2200℃ Kp 0.050 N2(g) + O2(g) ⇌ 2NO(g) 0.80 0.20 atm

2200℃ Kp 0.050 N2(g) + O2(g) ⇌ 2NO(g) 0.80 0.20 atm

 

 

평형에서 NO의 부분압력을 계산하시오.

 

 

At 2200°C, Kp = 0.050 for the reaction

N2(g) + O2(g) ⇌ 2NO(g)

What is the partial pressure of NO in equilibrium with N2 and O2

that were placed in a flask at initial pressures of 0.80 and 0.20 atm, respectively?

 

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ICE 도표를 작성하면,

 

........... N2(g) . + . O2(g) . ⇌ . 2NO(g)

초기(atm) . 0.80 ...... 0.20 ....... 0

변화(atm) . -x ........ -x ......... +2x

평형(atm) . 0.80-x .... 0.20-x ..... 2x

 

 

 

 

Kp = (P_NO)^2 / [(P_N2) (P_O2)]

 

Kp = (2x)^2 / [(0.80-x) (0.20-x)] = 0.050

 

4x^2 = 0.050(0.16 – x + x^2)

 

4x^2 = 0.0080 – 0.050x + 0.050x^2

 

3.95x^2 + 0.050x – 0.0080 = 0

 

 

 

근의 공식으로 x를 계산하면,

( 참고 https://ywpop.tistory.com/3302 )

 

x = 0.039

 

 

 

P_NO = 2x = 2(0.039) = 0.078 atm

 

 

 

답: 0.078 atm