Chapter 16: Problem 41
Is this reaction predicted to favor the products at low temperatures, at high temperatures, or both? Explain your answer briefly. \(\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{MgO}(\mathrm{s}) \quad \Delta_{\mathrm{r}} H^{\circ}=-601.70 \mathrm{~kJ} / \mathrm{mol}\)
Short Answer
Step by step solution
Analyze the Reaction Type
Apply the Concept of Exothermic Reactions
Consider Temperature Effects
Conclusion on Temperature Favorability
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Synthesis Reaction
These reactions are often depicted by the general formula \(A + B \rightarrow AB\), where \(A\) and \(B\) are reactants and \(AB\) is the product.
- Synthesis reactions are common in the formation of metal oxides from metals and oxygen.
- They are usually accompanied by energy changes, either releasing or absorbing heat.
- Synthesis reactions are instrumental in industries for forming necessary compounds.
Le Chatelier's Principle
Applying this principle to our synthesis reaction, we have an exothermic process, which means it releases heat.
- If the temperature is reduced, the system combats this change by producing more products to release more heat, enhancing product formation.
- Conversely, increasing the temperature pushes the equilibrium towards reactants, as the system absorbs extra heat favoring the reverse reaction.
Enthalpy Change
Exothermic reactions have a negative \(\Delta H\), demonstrating that they release energy to their surroundings. This release of energy makes the products more stable compared to the reactants.
- Reactions with large negative \(\Delta H\) values signify substantial energy release.
- Enthalpy change helps in understanding not only the energy dynamics but also the temperature suitability of the reaction.
- Analyzing \(\Delta H\) is crucial to industrial applications, as managing energy changes effectively can enhance reaction efficiency.