Problem 87 - Chemistry and Chemical Reactivity

Question

(a) Calculate the enthalpy change, \(\Delta_{r} H^{\circ},\) for the formation of 1.00 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements. $$ \mathrm{Sr}(\mathrm{s})+\mathrm{C}(\mathrm{s})+3 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{s}) $$ The experimental information available is \(\mathrm{Sr}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SrO}(\mathrm{s}) \quad \Delta_{i} H^{\circ}=-592 \mathrm{kJ} / \mathrm{mol}-\mathrm{pxn}\) \(\mathrm{SrO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{s})\) $$ \Delta_{\tau} H^{\circ}=-234 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} $$ C(graphite) \(+\mathbf{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})\) \(\Delta_{j} H^{\circ}=-394 \mathrm{kJ} / \mathrm{mol}-\mathrm{Dxn}\) (b) Draw an energy level diagram relating the energy quantities in this problem.

Step-by-Step Solution

Answer

The enthalpy change for the formation of strontium carbonate is -1220 kJ/mol.

Step 1: Understanding the Given Information

We have the formation reaction of strontium carbonate: \[ \mathrm{Sr}(\mathrm{s})+\mathrm{C}(\mathrm{s})+3 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{s}) \]and three given reactions with their respective enthalpies:1. \( \mathrm{Sr}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SrO}(\mathrm{s}) \) with \( \Delta_{i} H^{\circ}=-592 \, \mathrm{kJ/mol} \).2. \( \mathrm{SrO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{s}) \) with \( \Delta_{\tau} H^{\circ}=-234 \, \mathrm{kJ/mol} \).3. \( \mathrm{C} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}(\mathrm{g}) \) with \( \Delta_{j} H^{\circ}=-394 \, \mathrm{kJ/mol} \).Our goal is to use Hess's Law to find the enthalpy change for the formation of \( \mathrm{SrCO}_{3}(\mathrm{s}) \) from its elements.

Step 2: Applying Hess's Law

We need to arrange the given reactions so that they add up to the reaction for the formation of \( \mathrm{SrCO}_{3}(\mathrm{s}) \). - From reaction 1, the formation of \( \mathrm{SrO}(\mathrm{s}) \) involves strontium and oxygen.- From reaction 2, \( \mathrm{SrO}(\mathrm{s}) \) reacts with \( \mathrm{CO}_{2}(\mathrm{g}) \) to form \( \mathrm{SrCO}_{3}(\mathrm{s}) \).- Reaction 3 shows the formation of \( \mathrm{CO}_{2}(\mathrm{g}) \) from carbon and oxygen.Adding these reactions results in the desired equation:\( \mathrm{Sr}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) + 3/2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SrCO}_{3}(\mathrm{s}) \).

Step 3: Calculating the Total Enthalpy Change

Sum up the enthalpies from the provided reactions following Hess's Law:\[\begin{align*}\Delta_{r} H^{\circ} & = \Delta_{i} H^{\circ} + \Delta_{\tau} H^{\circ} + \Delta_{j} H^{\circ} \& = (-592) + (-234) + (-394) = -1220 \, \mathrm{kJ/mol}.\end{align*}\] So, the enthalpy change for the formation of \( \mathrm{SrCO}_{3}(\mathrm{s}) \) is \(-1220 \, \mathrm{kJ/mol} \).

Step 4: Drawing the Energy Level Diagram

Draw a diagram to represent the enthalpy changes of each step:1. Start with \( \mathrm{Sr}(\mathrm{s}) \), \( \mathrm{C}(\mathrm{s}) \), and \( \mathrm{O}_{2}(\mathrm{g}) \) at the bottom.2. Represent \( \mathrm{SrO}(\mathrm{s}) \) at a lower level by \(-592 \, \mathrm{kJ/mol} \).3. Show \( \mathrm{CO}_{2}(\mathrm{g}) \) at \(-394 \, \mathrm{kJ/mol} \) relative to \( \mathrm{C} \) and \( \mathrm{O}_{2} \).4. Finally, show \( \mathrm{SrCO}_{3}(\mathrm{s}) \) at \(-1220 \, \mathrm{kJ/mol} \), the lowest energy level in the diagram.Ensure the energy levels decrease reflecting the stability gained in each step.

Key Concepts

Enthalpy Change

Enthalpy is a measure of the total energy of a thermodynamic system, which includes both internal energy and the energy required to make space for it by displacing its surroundings. An enthalpy change, denoted as \( \Delta H \), occurs during a chemical reaction when reactants transform into products. This change in energy, which can be either absorbed or released, is crucial to understanding how a substance interacts with other materials.

In chemical reactions, the enthalpy change is defined as the difference in enthalpy between products and reactants. If the reaction releases heat, it is considered exothermic, and \( \Delta H \) will be negative, indicating that the system loses energy to the surroundings. Conversely, if the reaction absorbs heat, it is endothermic, and \( \Delta H \) will be positive, signaling an energy gain by the system.

Hess's Law states that the total enthalpy change of a reaction is the same, no matter whether the reaction occurs in one step or a series of steps. This allows the calculation of enthalpy change when direct measurement isn't feasible. By using given data for simpler reactions, such as the formation of \( \mathrm{SrO} \) or \( \mathrm{CO}_2 \), we can determine the enthalpy change for more complex reactions, like the formation of \( \mathrm{SrCO}_3 \), as demonstrated in the exercise solution.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, which often entails breaking and forming bonds between atoms. These reactions are usually accompanied by changes in energy. In the context of Hess's Law, understanding these energy changes helps in calculating the overall enthalpy change of complex reactions.

For example, the exercise discusses the reaction forming \( \mathrm{SrCO}_3 \) from its elements. Here, the reaction can be broken down into simpler reactions of strontium with oxygen to form strontium oxide (\( \mathrm{SrO} \)), and graphite with oxygen to produce carbon dioxide (\( \mathrm{CO}_2 \)). These intermediate reactions provide critical enthalpy values needed for the complete reaction.

By rearranging and combining these reactions, we can piece together the enthalpy changes of all the steps involved, using Hess's Law to ensure that the initial and final states of each element conform to the desired overall reaction. This methodology is advantageous for reactions that cannot be measured directly, thus highlighting the importance of intermediary reactions in calculating enthalpy changes.

Energy Level Diagram

An energy level diagram is a visual representation of the energy changes throughout a chemical reaction pathway. In the context of Hess's Law, it graphically illustrates how intermediate reactions and their associated energy changes contribute to the overall enthalpy change.

In the provided solution, the diagram starts with the reactants, \( \mathrm{Sr} \), \( \mathrm{C} \), and \( \mathrm{O}_2 \), at a high energy level. Each subsequent step represents a drop in energy, corresponding to reactions such as \( \mathrm{Sr} + 1/2 \mathrm{O}_2 \rightarrow \mathrm{SrO} \), with respective enthalpy changes like \(-592 \; \mathrm{kJ/mol}\). The final product, \( \mathrm{SrCO}_3 \), is placed at the lowest energy level, \(-1220 \; \mathrm{kJ/mol}\), showing the stabilization achieved after the complete reaction.

Such diagrams offer a clear understanding of how energy is released or absorbed at each step and aid in visually verifying Hess's Law. By accounting for the enthalpy changes individually, we can validate the derived total enthalpy change aligns with the sum of all energy levels, reinforcing comprehension of the reaction’s thermodynamics.