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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