500 K PCl5(g) ⇌ PCl3(g) + Cl2(g) Kp 0.497 1.66 atm PCl5(g)

500 K PCl5(g) ⇌ PCl3(g) + Cl2(g) Kp 0.497 1.66 atm PCl5(g)

For the reaction:

PCl5(g) ⇌ PCl3(g) + Cl2(g)

the equilibrium constant Kp is 0.497 at 500.0 K.

A gas cylinder at 500.0 K is charged with PCl5(g)

at an initial pressure of 1.66 atm.

What are the equilibrium pressures of

PCl5, PCl3, and Cl2 at this temperature?

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PCl5(g) ⇌ PCl3(g) + Cl2(g)

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

( 참고 https://ywpop.tistory.com/7136 )

( 관련 예제 https://ywpop.tistory.com/17886 )

Kp = (x) (x) / (1.66–x) = 0.497

x^2 + 0.497x – (1.66)(0.497) = 0

근의 공식으로 x를 계산하면,

( 참고 https://ywpop.tistory.com/3302 )

x = 0.693 atm

---> PCl3와 Cl2의 압력

1.66–x

= 1.66–0.693

= 0.967 atm

---> PCl5의 압력

[키워드] PCl5(g) ⇌ PCl3(g) + Cl2(g) Kp 기준