727℃ Kp 1.9 CO2 0.012 mol CO 0.025 mol

The equilibrium constant (Kp) for the reaction

C(s) + CO2(g) ⇌ 2CO(g) is 1.9 at 727℃.

What total pressure must be applied to the reacting system

to obtain 0.012 mole of CO2 and 0.025 mole of CO?

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평형에서 각 성분의 몰분율을 계산하면,

X_CO2 = 0.012 / (0.012 + 0.025) = 0.324

X_CO = 0.025 / (0.012 + 0.025) = 0.676

C(s) + CO2(g) ⇌ 2CO(g) 반응의

평형상수 식

Kp = (P_CO)^2 / (P_CO2)

( 참고 https://ywpop.tistory.com/7136 )

돌턴의 분압법칙

P_a = X_a × P_tot

( 참고 https://ywpop.tistory.com/48 )

Kp = (P_CO)^2 / (P_CO2)

= (X_CO × P_tot)^2 / (X_CO2 × P_tot)

= [(X_CO)^2 / (X_CO2)] × P_tot

P_tot = Kp / [(X_CO)^2 / (X_CO2)]

= 1.9 / [(0.676)^2 / (0.324)]

= 1.347

답: 1.3 atm

[참고]

> X_CO2 = 0.012 / (0.012 + 0.025) = 0.3243

> X_CO = 0.025 / (0.012 + 0.025) = 0.6757

> Kp = 1.9 / [(0.6757)^2 / (0.3243)] = 1.3496

[키워드] C(s) + CO2(g) ⇌ 2CO(g) 평형 기준