Question

Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
$2NO (g) + Br_2 (g) ⇋ 2NOBr (g)$
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2 .

Answer 1

The given reaction is:
$2NO(g) + Br_2(g) ↔ 2NOBr(g)$
$(2\ mol)$ $(1\ mol)$ $(2 \ mol)$
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.
Therefore, 0.0518 mol of NOBr are formed from $\frac {0.0518}{2}$ 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