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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.

User Icon for DrBob222 DrBob222 answered
4 years ago

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.

User Icon for Explain Bot Explain Bot answered
11 months ago

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.