Animated Solution for Chemistry - Chemical Thermodynamics: $(\Delta H-\Delta E)$ for the formation of carbon monoxide (CO) from its elements at $298\text{ K}$ is $(R=8.314{\text{ JK}}^{-1}{\text{ mol}}^{-1})$

In chemical thermodynamics, understanding the relationship between enthalpy ($\Delta H$) and internal energy ($\Delta E$ or $\Delta U$) is a fundamental skill. This problem asks us to find the difference $(\Delta H-\Delta E)$ for the formation of carbon monoxide (CO) from its constituent elements at standard conditions ($298\text{ K}$).

The first step in any thermodynamic problem involving a specific reaction is to write down the balanced chemical equation. Since we are dealing with the standard enthalpy of formation, we must write the reaction that forms exactly one mole of the product from its elements in their standard reference states.

For carbon monoxide, the elements are carbon and oxygen. The standard state of carbon is solid graphite, and for oxygen, it is diatomic gas ($O_2$).

Notice that we use a fractional coefficient for oxygen to ensure we only produce one mole of CO.

The First Law of Thermodynamics provides the bridge between enthalpy and internal energy for processes occurring at constant pressure:

Here, $\Delta n_g$ represents the change in the number of moles of gaseous substances during the reaction. By rearranging this equation, we can isolate the term we need to find:

This is where many students make a critical error. The term $\Delta n_g$ only accounts for species in the gaseous state. Solids and liquids do not expand significantly, so they do not contribute to the pressure-volume work term ($\Delta n_{g}RT$).

Let's calculate $\Delta n_g$ for our reaction:

Looking at our balanced equation: - Gaseous products: $1\text{ mole}$ of $CO(g)$ - Gaseous reactants: $\frac{1}{2}\text{ mole}$ of $O_2(g)$ - Solid reactants: Carbon $C(s)$ is ignored.

Now, we simply substitute the known values into our rearranged equation. We are given the universal gas constant $R=8.314{\text{ J K}}^{-1}{\text{ mol}}^{-1}$ and the temperature $T=298\text{ K}$.

Rounding to two decimal places, we get $1238.78{\text{ J mol}}^{-1}$. This perfectly matches option (b). The positive value indicates that the enthalpy change is greater than the internal energy change, which makes sense because the system does expansion work on the surroundings as the number of gaseous moles increases.