Question

For the reaction system, h2(g) + x2(g) ↔ 2 hx(g), kc = 24.4 at 300 k. a system made up from these components which is at equilibrium contains 0.200 moles of x2 and 0.600 moles of hx in a 4.00 liter container. calculate the number of moles of h2(g) present at equilibrium.

Answer 1

Answer: 0.0738 moles

Explanation:

1) Equilibrium chemical equation:

H₂(g) + X₂(g) ⇄ 2HX(g)

2) Equilibrium constant equation:

$Kc=\frac{[HX]^2}{[H_2][X_2]}$

3) Concentrations:

Molarity = number of moles of sulute / volume of solution in liters

[H₂] = 0.200 mol / 4.00 liter = 0.0500 M

[HX] = 0.600 mol / 4.00 liter = 0.150 M

4) Replace the known concentrations into the Kc equation:

$Kc=24.4=\frac{[0.150M]^2}{[0.0500M][X_2]}$

5) Solve for [X₂]

$[X_2]=\frac{0.150^2}{(24.4)(0.0500)} =0.0184M$

5) Convert the concentration to number of moles

number of moles = M×V = 0.0184M×4.00liter = 0.0738 moles

Answer 2

Kc=24.4=[HX]∧2/[H2]×[X2] =(0.6)∧2/(0.2)×[H2]
[H2] = 0.36/(24.4×0.2) = 0.07377 mole