1200 K 15.0 torr 36.3 torr 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)

1200 K 15.0 torr 36.3 torr 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)

 

 

1200 K에서 다음 반응을 조사한 결과

수증기의 평형 부분 압력이 15.0 torr, 전체 평형 압력은 36.3 torr.

3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)

1200 K에서 이 반응의 Kp를 계산하라.

힌트: 돌턴의 분압법칙을 사용하라.

 

 

In a study of the reaction at 1200 K

3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)

it was observed that when the equilibrium partial pressure of water vapor

is 15.0 torr, the total pressure at equilibrium is 36.3 torr.

Calculate the value of Kp for this reaction at 1200 K.

(Hint: Apply Dalton’s law of partial pressures.)

 

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▶ 참고: 돌턴의 분압법칙

[ https://ywpop.tistory.com/48 ]

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P_total = P_H2O + P_H2 = 36.3 torr

 

 

 

P_H2 = 36.3 – 15.0 = 21.3 torr

( Dalton’s law of partial pressures )

 

 

 

Kp = (P_H2)^4 / (P_H2O)^4

= (21.3)^4 / (15.0)^4

= 4.07

 

 

 

답: Kp = 4.07

 

 

 

 

[키워드] 1200 K water vapor 15.0 torr total 36.3 torr Kp 3Fe(s) + 4H2O(g) = Fe3O4(s) + 4H2(g)

 

[FAQ] [①23/06/20]