H2S 10.00 atm 800 K S2 0.020 atm Kp

H2S 10.00 atm 800 K S2 0.020 atm Kp

 

 

Suppose a tank initially contains H2S

at a pressure of 10.00 atm and a temperature of 800 K.

When the reaction has come to equilibrium,

the partial pressure of S2 vapor is 0.020 atm.

Calculate Kp.

 

2H2S(g) ⇌ 2H2(g) + S2(g)

 

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ICE 도표를 작성하면,

 

........... 2H2S(g) . ⇌ . 2H2(g) . + . S2(g)

초기(atm) . 10.00 ........ 0 .......... 0

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

평형(atm) . 10.00-2x ..... 2x ......... x

 

 

 

평형에서

x = P_S2 = 0.020 atm 이므로,

 

> P_H2S = 10.00 – 2(0.020) = 9.96 atm

> P_H2 = 2(0.020) = 0.040 atm

 

 

 

Kp = (P_H2)^2 (P_S2) / (P_H2S)^2

= (0.040)^2 (0.020) / (9.96)^2

= 3.23×10^(-7) atm