Problem 53
Question
When white phosphorus burns in air, it produces phosphorus(V) oxide. $$ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) ; \Delta H=-3010 \mathrm{~kJ} $$ What is \(\Delta H\) for the following equation? $$ \mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) $$
Step-by-Step Solution
Verified Answer
\( \Delta H = 3010 \text{ kJ} \) for the reverse reaction.
1Step 1: Understand the Reaction
First, note the given reaction for the formation of phosphorus(V) oxide: \( \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) \) with \( \Delta H = -3010 \text{ kJ} \). This means that forming phosphorus(V) oxide releases 3010 kJ of energy.
2Step 2: Identify the Reverse Reaction
The question asks for \( \Delta H \) of the reverse reaction: \( \mathrm{P}_{4} \mathrm{O}_{10}(s) \longrightarrow \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \). The reverse reaction involves breaking down phosphorus(V) oxide into white phosphorus and oxygen.
3Step 3: Apply the Concept of Reaction Reversibility
Remember that the enthalpy change of the reverse reaction is the opposite of the forward reaction. Since the °ΔH° for forming phosphorus(V) oxide is -3010 kJ, \( \Delta H \) for decomposing it will be the opposite: 3010 kJ.
Key Concepts
Enthalpy ChangeReversible ReactionsPhosphorus(V) Oxide
Enthalpy Change
The concept of enthalpy change (\( \Delta H \)) is a central part of thermochemistry. It represents the heat content change during a chemical reaction. When a substance undergoes a reaction, it either absorbs or releases energy. This energy exchange is measured as enthalpy change. For instance, when white phosphorus (\( \mathrm{P}_4 \)) burns and forms phosphorus(V) oxide (\( \mathrm{P}_4\mathrm{O}_{10} \)), it releases 3010 kJ of energy. This release of energy is indicated by the negative sign: \( \Delta H = -3010 \text{ kJ} \).For reactions:- **Exothermic reactions**: Energy is released, and \( \Delta H \) is negative.- **Endothermic reactions**: Energy is absorbed, and \( \Delta H \) is positive.Understanding \( \Delta H \) is crucial for predicting how a reaction will affect the surrounding temperature. In the exercise, the given reaction is exothermic, with energy released as heat when phosphorus combusts.
Reversible Reactions
Reversible reactions are processes where the reactants form products, which can then return to the reactants. Understanding reversible reactions involves acknowledging that reactions can proceed in both the forward and reverse direction. The enthalpy change (\( \Delta H \)) plays a crucial role here, as it indicates the magnitude and direction of energy difference in both processes.For the provided example of phosphorus(V) oxide formation:- Forward Reaction: \( \mathrm{P}_4(s) + 5 \mathrm{O}_2(g) \rightarrow \mathrm{P}_4\mathrm{O}_{10}(s) \), with \( \Delta H = -3010 \text{ kJ} \).- Reverse Reaction: \( \mathrm{P}_4\mathrm{O}_{10}(s) \rightarrow \mathrm{P}_4(s) + 5 \mathrm{O}_2(g) \), with \( \Delta H = 3010 \text{ kJ} \), energy is absorbed.Reversible reactions are essential in dynamic chemical systems and illustrate the importance of how energy changes impact chemical equilibria and reaction spontaneity.
Phosphorus(V) Oxide
Phosphorus(V) oxide, or diphosphorus pentoxide, is a commonly encountered compound of phosphorus, produced when white phosphorus burns. It is an important reagent in both laboratory and industrial chemistry, often used in applications where water removal is necessary, such as dehydrating acids or generating other phosphorus compounds.**Properties**- **Chemical Formula**: \( \mathrm{P}_4\mathrm{O}_{10} \)- **Physical State**: Solid- **Appearance**: White crystallineThe given exercise illustrates an important reaction involving phosphorus(V) oxide, demonstrating its formation and subsequent decomposition. These reactions highlight both the practical use in energetics and theoretical importance in reaction mechanics. Efficient handling of \( \mathrm{P}_4\mathrm{O}_{10} \) showcases its versatility in chemical transformations.
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