Question

Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:

\[\ce{2NO(g) + Br2 (g) ⇌ 2NOBr (g)}\]

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 the equilibrium amount of NO and Br₂.

Answer 1

The given reaction is:

\[\ce{\underset{\text{2 mol}}{2NO_{(g)}} + \underset{1 mol}{Br_{2(g)}} ⇌ \underset{2 mol}{2NOBr_{(g)}}}\]

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, or

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