2H2S(g) ⇌ 2H2(g) + S2(g) 평형상수. 9.28×10^-3 mol H2S

2H2S(g) ⇌ 2H2(g) + S2(g) 평형상수. 9.28×10^-3 mol H2S

 

 

A 1.00 L reaction container initially contains 9.28×10^-3 moles of H2S.

At equilibrium the concentration of H2S is 7.06×10^-3 moles.

Calculate the equilibrium constant for the reaction

2H2S(g) ⇌ 2H2(g) + S2(g)

 

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용기의 부피 = 1.00 L 이므로,

[H2S] = 9.28×10^(-3) M = 0.00928 M

 

 

ICE 도표를 작성하면,

 

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

초기(M) ... 0.00928 ....... 0 .......... 0

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

평형(M) ... 0.00928-2x .... 2x ......... x

 

 

평형에서,

0.00928 – 2x = 0.00706 이므로,

x = (0.00928 – 0.00706) / 2 = 0.00111

 

 

K = [H2]^2 [S2] / [H2S]^2

= [(2×0.00111)^2 (0.00111)] / (0.00706)^2

= 0.000109754

= 1.10×10^(-4)

 

 

답: 1.10×10^(-4)