A sample of gaseous PCl5 was introduced into an evacuated flask so that the pressure of pure PCl5 would be 0.52 atm at 436 K.?

[匿名]

2019-02-19 2:23 pm

However, PCl5 decomposes to gaseous PCl3 and Cl2, and the actual pressure in the flask was found to be 0.88 atm. Calculate Kp for the decomposition reaction below at 436 K.

Also calculate K at this temperature.

回答 (1)

wanszeto

2019-02-19 8:50 pm

✔ 最佳答案

_______________ PCl₅(g) ____ ⇌ ____ PCl₃(g) _____ + _____ Cl₂(g) …… Kp
Initial: _________ 0.52 atm ___________ 0 atm ____________ 0 atm
Change: ________ -y atm ___________ +y atm ___________ +y atm
Equilibrium: __ (0.52 - y) atm __________ y atm ____________ y atm

Total pressure in atm at equilibrium:
(0.52 - y) + y + y = 0.88
0.52 + y = 0.88
y = 0.36

Kp = P(PCl₃) P(Cl₂) / P(PCl₅) = y² / (0.52 - y) = 0.36²/ (0.52 - 0.36) = 0.81

PV = nRT
n/V = P/(RT)
Hence, concentration = P/(RT)

K = [PCl₃][Cl₂]/[PCl₅]
= {P(PCl₃)/(RT)} {P(Cl₂)/RT}/{P(PCl₅)/RT}
= {P(PCl₃) P(Cl₂) / P(PCl₅)} / (RT)
= Kp / (RT)
= 0.81 / (0.08206 × 436)
= 0.023 (to 2 sig. fig.)

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