Nitrosyl chloride (NOCl) is a highly toxic gas that decomposes according to the equation $$2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$$ 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

To determine the equilibrium concentration of NOCl(g), we can use the stoichiometry of the reaction. According to the balanced equation:

$2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$

From the reaction, it can be inferred that for every 2 moles of NOCl that decompose, 1 mole of Cl₂ is produced.

1. Calculate the moles of Cl₂ produced:

   • Given [Cl₂] = 0.20 M in a 1.0 L flask, the number of moles of Cl₂ at equilibrium is:

   $\text{moles of } Cl_2 = [Cl_2] \times \text{Volume} = 0.20 \text{ mol/L} \times 1\text{ L} = 0.20 \text{ mol}$

2. Relate moles of NOCl and Cl₂:

   • From the balanced equation, for each 1 mole of Cl₂ produced, 2 moles of NOCl are consumed. Therefore, the moles of NOCl that have decomposed:

   $\text{moles of } NOCl \text{ decomposed} = 2 \times \text{moles of } Cl_2 = 2 \times 0.20 = 0.40 \text{ mol}$

3. Calculate the initial moles of NOCl:

   • Initial moles of NOCl = 1.2 mol
   • Remaining moles of NOCl at equilibrium:

   $\text{Remaining moles of } NOCl = \text{Initial moles} - \text{moles decomposed} = 1.2 - 0.40 = 0.80 \text{ mol}$

4. Find the equilibrium concentration of NOCl:

   • The concentration of NOCl at equilibrium:

   $[NOCl] = \frac{\text{Remaining moles}}{\text{Volume}} = \frac{0.80 \text{ mol}}{1.0 \text{ L}} = 0.80 \text{ M}$

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