The equilibrium constant for the reaction given below at 400 K, if ∆H° = 77.2KJ/mol and ∆S° = 122JK⁻¹ mol⁻¹, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
The equilibrium constant for the reaction given below at 400 K, if ∆H° = 77.2KJ/mol and ∆S° = 122JK⁻¹ mol⁻¹, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
From ΔG° = ΔH° – TΔS°
⇒ – 2.303RTlog Kc = ΔH° – TΔS° = 77.2 * 10³ – 400 * 122 = 28400 J
So log K? = (28400 / (2.303*8.314*400) ∴ [Kc = 1.958 * 10? ]
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