Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
2NO (g) + Br2 (g) ⇌ 2NOBr (g)
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at the constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2 .
The given reaction is:
`2NO_(g) + Br_(2(g)) ↔ 2NOBr_(g)`
2mol 1mol 2mol
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.
Therefore, 0.0518 mol of NOBr are formed from `0.0518/2 mol` of Br, ot
0.0259 mol of NO.
The amount of NO and Br present initially is as follows:
[NO] = 0.087 mol [Br2] = 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:
[Br2] = 0.0437 – 0.0259
= 0.0178 mol
The balanced chemical equation for the reaction is:
According to the equation, 2 moles of NO (g) react with 1 mole of Br2 (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 Br2 (g) taking part in reaction = 1/2 x 0.0518 = 0.0259 mol
No. of moles of Br2 (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) + Br2(g) ——————> 2NOBr(g)
Initial moles 0.087 0.0437 0
Moles at eqm. point: 0.0352 0.0178 0.0518