Nitrosyl bromide decomposes according to the following equation - HSC - SSCE Chemistry - Question 13 - 2022 - Paper 1

At equilibrium, the concentration of NOBR is found to be 0.46 mol L$^{-1}$. The change in moles of NOBR is:

$\Delta n = 0.64 - 0.46 = 0.18 \text{ mol}$

From the balanced equation, for every 2 moles of NOBR that decompose, 2 moles of NO and 1 mole of Br2 are produced. Therefore, the relationship between NOBR decomposition and product formation is:

$\frac{2 ext{ moles NO}}{2 ext{ moles NOBR}} = \frac{1 ext{ mole Br2}}{2 ext{ moles NOBR}}$

Thus, the moles of NO produced is equal to the change in moles of NOBR: 0.18 mol produces 0.18 mol of NO. The moles of Br2 is half that value:

\text{Moles of Br2} = \frac{0.18}{2} = 0.09 \text{ mol}$$

We can now calculate the equilibrium concentrations of NO and Br2 in the flask:

• For NO: $[NO] = \frac{0.18 \text{ mol}}{1.00 \text{ L}} = 0.18 \text{ mol L}^{-1}$

• For Br2: $[Br2] = \frac{0.09 \text{ mol}}{1.00 \text{ L}} = 0.09 \text{ mol L}^{-1}$