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At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H2 (g) + Br2 (g) ↔ 2HBr (g) A mixture of 0.682 mol of H2 and 0.440 mol of Br2 is combined in a reaction vessel with a volume of 2.00 L. At equilibrium at 700 K, there are 0.516 mol of H2 present. At equilibrium, there are __________ mol of Br2 present in the reaction vessel.

Respuesta :

Answer : The moles of [tex]Br_2[/tex] at equilibrium is 0.274 mole.

Solution :

First we have to calculate the concentration of [tex]H_2[/tex] and [tex]Br_2[/tex].

Concentration of [tex]H_2[/tex] = [tex]\frac{Moles}{Volume}=\frac{0.682mole}{2.00L}=0.341M[/tex]

Concentration of [tex]Br_2[/tex] = [tex]\frac{Moles}{Volume}=\frac{0.440mole}{2.00L}=0.220M[/tex]

Concentration of [tex]H_2[/tex] at equilibrium = [tex]\frac{Moles}{Volume}=\frac{0.516mole}{2.00L}=0.258M[/tex]

The given equilibrium reaction is,

                            [tex]H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)[/tex]

Initially conc.     0.341   0.220            0

At equilibrium.  (0.341-x) (0.220-x)     2x

The expression of [tex]K_c[/tex] will be,

[tex]K_c=\frac{[HBr]^2}{[H_2][Br_2]}[/tex]

As, the concentration of [tex]H_2[/tex] at equilibrium = 0.258 M

That means,

0.341 - x = 0.258

x = 0.341 - 0.258

x = 0.083 M

The concentration of [tex]Br_2[/tex] at equilibrium = (0.220-x) = (0.220-0.083) = 0.137 M

Now we have to calculate the moles of [tex]Br_2[/tex] at equilibrium.

[tex]\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}[/tex]

[tex]0.137M=\frac{\text{Moles of }Br_2}{2.00L}[/tex]

[tex]\text{Moles of }Br_2=0.137M\times 2.00L[/tex]

[tex]\text{Moles of }Br_2=0.274mole[/tex]

Therefore, the moles of [tex]Br_2[/tex] at equilibrium is 0.274 mole.

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