IBr 0.025 atm이 2.0 L 용기에 들어있다면

IBr 0.025 atm이 2.0 L 용기에 들어있다면

 

 

온도 150℃에서 평형 반응 2IBr(g) ⇌ I2(g) + Br2(g) 의 Kp = 8.5×10^-3 이다.

IBr 0.025 atm이 2.0 L 용기에 들어있다면, 평형에 도달한 후 모든 물질의 부분 압력은 얼마인가?

 

 

For the equilibrium 2IBr(g) ⇌ I2(g) + Br2(g) Kp = 8.5×10^-3 at 150℃. If 0.025 atm of IBr is placed in a 2.0 L container, what is the partial pressures of all substances after equilibrium is reached?

 

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ICE 도표를 작성하면,

 

........... 2IBr(g) . ⇌ . I2(g) + Br2(g)

초기(atm) . 0.025 ........ 0 ..... 0

변화(atm) . -2x ...........+x .... +x

평형(atm) . 0.025-2x ..... x ..... x

 

 

Kp = (P_I2) (P_Br2) / (P_IBr)^2

8.5×10^-3 = (x) (x) / (0.025-2x)^2

 

8.5×10^-3 = (x)^2 / (0.025-2x)^2

8.5×10^-3 = [x / (0.025-2x)]^2

x / (0.025-2x) = [8.5×10^(-3)]^(1/2)

x = [8.5×10^(-3)]^(1/2) × 0.025 - [8.5×10^(-3)]^(1/2) × 2x

x + [8.5×10^(-3)]^(1/2) × 2x = [8.5×10^(-3)]^(1/2) × 0.025

x(1 + [8.5×10^(-3)]^(1/2) × 2) = [8.5×10^(-3)]^(1/2) × 0.025

 

x = ([8.5×10^(-3)]^(1/2) × 0.025) / (1 + [8.5×10^(-3)]^(1/2) × 2)

= 0.001946 atm = P_I2 = P_Br2

 

P_IBr = 0.025 – 2(0.001946) = 0.0211 atm

 

 

답: P_IBr = 0.021 atm, P_I2 = P_Br2 = 0.0019 atm