Degree Of Dissociation

Question

At 500 K, for the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), if the degree of dissociation is 0.4 and initial concentration of PCl₅ is 1 M, the value of Kc is:

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Reaction: $PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g)$
- Initial concentrations: $[PCl_5]_0 = 1.0 	ext{ M}, [PCl_3] = 0, [Cl_2] = 0$
- Degree of dissociation $(\alpha) = 0.4$
At equilibrium:
- $[PCl_5] = C(1-\alpha) = 1.0(1 - 0.4) = 0.6 	ext{ M}$
- $[PCl_3] = C\alpha = 1.0 	imes 0.4 = 0.4 	ext{ M}$
- $[Cl_2] = C\alpha = 1.0 	imes 0.4 = 0.4 	ext{ M}$
Calculation of $K_c$:
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.4 	imes 0.4}{0.6} = \frac{0.16}{0.6} = 0.2667 \approx 0.267$

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Degree Of Dissociation

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