75.0 g PCl5(g) 3.00 L 250℃ Kp 1.80

75.0 g PCl5(g) 3.00 L 250℃ Kp 1.80

75.0 g of PCl5(g) is introduced into an evacuated 3.00 L vessel

and allowed to reached equilibruim at 250℃.

PCl5(g) ⇌ PCl3(g) + Cl2(g)

If Kp = 1.80 for this reaction,

what is the total pressure inside the vessel at equilibrium?

a) 2.88 atm

b) 2.27 atm

c) 4.54 atm

d) 7.42 atm

e) 9.69 atm

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PCl5의 몰질량 = 208.24 g/mol 이므로,

75.0 g / (208.24 g/mol) = 0.36016 mol

( 참고 https://ywpop.tistory.com/7738 )

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

P = nRT / V

= (0.36016) (0.08206) (273.15 + 250) / (3.00)

= 5.154 atm

---> PCl5의 초기 압력

Kp = (P_PCl3) (P_Cl2) / (P_PCl5)

1.80 = (x) (x) / (5.154 – x)

( 참고 https://ywpop.tistory.com/17886 )

x^2 + 1.80x – (1.80)(5.154) = 0

근의 공식으로 x를 계산하면,

( 참고 https://ywpop.tistory.com/3302 )

x = 2.276

평형에서,

> PCl3의 압력 = C2의 압력 = 2.276 atm

> PCl5의 압력 = 5.154 – 2.276 = 2.878 atm

전체 압력 = 2(2.276) + 2.878 = 7.43 atm

답: d) 7.42 atm

[키워드] PCl5(g) ⇌ PCl3(g) + Cl2(g) 기준