25℃ 152.0 kPa N2O4(g) 101.3 kPa NO2(g) 51.9 kPa
A flask is charged with
152.0 kPa of N2O4(g) and 101.3 kPa NO2(g) at 25℃,
and the following equilibrium is achieved
N2O4(g) ⇌ 2NO2(g)
After equilibrium is reached,
the partial pressure of NO2 is 51.9 kPa.
(a) What is the equilibrium partial pressure of N2O4?
(b) Calculate the value of Kp for the reaction.
(c) Calculate Kc for the reaction.
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(a) What is the equilibrium partial pressure of N2O4?
ICE 도표를 작성하면,
( 참고 https://ywpop.tistory.com/22892 )
평형에서, NO2 is 51.9 kPa 이므로,
101.3 + 2x = 51.9
x = (51.9 – 101.3) / 2 = –24.7
152.0 – x = 152.0 – (–24.7)
= 176.7 kPa N2O4(g)
---> equilibrium partial pressure of N2O4
(b) Calculate the value of Kp for the reaction.
Kp = (P_NO2)^2 / (P_N2O4)
( 참고: 평형상수식 https://ywpop.tistory.com/7136 )
= (51.9)^2 / (176.7)
= 15.24
(c) Calculate Kc for the reaction.
Kc = Kp / (RT)^Δn
( 참고 https://ywpop.tistory.com/6523 )
Δn = 기체 생성물들의 몰수 – 기체 반응물들의 몰수
= 2 – 1 = 1
Kc = Kp / (RT)^Δn
= 15.24 / [0.08206 × (273.15 + 25)]
= 0.6229
[키워드] N2O4(g) ⇌ 2NO2(g) 평형 기준, kc kp 기준, kp kc 기준
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