In the reaction 2 H2(g) + S2(g) reverse reaction arrow 2 H2S(g), Kc = 6.28 multiplied by 103 at 900 K. What is the equilibrium value of [H2] if at equilibrium [H2S] = [S2]1/2?
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I don't get this reverse reaction arrow bit at all. Do you mean the reaction is really 2 H2S ==>2H2 + S2 or are you trying to communicate that this is a double arrow like this <---->
First calculate the molarity (M) of each species.
M H2 = moles H2 / L = 2.50 / 12.0 = 0.208 M
M S2 = moles S2 / L = 1.35 x 10^-5 / 12.0 = 1.13 x 10^-6 M
M H2S = moles H2S / L = 8.70 / 12.0 = 0.725 M
Kc = [H2S]^2 / ([H2]^2[S2]) = (0.725^2) / ((0.208)^2(1.13 x 10^-6)) = 1.08 x 10^7
M H2 = moles H2 / L = 2.50 / 12.0 = 0.208 M
M S2 = moles S2 / L = 1.35 x 10^-5 / 12.0 = 1.13 x 10^-6 M
M H2S = moles H2S / L = 8.70 / 12.0 = 0.725 M
Kc = [H2S]^2 / ([H2]^2[S2]) = (0.725^2) / ((0.208)^2(1.13 x 10^-6)) = 1.08 x 10^7
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