Q.Given the standard enthalpies of combustion: $\text{C}(s) = -393.5\ \text{kJ mol}^{-1}$, $\text{H}_2(g) = -285.8\ \text{kJ mol}^{-1}$, and $\text{C}_2\text{H}_6(g) = -1560\ \text{kJ mol}^{-1}$, use Hess's law to calculate the standard enthalpy of formation of ethane, $2\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)$.

[!TLDR] Combine the combustion enthalpies of carbon and hydrogen (as elements) with the combustion enthalpy of ethane to isolate the formation reaction. [!ANSWER …

The target reaction is $2\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)$. Multiply the combustion of carbon by 2: $2\text{C}(s) + 2\text{O}_2(g) \rightarrow 2\text{CO}_2(g)$, $\Delta H = 2(-393.5) = -787.0\ \text{kJ}$. Multiply the combustion of hydrogen by 3: $3\text{H}_2(g) + \tfrac{3}{2}\text{O}_2(g) \rightarrow 3\text{H}_2\text{O}(l)$, $\Delta H = 3(-285.8) = -857.4\ \text{kJ}$. Adding these gives the formation of $2\text{CO}_2 + 3\text{H}_2\text{O}$ from $2\text{C} + 3\text{H}_2$ (plus $\tfrac{7}{2}\text{O}_2$), total $\Delta H = -787.0 + (-857.4) = -1644.4\ \text{kJ}$. The combustion of ethane itself is $\text{C}_2\text{H}_6(g) + \tfrac{7}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$, $\Delta H_c = -1560\ \text{kJ mol}^{-1}$. Subtracting this combustion equation from the combined elemental-combus …

Scale the elemental combustion equations to match the target's stoichiometry, sum them, then subtract the ethane combustion equation to cancel the common combustion products and …

Forgetting to multiply the combustion enthalpies of C and H2 by their stoichiometric coefficients (2 and 3) before combining; adding instead of …