Question

Unit 5 Activity 1: Intro to Equilibrium and Keq [5 marks]
1. 0.32 moles of NOCl(g) was placed in a 2 L reaction vessel and found to contain a total of 0.12 mol of NOCl
at equilibrium. Calculate the equilibrium concentrations of NOCl, NO and Cl2
. [5]
NOCl(g) ⇌ NO(g) + Cl2(g)

Answer 1

To determine the equilibrium concentrations of NOCl, NO, and Cl2, we can start with the following information:

**Initial moles of NOCl**: 0.32 moles
**Volume of the reaction vessel**: 2 L
**Equilibrium moles of NOCl**: 0.12 moles

First, we can calculate the initial concentration of NOCl:
\[
\text{Initial concentration of NOCl} = \frac{\text{moles}}{\text{volume}} = \frac{0.32 \, \text{mol}}{2 \, \text{L}} = 0.16 \, \text{M}
\]

Next, we find the change in the number of moles of NOCl as it reaches equilibrium:
\[
\text{Change in moles of NOCl} = 0.32 \, \text{mol} - 0.12 \, \text{mol} = 0.20 \, \text{mol}
\]

This change indicates that 0.20 moles of NOCl dissociated into NO and Cl2. According to the stoichiometry of the reaction:
\[
\text{NOCl}(g) ⇌ \text{NO}(g) + \text{Cl}_2(g)
\]
For every 1 mole of NOCl that dissociates, 1 mole of NO and 1 mole of Cl2 are produced.

Thus, at equilibrium:
- Moles of NO produced = 0.20 moles
- Moles of Cl2 produced = 0.20 moles

Now we can find the equilibrium concentrations.

**Equilibrium concentration of NOCl**:
\[
[\text{NOCl}]_{eq} = \frac{\text{moles of NOCl at equilibrium}}{\text{volume}} = \frac{0.12 \, \text{mol}}{2 \, \text{L}} = 0.06 \, \text{M}
\]

**Equilibrium concentration of NO**:
\[
[\text{NO}]_{eq} = \frac{\text{moles of NO at equilibrium}}{\text{volume}} = \frac{0.20 \, \text{mol}}{2 \, \text{L}} = 0.10 \, \text{M}
\]

**Equilibrium concentration of Cl2**:
\[
[\text{Cl}_2]_{eq} = \frac{\text{moles of Cl2 at equilibrium}}{\text{volume}} = \frac{0.20 \, \text{mol}}{2 \, \text{L}} = 0.10 \, \text{M}
\]

To summarize, the equilibrium concentrations are:
- \([\text{NOCl}]_{eq} = 0.06 \, \text{M}\)
- \([\text{NO}]_{eq} = 0.10 \, \text{M}\)
- \([\text{Cl}_2]_{eq} = 0.10 \, \text{M}\)