Problem 22 - Basic Principles of Organic Chemistry

Question

A \(C-F\) bond energy can be computed from thermochemical studies of the vapor- phase reaction $$ \mathrm{CH}_{4}+4 \mathrm{~F}_{2} \rightarrow \mathrm{CF}_{4}+4 \mathrm{HF} \quad \Delta H^{0}=-460 \mathrm{kcal} $$ Show how the \(\Delta H^{0}\) value for this reaction may be used to calculate the energy of the \(\mathrm{C}-\mathrm{F}\) bond if all the other bond energies are known.

Step-by-Step Solution

Answer

The C-F bond energy is 114 kcal/mol.

Step 1: Identify Known Bond Energies

We need to first identify all the known bond energies in the molecules involved. For methane (\( \mathrm{CH}_4 \)), we have four C-H bonds. For fluorine (\( \mathrm{F}_2 \)), there are four F-F bonds. In tetrafluoromethane (\( \mathrm{CF}_4 \)), there are four C-F bonds, and in hydrogen fluoride (\( \mathrm{HF} \)), there are four H-F bonds.

Step 2: Write the Equation for Reaction Enthalpy

We express the reaction enthalpy, \( \Delta H^0 \), in terms of bond enthalpies. \[ \Delta H^0 = \sum (\text{Bond Energies of Reactants}) - \sum (\text{Bond Energies of Products}) \] So, \( \Delta H^0 = (4E_{\mathrm{C-H}} + 4E_{\mathrm{F-F}}) - (4E_{\mathrm{C-F}} + 4E_{\mathrm{H-F}}) \).

Step 3: Solve for C-F Bond Energy

Rearrange the equation from Step 2 to solve for the C-F bond energy, \( E_{\mathrm{C-F}} \). \[ 4E_{\mathrm{C-F}} = 4E_{\mathrm{C-H}} + 4E_{\mathrm{F-F}} - 4E_{\mathrm{H-F}} - \Delta H^0 \] \[ E_{\mathrm{C-F}} = \frac{4E_{\mathrm{C-H}} + 4E_{\mathrm{F-F}} - 4E_{\mathrm{H-F}} - \Delta H^0}{4} \]

Step 4: Substitute Known Values

We substitute the known bond energies into the equation: \( E_{\mathrm{C-H}} = 99 \mathrm{kcal/mol} \), \( E_{\mathrm{F-F}} = 36 \mathrm{kcal/mol} \), \( E_{\mathrm{H-F}} = 136 \mathrm{kcal/mol} \), and \( \Delta H^0 = -460 \mathrm{kcal} \). Now calculate \( E_{\mathrm{C-F}} \). \[ E_{\mathrm{C-F}} = \frac{4 \times 99 + 4 \times 36 - 4 \times 136 - (-460)}{4} \]

Step 5: Perform the Calculation

Calculate the energy value: \( E_{\mathrm{C-F}} = \frac{396 + 144 - 544 + 460}{4} = \frac{456}{4} = 114 \mathrm{kcal/mol} \).

Key Concepts

Thermochemical Calculations

Thermochemical calculations are a crucial tool in chemistry used to determine the heat exchange associated with chemical reactions. This concept is particularly useful for evaluating reaction enthalpies and bond energies. When reacting substances, atoms and molecules undergo reconfiguration, often requiring or releasing energy. The calculation of this energy is fundamental to predicting reaction behavior.

To perform thermochemical calculations, you should understand the principles of energy conservation and explicitly determine the enthalpy change, denoted as \( \Delta H^0 \).

• It involves calculating the balance between the energy of bonds broken in reactants and formed in products.
• Subtract the sum of bond energies of products from those of reactants, leading to the equation: \( \Delta H^0 = \sum(\text{Bond Energies of Reactants}) - \sum(\text{Bond Energies of Products}) \).
• Remember that enthalpy is typically measured in kilocalories per mole or kilojoules per mole.

Understanding and applying these principles can be effortlessly done with consistent practice and familiarization with the bond energies involved.

Bond Enthalpy

Bond enthalpy, also known as bond dissociation energy, is a measure of the strength of a chemical bond. It represents the amount of energy required to break one mole of a specific bond in a gaseous substance.

• The bond enthalpy values are typically averages derived from experiments on similar compounds.
• These values are essential for predicting the stability of molecules and the energy change in chemical reactions.
• Factors influencing bond enthalpy include the type of bonded atoms and their electronegativity, as well as the bond length.

For the given exercise, bond enthalpy information helps in calculating specific bond energies like that for the \( \text{C-F} \) bond.Consider the sample calculation in the solution steps:- Given bond energies for \( \text{C-H} \), \( \text{F-F} \), and \( \text{H-F} \).- Use these known values and reaction enthalpy to solve for the unknown \( \text{C-F} \) bond enthalpy.

This systematic approach ensures accurate calculation of bond energies using enthalpy data from known reactions and compounds.

Reaction Enthalpy

Reaction enthalpy is the overall enthalpic change during a chemical reaction, providing insight into the energy dynamics of the process. It can indicate whether a reaction is exothermic (releases energy) or endothermic (absorbs energy).

An exothermic reaction, like the one in the exercise with \( \Delta H^0 = -460 \text{kcal/mol} \), releases energy, thus having a negative \( \Delta H^0 \). In contrast, a positive \( \Delta H^0 \) suggests an endothermic nature.

• Understanding this change is vital for calculating reaction feasibility and designing energy-efficient processes.
• The reaction enthalpy also helps in determining how bond energies transform during chemical changes.
• To compute a specific bond energy from reaction enthalpy, manipulate the equation to highlight the desired unknown value.

In chemical synthesis and industrial applications, reaction enthalpy guides decisions on reaction conditions and optimization.

This concept, structured around bond energy contribution, forms the basis for more advanced studies and practical applications in thermochemistry.