Q.Given the following data: $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g),\ \Delta H_1 = -393.5\ \text{kJ mol}^{-1}$; and $\text{CO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g),\ \Delta H_2 = -283.0\ \text{kJ mol}^{-1}$. Use Hess's law to calculate the enthalpy of formation of $\text{CO}(g)$ from its elements.

[!TLDR] Subtract the CO-to-CO2 combustion equation from the C-to-CO2 combustion equation to isolate the target formation reaction. [!ANSWER] …

The target reaction is $\text{C}(s) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)$. Given: (i) $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$, $\Delta H_1 = -393.5\ \text{kJ mol}^{-1}$; (ii) $\text{CO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g)$, $\Delta H_2 = -283.0\ \text{kJ mol}^{-1}$. Subtracting equation (ii) from equation (i): $\left[\text{C}(s)+\text{O}_2(g)\right] - \left[\text{CO}(g)+\tfrac{1}{2}\text{O}_2(g)\right] \rightarrow \text{CO}_2(g) - \text{CO}_2(g)$, which simplifies (the $\text{CO}_2(g)$ cancels on the product side, and $\text{O}_2(g) - \tfrac12\text{O}_2(g) = \tfrac12\text{O}_2(g)$ remains on the reactant side, while $-\text{CO}(g)$ moves to become $+\text{CO}(g)$ on the product side) to exactly the target equation: $\text{C}(s) + \tfrac{1}{ …

Manipulate the two given equations algebraically (here, subtracting one from the other) until they combine to give the target equation, then …

Adding the two enthalpies instead of subtracting; making a sign error when su …