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Question: For the reaction \(PCl_{3}(g) + Cl_{2}(g)\) ⇌\(PCl_{5}\) at \(250^{o}C\), the value of \(K_{c}\) is ...

For the reaction PCl3(g)+Cl2(g)PCl_{3}(g) + Cl_{2}(g) ⇌PCl5PCl_{5} at 250oC250^{o}C, the value of KcK_{c} is 26, then the value of KpK_{p} on the same temperature will be

A

0.61

B

0.57

C

0.83

D

0.46

Answer

0.61

Explanation

Solution

∆ng = 1 – 2 = –1

∴Kp=Kc(RT)Δn\therefore K_{p} = K_{c}(RT)^{\Delta n}; ∵Kp=Kc(RT)−1\because K_{p} = K_{c}(RT)^{- 1}

Since R = 0.0821 litre atm k–1 mol–1, T = 250oC = 250 + 273

= 523 Kp=26(0.0821×523)−1=0.61K_{p} = 26(0.0821 \times 523)^{- 1} = 0.61