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puteri [66]
3 years ago
9

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 70

0 K, there are 0.536 mol of H2 present. At equilibrium, there are ________ mol of Br2 present in the reaction vessel.
Chemistry
1 answer:
ahrayia [7]3 years ago
7 0

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.

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