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When white phosphorus burns in air, it produces phosphorus (V) oxide.
[tex]\[
P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s); \quad \Delta H = -3010 \, \text{kJ}
\][/tex]

What is [tex]$\Delta H$[/tex] for the following equation?
[tex]\[
P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g)
\][/tex]

Sagot :

GinnyAnswer
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]
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