2(g)

0.8 mole 0.2 mole 0

Equilibrium 0.8 – x 0.2 + x x

= 0.6 = 0.4 = 0.2

∴ x = 0.2

Now, $K_{C} = \frac{\left\lbrack PCl_{3} \right\rbrack\left\lbrack Cl_{2} \right\rbrack}{\left\lbrack PCl_{5} \right\rbrack} = \frac{\left( \frac{0.4}{1} \right) \times \left( \frac{0.2}{1} \right)}{\left( \frac{0.6}{1} \right)} = 0.133\text{ M}$

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