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: $$ \begin{aligned} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ} &=-890.3 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ} &=-571.6 \mathrm{~kJ} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ} &=-3120.8 \mathrm{~kJ} \end{aligned} $$

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

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: \( \textrm{C}_{2} \textrm{H}_{6}(g) + \frac{7}{2} \textrm{O}_{2}(g) \rightarrow 2\textrm{CO}_{2}(g) + 3 \textrm{H}_{2} \textrm{O}(l) \quad \Delta H^{\circ} = -1560.4 \textrm{~kJ}\)

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: \(\textrm{2 CH}_4(g) \rightarrow \textrm{C}_2 \textrm{H}_6(g) + \textrm{H}_2(g) \quad \Delta H^{\circ} = (890.3 + 285.8 - 1560.4) \textrm{~kJ}\)

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: \(\Delta H^{\circ} = -384.3\textrm{~kJ}\)