Question

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 0.536 mol of H2 present. At equilibrium, there are ________ mol of Br2 present in the reaction vessel.

Answer 1

Answer: 0.294 mol of $Br_2$ present in the reaction vessel.

Explanation:

Initial moles of  $H_2$ = 0.682 mole

Initial moles of  $Br_2$ = 0.440 mole

Volume of container = 2.00 L

Initial concentration of $H_2=\frac{moles}{volume}=\frac{0.682moles}{2.00L}=0.341M$

Initial concentration of $Br_2=\frac{moles}{volume}=\frac{0.440moles}{2.00L}=0.220M$

equilibrium concentration of $H_2=\frac{moles}{volume}=\frac{0.536mole}{2.00L}=0.268M$

The given balanced equilibrium reaction is,

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

Initial conc.              0.341 M     0.220 M         0  M

At eqm. conc.    (0.341-x) M      (0.220-x) M     (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[HBr]^2}{[Br_2]\times [H_2]}$

we are given : (0.341-x) = 0.268 M

x= 0.073 M

Thus equilibrium concentration of $Br_2$ = (0.220-x) M = (0.220-0.073) M = 0.147 M

$[Br_2]=\frac{moles}{volume}\\0.147=\frac{xmole}{2.00L}\\\\x=0.294 mole$

Thus there are 0.294  mol of $Br_2$ present in the reaction vessel.

Answer 2

There are 0.294 mol of Br2 present in the reaction vessel when there is the mixture of 0.682 mol of H2 and 0.440 mol of Br_2.

Calculation of the number of moles:

Since

Initial moles of H2 is 0.682 moles

Initial moles of Br_2 = 0.440  moles

And, Volume of container = 2.00 L

So here the initial concentration of H2 is = 0.682 moles / 2 = 0.341 M

The initial concentration of Br2 = 0.440/2  = 0.220 M

And, the equilibrium concentration should be = 0.536 / 2 = 0.268 M

Now

The moles should be

= 0.147 * 2

= 0.294 moles

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