A 0.15 mol sample of H2S is placed in a10 L reaction vessel and heated to 1132◦C. At equilibrium, 0.03 mol H2is present. Calculate the value of Kc for the reaction2 H2S(g)⇀↽2 H2(g) + S2(g)at 1132◦C.
To calculate the value of Kc for the given reaction, we first need to determine the concentrations of each species at equilibrium.
Given:
Initial moles of H2S = 0.15 mol
Moles of H2 at equilibrium = 0.03 mol
We can assume that the volume of the reaction vessel remains constant (10 L). Therefore, the concentrations of H2S and H2 can be calculated as follows:
Concentration of H2S = Moles of H2S / Volume
= 0.15 mol / 10 L
= 0.015 mol/L
Concentration of H2 = Moles of H2 / Volume
= 0.03 mol / 10 L
= 0.003 mol/L
Since the stoichiometry of the reaction is 2 H2S ↔ 2 H2 + S2, we can determine the concentration of S2 using the stoichiometry and the concentration of H2S:
Concentration of S2 = (Concentration of H2S)^2
= (0.015 mol/L)^2
= 0.000225 mol^2/L^2
Now, we can use the concentrations of the three species to calculate Kc using the equilibrium constant expression:
Kc = [H2]^2 * [S2] / [H2S]^2
Substituting the values we calculated:
Kc = (0.003 mol/L)^2 * 0.000225 mol^2/L^2 / (0.015 mol/L)^2
Simplifying the expression:
Kc = 0.00000000405 mol^2/L^2 / 0.000225 mol^2/L^2
= 0.000018
Therefore, the value of Kc for the given reaction at 1132°C is 0.000018.
0.15 mols/10 L = 0.015 mols/L = 0.015 M
0.03 mols/10 L = 0.003 M
...............2H2S(g)⇀↽ 2H2(g) + S2(g)
I..................0.015...........0..............0
C.................-2x.............2x...............x
E................0.015-2x......2x...............x
But the problem tells you (H2) at equilibrium is 0.003. Therefore, use that to evaluate x, (from 0.015-2x =0.003) then 2x, then place those numbers into the Kc expression and solve for Kc. Post your work if you get stuck.