Question

It is interesting to compare the "fuel value" of a hydrocarbon 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? $$ \begin{aligned} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{CH}_{4}(g)+4 \mathrm{~F}_{2}(g) & \longrightarrow \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) \end{aligned} $$

Short answer

The second reaction involving methane and fluorine has a more exothermic enthalpy change, \(-1824.3 \, \mathrm{kJ/mol}\), as compared to the first reaction involving methane and oxygen with an enthalpy change of \(-802.3 \, \mathrm{kJ/mol}\).

Step-by-step solution

Write the known enthalpies of formation

We are given the enthalpy of formation of \(\mathrm{CF}_{4}(g)\) as -679.9 kJ/mol. We need to find the enthalpy change of both reactions. To do this, we need standard enthalpy of formation values for the other substances involved in each reaction. Remember that the standard enthalpy of formation for an element in its natural state is zero. We should be able to find the standard enthalpy of formation values for the other substances in a textbook or online. Here are the \(\Delta H_{\text{f}}^\circ\) for the required substances: \(\Delta H_{\text{f}}^\circ(\mathrm{CH}_4) = -74.9 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{O}_2) = 0 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{CO}_2) = -393.5 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{H}_2\mathrm{O}) = -285.8 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{F}_2) = 0 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{f}}^\circ(\mathrm{HF}) = -273.3 \, \mathrm{kJ/mol}\)

Calculate the enthalpy change of the first reaction

We can calculate the enthalpy change of the first reaction using the formula: \(\Delta H_{\text{rxn}} = \Sigma n \Delta H_{\text{f}}^{\circ}(\text{products}) - \Sigma m \Delta H_{\text{f}}^\circ(\text{reactants})\) For the first reaction: \(\Delta H_{\text{rxn}1} = [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CO}_2) + 2 \times \Delta H_{\text{f}}^\circ(\mathrm{H}_2\mathrm{O})] - [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CH}_4) + 2 \times \Delta H_{\text{f}}^\circ(\mathrm{O}_2)]\) By substituting the values obtained in Step 1, we get: \(\Delta H_{\text{rxn}1} = [1 \times (-393.5) + 2 \times (-285.8)] - [1 \times (-74.9) + 2 \times (0)] = -802.3 \, \mathrm{kJ/mol}\)

Calculate the enthalpy change of the second reaction

Now, let's calculate the enthalpy change of the second reaction using the same formula used above. For the second reaction: \(\Delta H_{\text{rxn}2} = [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CF}_4) + 4 \times \Delta H_{\text{f}}^\circ(\mathrm{HF})] - [1 \times \Delta H_{\text{f}}^\circ(\mathrm{CH}_4) + 4 \times \Delta H_{\text{f}}^\circ(\mathrm{F}_2)]\) By substituting the values, we get: \(\Delta H_{\text{rxn}2} = [1 \times (-679.9) + 4 \times (-273.3)] - [1 \times (-74.9) + 4 \times (0)] = -1824.3 \, \mathrm{kJ/mol}\)

Compare the enthalpy changes

Now that we have both enthalpy changes, let's compare them to find which reaction is more exothermic. \(\Delta H_{\text{rxn}1} = -802.3 \, \mathrm{kJ/mol}\) \(\Delta H_{\text{rxn}2} = -1824.3 \, \mathrm{kJ/mol}\) Comparing them, it is clear that \(\Delta H_{\text{rxn}2} < \Delta H_{\text{rxn}1}\), which means that the second reaction is more exothermic \((\)gaining more heat\()\) than the first one.

Key concepts

enthalpy of formation

The enthalpy of formation is a crucial concept in chemistry that tells us about the energy change when one mole of a substance is formed from its elements in their standard states. This value is typically given in kilojoules per mole (kJ/mol), and it is measured under standard conditions (usually 1 atm of pressure and 25°C). The enthalpy of formation can be either positive or negative, depending on whether the formation of the substance releases or absorbs energy.

• A negative enthalpy of formation indicates that the formation process releases energy, thus stabilizing the compound, and often signifies that the reaction is exothermic.
• A positive enthalpy of formation suggests that the formation process requires energy input, making it less favorable under standard conditions.

In the exercise provided, the enthalpy of formation is used to calculate the enthalpy changes of different chemical reactions by using given values of other substances involved.

exothermic reactions

Exothermic reactions are chemical reactions that release energy, usually in the form of heat, to their surroundings. This release of energy often results in an increase in temperature or visible light or sound. The enthalpy change (\(\Delta H \)) for an exothermic reaction is negative because the energy required to break the bonds in the reactants is less than the energy released when new bonds are formed in the products.

• Exothermic reactions are spontaneous since they release energy, which often drives the reaction forward.
• The measure of how exothermic a reaction is can be used to compare reactions, as seen in the exercise where two reactions were compared based on their enthalpy changes to determine which was more exothermic.

Examples of exothermic reactions include combustion and many oxidation-reduction reactions, such as the burning of methane with oxygen.

chemical reactions

Chemical reactions involve the transformation of substances through the breaking of chemical bonds in reactants and the formation of new bonds in products. In the analysis of chemical reactions, it is important to consider energy changes, as these determine the feasibility and type of reaction.

• The reaction can be classified as exothermic or endothermic based on whether it releases or absorbs heat, respectively.
• The enthalpy change gives an idea about the energy involved in the reaction, and it is calculated from the enthalpies of formation of the reactants and products.
• Understanding the type of chemical reactions helps predict the behavior and outcome of reactions, making it easier to control them in industrial and laboratory settings.

In the given exercise, we observe how enthalpy changes help compare two different pathways of converting methane into other compounds, demonstrating the significant role of energy considerations in chemical reactions.

calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. This helps in understanding the energy dynamics of any chemical process. In laboratory settings, calorimetry measures the enthalpy change of chemical reactions by observing temperature changes in a solution or calorimeter surroundings.

• By knowing the specific heat capacity of a system and the mass of the reactants, calorimetric measurements can determine the enthalpy change associated with a reaction.
• Calorimetry can be used for various applications, including combustion reactions, such as those involving hydrocarbons like methane.

In the exercise given, even though specific calorimetric data isn't mentioned, understanding the enthalpy changes of reactions is a fundamental step facilitated through calorimetric measurements in real scenarios.