Question

For the reaction system, H2(g) + X2(g) <--> 2HX(g), Kc = 24.4 at 300 K. A system made up from these components which is at equilibrium contains 0.150 moles of H2 and 0.600 moles of HX in a 3.00 liter container. Calculate the number of moles of X2(g) present at equilibrium.

Answer 1

The number of moles of X2(g) present at equilibrium is 0.0981.

Step-by-step explanation:

Being:

aA + bB ⇔ cC + dD

the reaction constant Kc is defined as:

`Kc=([C]^(c) *[D]^(d) )/([A]^(a)*[B]^(b) )`

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Being the balanced reaction:

H₂(g) + X₂(g) ⇔ 2 HX (g)

the constant Kc is:

`Kc=([HX]^(2) )/([H_(2) ]*[X_(2) ])`

In this case, Kc = 24.4 and being
`Concentration=(number of moles of solute)/(volume)`:

• 
  `[H_(2) ]=(0.150 moles)/(3 liters)=0.05 (moles)/(liter)`

• 
  `[HX]=(0.600 moles)/(3 liters)=0.2 (moles)/(liter)`

Replacing in the definition of the constant of Kc:

`24.4=(0.2^(2) )/(0.05*[X_(2) ])`

Solving:

`[X_(2) ]=(0.2^(2) )/(0.05*24.4)`

[X₂]= 0.0327

Applying the definition of concentration
`[X_(2) ]=(number of moles of X_(2) )/(volume)`, and the volume being 3 liters:

`0.0327=(number of moles of X_(2) )/(3 liters)`

Solving:

number of moles of X₂= 0.0327* 3 liters

number of moles of X₂= 0.0981

The number of moles of X2(g) present at equilibrium is 0.0981.