0.500 mol PCl5 was placed into a 0.250 L

0.500 mol PCl5 was placed into a 0.250 L

 

 

Consider the reaction

PCl3(g) + Cl2(g) ⇌ PCl5(g)

At a certain temperature,

0.500 mol of PCl5 was placed into a 0.250 L container

and allowed to react.

At equilibrium, the container held 0.100 mol of PCl5.

What is the value of K for this reaction?

 

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초기 [PCl5]

= 0.500 mol / 0.250 L = 2.00 M

 

 

 

ICE 도표를 작성하면,

 

..............PCl3(g) + Cl2(g) ⇌ PCl5(g)

초기(M): 0 ............. 0 ........... 2.00

변화(M): +x ........... +x ......... –x

평형(M): x ............. x ........... 2.00–x

 

 

 

[참고] 화학 반응식 상의

생성물인 PCl5의 초기농도가 주어졌고,

평형에서 PCl5의 농도가 감소했으므로,

역반응이 일어난 것이다.

이럴 경우, ICE 도표에서 변화(M)를 작성할 때,

반응물은 (+), 생성물은 (–)를 붙이면 된다.

 

 

 

평형 [PCl5]

= 0.100 mol / 0.250 L = 0.400 M

 

 

 

2.00 – x = 0.400

 

x = 2.00 – 0.400 = 1.60

 

 

 

K = [PCl5] / [PCl3] [Cl2]

= (2.00–x) / (x) (x)

= (0.400) / (1.60)^2

= 0.156

 

 

 

답: K = 0.156

 

 

 

 

[키워드] PCl3(g) + Cl2(g) ⇌ PCl5(g) ICE table 기준문서, 역반응 ICE table 기준문서