Question

Using the information in the following table, determine \(\Delta \mathrm{H}^{\circ}\) for the reactions: a) \(3 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{4}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})\) b) \(\mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+\mathrm{CO}(\mathrm{g}) \rightarrow 3 \mathrm{FeO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) c) \(\mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) $$ \begin{array}{|c|c|} \hline {\text { Heats of Formation }} \\ \hline \text { Compound } & \Delta \mathrm{H}^{\circ}(\mathrm{Kcal} / \mathrm{mole}) \\ \hline \mathrm{CO}(\mathrm{g}) & -26.4 \\ \hline \mathrm{CO}_{2}(\mathrm{~g}) & -94.1 \\ \hline \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) & -197 \\ \hline \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s}) & -267 \\ \hline \mathrm{FeO}(\mathrm{s}) & -63.7 \\ \hline \end{array} $$

Short answer

For the given reactions: a) \(3Fe_{2}O_{3}(s)+CO(g) \rightarrow 2Fe_{2}O_{4}(s)+CO_{2}(g)\): \[\Delta H^{\circ} = -36.5 \, Kcal/mol\] b) \(Fe_{3}O_{4}(s)+CO(g) \rightarrow 3FeO(s)+CO_{2}(g)\): \[\Delta H^{\circ} = 22.3 \, Kcal/mol\] c) \(FeO(s)+CO(g) \rightarrow Fe(s)+CO_{2}(g)\): \[\Delta H^{\circ} = -41 \, Kcal/mol\]

Step-by-step solution

Identify the heats of formation of the reactants and products

Use the provided table to find the heats of formation for the reactants and products in each equation. The reactants are on the left side of the equation, and the products are on the right side. The provided heats of formation are in Kcal/mole.

Apply Hess's Law for each reaction

For each reaction, calculate \(\Delta H^{\circ}\) using the expression: \[\Delta H^{\circ} = \Sigma \Delta H_f^{\circ}(\text{products}) - \Sigma \Delta H_f^{\circ}(\text{reactants})\]

Determination of \(\Delta H^{\circ}\) for reaction (a)

The heat of reaction for the equation: \(3 Fe_{2}O_{3}(s)+CO(g) \rightarrow 2 Fe_{2}O_{4}(s)+CO_{2}(g)\) can be calculated as follows: \[\Delta H^{\circ} = 2 \times \Delta H_f^{\circ}(Fe_{2}O_{4}) + \Delta H_f^{\circ}(CO_{2}) - [3 \times \Delta H_f^{\circ}(Fe_{2}O_{3}) + \Delta H_f^{\circ}(CO)]\]

Calculation for Reaction (a)

Here, we substitute the values for the formation heats from the table into the equation and solve: \[\Delta H^{\circ} = 2 \times -267 + -94.1 - [3 \times -197 + -26.4]\] \[\Delta H^{\circ} = -628.1 + 591.6 = -36.5 \, Kcal/mol\]

Repeat for Reactions (b) and (c)

Following the procedure above, calculate \(\Delta H^{\circ}\) for reactions (b) and (c). For Reaction (b): \(Fe_{3}O_{4}(s)+CO(g) \rightarrow 3 FeO(s)+CO_{2}(g)\) \[\Delta H^{\circ} = 3 \times \Delta H_f^{\circ}(FeO) + \Delta H_f^{\circ}(CO_{2}) - [\Delta H_f^{\circ}(Fe_{3}O_{4}) + \Delta H_f^{\circ}(CO)]\] For Reaction (c): \(FeO(s)+CO(g) \rightarrow Fe(s)+CO_{2}(g)\) \[\Delta H^{\circ} = \Delta H_f^{\circ}(Fe) + \Delta H_f^{\circ}(CO_{2}) - [\Delta H_f^{\circ}(FeO) + \Delta H_f^{\circ}(CO)]\] Substitute the respective heats from the table into equations for reactions (b) and (c) to solve for \(\Delta H^{\circ}\). The heat of reaction for metallic iron Fe(s) is zero because it is a pure element in its standard state at 25°C. By following these steps, we can calculate the heat of a reaction using the heats of formation for each reaction.

Key concepts

Hess's Law

Hess's Law is a fundamental principle in thermochemistry that states that the total enthalpy change for a chemical reaction is the same no matter how the reaction takes place, as long as the initial and final conditions are the same. In simpler terms, the heat released or absorbed in a chemical process is consistent irrespective of the multiple stages or steps the reaction might undergo. This principle is incredibly useful when direct measurement of heat in a reaction is difficult, as it allows us to use known enthalpy changes of other reactions to calculate the enthalpy change of the reaction in question.

For students having difficulties with understanding Hess's Law, a useful analogy is to think of it as a financial transaction. No matter which route you take to pay for an item, whether you pay by cash, credit, or through a series of smaller transactions, the total amount spent will be the same. Similarly, in a chemical reaction, the total enthalpy change is independent of the path taken by the reaction. Knowing this helps you relate various reactions and form a 'thermochemical equation' that will yield the enthalpy change of the target reaction.

Heats of Formation

The heat of formation, also known by its symbol \( \Delta H_f^\circ \), is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. These values are fundamental to calculating the overall enthalpy changes in reactions using Hess's Law. The standard state refers to the most stable physical form of the element at 1 atmosphere pressure and at a specified temperature, usually 25°C (298 K).

For instance, when we calculate the enthalpy change in a given reaction, we'll need to find the difference between the sum of heats of formation for the products and the sum for the reactants. If you're trying to visualize these concepts, you might imagine the heat of formation as the 'price tag' of creating a compound from its basic, natural ingredients. These 'price tags' allow us to understand the energy economics within a chemical reaction. It's essential for students to know that the heat of formation for an element in its standard state is zero, which reflects the fact that there is no energy change when an element is already in its most stable form.

Thermochemistry

Thermochemistry is the study of energy and heat associated with chemical reactions and/or physical transformations. It involves understanding how energy is transferred within systems as heat and how this affects the properties and behavior of matter. This branch of chemistry is critical in determining which reactions are feasible from an energetic standpoint and is used extensively in both academic research and industrial applications.

As part of learning thermochemistry, students explore key concepts like enthalpy (\(H\)), entropy (\(S\)), free energy (\(G\)), and the laws of thermodynamics, which govern the principles of energy transfer. It is important for students to grasp that factors such as temperature, pressure, and the nature of the reactants and products can influence the direction and extent to which a reaction occurs. Discussing real-life applications, such as energy production or the environmental impact of reactions, can aid in comprehending the relevance of thermochemistry.

Chemical Reactions

Chemical reactions involve the transformation of one set of chemical substances to another. These reactions are governed by the laws of conservation of mass and energy. During a chemical reaction, bonds between atoms are broken and new bonds are formed, which may result in the absorption or release of energy. Understanding the different types of chemical reactions – synthesis, decomposition, single displacement, double displacement, combustion, and acid-base, among others – helps students to predict the outcome of reactions and to grasp the science behind the changes they observe in lab experiments or in the world around them.

It's critical for students to realize that although atoms rearrange during chemical reactions, they are neither created nor destroyed; they simply shift from one type of molecule to another. This is why balancing chemical equations is a crucial skill in studying chemistry. Mastery of this skill not only ensures adherence to the conservation laws but also enhances comprehension of reaction stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction.