Nitrosyl chloride (NOCl) is a highly toxic gas that decomposes according to the equation $$2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g)$$ To investigate the reaction, 1.2 mol of NOCl(g) is placed in an empty 1.0 L flask and allowed to reach equilibrium - VCE - SSCE Chemistry - Question 6 - 2010 - Paper 1

From the equilibrium equation: $2\text{NOCl}(g) \rightleftharpoons 2\text{NO}(g) + \text{Cl}_2(g),$ it can be seen that 1 mole of Cl$_2$ is produced for every 2 moles of NOCl that decompose.

1. Initial moles of NOCl = 1.2 mol. Since it’s in a 1.0 L flask, the initial concentration of NOCl is: $[\text{NOCl}]_{initial} = \frac{1.2 \text{ mol}}{1.0 \text{ L}} = 1.2 \text{ M}$

2. At equilibrium, [Cl$_2$] = 0.20 M implies:

   • From the stoichiometry: 0.20 moles of Cl$_2$ produced means (0.20 \times 2 = 0.40) moles of NOCl will decompose.

3. Calculate the equilibrium concentration of NOCl:

   • Change in concentration of NOCl: $[\text{NOCl}]_{equilibrium} = [\text{NOCl}]_{initial} - (0.40 \text{ M})$ $[\text{NOCl}]_{equilibrium} = 1.2 \text{ M} - 0.40 \text{ M} = 0.80 \text{ M}$

Thus, the equilibrium concentration of NOCl(g) is 0.80 M, which corresponds to option A.