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 a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate the equilibrium amount of NO and Br2.
The given reaction is:
2NO(g)+Br2(g) ⇌ 2NOBr(g)
2mol 1mol 2mol
Now, 2 mol of NOBr is formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr is formed from 0.0518 mol of NO.
Again, 2 mol of NOBr is formed from 1 mol of Br.
Therefore, 0.0518 mol of NOBr is formed from \( \dfrac{0.0518}{2}\) mol or 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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