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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.
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Solution
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
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