2H2S(g)↔2H2(g)+S2(g)
Make ICE table
[H2S] [H2] [S2]
Initial 3.60 0 0
Change -x +x +0.5x
Equilibrium 3.60-x x 0.5 x
Kc=[H2S]2[H2]2[S2]
2.2×10−4=(3.60−x)2(x2)(0.5x)
make an assumption that 3.60-x = 3.60
2.2×10−4=(3.60)2(x2)(0.5x)
x=0.179
[H2S]=3.6−0.179=3.421M
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