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A reaction N(2)+ 3H(2) hArr 2NH(3) + 92k...

A reaction `N_(2)+ 3H_(2) hArr 2NH_(3) + 92`k.j is at equilibrium. If the concentration of `N_(2)` is increased the temperature of the system

A

decreases

B

increases

C

remains constant

D

becomes half

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the equilibrium reaction \( N_2 + 3H_2 \rightleftharpoons 2NH_3 + 92 \, \text{kJ} \), we need to analyze the effects of increasing the concentration of \( N_2 \) on the system's temperature. ### Step-by-Step Solution: 1. **Identify the Reaction**: The reaction given is: \[ N_2 + 3H_2 \rightleftharpoons 2NH_3 + 92 \, \text{kJ} \] This indicates that the formation of ammonia (\( NH_3 \)) is an exothermic reaction, as energy (92 kJ) is released. 2. **Le Chatelier's Principle**: According to Le Chatelier's Principle, if a change is made to a system at equilibrium, the system will adjust to counteract that change and restore a new equilibrium. 3. **Effect of Increasing \( N_2 \)**: If the concentration of \( N_2 \) is increased, the system will respond by shifting the equilibrium to the right to consume the added \( N_2 \). This means more \( NH_3 \) will be produced. 4. **Temperature Consideration**: Since the reaction is exothermic, producing more \( NH_3 \) will release more heat. To counteract this increase in heat (due to the exothermic nature of the forward reaction), the system will adjust by decreasing the temperature. 5. **Conclusion**: Therefore, when the concentration of \( N_2 \) is increased, the temperature of the system will decrease in order to maintain equilibrium. ### Final Answer: The temperature of the system will decrease. ---
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