Animated Solution for Chemistry - Chemical Thermodynamics: The heats of combustion of carbon and carbon monoxide are $-393.5$ and $-283.5{\text{ kJ mol}}^{-1}$, respectively. The heat of formation (in kJ) of carbon monoxide per mole is

Imagine we are mapping out the energy states of carbon and its oxides. We always start by defining a baseline, a zero-energy reference line. In thermochemistry, this baseline is formed by elements in their standard states. For our system, this means solid carbon, $C(s)$, and oxygen gas, $O_2(g)$, sit right at the $H=0$ mark.

When carbon undergoes complete combustion, it reacts with oxygen to form carbon dioxide, $C{O}_2(g)$. The problem tells us this process releases $393.5{\text{ kJ mol}}^{-1}$ of energy. Because energy is released, the enthalpy change is negative, meaning $C{O}_2(g)$ sits deep down in our energy well at $-393.5{\text{ kJ mol}}^{-1}$.

Now, let's look at carbon monoxide, $CO(g)$. We are given its heat of combustion as well. When $CO(g)$ burns, it also forms $C{O}_2(g)$, releasing $283.5{\text{ kJ mol}}^{-1}$.

This is a crucial piece of information! It tells us that carbon monoxide is an intermediate state. It has less energy than pure carbon and oxygen, but more energy than fully oxidized carbon dioxide. It sits on a 'shelf' partway down our energy well.

Our ultimate goal is to find the heat of formation of carbon monoxide. This is the enthalpy change for the direct reaction from our standard elements to $CO(g)$:

According to Hess's Law, the total enthalpy change of a reaction is independent of the pathway taken. It is simply a manifestation of the conservation of energy. Therefore, the energy drop from the top level directly to the $CO(g)$ shelf must equal the total drop to the bottom minus the drop from the shelf to the bottom.

Mathematically, we can set up the equations:

1. $C(s)+O_2(g)\to C{O}_2(g)\Delta H_1=-393.5{\text{ kJ mol}}^{-1}$ 2. $CO(g)+\frac{1}{2}{O}_2(g)\to C{O}_2(g)\Delta H_2=-283.5{\text{ kJ mol}}^{-1}$

To get our target equation, we simply subtract equation (2) from equation (1):

Rearranging this gives us exactly what we want:

Since we subtracted the chemical equations, we must do the exact same operation to their enthalpy values:

The heat of formation of carbon monoxide is $-110.5{\text{ kJ mol}}^{-1}$ (accounting for slight variations in standard data, the closest option is $-110.5$). This tells us that while forming $CO(g)$ releases energy, it still holds a significant amount of chemical potential energy compared to $C{O}_2(g)$, making it an excellent industrial fuel!