Consider the following equilibrium: PCI5(g) → PC13(g) + Cl2(g) ΔΗ 92 k The concentration of Cl2 at equilibrium may be increased by a. decreasing the temperature. adding PC13 to the system. adding PCI5 to the system. c. increasing the pressure. е. adding a catalyst.

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### Equilibrium Analysis in Chemical Reactions

#### Consider the following equilibrium reaction:

\[ \text{PCl}_5(g) \leftrightarrow \text{PCl}_3(g) + \text{Cl}_2(g) \quad \Delta H = 92 \text{kJ} \]

The concentration of \( \text{Cl}_2 \) at equilibrium can be increased by:

a. Decreasing the temperature.

b. Adding \( \text{PCl}_3 \) to the system.

c. Adding \( \text{PCl}_5 \) to the system.

d. Increasing the pressure.

e. Adding a catalyst.

#### Explanation of Options:

- **Decreasing the temperature:** This affects the equilibrium position if the reaction is exothermic or endothermic. Since the reaction is endothermic (ΔH = 92 kJ), decreasing the temperature will shift the equilibrium to the left, decreasing \( \text{Cl}_2 \) concentration.
  
- **Adding \( \text{PCl}_3 \):** According to Le Chatelier’s principle, adding a product will shift the equilibrium to the left, decreasing \( \text{Cl}_2 \).

- **Adding \( \text{PCl}_5 \):** This will shift the equilibrium to the right, increasing the concentration of \( \text{Cl}_2 \).

- **Increasing the pressure:** This depends on the number of moles of gas on each side of the equilibrium. Here, there is an equal number of moles of gas (1 mole of \( \text{PCl}_5 \) equals 1 mole of \( \text{PCl}_3 \) + 1 mole of \( \text{Cl}_2 \)), so pressure change won’t significantly affect the equilibrium position.

- **Adding a catalyst:** Catalysts speed up the reaction rate to reach equilibrium faster but do not affect the equilibrium concentrations of reactants and products.

In summary, the concentration of \( \text{Cl}_2 \) at equilibrium can be increased by option c: **adding \( \text{PCl}_5 \) to the system**.
Transcribed Image Text:### Equilibrium Analysis in Chemical Reactions #### Consider the following equilibrium reaction: \[ \text{PCl}_5(g) \leftrightarrow \text{PCl}_3(g) + \text{Cl}_2(g) \quad \Delta H = 92 \text{kJ} \] The concentration of \( \text{Cl}_2 \) at equilibrium can be increased by: a. Decreasing the temperature. b. Adding \( \text{PCl}_3 \) to the system. c. Adding \( \text{PCl}_5 \) to the system. d. Increasing the pressure. e. Adding a catalyst. #### Explanation of Options: - **Decreasing the temperature:** This affects the equilibrium position if the reaction is exothermic or endothermic. Since the reaction is endothermic (ΔH = 92 kJ), decreasing the temperature will shift the equilibrium to the left, decreasing \( \text{Cl}_2 \) concentration. - **Adding \( \text{PCl}_3 \):** According to Le Chatelier’s principle, adding a product will shift the equilibrium to the left, decreasing \( \text{Cl}_2 \). - **Adding \( \text{PCl}_5 \):** This will shift the equilibrium to the right, increasing the concentration of \( \text{Cl}_2 \). - **Increasing the pressure:** This depends on the number of moles of gas on each side of the equilibrium. Here, there is an equal number of moles of gas (1 mole of \( \text{PCl}_5 \) equals 1 mole of \( \text{PCl}_3 \) + 1 mole of \( \text{Cl}_2 \)), so pressure change won’t significantly affect the equilibrium position. - **Adding a catalyst:** Catalysts speed up the reaction rate to reach equilibrium faster but do not affect the equilibrium concentrations of reactants and products. In summary, the concentration of \( \text{Cl}_2 \) at equilibrium can be increased by option c: **adding \( \text{PCl}_5 \) to the system**.
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