NOCl의 해리백분율과 평형상수

NOCl의 해리백분율과 평형상수

 

 

2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400 ℃. After equilibrium was established, it was found that 28% of the NOCl had dissociated according to the equation;

2NOCl(g) ⇌ 2NO(g) + Cl2(g)

 

Calculate the equilibrium constant, Kc, for the reaction.

 

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NOCl의 초기농도 = 2.50 mol / 2.50 L = 1.00 M

 

 

ICE 도표를 작성하면,

 

	2NOCl(g)	⇌	2NO(g)	+	Cl2(g)
초기(M) :	1.00		0		0
변화(M) :	-2x		+2x		+x
평형(M) :	1.00-2x		2x		x

 

 

NOCl이 28.0% 해리되었으므로, 평형에서 남아있는 NOCl의 양은

100 – 28.0 = 72.0% 이고, 이것을 몰농도로 나타내면,

1.00 M * 72.0% = 0.720 M

 

평형에서,

1.00-2x = 0.720

2x = 1.00 – 0.720 = 0.28

x = 0.28 / 2 = 0.14

 

2NOCl(g) ⇌ 2NO(g) + Cl2(g)

Kc = ([NO]^2 [Cl2]) / [NOCl]^2

= ((2*0.14)^2 (0.14)) / (0.720)^2

= 0.021

 

 

답: 0.021