Calculate Reaction Enthalpy: Reversing And Scaling

Hey there, chemistry enthusiasts! Today, we're diving into a fascinating problem that involves manipulating chemical reactions and understanding their associated energy changes. Specifically, we'll be tackling how to determine the molar enthalpy for a reaction when it's been reversed and scaled. This concept is fundamental in thermochemistry, helping us predict how much energy is absorbed or released during chemical transformations. Our starting point is the formation of calcium oxide (CaO) from its constituent elements, calcium (Ca) and oxygen (O₂), which is an exothermic process, releasing a significant amount of energy. The given reaction is: $Ca(s) + \frac{1}{2} O_2(g) \rightarrow CaO(s) \quad \Delta H^{\circ} = -635 \text{ kJ/mol}$. This value, $\Delta H^{\circ}$, represents the standard enthalpy change for the formation of one mole of CaO. The negative sign indicates that the reaction releases heat into the surroundings, making it exothermic. It's crucial to remember that enthalpy changes are state functions, meaning they depend only on the initial and final states, not the path taken. This principle, known as Hess's Law, is the backbone of manipulating thermochemical equations. It allows us to combine known reactions to determine the enthalpy change of an unknown reaction. In this case, we are given a direct formation reaction and asked to find the enthalpy for the reverse and scaled decomposition reaction. Let's break down the process step-by-step, ensuring we grasp every nuance of enthalpy manipulation. Understanding these fundamental principles will not only help us solve this specific problem but also equip us with the tools to analyze a wide array of chemical reactions and their energetic implications. Whether you're a student grappling with thermochemistry homework or a curious mind wanting to understand the energy dynamics of chemical processes, this article aims to provide a clear and comprehensive explanation.

Understanding Enthalpy Changes in Chemical Reactions

Before we jump into manipulating the given reaction, it's vital to solidify our understanding of what molar enthalpy truly signifies in the context of chemical reactions. The symbol $\Delta H^{\circ}$ denotes the standard enthalpy change, where the '°' indicates standard conditions (usually 298 K and 1 atm pressure). The 'kJ/mol' unit tells us the amount of energy transferred per mole of reaction as written. Our initial reaction, $Ca(s) + \frac{1}{2} O_2(g) \rightarrow CaO(s) \quad \Delta H^{\circ} = -635 \text{ kJ/mol}$, describes the process where one mole of solid calcium reacts with half a mole of gaseous oxygen to produce one mole of solid calcium oxide. The $-635 \text{ kJ/mol}$ means that for every mole of CaO formed under standard conditions, $635 \text{ kJ}$ of energy is released. Now, let's consider the key principles of Hess's Law that are relevant here. Firstly, if a reaction is reversed, the sign of its enthalpy change is also reversed. This makes intuitive sense: if forming a compound releases energy, then breaking that compound back down into its elements must require the same amount of energy input. Secondly, if a reaction is multiplied by a stoichiometric coefficient, its enthalpy change is multiplied by the same coefficient. This is because enthalpy is an extensive property, meaning it scales with the amount of substance involved. If we want to make twice as much product, we'll need twice the energy input or output. With these principles in mind, we can confidently approach the target reaction and derive its enthalpy change from the given information. It’s like solving a puzzle where each piece (known reaction) has a specific value, and we need to arrange them correctly to find the value of the final picture (target reaction). The beauty of thermochemistry lies in this predictability and consistency, allowing us to calculate energy changes even for reactions that are difficult or impossible to measure directly in a laboratory setting. So, let's roll up our sleeves and apply these powerful concepts.

Manipulating the Given Reaction

We are provided with the following thermochemical equation: $Ca(s) + \frac{1}{2} O_2(g) \rightarrow CaO(s) \quad \Delta H^{\circ} = -635 \text{ kJ/mol}$. Our goal is to determine the molar enthalpy for the reaction: $2 CaO(s) \rightarrow 2 Ca(s) + O_2(g)$. Let's analyze the differences between the given reaction and our target reaction. First, notice that the roles of reactants and products are swapped. In the given reaction, Ca and O₂ are reactants, forming CaO. In our target reaction, CaO is the reactant, breaking down into Ca and O₂. This means we need to reverse the given reaction. When we reverse a reaction, we must also reverse the sign of its enthalpy change. So, reversing the formation of CaO gives us:

$CaO(s) \rightarrow Ca(s) + \frac{1}{2} O_2(g)$

And the enthalpy change for this reversed reaction becomes: $\Delta H^{\circ} = +635 \text{ kJ/mol}$.

Now, let's compare the stoichiometric coefficients. In our target reaction, we have $2 \text{ CaO(s)}$ on the reactant side, while in the reversed reaction we just wrote, we have $1 \text{ CaO(s)}$. Similarly, on the product side, we have $2 \text{ Ca(s)}$ and $1 \text{ O}_2(g)$ in the target reaction, compared to $1 \text{ Ca(s)}$ and $\frac{1}{2} \text{ O}_2(g)$ in our reversed reaction. This indicates that we need to scale the reversed reaction by a factor of 2. According to Hess's Law, when we multiply a reaction by a certain factor, we must also multiply its enthalpy change by the same factor. Therefore, we multiply both the reversed reaction and its enthalpy change by 2:

$2 \times (CaO(s) \rightarrow Ca(s) + \frac{1}{2} O_2(g))$

This gives us:

