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.

Asked by Abhisek | 1 year ago |  90

#### 1 Answer

##### Solution :-

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

Answered by Pragya Singh | 1 year ago

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