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?
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
To find the equilibrium value of [H2], we first need to understand how the equilibrium constant (Kc) relates to the concentrations of the reactants and products at equilibrium.
The given reaction is: 2 H2(g) + S2(g) ⇌ 2 H2S(g)
According to the law of mass action, the equilibrium constant expression (Kc) for this reaction is:
Kc = [H2S]^2 / ([H2]^2 × [S2])
Next, we are given that at equilibrium, [H2S] = [S2]^(1/2). We can substitute this expression into the Kc expression:
Kc = ([S2]^(1/2))^2 / ([H2]^2 × [S2])
Simplifying the expression:
Kc = [S2] / ([H2]^2 × [S2])
Since [S2] appears in both the numerator and denominator, it cancels out:
Kc = 1 / [H2]^2
We can rearrange the equation to solve for [H2]:
[H2]^2 = 1 / Kc
Taking the square root of both sides:
[H2] = √(1 / Kc)
Plugging in the value of Kc = 6.28 × 10^3:
[H2] = √(1 / 6.28 × 10^3)
Calculating the result:
[H2] ≈ 0.0252
Therefore, the equilibrium value of [H2] is approximately 0.0252 (mol/L).