Nitric oxide reacts with Br_2 and gives nitrosyl bromide as per r… - Vidyadip

Question

Nitric oxide reacts with $Br_2$ and gives nitrosyl bromide as per reaction given below:
$2\text{NO (g) + Br}_2\text{ (g)}\leftrightharpoons2\text{NOBr (g)}$
When $0.087$ mol of $NO$ and $0.0437$ mol of $Br _2$ are mixed in a closed container at constant temperature, $0.0518$ mol of $NOBr$ is obtained at equilibrium. Calculate equilibrium amount of $NO$ and $Br _2$.

✓

Answer

The given reaction is:
$\begin{matrix}2\text{NO}_{\text{(g)}}&+&\text{Br}_{2\text{(g)}}&\leftrightarrow&2\text{NOBr}_\text{(g)}\\2\text{ mol}&&\text{1 mol}&&\text{2 mol}\end{matrix}$
Now, 2mol of NOBr are formed from 2mol of NO. Therefore, 0.0518mol of NOBr are formed from 0.0518mol of NO.
Again, 2mol of NOBr are formed from 1mol of Br.
Therefore, 0.0518mol of NOBr are formed from $\frac{0.0518}{2}$mol of Br, or 0.0259mol of NO.
The amount of NO and Br present initially is as follows:
$[NO] = 0.087mol [Br_2] = 0.0437mol$
Therefore, the amount of $NO$ present at equilibrium is:
$[NO] = 0.087 – 0.0518$
$= 0.0352mol$
And, the amount of Br present at equilibrium is:
$[Br_2] = 0.0437 – 0.0259$
$= 0.0178mol$

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