Question

The methane molecule, \(\mathrm{CH}_{4}\), has the geometry shown in Figure 2.21. Imagine a hypothetical process in which the methane molecule is "expanded," by simultaneously extending all four \(\mathrm{C}-\mathrm{H}\) bonds to infinity. We then have the process $$ \mathrm{CH}_{4}(g) \cdots \mathrm{C}(g)+4 \mathrm{H}(g) $$ (a) Compare this process with the reverse of the reaction that represents the standard enthalpy of formation. (b) Calculate the enthalpy change in each case. Which is the more endothermic process? What accounts for the difference in \(\Delta H^{\circ}\) values? (c) Suppose that \(3.45 \mathrm{~g} \mathrm{CH}_{4}(g)\) is reacted with \(1.22 \mathrm{~g} \mathrm{~F}_{2}(\mathrm{~g})\), forming \(\mathrm{CF}_{4}(\mathrm{~g})\) and \(\mathrm{HF}(\mathrm{g})\) as sole products. What is the limiting reagent in this reaction? If the reaction occurs at constant pressure, what amount of heat is evolved?

Short answer

The hypothetical process of expanding a methane molecule involves breaking all \(\mathrm{C}-\mathrm{H}\) bonds and is endothermic, while the standard enthalpy of formation of methane involves forming these bonds and is exothermic. The hypothetical process is more endothermic. In the reaction between methane and fluorine gas, fluorine is the limiting reagent. The heat evolved in the reaction can be calculated as the product of the moles of the limiting reagent (fluorine) and the enthalpy change for the reaction. The process results in a net release of energy due to the breaking of C-H bonds and the formation of C-F and H-F bonds.

Step-by-step solution

Compare the hypothetical process with the reverse of the reaction that represents the standard enthalpy of formation

The hypothetical process is given as: \[ \mathrm{CH}_{4}(g) \cdots \mathrm{C}(g)+4 \mathrm{H}(g) \] The standard enthalpy of formation for methane is the reverse of the reaction in which methane is formed from its constituent elements in their standard states: \[ \mathrm{C}(s, \text{graphite}) + 2\mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g) \] Comparing the hypothetical process with the standard enthalpy of formation reaction, we can see that the hypothetical process involves the breaking of all \(\mathrm{C}-\mathrm{H}\) bonds, whereas the standard enthalpy of formation involves the formation of these bonds.

Calculate the enthalpy change in both cases

For both cases, we can calculate the enthalpy changes from the standard enthalpies of formation of the respective reactants and products. For the hypothetical process, we have: \[ \Delta H^{\circ}_{\text{hyp}} = [\mathrm{C}(g)+4 \mathrm{H}(g)] - [\mathrm{CH}_{4}(g)] = [\Delta H_f^{\circ}(\mathrm{C}(g)) + 4\Delta H_f^{\circ}(\mathrm{H}(g))] - [\Delta H_f^{\circ}(\mathrm{CH}_{4}(g))] \] Since the standard enthalpies of formation for elements in their standard state are zero, the change in enthalpy due to breaking the bonds is equal to the negative of the standard enthalpy of formation of methane: \[ \Delta H^{\circ}_{\text{hyp}} = -\Delta H_f^{\circ}(\mathrm{CH}_{4}(g)) \] For the reverse reaction that represents the standard enthalpy of formation, we have: \[ \Delta H^{\circ}_{\text{reverse}} = [\mathrm{CH}_{4}(g)] - [\mathrm{C}(s,\text{graphite}) + 2\mathrm{H}_{2}(g)] = [\Delta H_f^{\circ}(\mathrm{CH}_{4}(g))] - [\Delta H_f^{\circ}(\mathrm{C}(s,\text{graphite})) + 2\Delta H_f^{\circ}(\mathrm{H}_{2}(g))] \] Since the standard enthalpies of formation for elements in their standard state are zero, the change in enthalpy for this process is equal to the standard enthalpy of formation of methane: \[ \Delta H^{\circ}_{\text{reverse}} = \Delta H_f^{\circ}(\mathrm{CH}_{4}(g)) \] The standard enthalpy of formation of methane is available in standard reference texts and is approximately \(-74.8\mathrm{~kJ/mol}\). Comparing both processes, the hypothetical process is more endothermic, as it involves the breaking of the \(\mathrm{C}-\mathrm{H}\) bonds.

