Question

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.

Answer 1

Answer : The moles of
`Br_2` at equilibrium is 0.274 mole.

Solution :

First we have to calculate the concentration of
`H_2` and
`Br_2`.

Concentration of
`H_2` =
`(Moles)/(Volume)=(0.682mole)/(2.00L)=0.341M`

Concentration of
`Br_2` =
`(Moles)/(Volume)=(0.440mole)/(2.00L)=0.220M`

Concentration of
`H_2` at equilibrium =
`(Moles)/(Volume)=(0.516mole)/(2.00L)=0.258M`

The given equilibrium reaction is,

`H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)`

Initially conc. 0.341 0.220 0

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

The expression of
`K_c` will be,

`K_c=([HBr]^2)/([H_2][Br_2])`

As, the concentration of
`H_2` at equilibrium = 0.258 M

That means,

0.341 - x = 0.258

x = 0.341 - 0.258

x = 0.083 M

The concentration of
`Br_2` at equilibrium = (0.220-x) = (0.220-0.083) = 0.137 M

Now we have to calculate the moles of
`Br_2` at equilibrium.

`\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}`

`0.137M=\frac{\text{Moles of }Br_2}{2.00L}`

`\text{Moles of }Br_2=0.137M* 2.00L`

`\text{Moles of }Br_2=0.274mole`

Therefore, the moles of
`Br_2` at equilibrium is 0.274 mole.