The heats of combustion of carbon and carbon monoxide are $$-393.5$$ and $$-283.5$$ kJ mol$$^{-1}$$, respectively. The heat of formation (in kJ) of carbon monoxide per mole is:

First, let us recall what is meant by “heat of combustion”. It is the enthalpy change when one mole of a substance burns completely in oxygen, forming the most stable oxide(s). Mathematically we write the combustion reactions as:

$$\text{(1) } C\;(s) + O_2\;(g) \rightarrow CO_2\;(g), \qquad \Delta H_1 = -393.5\ \text{kJ mol}^{-1}$$

$$\text{(2) } CO\;(g) + \tfrac12 O_2\;(g) \rightarrow CO_2\;(g), \qquad \Delta H_2 = -283.5\ \text{kJ mol}^{-1}$$

The quantity we are asked to find is the standard heat of formation of carbon monoxide. By definition, the heat (enthalpy) of formation of a compound is the enthalpy change for forming one mole of that compound from its constituent elements in their standard states. For carbon monoxide the formation reaction is:

$$\text{(target) } C\;(s) + \tfrac12 O_2\;(g) \rightarrow CO\;(g), \qquad \Delta H_f = ?$$

To obtain this unknown ΔHf, we apply Hess’s law. Hess’s law states that if a reaction can be expressed as the algebraic sum of two or more other reactions, then its enthalpy change is the corresponding algebraic sum of the enthalpy changes of those reactions.

We already have two reactions, (1) and (2). If we can combine them algebraically to give the target reaction, their ΔH values can be combined in the same way to yield ΔHf.

Observe that reaction (2) has $$CO$$ on the reactant side, but we want $$CO$$ on the product side in the target equation. Therefore, we reverse reaction (2). When a reaction is reversed, the sign of ΔH is also reversed. So, reversing (2) gives:

$$\text{(2ʹ) } CO_2\;(g) \rightarrow CO\;(g) + \tfrac12 O_2\;(g), \qquad \Delta H_{2ʹ} = +283.5\ \text{kJ mol}^{-1}$$

Now we add reaction (1) and reaction (2ʹ) term by term:

$$ \begin{aligned} &C\;(s) + O_2\;(g) &\rightarrow&\; CO_2\;(g) &&\quad \Delta H_1 = -393.5\ \text{kJ} \\ +&\; CO_2\;(g) &\rightarrow&\; CO\;(g) + \tfrac12 O_2\;(g) &&\quad \Delta H_{2ʹ} = +283.5\ \text{kJ} \\ \hline &C\;(s) + \cancel{O_2\;(g)} + \cancel{CO_2\;(g)} &\rightarrow&\; CO\;(g) + \tfrac12 \cancel{O_2\;(g)} && \\ \end{aligned} $$

After canceling the common species $$CO_2\;(g)$$ and $$\tfrac12 O_2\;(g)$$, we are left exactly with the target formation reaction:

$$C\;(s) + \tfrac12 O_2\;(g) \rightarrow CO\;(g)$$

According to Hess’s law, the enthalpy change for the target reaction is the algebraic sum of $$\Delta H_1$$ and $$\Delta H_{2ʹ}$$:

$$\Delta H_f = \Delta H_1 + \Delta H_{2ʹ} = (-393.5\$$ kJ $$) + (+283.5\$$ kJ $$) = -110.0\$$ kJ

The numerical value matches, and the negative sign indicates that the formation of carbon monoxide from its elements releases heat (it is exothermic).

Hence, the correct answer is Option B.