for the equilibrium 2H2S(g)=2H2(g)+S2(g) Kc=9.0x10^-8 at 700 degree cl the initial concerntration of the three gases are 0.270M H2S,0.270 H2 and 0.135 S2. determine the equilibrium concentrations of the three gases

Answer 1

..................2H2S(g) = => 2H2(g)+S2(g)
I..................0.270...........270......0.135
C..................+2x.............-2x..........-x
E..............0.270+2x........0.370-2x...-x

You will notice that the equilibrium tells you that the reaction must go to the left. Why? Because Qeq is too large as follows:
Qeq = (H2)^2(S2)/(H2S)^2 or
Qeq = (0.270)^2(0.135)/(0.270)^2 = 0.135
Keq = 9.0E-8. Since Qeq is so much larger than Keq that means the products are too large ad the reactants are too small so the rxn must move to the left in order to decrease the products and increase the reactants.
Plug the E line into the Keq expression above and solve for each of the quantities. Post your work if you get stuck.