Animated Solution for Chemistry - Chemical Thermodynamics: The standard enthalpy of formation ($\Delta H_f^{\circ }$) at $298\text{ K}$ for methane, ${\text{CH}}_4(g)$ is $-74.8{\text{ kJ mol}}^{-1}$. The addition information required to determine the average energy for C—H bond formation would be

Imagine you are a chemical detective trying to figure out exactly how much energy is stored in a single carbon-hydrogen bond in methane (${\text{CH}}_4$). You are given a crucial clue: the standard enthalpy of formation of methane is $-74.8{\text{ kJ mol}}^{-1}$. But is this enough?

To solve this, we need to understand what the "standard enthalpy of formation" actually means. It is the energy change when one mole of a compound is formed from its constituent elements in their most stable standard states at $298\text{ K}$ and $1\text{ bar}$.

For methane, the elements are carbon and hydrogen. At room temperature, carbon exists as a solid (graphite), and hydrogen exists as a diatomic gas (${\text{H}}_2$). So, the formation reaction is:

Here is the catch: bond energy is strictly defined as the energy required to break one mole of bonds in a gaseous molecule to form gaseous atoms.

Our starting materials are solid carbon and diatomic hydrogen gas. We cannot directly measure the C-H bond energy from these states. We must first convert our reactants into isolated gaseous atoms. This is where Hess's Law comes to our rescue, allowing us to build a thermodynamic cycle.

Step 1: Sublimation of Carbon We must first vaporize the solid carbon into gaseous carbon atoms. This process requires energy, known as the enthalpy of sublimation ($\Delta H_{\text{sub}}$).

Step 2: Dissociation of Hydrogen Next, we have two moles of hydrogen gas (${\text{H}}_2$). We need to break the H-H bonds to get four individual gaseous hydrogen atoms. This requires the bond dissociation energy of hydrogen ($\Delta H_{\text{diss}}$). Since we have two moles of ${\text{H}}_2$, the energy needed is $2\times \Delta H_{\text{diss}}$.

Step 3: Formation of C-H Bonds Finally, the isolated gaseous carbon and hydrogen atoms combine to form methane. In this process, four C-H bonds are formed, releasing energy equal to four times the C-H bond energy ($4\times \Delta H_{\text{C-H}}$).

According to Hess's Law, the total enthalpy change of the direct formation reaction must equal the sum of the enthalpy changes of these three steps:

Looking at this master equation, it becomes crystal clear. To find the average C-H bond energy ($\Delta H_{\text{C-H}}$), knowing just the enthalpy of formation ($\Delta H_f^{\circ }$) is not enough. We absolutely must know the enthalpy of sublimation of carbon and the dissociation energy of hydrogen.

This elegant application of Hess's Law shows how macroscopic thermodynamic properties are intimately connected to the microscopic world of chemical bonds!