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7.9     Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

              2NO_{(g)}+Br_{2}_{(g)}\rightleftharpoons 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  .

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The initial concentration of NO and Br_2  is  0.087 mol and 0.0437 mol resp.

The given chemical reaction is-
2NO_{(g)}+Br_{2}_{(g)}\rightleftharpoons 2NOBr_{(g)}

Here, 2 mol of NOBr produces from 2 mol of NO. So, 0.0518 mol of NOBr  is obtained from 0.518 mol of NO.

Again, From 1 mol of Br_2  two mol of NOBr produced. So, to produce 0.518 mol of NOBr we need \frac{0.518}{2} = 0.0259 mol of Br_2.

Thus, the amount of NO present at equilibrium = 0.087 - 0.0518 = 0.0352 mol

and the amount of Br_2 present at the equilibrium = 0.0437-0.0259 = 0.0178 mol

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