Animated Solution for Chemistry - Chemical Thermodynamics: The difference between $\Delta H$ and $\Delta U$ ($\Delta H-\Delta U$), when the combustion of one mole of heptane (l) is carried out at a temperature $T$, is equal to

Mastering $\Delta H$ and $\Delta U$

Thermochemistry often feels like a maze of abstract variables, but at its core, it is just the accounting of energy. When we burn a fuel like heptane, energy is released. But how we measure that energy depends on the conditions of our experiment. This brings us to the classic relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$).

The relationship between the heat exchanged at constant pressure ($\Delta H$) and the heat exchanged at constant volume ($\Delta U$) is governed by the work done by the gases expanding or contracting during the reaction. The master equation is:

Here, $\Delta n_g$ is the absolute star of the show. It represents the change in the number of moles of gaseous substances. Why only gases? Because the volume occupied by solids and liquids is practically negligible compared to gases. Therefore, any significant expansion work ($P\Delta V$) is entirely due to the creation or consumption of gas molecules.

To find $\Delta n_g$, we must first write a perfectly balanced chemical equation for the combustion of heptane ($C_7{H}_{16}$). Combustion means reacting the hydrocarbon with oxygen gas ($O_2$) to produce carbon dioxide ($C{O}_2$) and water ($H_{2}O$).

The general formula for burning any hydrocarbon $C_x{H}_y$ is incredibly useful here:

For heptane, $x=7$ and $y=16$. Substituting these values, we get:

Which simplifies to our balanced equation:

This is where many students make a fatal error. You must pay extreme attention to the physical states of the reactants and products. Heptane is a liquid at room temperature. Oxygen and carbon dioxide are gases. But what about water?

Unless the problem explicitly states that the reaction occurs at a high temperature (above $100^{\circ }\text{C}$), we assume standard conditions where water is a liquid.

Now, let's calculate $\Delta n_g$ by counting only the gaseous moles:

With $\Delta n_g$ securely in hand, we return to our master equation to find the difference between $\Delta H$ and $\Delta U$:

Substituting our value of $\Delta n_g=-4$, we arrive at the final answer:

This negative sign tells us a physical story: because the number of gas moles decreased during the reaction, the system contracted. The atmosphere did work on the system, meaning the heat released at constant pressure ($\Delta H$) is slightly more negative (more exothermic) than the heat released at constant volume ($\Delta U$). Always respect the physical states, and the math will naturally fall into place!