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

Step-by-step explanation:

1) Equilibrium chemical equation:

H₂(g) + X₂(g) ⇄ 2HX(g)

2) Equilibrium constant equation:


`Kc=([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=([0.150M]^2)/([0.0500M][X_2])`

5) Solve for [X₂]


`[X_2]=(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