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7.19     A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl was found to be 0.5 × 10-1 mol L-1. If value of Kc is 8.3 × 10-3, what are the concentrations of PCl3 and Cl2 at equilibrium?

                PCl_{5}_{(g)}\rightleftharpoons PCl_{3}_{(g)}+Cl_{2}_{(g)}

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We have,

concentration of PCl_5 = 0.05 mol/L
and K_c = 8.3\times 10^{-3}

Suppose the concentrations of both PCl_3 and Cl_2 at equilibrium be x mol/L. The given reaction is:

                           PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2(g)

at equilibrium       0.05                    x                     x

it is given that the value of the equilibrium constant, K_c = 8.3\times 10^{-3}

Now we can write the expression for equilibrium as:

K_c = \frac{x^2}{0.05}=8.3\times 10^{-3}
\Rightarrow x = \sqrt{4.15\times 10^{-4}}
            =0.0204 (approx)

Hence the concentration of PCl_3 and Cl_2  is 0.0204 mol / L

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