25℃ 평형상수(Kp) 계산. CuS(s) + H2(g) → H2S(g) + Cu(s)

25℃ 평형상수(Kp) 계산. CuS(s) + H2(g) → H2S(g) + Cu(s)

 

 

For the reaction CuS(s) + H2(g) → H2S(g) + Cu(s),

G_f (CuS) = –53.6 kJ/mol

G_f (H2S) = –33.6 kJ/mol

H_f (CuS) = –53.1 kJ/mol

H_f (H2S) = –20.6 kJ/mol

Calculate the value of the equilibrium constant (Kp)

for this reaction at 298 K.

 

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ΔG°_rxn = [생성물들의 ΔG°_f 합] – [반응물들의 ΔG°_f 합]

( 참고 https://ywpop.tistory.com/7393 )

 

= [(–33.6) + (0)] – [(–53.6) + (0)]

= +20 kJ

= +20000 J

 

 

 

K = e^[–ΔG°/RT]

( 참고 https://ywpop.tistory.com/10336 )

 

= e^[–(+20000) / ((8.314) (273+25))]

= 0.000312

 

 

 

답: 3.12×10^(-4)

 

 

 

 

[참고] Δn = 0 이므로,

Kc = Kp

( 참고 https://ywpop.tistory.com/6523 )

 

 

 

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