Animated Solution for Chemistry - Chemical Thermodynamics: The value of enthalpy change ($\Delta H$) for the reaction ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}(\mathrm{l})+3{\mathrm{O}}_2(\mathrm{g})\to 2C{\mathrm{O}}_2(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$ at $27^{\circ }\mathrm{C}$ is $-1366.5{\text{ kJ mol}}^{-1}$. The value of internal energy change for the above reaction at this temperature will be

The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is one of the most fundamental concepts in chemical thermodynamics. It bridges the gap between the heat exchanged at constant pressure and the heat exchanged at constant volume.

The reaction produces carbon dioxide gas ($\mathrm{C}{\mathrm{O}}_2$) and liquid water (${\mathrm{H}}_2\mathrm{O}$).

The problem states that the enthalpy change ($\Delta H$) is $-1366.5{\text{ kJ mol}}^{-1}$. The negative sign indicates that this is an exothermic reaction—heat is being released into the surroundings. But what about the internal energy change ($\Delta U$)? To find that, we need to account for the work done by or on the gases during the reaction.

The first law of thermodynamics gives us the beautiful relationship:

Here, $\Delta n_g$ represents the change in the number of moles of gaseous substances. Why only gases? Because the volume occupied by solids and liquids is negligible compared to gases, so they don't contribute significantly to expansion or compression work ($P\Delta V$).

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

Looking at the balanced chemical equation:

On the product side, we have $2$ moles of $\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$. On the reactant side, we have $3$ moles of ${\mathrm{O}}_2(\mathrm{g})$. Notice how we completely ignore the liquid ethanol and liquid water!

A negative $\Delta n_g$ means the number of gas moles decreases. The system is effectively being compressed by the atmosphere, meaning work is done on the system.

Now, we substitute our known values into the master equation. Crucial Trap: We must ensure all energy units match! Since $\Delta H$ is in kilojoules ($\text{kJ}$), we must convert the universal gas constant $R$ from Joules to kilojoules by multiplying by $10^{-3}$.

Substituting these into the equation:

Let's evaluate the work term ($\Delta n_{g}RT$):

Now, isolate $\Delta U$:

Rounding to one decimal place to match our options, we get:

This perfectly matches option (c). The internal energy change is slightly less negative than the enthalpy change because the atmosphere did some work on the system, adding a bit of energy back into it!