Problem 97 - Chemistry The Molecular Science

Question

From these enthalpies of reaction, $$ \begin{array}{ll} \mathrm{CaCO}_{3}(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) & \Delta_{\mathrm{r}} H^{\circ}=178.3 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}) & \Delta_{\mathrm{r}} H^{\circ}=-65.2 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}) \longrightarrow \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})\) $$ \Delta_{\mathrm{r}} H^{\circ}=-16.7 \mathrm{~kJ} / \mathrm{mol} $$ calculate \(\Delta_{\mathrm{r}} H^{\circ}\) for \(\mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CaCO}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

\(-129.8 \, \mathrm{kJ/mol}\).

Step 1: Identify Given Reactions

List all given reactions and their enthalpy changes.1. \( \mathrm{CaCO}_{3} \rightarrow \mathrm{CaO} + \mathrm{CO}_{2} \), \( \Delta_{\mathrm{r}} H^{\circ} = 178.3 \, \mathrm{kJ/mol} \)2. \( \mathrm{CaO} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca(OH)}_{2} \), \( \Delta_{\mathrm{r}} H^{\circ} = -65.2 \, \mathrm{kJ/mol} \)3. \( \mathrm{Ca(OH)}_{2} \rightarrow \mathrm{Ca}^{2+} + 2 \mathrm{OH}^{-} \), \( \Delta_{\mathrm{r}} H^{\circ} = -16.7 \, \mathrm{kJ/mol} \).

Step 2: Combine Reactions to Target Equation

Arrange the given reactions to match the target equation: \( \mathrm{Ca}^{2+} + 2\mathrm{OH}^{-} + \mathrm{CO}_{2} \rightarrow \mathrm{CaCO}_{3} + \mathrm{H}_{2} \mathrm{O} \). Reverse and combine reactions as necessary.

Step 3: Reverse Reactions 1 and 2

Reverse the reactions where needed, as the products in the target reaction become reactants, and vice versa.1. \( \mathrm{CaO} + \mathrm{CO}_{2} \rightarrow \mathrm{CaCO}_{3} \), \( \Delta_{\mathrm{r}} H^{\circ} = -178.3 \, \mathrm{kJ/mol} \) (reversed)2. \( \mathrm{Ca(OH)}_{2} \rightarrow \mathrm{CaO} + \mathrm{H}_{2} \mathrm{O} \), \( \Delta_{\mathrm{r}} H^{\circ} = 65.2 \, \mathrm{kJ/mol} \) (reversed)

Step 4: Combine to Find Overall Enthalpy

Add all three enthalpy changes accordingly:\[ \Delta_{\mathrm{r}} H^{\circ} = (-178.3) + 65.2 - 16.7 = -129.8 \, \mathrm{kJ/mol} \]

Step 5: Final Result

The enthalpy change for the target reaction is \(-129.8 \, \mathrm{kJ/mol}\).

Key Concepts

Enthalpy of Reaction

Enthalpy of reaction, often denoted as \(\Delta_r H^{\circ}\), is a crucial concept in chemical thermodynamics. It refers to the heat change that occurs during a chemical reaction at standard conditions, often measured at 1 bar and at a specified temperature, typically 298 K. Enthalpy change can either be exothermic, indicating that heat is released by the reaction (\(\Delta_r H^{\circ} < 0\)), or endothermic, meaning heat is absorbed from the surroundings (\(\Delta_r H^{\circ} > 0\)).
Understanding the enthalpy of reaction allows chemists to predict how a reaction will behave energetically. In the problem outlined, we are dealing with reactions involving calcium compounds and carbon dioxide, each with its own associated enthalpy change. By calculating the overall enthalpy of reaction for a series of reactions, we can determine the net heat effect of the processes involved.
Key points to remember about enthalpy of reaction include:

• Standard enthalpies reflect changes in energy content per mole of substance.
• Enthalpy changes depend on the physical states of the reactants and products.
• They are path-independent; only the initial and final states matter in calculations.

Reaction Enthalpy Calculations

Reaction enthalpy calculations involve finding the overall enthalpy change for a series of chemical reactions. These reactions can be individual steps in a multi-step process.
The principle tool in these calculations is Hess's Law, which states that if a reaction can be expressed as a series of steps, the total enthalpy change is the sum of the enthalpy changes for each step.
To calculate the enthalpy change for the exercise at hand, we need to:

• Identify all given reactions and associate their respective enthalpy changes.
• Manipulate the reactions by reversing or combining them to arrive at the target equation.
• Reverse reactions will change the sign of their enthalpy values, as observed in the solution process.

This systematic combination of reactions allows us to ascertain the desired enthalpy change for the final reaction of interest, demonstrating the power of Hess's Law in understanding complex reaction networks.
In our example, once the reactions were aligned to match the target equation, their enthalpy changes were simply added to yield a final result of \(-129.8 \, \mathrm{kJ/mol}\).

Chemical Thermodynamics

Chemical thermodynamics is the study of energy transformations in chemical processes. It provides insights into how energy is absorbed or released during chemical reactions, helping predict reaction behavior under various conditions.
Central to chemical thermodynamics is the concept of enthalpy, alongside other terms like entropy and free energy. These terms can help assess the spontaneity and feasibility of reactions.
Within our problem context, chemical thermodynamics helps explain how calcium carbonate undergoes transformations with substances like water and carbon dioxide, focusing specifically on heat changes.
Understanding these transformations assists in:

• Predicting equilibrium positions and directions of reactions.
• Determining the energy efficiency of reactions, relevant in industrial applications.
• Evaluating environmental implications, as chemical energy changes relate to broader ecological balances.

By exploring these thermodynamic principles, students can develop a profound understanding of energy changes in chemical processes, reflecting real-world scenarios and technological applications.