------------------ 2H2S(g) <======> 2H2(g) + S2(g) : Kc = 1.67x10-7
Initl.conc(M): 0.100 ------------------- 0.100, ----- 0.00
change (M): - 2y, --------------------- +2y, -------- +y
eqm.conc(M): (0.100 - 2y), --------- (0.100+2y), y
Kc = 1.67x10-7 = [H2(g)]2x[S2(g)] / [H2S(g)]2
=> 1.67x10-7 = (0.100+2y)2 x y / (0.100 - 2y)2
=> 4y3 + 0.4y2 + 0.01y - 1.67x10-9 = 0
On solving the above equation we get
y = 1.67x10-7 M
Hence equilibrium concentration ofS2(g), [S2(g) = y = 1.67x10-7 M (answer)
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