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For the gas phase exothermic reaction, A...

For the gas phase exothermic reaction,` A(g) + 2B(g)

A

Decreasing the temperature

B

Increasing the pressure

C

Adding inert yas at constant pressure

D

Removing C(g) at equilibrium

Text Solution

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The correct Answer is:
To solve the question regarding the gas phase exothermic reaction \( A(g) + 2B(g) \rightleftharpoons C(g) \) and how to increase the equilibrium moles of \( A \), we can follow these steps: ### Step 1: Understand the Reaction The given reaction is exothermic, meaning it releases heat. The reaction can be represented as: \[ A(g) + 2B(g) \rightleftharpoons C(g) \] ### Step 2: Apply Le Chatelier's Principle Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift in the direction that counteracts the change. ### Step 3: Analyze the Options 1. **Decreasing the Temperature**: For an exothermic reaction, decreasing the temperature shifts the equilibrium to the right (towards the products). This would not increase the moles of \( A \). 2. **Increasing the Pressure**: Increasing the pressure shifts the equilibrium towards the side with fewer moles of gas. In this case, there are 3 moles of reactants (1 mole of \( A \) and 2 moles of \( B \)) and 1 mole of product (\( C \)). Thus, increasing the pressure would shift the equilibrium to the right, again not increasing the moles of \( A \). 3. **Adding Inert Gas at Constant Pressure**: Adding an inert gas at constant pressure does not affect the partial pressures of the reactants and products. Therefore, it does not shift the equilibrium position and would not increase the moles of \( A \). 4. **Removing \( C \) at Equilibrium**: If \( C \) is removed from the system, the concentration of \( C \) decreases. According to Le Chatelier's Principle, the equilibrium will shift to the right to produce more \( C \), which means it will consume \( A \) and \( B \). This option does not increase the moles of \( A \). ### Conclusion After analyzing all the options, the only way to increase the moles of \( A \) is by removing \( C \) at equilibrium, which shifts the equilibrium to the right. However, this option does not actually increase the moles of \( A \) but rather decreases it. Therefore, none of the options provided would increase the moles of \( A \) directly. ### Final Answer None of the options effectively increase the equilibrium moles of \( A \).
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