$2 CaO(s) \rightarrow 2 Ca(s) + 2 \times \frac{1}{2} O_2(g)$

Simplifying the oxygen part, we get:

$2 CaO(s) \rightarrow 2 Ca(s) + O_2(g)$

This is exactly our target reaction! Now, let's apply the same scaling to the enthalpy change:

$

\Delta H^{\circ} = 2 \times (+635 \text{ kJ/mol})$

$

\Delta H^{\circ} = +1270 \text{ kJ/mol}$

So, the molar enthalpy for the reaction $2 CaO(s) \rightarrow 2 Ca(s) + O_2(g)$ is $+1270 \text{ kJ/mol}$. The positive sign indicates that this reaction is endothermic; it requires $1270 \text{ kJ}$ of energy to decompose 2 moles of CaO into its elements.

The Significance of Reversing and Scaling Reactions

Understanding how to reverse and scale chemical reactions is a cornerstone of chemical thermodynamics, enabling us to predict and control energy transformations. The process we just completed—reversing and scaling the formation of calcium oxide to find the enthalpy for its decomposition—illustrates a powerful application of Hess's Law. Hess's Law states that the total enthalpy change for a chemical reaction is independent of the pathway taken, meaning it only depends on the initial and final states. This principle allows us to construct complex reaction pathways from simpler, known reactions. When we are given a reaction and its enthalpy change, say $A \rightarrow B \quad \Delta H = x$, and we need the enthalpy for the reverse reaction $B \rightarrow A$, we simply change the sign of the enthalpy: $\Delta H = -x$. This is because the energy released in forming B from A must be absorbed to break B back down into A. Similarly, if we need to adjust the quantities of reactants and products, we multiply the entire reaction equation by a stoichiometric factor, say $n$. For example, if we have $A \rightarrow B \quad \Delta H = x$, then $nA \rightarrow nB \quad \Delta H = nx$. This is because enthalpy is an extensive property; it scales linearly with the amount of substance. In our specific problem, the initial reaction $Ca(s) + \frac{1}{2} O_2(g) \rightarrow CaO(s) \quad \Delta H^{\circ} = -635 \text{ kJ/mol}$ represents the formation of one mole of CaO. We wanted to find the enthalpy for the reaction $2 CaO(s) \rightarrow 2 Ca(s) + O_2(g)$. By reversing the initial reaction, we got $CaO(s) \rightarrow Ca(s) + \frac{1}{2} O_2(g)$ with $\Delta H^{\circ} = +635 \text{ kJ/mol}$. Then, by multiplying this reversed reaction by 2, we obtained the target reaction with an enthalpy change of $2 \times (+635 \text{ kJ/mol}) = +1270 \text{ kJ/mol}$. This ability to manipulate enthalpy changes is crucial for many areas of chemistry, including calculating bond energies, determining the feasibility of reactions, and designing industrial chemical processes. It provides a theoretical framework to predict energy requirements and outputs without needing to conduct potentially dangerous or expensive experiments. The consistent application of these rules ensures accuracy and reliability in our thermodynamic calculations.

Conclusion: Mastering Enthalpy Calculations

In summary, we've successfully determined the molar enthalpy for the reaction $2 CaO(s) \rightarrow 2 Ca(s) + O_2(g)$ by applying the fundamental principles of thermochemistry, specifically Hess's Law. We started with the given enthalpy of formation for calcium oxide: $Ca(s) + \frac{1}{2} O_2(g) \rightarrow CaO(s) \quad \Delta H^{\circ} = -635 \text{ kJ/mol}$. To arrive at our target reaction, we first reversed the given reaction. Reversing a reaction negates the sign of its enthalpy change. Thus, the decomposition of CaO into its elements initially becomes $CaO(s) \rightarrow Ca(s) + \frac{1}{2} O_2(g)$ with an enthalpy change of $\Delta H^{\circ} = +635 \text{ kJ/mol}$. Next, we observed that the target reaction involves twice the amount of CaO and its constituent elements compared to the reversed reaction. Therefore, we scaled the reversed reaction by a factor of 2. Multiplying the reaction by 2 gave us $2 CaO(s) \rightarrow 2 Ca(s) + O_2(g)$, which is our desired reaction. Consequently, we multiplied the enthalpy change by the same factor: $2 \times (+635 \text{ kJ/mol}) = +1270 \text{ kJ/mol}$. The final molar enthalpy for the decomposition of 2 moles of CaO is $+1270 \text{ kJ/mol}$. This positive value signifies that the reaction is endothermic, meaning it requires energy input to occur. Mastering these techniques of reversing and scaling chemical equations is essential for solving a wide range of problems in chemical thermodynamics. It empowers you to calculate enthalpy changes for reactions that may not be directly measurable, providing crucial insights into the energetic landscape of chemical transformations. These skills are foundational for anyone studying chemistry, from introductory courses to advanced research.

For further exploration into the fascinating world of chemical thermodynamics and enthalpy calculations, you can refer to reliable resources such as:

• The American Chemical Society (ACS): Their website offers a wealth of educational materials and research on various chemistry topics, including thermochemistry. You can find it at American Chemical Society.
• Khan Academy: This platform provides comprehensive and free educational videos and articles on chemistry, making complex concepts like enthalpy easy to understand. Explore their chemistry section at Khan Academy Chemistry.