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Sagot :
To solve the problem of finding the \(\Delta H\) in \( \text{kJ} \) of heat released per mole of \( \text{NH}_3 \) formed in the Haber Process, let's break it down step by step.
Given:
[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) + 100.4 \; \text{kJ} \][/tex]
This tells us that the reaction yields 100.4 kJ of energy when 2 moles of \( \text{NH}_3 \) are produced. We need to find the heat released per mole of \( \text{NH}_3 \).
Step-by-Step Solution:
1. Determine the total heat released for the reaction:
The overall reaction releases 100.4 kJ of energy.
2. Identify the amount of \( \text{NH}_3 \) produced:
According to the equation, 100.4 kJ is released when 2 moles of \( \text{NH}_3 \) are formed.
3. Calculate the heat released per mole of \( \text{NH}_3 \):
To find the heat released per mole, we divide the total heat by the number of moles of \( \text{NH}_3 \) produced.
[tex]\[ \Delta H_{\text{per mole of NH}_3} = \frac{100.4 \; \text{kJ}}{2 \; \text{moles of NH}_3} = 50.2 \; \text{kJ per mole of NH}_3 \][/tex]
4. Consider the direction of the heat flow:
The question specifies that the heat is released, meaning it is an exothermic reaction. Thus, the enthalpy change \(\Delta H\) should be negative when considering the perspective of the reaction.
[tex]\[ \Delta H_{\text{per mole of NH}_3} = -50.2 \; \text{kJ per mole of NH}_3 \][/tex]
Therefore, the \(\Delta H\) of heat released per mole of \( \text{NH}_3 \) formed is \(-50.2 \; \text{kJ}\). The correct answer is:
[tex]\[ \boxed{-50.2 \; \text{kJ}} \][/tex]
Given:
[tex]\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) + 100.4 \; \text{kJ} \][/tex]
This tells us that the reaction yields 100.4 kJ of energy when 2 moles of \( \text{NH}_3 \) are produced. We need to find the heat released per mole of \( \text{NH}_3 \).
Step-by-Step Solution:
1. Determine the total heat released for the reaction:
The overall reaction releases 100.4 kJ of energy.
2. Identify the amount of \( \text{NH}_3 \) produced:
According to the equation, 100.4 kJ is released when 2 moles of \( \text{NH}_3 \) are formed.
3. Calculate the heat released per mole of \( \text{NH}_3 \):
To find the heat released per mole, we divide the total heat by the number of moles of \( \text{NH}_3 \) produced.
[tex]\[ \Delta H_{\text{per mole of NH}_3} = \frac{100.4 \; \text{kJ}}{2 \; \text{moles of NH}_3} = 50.2 \; \text{kJ per mole of NH}_3 \][/tex]
4. Consider the direction of the heat flow:
The question specifies that the heat is released, meaning it is an exothermic reaction. Thus, the enthalpy change \(\Delta H\) should be negative when considering the perspective of the reaction.
[tex]\[ \Delta H_{\text{per mole of NH}_3} = -50.2 \; \text{kJ per mole of NH}_3 \][/tex]
Therefore, the \(\Delta H\) of heat released per mole of \( \text{NH}_3 \) formed is \(-50.2 \; \text{kJ}\). The correct answer is:
[tex]\[ \boxed{-50.2 \; \text{kJ}} \][/tex]
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