PCl3 (g) + Cl2 (g) ⇌ PCl5 (g) this equation Kp and Kc solve
PCl3 (g) + Cl2 (g) ⇌ PCl5 (g) this equation Kp and Kc solve
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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)]
Kp = 1.24atm
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