Problem 63 - Chemistry and Chemical Reactivity

Question

The enthalpy change for the oxidation of naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8},\) is measured by calorimetry. $$ \begin{aligned} \mathrm{C}_{10} \mathrm{H}_{g}(\mathrm{s})+12 \mathrm{O}_{2}(\mathrm{g}) \rightarrow & 10 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell) \\\ \Delta_{i} H^{\circ} &=-5156.1 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} \end{aligned} $$ Use this value, along with the standard enthalpies of formation of \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell),\) to calculate the enthalpy of formation of naphthalene, in kJ/mol.

Step-by-Step Solution

Answer

The enthalpy of formation for naphthalene is 77.9 kJ/mol.

Step 1: Understand the Reaction Enthalpy Change

The given reaction shows the complete oxidation of naphthalene. The enthalpy change for the reaction, \( \Delta_i H^\circ \), is -5156.1 kJ/mol, which means the reaction releases 5156.1 kJ of energy per mole of naphthalene oxidized.

Step 2: Write the Enthalpy Change Equation

The enthalpy change for the reaction can be expressed in terms of the enthalpy of formation of the reactants and products:\[ \Delta_r H^\circ = \sum \Delta_f H^\circ (\text{products}) - \sum \Delta_f H^\circ (\text{reactants}) \]

Step 3: Plug Known Values into the Equation

Using the standard enthalpies of formation of the products:- For \( \text{CO}_2(g) \), \( \Delta_f H^\circ = -393.5 \, \text{kJ/mol} \)- For \( \text{H}_2O(l) \), \( \Delta_f H^\circ = -285.8 \, \text{kJ/mol} \)Plug these into the enthalpy equation:\[\Delta_i H^\circ = [10(-393.5) + 4(-285.8)] - [\Delta_f H^\circ (\text{C}_{10}\text{H}_8) + 12(0)]\]\[-5156.1 = [-3935 - 1143.2] - \Delta_f H^\circ (\text{C}_{10}\text{H}_8)\]

Step 4: Simplify and Solve the Equation

Calculate the sum of enthalpies of the products:\[ -3935 - 1143.2 = -5078.2 \, \text{kJ/mol} \]Now, simplify the equation:\[-5156.1 = -5078.2 - \Delta_f H^\circ (\text{C}_{10}\text{H}_8)\]Rearranging:\[\Delta_f H^\circ (\text{C}_{10}\text{H}_8) = -5078.2 + 5156.1\]\[\Delta_f H^\circ (\text{C}_{10}\text{H}_8) = 77.9 \, \text{kJ/mol} \]

Step 5: Conclusion

The enthalpy of formation for naphthalene, \( \text{C}_{10}\text{H}_8 \), is calculated to be 77.9 kJ/mol.

Key Concepts

Calorimetry

Calorimetry is an important experimental technique used to measure the amount of heat exchanged in chemical reactions or physical changes. It offers insights into reaction energetics by quantifying the heat transfer. In the context of our problem, the calorimetry method is utilized to understand the heat released during the oxidation of naphthalene.

To perform calorimetry, a calorimeter, a device that can absorb or release heat, is used. The change in temperature measured inside the calorimeter, alongside specific heat capacities, allows us to determine the enthalpy change of the reaction.

• Calorimeter: A device for measuring heat transfer.
• Specific heat capacity: The heat required to raise the temperature of a substance.
• Temperature change: Directly measured to assess energy change.

Understanding calorimetry equips us with the foundational tools to gauge how much energy, in terms of heat, is involved in various reactions, enabling calculations like those involved in determining the enthalpy change of naphthalene's oxidation.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a core concept in thermodynamics that describes the heat content change in a system during a reaction at constant pressure. For the naphthalene oxidation reaction, this change is represented by \( \Delta_i H^\circ = -5156.1 \) kJ/mol. This negative value indicates that the reaction releases energy, making it exothermic.

When considering enthalpy changes:

• Endothermic processes absorb heat (positive \( \Delta H \)).
• Exothermic processes release heat (negative \( \Delta H \)).

The enthalpy change helps to understand the energy profile of a chemical reaction by indicating how energy is absorbed or released through the formation and breaking of bonds.
Understanding enthalpy change allows us to further apply this concept to solve for related thermodynamic values, such as standard enthalpies of formation.

Standard Enthalpies of Formation

Standard enthalpies of formation, represented as \( \Delta_f H^\circ \), are the changes in enthalpy when one mole of a compound is formed from its elements in their standard states. This measure is critical for calculating the overall enthalpy change in complex reactions through Hess's law.

Given standards, like for COelements: Oxygen and Carbon, help establish the reaction's thermodynamic landscape.

• For CO\(_2\)(g), \( \Delta_f H^\circ = -393.5 \) kJ/mol.
• For H\(_2\)O(l), \( \Delta_f H^\circ = -285.8 \) kJ/mol.

By using these known values, along with Hess's law, we connect the dots in understanding and calculating the enthalpy of less straightforward reactions, such as the formation of naphthalene.

Chemical Reaction

Chemical reactions involve the transformation of reactants into products, with accompanying energy changes. This energy change drives the reaction's final state, determining whether it is energetically favorable or not.

The oxidation of naphthalene is given by the reaction:

• \( \mathrm{C}_{10} \mathrm{H}_{8} + 12 \mathrm{O}_{2} \rightarrow 10 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O} \)
• The associated energy change, \( \Delta_i H^\circ \), is noted as -5156.1 kJ/mol.

Reactions can be influenced by different factors, such as catalysts, temperature, and pressure. However, understanding their general makeup and products, like in the case of the naphthalene oxidation reaction, aids in predicting energy outputs and overall reaction feasibility.
These foundations in reaction chemistry enrich comprehension, which is crucial in applying concepts such as calculating the enthalpy of formation and understanding calorimetric data for detailed thermodynamic studies.