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

Correct Answer - `0.0352"mol"` of `NO` and `0.0178"mol"` of `Br_(2)`
The given reaction
`{:(2NO_(g),,Br_(2(g)),harr,2NOBr_((g))),(2"mol",,1"mol",,2"mol"):}`
Now, 2 mol of NoBr are formed from 2 mol of NO. Therefore, `0.0518` mol of `NOBr` are formed from `0.0518` mol of `NO`.
Again, `2` mol of `NOBr` are formed from 1 mol of Br.
Again, 2 mol of NOBr are formed from 1 mol of Br.
Therefore, `0.0518` mol of NOBr are formed from
`(0.0518)/(2)` mol of `Br`, or
`0.0259` mol of `NO`.
The amount of `NO` and `Br` present initially is as follows:
`[NO] = 0.087 "mol"[Br_(2)] = 0.0437` mol
Therefore, the amount of `NO` present at equilibrium is :
`[NO] = 0.087 - 0.0518`
`= 0.0352` mol
And, the amount of Br present at equilibrium is :
`[Br_(2)] = 0.0437 - 0.0259`
`= 0.0178` mol