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 : The moles of at equilibrium is 0.274 mole.

Solution :

First we have to calculate the concentration of and .

Concentration of =

Concentration of =

Concentration of at equilibrium =

The given equilibrium reaction is,

Initially conc.     0.341   0.220            0

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

The expression of will be,

As, the concentration of at equilibrium = 0.258 M

That means,

0.341 - x = 0.258

x = 0.341 - 0.258

x = 0.083 M

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

Now we have to calculate the moles of at equilibrium.

Therefore, the moles of at equilibrium is 0.274 mole.

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