Question

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below:
`2NO_((g))+Br_(2(g))hArr2NOBr_((g))`.
When `0.087 "mole"` of `NO` and `0.0437 "mole"` of `Br_(2)` are mixed in a closed container at constant temperature, `0.0518 "mole"` of `NOBr` is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

Answer 1

The balanced chemical equation for the reaction is :
`2NO(g) + Br_(2)(g) hArr 2NOBr(g)`
According to the equation , 2 moles of NO (g) react with 1 mole of `Br_(2)(g)` to form 2 moles of `NOBr(g)`. The composition of the equilibrium mixture can be calculated as follows :
No. of moles of NOBr(g) formed at equilibrium =0.0518 mol (given)
No. of moles of NO(g) taking part in reaction =0.0518 mol
No. of moles of NO(g) left at equilibrium =0.087 - 0.0518 =0.0352 mol
No. of moles of `Br_(2)(g)` taking part in reaction `=1//2 xx 0.0518 =0.0259`mol
No. moles of `Br_(2) (g)` left at equilibrium =0.0437 -0.0259 =0.0178 mol
the initial molar concentration and equilibrium molar concentration of different species may be represented as :
`{:(,2NO(g),+,Br_(2)(g),hArr,2NOBr(g)),("Initial moles" ,0.087,,0.0437,,0),("Moles at eqm. point" ,0.0352,,0.0178,,0.0518):}`