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:
- \([\text{NOCl}]_{eq} = 0.06 , \text{M}\)
- \([\text{NO}]_{eq} = 0.10 , \text{M}\)
- \([\text{Cl}2]{eq} = 0.10 , \text{M}\)
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