Determine the limiting reagent and amount of heat evolved in a reaction between methane and fluorine gas

First, let's write down the balanced chemical equation for the reaction between methane and fluorine gas: \[ \mathrm{CH}_{4}(g) + 4\mathrm{F}_{2}(g) \rightarrow \mathrm{CF}_{4}(g) + 4\mathrm{HF}(g) \] Now, let's determine the moles of methane and fluorine using their respective masses and molar masses: Moles of methane = \(\frac{3.45 \mathrm{~g}}{16.04 \mathrm{~g/mol}} = 0.215 \mathrm{~mol}\) Moles of fluorine = \(\frac{1.22 \mathrm{~g}}{38 \mathrm{~g/mol} \times 2} = 0.0161 \mathrm{~mol}\) By comparing the moles of the reactants with their stoichiometric coefficients, we can see that fluorine is the limiting reagent, as 0.0161 moles of fluorine are enough to react with only 0.00403 moles of methane. Heat evolved in the reaction can be calculated as the product of the moles of the limiting reagent (fluorine) and the enthalpy change for the reaction: q (heat evolved) = 0.0161 mol \(\times \Delta H_{\text{rxn}}\). The value for \(\Delta H_{\text{rxn}}\) is not provided in the problem. However, based on the given information, it is possible to estimate the amount of heat evolved. Since the reaction between methane and fluorine involves the breaking of C-H bonds and the formation of C-F and H-F bonds, which are generally more exothermic, this process will likely result in a net release of energy.

Key concepts

Methane Molecule

The methane molecule, known chemically as \(\mathrm{CH}_4\), is a simple hydrocarbon compound made up of one carbon atom bonded to four hydrogen atoms. It is the main component of natural gas and plays a vital role in the energy sector. Methane is of significant interest in chemistry due to its structure and interactions. In the methane molecule, the bonds are arranged in a three-dimensional tetrahedral shape. This configuration minimizes repulsion between the bonds, making it a stable and common molecule in organic chemistry.

• The carbon-hydrogen bonds in methane are covalent, meaning they involve the sharing of electrons between carbon and hydrogen.
• The geometry of methane is critical as it determines how the molecule interacts with others, affecting how it reacts under different conditions.
• The bond angle between the hydrogen atoms in methane is approximately 109.5 degrees, a characteristic of the tetrahedral shape.

Understanding the methane molecule helps in grasping broader concepts like hydrocarbons, molecular geometry, and chemical bonding.

Endothermic Process

An endothermic process is a chemical reaction or change that absorbs energy from its surroundings, typically in the form of heat. In the context of breaking down methane (\(\mathrm{CH}_4\)), the process involves breaking all the \(\mathrm{C}-\mathrm{H}\) bonds, requiring energy input. This contrasts with exothermic processes that release energy.

• The breaking of \(\mathrm{C}-\mathrm{H}\) bonds in methane is an endothermic process because it requires energy to overcome the bond energy holding the atoms together.
• Such processes are crucial in various industrial and chemical applications, where controlling the absorption and release of energy is necessary.
• In thermodynamics, endothermic reactions result in a positive change in enthalpy (\(\Delta H\)), indicating that the system has absorbed energy.

This concept is essential for understanding reaction energetics and the conditions under which certain reactions will occur.

Limiting Reagent

The concept of a limiting reagent plays a crucial role in understanding chemical reactions. The limiting reagent is the reactant that is entirely consumed first during a chemical reaction, determining the maximum amount of product that can be formed. When methane and fluorine react, as in the exercise, identifying the limiting reagent allows for accurate prediction of reaction outcomes.

• In the given reaction, \(\mathrm{CH}_4 + 4\mathrm{F}_2 \rightarrow \mathrm{CF}_4 + 4\mathrm{HF}\), calculating the moles of each reactant helps determine which is the limiting reagent.
• Fluorine, with fewer moles available relative to methane, becomes the limiting reagent.
• Recognizing the limiting reagent is vital in industrial processes to optimize reactant use and reduce waste.

Understanding this concept is foundational for students to advance in stoichiometry and efficiently carry out chemical reactions.