압력 평형상수, Kp. Br2(g) + Cl2(g) ⇌ 2BrCl(g)

압력 평형상수, Kp. Br2(g) + Cl2(g) ⇌ 2BrCl(g)

 

 

온도 400 K에서 반응 Br2(g) + Cl2(g) ⇌ 2BrCl(g) 이고, 평형 상수 Kp는 7.0이다. BrCl(g)의 초기 압력이 1.00 atm일 때 평형에 도달한다면 BrCl의 최종 압력은 얼마인가?

 

For the equilibrium Br2(g) + Cl2(g) ⇌ 2BrCl(g), the equilibrium constant Kp is 7.0 at 400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is allowed to come to equilibrium, what is the final (equilibrium) pressure of BrCl?

 

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ICE table

 

	Br2(g)	+	Cl2(g)	⇌	2BrCl(g)
초기, I(atm) :	0		0		1
변화, C(atm) :	+x		+x		-2x
평형, E(atm) :	x		x		1-2x

 

 

Kp = (P_BrCl)^2 / (P_Br2) (P_Cl2)

= (1.00-2x)^2 / x^2 = 7.0

 

(1.00-2x) / x = 2.65

1.00-2x = 2.65x

4.65x = 1.00

x = 0.215

 

따라서 평형에서,

Br2 = Cl2 = 0.215 atm

BrCl = 1.00 – 2(0.215) = 0.57 atm

 

 

답: 0.57 atm