Question

We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethane: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$ \begin{aligned} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ} &=-890.3 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ} &=-571.6 \mathrm{~kJ} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ} &=-3120.8 \mathrm{~kJ} \end{aligned} $$

Short answer

The enthalpy change for the conversion of methane to ethane can be calculated using Hess's Law and the given thermochemical data. By rearranging and modifying the thermochemical reactions, we can obtain the target reaction: \(2 \textrm{CH}_4(g) \rightarrow \textrm{C}_2 \textrm{H}_6(g) + \textrm{H}_2(g)\). Summing up the enthalpy changes of the rearranged reactions, we find the enthalpy change for the target reaction to be \(\Delta H^{\circ} = -384.3\textrm{~kJ}\).

Step-by-step solution

Write down the target reaction.

First, write down the target reaction whose enthalpy change needs to be found: \(2 \textrm{CH}_4(g) \rightarrow \textrm{C}_2 \textrm{H}_6(g) + \textrm{H}_2(g)\)

Rearrange and modify the given thermochemical reactions.

Now, we need to rearrange the given thermochemical equations to match the target reaction. We might need to multiply or divide the equations to achieve this. Keep in mind that if you reverse a reaction, the sign of the enthalpy change will also reverse. If you multiply or divide a reaction by any factor, the enthalpy change should also be multiplied or divided by the same factor accordingly. 1. In the first given reaction, the methane and water molecules are on opposite sides of the equation compared to the target reaction, so to fix it, reverse the first reaction and change the sign of its enthalpy change: \( \textrm{CO}_2(g) + 2 \textrm{H}_2 \textrm{O}(l) \rightarrow \textrm{CH}_4(g) + 2 \textrm{O}_2(g) \quad \Delta H^{\circ} = 890.3 \textrm{~kJ}\) 2. In the second given reaction, we have \(\textrm{H}_2(g)\) on the left side, while it is on the right side in the target equation. Therefore, we need to reverse this reaction as well and change the sign of its enthalpy change: \(\textrm{H}_{2} \textrm{O}(l) \rightarrow \frac{1}{2} \textrm{O}_{2}(g) + \textrm{H}_{2}(g) \quad \Delta H^{\circ} = 285.8 \textrm{~kJ}\) 3. The third reaction has ethane molecules on the left side like the target equation, but it has two ethane molecules, while the target reaction has only one. To fix this, divide the third reaction by 2 and also divide its enthalpy change by 2: \( \textrm{C}_{2} \textrm{H}_{6}(g) + \frac{7}{2} \textrm{O}_{2}(g) \rightarrow 2\textrm{CO}_{2}(g) + 3 \textrm{H}_{2} \textrm{O}(l) \quad \Delta H^{\circ} = -1560.4 \textrm{~kJ}\)

Add the rearranged thermochemical equations and enthalpy changes to get the target reaction.

Now, we can sum up these rearranged thermochemical equations and their enthalpy changes to obtain the target reaction: \( \textrm{CO}_2(g) + 2 \textrm{H}_2\textrm{O}(l) \rightarrow \textrm{CH}_4(g) + 2 \textrm{O}_2(g) \quad \Delta H^{\circ} = 890.3 \textrm{~kJ}\) \(+ \textrm{H}_{2} \textrm{O}(l) \rightarrow \frac{1}{2} \textrm{O}_{2}(g) + \textrm{H}_{2}(g) \quad \Delta H^{\circ} = 285.8 \textrm{~kJ}\) \(- \textrm{C}_{2} \textrm{H}_{6}(g) + \frac{7}{2} \textrm{O}_{2}(g) \rightarrow 2\textrm{CO}_{2}(g) + 3 \textrm{H}_{2} \textrm{O}(l) \quad \Delta H^{\circ} = -1560.4 \textrm{~kJ}\) The result is: \(\textrm{2 CH}_4(g) \rightarrow \textrm{C}_2 \textrm{H}_6(g) + \textrm{H}_2(g) \quad \Delta H^{\circ} = (890.3 + 285.8 - 1560.4) \textrm{~kJ}\)

Calculate the enthalpy change for the target reaction.

Finally, calculate the enthalpy change for the target reaction by summing up the enthalpy changes of the rearranged thermochemical reactions: \(\Delta H^{\circ} = 890.3 \textrm{~kJ} + 285.8\textrm{~kJ} - 1560.4\textrm{~kJ} = -384.3\textrm{~kJ}\) Thus, the enthalpy change for the conversion of methane to ethane is: \(\Delta H^{\circ} = -384.3\textrm{~kJ}\)

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a fundamental concept in thermodynamics that refers to the heat absorbed or released during a chemical reaction at constant pressure. When discussing chemical reactions, this measurement helps us understand how much energy is involved in breaking and forming bonds. This value can be either negative, indicating an exothermic reaction where heat is released, or positive, indicative of an endothermic reaction where heat is absorbed.

• Exothermic Reaction: \( \Delta H < 0 \)
• Endothermic Reaction: \( \Delta H > 0 \)

Enthalpy change is important for calculating the viability and spontaneity of chemical processes. It shows whether a reaction releases energy that can be harnessed or needs additional energy to proceed. In Hess's Law, we use enthalpy change to find the overall energy change when converting reactants to products, even if the step-by-step energy changes are not directly measurable. By adding up enthalpy changes of a series of reactions that lead to the overall reaction, we can calculate the total enthalpy change for complex reactions.

Thermochemical Equations

A thermochemical equation is a chemical equation that includes not just the chemical changes but also the enthalpy change associated with the reaction. When writing these equations, it is important to remember:

• All states of reactants and products must be given (solid, liquid, gas, aqueous).
• The enthalpy change, \( \Delta H \), is included, often next to the equation, to indicate if the reaction absorbs or releases heat.

Thermochemical equations help track the energy flow in reactions and are crucial when applying Hess's Law. When rearranging or balancing these equations to match desired reactions, changes must reflect in \( \Delta H \) values. If a reaction is reversed, the sign of \( \Delta H \) also reverses. If you adjust the coefficients by any factor, you must apply the same factor to \( \Delta H \). This adjustment ensures the calculated enthalpy change represents the actual energy change occurring in that reaction scenario.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and formation of chemical bonds, resulting in new products. The study of these reactions often focuses on the exchange of energy and the identification of reactants and products over three main phases: gaseous, liquid, and solid states. This exchange of energy is what we are particularly interested in when studying enthalpy changes.

• Reactants: The starting substances in a reaction.
• Products: The new substances formed as a result of the reaction.
• Energy: Can be absorbed or released during the reaction process.

Hess's Law is essential for constructing a pathway of reactions that lead to an overall reaction when the direct measurement is complex. Through a series of calculations involving known thermochemical equations, Hess's Law allows us to deduce \( \Delta H \) for reactions otherwise not easily measured. Understanding how reactions occur and the energy involved helps chemists design more efficient, safer, and less energy-intensive processes.