Question

It is interesting to compare the "fuel value" of a hydro- carbon in a hypothetical world where oxygen is not the combustion agent. The enthalpy of formation of \(\mathrm{CF}_{4}(g)\) is \(-679.9 \mathrm{kJ} / \mathrm{mol} .\) Which of the following two reactions is the more exothermic? $$\begin{array}{l}{\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)} \\ {\mathrm{CH}_{4}(g)+4 \mathrm{F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g)}\end{array}$$

Short answer

The combustion of methane with fluorine (CH4(g) + 4F2(g) -> CF4(g) + 4HF(g)) is more exothermic than the combustion with oxygen, as it has a more negative change in enthalpy (∆H = -1698.5 kJ/mol) compared to the combustion with oxygen (CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)) with a ∆H value of -802.5 kJ/mol.

Step-by-step solution

Find ∆H for reaction 1

For this step, we will need the standard enthalpies of formation for each of the products and reactants in reaction 1. The following values are needed: - Methane (CH4): -74.6 kJ/mol - Oxygen (O2): 0 kJ/mol (by definition, since it's a diatomic element) - Carbon dioxide (CO2): -393.5 kJ/mol - Water (H2O, gas): -241.8 kJ/mol The change in enthalpy for the reaction (∆H_reaction1) can be calculated as follows: \[\Delta H_{\text{reaction1}} = \sum H_{\text{products}} - \sum H_{\text{reactants}} \] \[\Delta H_{\text{reaction1}} = [1 (-393.5) + 2 (-241.8)] - [1 (-74.6) + 2 (0)]\]

Calculate ∆H for reaction 1

Now, let's plug in the known values and calculate ∆H for reaction 1: \[\Delta H_{\text{reaction1}} = [-393.5 - 2(241.8)] - (-74.6)\] \[\Delta H_{\text{reaction1}} = [-393.5 - 483.6 + 74.6 ]\] \[\Delta H_{\text{reaction1}} = -802.5 \, \text{kJ/mol}\]

Find ∆H for reaction 2

For this step, we will need the standard enthalpies of formation for each of the products and reactants in reaction 2. The following values are needed: - Methane (CH4): -74.6 kJ/mol - Carbon tetrafluoride (CF4): -679.9 kJ/mol - Hydrogen fluoride (HF): -273.3 kJ/mol - Fluorine (F2): 0 kJ/mol (by definition, since it's a diatomic element) The change in enthalpy for the reaction (∆H_reaction2) can be calculated as follows: \[\Delta H_{\text{reaction2}} = \sum H_{\text{products}} - \sum H_{\text{reactants}} \] \[\Delta H_{\text{reaction2}} = [1 (-679.9) + 4 (-273.3)] - [1 (-74.6) + 4 (0)]\]

Calculate ∆H for reaction 2

Now, let's plug in the known values and calculate ∆H for reaction 2: \[\Delta H_{\text{reaction2}} = [-679.9 - 4(273.3)] - (-74.6)\] \[\Delta H_{\text{reaction2}} = [-679.9 - 1093.2 + 74.6]\] \[\Delta H_{\text{reaction2}} = -1698.5 \, \text{kJ/mol}\]

Compare ∆H values for both reactions

Now we have the ∆H values for both reactions: - Reaction 1: ∆H = -802.5 kJ/mol - Reaction 2: ∆H = -1698.5 kJ/mol As the ∆H value for reaction 2 is more negative than that for reaction 1, reaction 2 is more exothermic. So, the combustion of methane with fluorine is more exothermic than combustion with oxygen.

Key concepts

Exothermic Reaction

Exothermic reactions are chemical processes where energy is released to the surroundings, usually in the form of heat. In these reactions, the enthalpy change (\( \Delta H \) is negative, indicating that the products possess less energy than the reactants. This release of energy is what typically makes exothermic reactions feel hot.
Exothermic reactions are prevalent in various everyday phenomena:

• Combustion reactions, like burning wood or gasoline, release heat, keeping homes warm or engines running.
• Chemical hand warmers use exothermic reactions to provide warmth in cold environments.

In the exercise, two reactions were compared for their exothermicity. Reaction 2, involving methane and fluorine, was more exothermic owing to its larger negative \( \Delta H \) value, demonstrating a larger energy release to the surroundings.

Standard Enthalpy of Formation

The standard enthalpy of formation is a key concept in calculating the enthalpy changes of reactions. It refers to the change in enthalpy when one mole of a compound is formed from its elements under standard conditions (usually 1 atm pressure and 25°C).
For any element in its standard state (like \( O_2 \) gas or \( F_2 \) gas), the standard enthalpy of formation is zero. This serves as a baseline for calculating the energy released or absorbed during chemical reactions.
In the problem, this data was used to calculate the total enthalpy change in both reactions. By utilizing standard enthalpies of formation for each compound, it is possible to determine which reaction releases more energy and is more exothermic.

Combustion

Combustion is a class of exothermic reactions where a substance reacts with an oxidant, often releasing energy as heat and light. It's a crucial reaction in both daily life and industrial applications.
Conventionally, combustion involves hydrocarbon fuels like methane (\( CH_4 \)), with oxygen as the oxidant, producing carbon dioxide and water. This process powers engines, heats buildings, and facilitates cooking.
In the given exercise, a comparison is made between the traditional combustion of methane with oxygen and an atypical combustion using fluorine. While both are exothermic, the use of fluorine resulted in a greater release of energy, reflecting in a more negative \( \Delta H \).
The differing heat outputs in such reactions can be utilized to optimize energy efficiency in various applications, from creative industrial uses to novel technological designs.