Question

Chloroform, \(\mathrm{CHCl}_{3},\) is formed from methane and chlorine in the following reaction. $$\mathrm{CH}_{4}(\mathrm{g})+3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{HCl}(\mathrm{g})+\mathrm{CHCl}_{3}(\mathrm{g})$$ Calculate \(\Delta H_{\mathrm{rxn}}^{\circ}\), the enthalpy change for this reaction, using the enthalpy of formation of \(\mathrm{CHCl}_{3}(\mathrm{g}), \Delta \mathrm{H}_{f}^{\circ}=\) \(-103.1 \mathrm{kJ} / \mathrm{mol}),\) and the enthalpy changes for the following reactions: $$\begin{aligned}\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) & \\\\\Delta H_{\mathrm{rxn}}^{\circ} &=-890.4 \mathrm{kJ}\end{aligned}$$ $$\begin{array}{ll}2 \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) & \Delta H_{\mathrm{rxn}}^{\circ}=+184.6 \mathrm{kJ} \\\\\mathrm{C}(\text { graphite })+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) & \Delta H_{f}^{\circ}=-393.5 \mathrm{kJ} \\\\\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell) & \Delta H_{f}^{\circ}=-285.8 \mathrm{kJ}\end{array}$$

Short answer

The enthalpy change for the reaction is \(-305.2 \, \mathrm{kJ}\).

Step-by-step solution

Write the Equation Using Enthalpies

Use Hess's law to find the enthalpy change \( \Delta H_{\mathrm{rxn}}^{\circ} \) for the overall reaction. We are provided with several reactions and their enthalpies. We want the target reaction: \( \mathrm{CH}_{4}(\mathrm{g}) + 3 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{HCl}(\mathrm{g}) + \mathrm{CHCl}_{3}(\mathrm{g}) \). We have \( \Delta H_{f}^{\circ} \) for \( \mathrm{CHCl}_{3} \), among other reactions.

Calculate Formation of Methane

To work with Hess's law effectively, calculate the formation enthalpy of methane \( \mathrm{CH}_{4} \) \((\Delta H_{f}^{\circ}) \). Rearrange the provided reaction: \( \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g}) \) with \( \Delta H_{\mathrm{rxn}}^{\circ} = -890.4 \), and use the given \( \Delta H_{f}^{\circ} \'s \) for water and carbon dioxide: \(\Delta H_{f}(\mathrm{H}_{2}\mathrm{O}) = -285.8\,\mathrm{kJ}/\mathrm{mol}\) and \(\Delta H_{f}(\mathrm{CO}_{2}) = -393.5\,\mathrm{kJ}/\mathrm{mol}\).\[\Delta H_{f}(\mathrm{CH}_{4}) = \Delta H_{\mathrm{rxn}} - \left( 2 \times \Delta H_{f}(\mathrm{H}_{2}\mathrm{O}) + \Delta H_{f}(\mathrm{CO}_{2}) \right) \\Delta H_{f}(\mathrm{CH}_{4}) = -890.4 - \left( 2 \times -285.8 + (-393.5) \right) = -74.8\,\mathrm{kJ}/\mathrm{mol}\]

Calculate Decomposition of HCl

Use the provided equation \( 2 \mathrm{HCl}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \) with \( \Delta H_{\mathrm{rxn}} = +184.6\,\mathrm{kJ} \) to find \( \Delta H_{f}(\mathrm{HCl}) \). Reverterse the reaction's direction changing sign of \( \Delta H \).\[\mathrm{H}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g}) \\Delta H_{\mathrm{rxn}} = -184.6\,\mathrm{kJ}\] Then, divide by 2: \[\Delta H_{f}(\mathrm{HCl}) = \frac{-184.6}{2}\,\mathrm{kJ}/\mathrm{mol} = -92.3\,\mathrm{kJ}/\mathrm{mol}\]

Combine to Find Overall Reaction

To find \( \Delta H_{\mathrm{rxn}} \) of the requested reaction, sum the enthalpies of formation of products minus those of reactants, using coefficients from the balanced equation. (Note: The chlorine molecule \( \mathrm{Cl}_{2} \) is an element in its standard state, so \( \Delta H_{f} = 0 \) for \( \mathrm{Cl}_{2} \)) \[\Delta H_{\mathrm{rxn}} = \left( 3 \times -92.3 + (-103.1) \right) - \left( (-74.8) + 0 \right) \] Calculate: \[\Delta H_{\mathrm{rxn}} = \left(-276.9 -103.1 \right) + 74.8 = -305.2 \,\mathrm{kJ}\]

Conclusion: Result

The calculated enthalpy change for the overall reaction \( \mathrm{CH}_{4}(\mathrm{g}) + 3 \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{HCl}(\mathrm{g}) + \mathrm{CHCl}_{3}(\mathrm{g}) \) is \( \Delta H_{\mathrm{rxn}}^{\circ} = -305.2\,\mathrm{kJ}\).

Key concepts

Enthalpy of Formation

Enthalpy of formation is a central idea in thermochemistry. It refers to the heat change involved in the formation of a compound from its elements in their standard states. Consider a compound like chloroform (\( \text{CHCl}_3 \)). The enthalpy of formation, denoted as \( \Delta H_f^{\circ} \), represents the energy change when 1 mole of chloroform is formed from its elements, carbon, hydrogen, and chlorine, in standard conditions (1 atm pressure and 298 K temperature).

• Elements: Pure substances like \( \text{C} \) (graphite) and \( \text{Cl}_2 \) (\( \text{g} \))
• Standard state: The most stable state of the element at 1 atm and 298 K

For \( \text{CHCl}_3 \), the given \( \Delta H_f^{\circ} = -103.1 \, \text{kJ/mol} \). This negative sign indicates it's exothermic, meaning energy is released during its formation.

Enthalpy Change

Enthalpy change, represented as \( \Delta H_{\text{rxn}} \), is the energy change during a chemical reaction. It's vital for understanding whether a reaction absorbs or releases heat.In the context of chloroform formation, the total enthalpy change is calculated using Hess's Law, which states that enthalpy change is the same, regardless of the pathway from reactants to products. :

• Combining known reactions: Utilize provided reactions to compute \( \Delta H_{\text{rxn}} \)
• Energy balance: Sum the enthalpies of products and subtract reactants' enthalpies

In our exercise, multiple equations with known enthalpy values assist in calculating the overall \( \Delta H_{\text{rxn}} \) of the chloroform formation:\[ \Delta H_{\text{rxn}} = \text{(products)} - \text{(reactants)}\]For \( \text{CH}_4 \):\[ \Delta H_{f}^{\circ} \text{(CH}_4\text{)} = -74.8 \, \text{kJ/mol}\]

Chemical Reactions

Chemical reactions involve breaking and forming bonds, transforming reactants into products. Understanding reactions is key to calculating enthalpy changes.Take the reaction:\[ \text{CH}_4 (g) + 3 \text{Cl}_2 (g) \rightarrow 3 \text{HCl} (g) + \text{CHCl}_3 (g)\]Here, methane reacts with chlorine gas to produce hydrogen chloride and chloroform.

• Reactants: \( \text{CH}_4 \) and \( \text{Cl}_2 \) are the starting chemicals
• Products: \( \text{HCl} \) and \( \text{CHCl}_3 \) are the resulting substances

During this transformation, bonds in methane and chlorine molecules break, which requires energy. Simultaneously, new bonds form in hydrogen chloride and chloroform, releasing energy. Overall, understanding these processes helps solve the enthalpy change calculations thoroughly, as highlighted in this exercise, emphasizing the concept of state standardization and balanced reactions.