Answer :
Certainly! Let's begin by understanding the relationship between the given reaction and the one for which we need to determine the enthalpy change ([tex]\(\Delta H\)[/tex]).
1. Given Reaction and Its Enthalpy Change:
The given reaction is:
[tex]\[ P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
with an enthalpy change of:
[tex]\[ \Delta H = -3010 \, \text{kJ} \][/tex]
This indicates that the reaction is exothermic, releasing 3010 kJ of energy.
2. Reverse Reaction:
We need to find the [tex]\(\Delta H\)[/tex] for the reverse reaction:
[tex]\[ P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g) \][/tex]
The enthalpy change for a reverse reaction is the opposite of the forward reaction.
3. Calculating [tex]\(\Delta H\)[/tex] for the Reverse Reaction:
Since the forward reaction releases energy ([tex]\(\Delta H = -3010 \, \text{kJ}\)[/tex]), the reverse reaction will absorb the same amount of energy. Therefore, the enthalpy change for the reverse reaction will be:
[tex]\[ \Delta H = +3010 \, \text{kJ} \][/tex]
This indicates that the reverse reaction is endothermic.
4. Conclusion:
The enthalpy change ([tex]\(\Delta H\)[/tex]) for the reverse reaction,
[tex]\[ P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g) \][/tex]
is:
[tex]\[ \boxed{3010 \, \text{kJ}} \][/tex]
1. Given Reaction and Its Enthalpy Change:
The given reaction is:
[tex]\[ P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
with an enthalpy change of:
[tex]\[ \Delta H = -3010 \, \text{kJ} \][/tex]
This indicates that the reaction is exothermic, releasing 3010 kJ of energy.
2. Reverse Reaction:
We need to find the [tex]\(\Delta H\)[/tex] for the reverse reaction:
[tex]\[ P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g) \][/tex]
The enthalpy change for a reverse reaction is the opposite of the forward reaction.
3. Calculating [tex]\(\Delta H\)[/tex] for the Reverse Reaction:
Since the forward reaction releases energy ([tex]\(\Delta H = -3010 \, \text{kJ}\)[/tex]), the reverse reaction will absorb the same amount of energy. Therefore, the enthalpy change for the reverse reaction will be:
[tex]\[ \Delta H = +3010 \, \text{kJ} \][/tex]
This indicates that the reverse reaction is endothermic.
4. Conclusion:
The enthalpy change ([tex]\(\Delta H\)[/tex]) for the reverse reaction,
[tex]\[ P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g) \][/tex]
is:
[tex]\[ \boxed{3010 \, \text{kJ}} \][/tex]
