Question

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

Answer 1

Answer: At equilibrium , there are 0.274 moles of
`Br_2`

Step-by-step explanation:

Moles of
`H_2` = 0.682 mole

Moles of
`Br_2` = 0.440 mole

Volume of solution = 2.00 L

Initial concentration of
`H_2` =
`(0.682)/(2.00)=0.341 M`

Initial concentration of
`Br_2` =
`(0.440)/(2.00)=0.220 M`

Equilibrium concentration of
`H_2` =
`(0.516)/(2.00)=0.258 M`

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

Given : (0.341-x) M = 0.258 M

x= 0.083 M

Thus equilibrium concentartion of
`Br_2` = (0.220-0.083) M = 0.137 M

Thus moles of
`Br_2` at equilibrium =
`0.137M* 2.00L=0.274mol`

At equilibrium , there are 0.274 moles of
`Br_2`