Question

# Keq= 798 for the reaction: 2SO2 (g) + O2 (g) ⇄ 2SO3 (g). In a particular...

Keq= 798 for the reaction: 2SO2 (g) + O2 (g) ⇄ 2SO3 (g). In a particular mixture at equilibrium, [SO2] = 4.20 M and [SO3] = 11.0M. Calculate the equilibrium [O2] in this mixture.

H2(g) + S(s) ⇄ H2S(g) Keq= 14 0.60 moles of H2 and 1.4 moles of S are placed into a 2.0L flask and allowed to reach equilibrium. Calculate the [H2] at equilibrium.

Given,

Keq = 798

[SO2] = 4.20 M

[SO3] = 11.0 M

[O2] = ?

2 SO2 (g) + O2 (g) ⇄ 2 SO3 (g) [BALANCED]

Keq = [SO3]2 / [SO2]2[O2]

Put the values,

798 = (11.0)2 / (4.20)2[O2]

[O2] = (11.0)2 / (4.20)2 * 798

[O2] = 0.00859575241 M or 8.6 * 10-3 M [ANSWER]

Given,

Keq = 14

moles of H2 = 0.60 mol

moles of S = 1.4 mol

Volume of flask = 2.0 L

[H2] = ?

H2 (g) + S (s) ⇄ H2S (g) [BALANCED]

Keq = [H2S]/[H2][S] ------------------X

Also,

Molarity = Moles/ Volume

So, Molarity of H2S = 0.60 mol/2.0 L = 0.3 M --------A

Molarity of S = 1.4 mol /2.0 L = 0.7 M ------------B

Put Keq , A and B in X,

Keq = [H2S]/[H2][S]

14 = 0.3/[H2]*0.7

[H2] = 0.3/(14*0.7)