Question

Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:
\(2NO (g) + Br_2 (g) ⇋ 2NOBr (g)\)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂ .

Answer 1

The given reaction is:
                                                                           \(2NO(g) + Br_2(g) ↔ 2NOBr(g)\)
                                                                            \((2\ mol)\)      \((1\ mol)\)     \((2 \ mol)\)
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 \(\frac {0.0518}{2}\) 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

Answer 2

\(2NO(g)+Br2​(g)⇌2NOBr(g)\)
Starting moles of \(NO=0.087\) and \(Br_2 = 0.0437.\)
At equilibrium, moles of \(NO = 2a\) and moles of \(Br_2 = a.\)
So, moles of NO at equilibrium \(=(0.087−2a)\) and moles of \(Br_2\)​ at equilibrium \(=(0.0437−a)\)
Given: \(2a=0.0518\), thus \(a = \frac{0.0518}{2} = 0.0259.\)
At equilibrium:
Moles of \(NO=0.0870−0.0518=0.0352\) mole.
Moles of \(Br_2 = 0.0437 - 0.0259 = 0.0178\) mole.