PCl3 (g) + Cl2 (g)  ⇌  PCl5 (g) this equation Kp and Kc solve

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2 Answers

I think your question is not complete

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PCl 3(g)+ Cl2(g)PCl5
Kc = 3.26 x 10-2 at 191 k

Kp = Kc(RT)∆n  ......where ∆n denotes (no. of moles in product - no. of moles in reactant)
∆n =2-1 = 1 
TK = 191 + 273 = 464 K

Kp = (3.26 x 10-2 M) [(0.0821 L atm/k mol ) ( 464 K)]
K= 1.24atm

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