Animated Solution for Chemistry - Chemical Thermodynamics: For the complete combustion of ethanol, $C_2{H}_{5}OH(l)+3O_2(g)⟶2C{O}_2(g)+3H_{2}O(l)$ the amount of heat produced as measured in bomb calorimeter is $1364.47{\text{ kJ mol}}^{-1}$ at $25^{\circ }\text{C}$. Assuming ideality, the enthalpy of combustion, ${\Delta }_{c}H$ for the reaction will be $[R=8.314{\text{ J K}}^{-1}{\text{mol}}^{-1}]$

Inside the Bomb Calorimeter

Imagine a bomb calorimeter. It is a rigid, heavily reinforced steel container submerged in a precisely monitored water bath. When we ignite the ethanol inside this "bomb," the reaction occurs at a strictly constant volume. Because the volume cannot change, the system cannot do any expansion work on its surroundings ($P\Delta V=0$).

According to the First Law of Thermodynamics, the heat exchanged at constant volume ($q_v$) is exactly equal to the change in the system's internal energy. Therefore, the heat measured by the calorimeter directly gives us:

Connecting $\Delta U$ and $\Delta H$

While the bomb calorimeter gives us the internal energy change ($\Delta U$), the question asks for the enthalpy of combustion (${\Delta }_{c}H$). Enthalpy is defined as the heat exchanged at constant pressure ($q_p$), which is how most reactions occur in the real world (like burning fuel in an open beaker).

To bridge the gap between constant volume and constant pressure, we use the fundamental thermodynamic relation for chemical reactions:

Here, the term $\Delta n_{g}RT$ represents the expansion work the gases would do if the reaction were allowed to expand against atmospheric pressure.

Calculating $\Delta n_g$

To use our master equation, we must calculate $\Delta n_g$, the change in the number of moles of gaseous species. This is where many students make a critical error. You must look exclusively at the gases in the balanced chemical equation:

Notice the physical states! Ethanol and water are liquids, so their volumes are negligible compared to the gases. We only count the oxygen and carbon dioxide:

Taming the Gas Constant

Now, we gather our variables. The temperature is $25^{\circ }\text{C}$, which we must convert to Kelvin:

The gas constant $R$ is given as $8.314{\text{ J K}}^{-1}{\text{mol}}^{-1}$. Warning! Our internal energy $\Delta U$ is in kilo-Joules ($\text{kJ}$), but $R$ is in Joules ($\text{J}$). If you add them directly, your answer will be catastrophically wrong. We must divide $R$ by $1000$ to synchronize the units:

With all our traps avoided, we substitute the values into the master equation:

First, we evaluate the work term:

Finally, we add this to the internal energy:

Rounding to two decimal places, we get $-1366.95{\text{ kJ mol}}^{-1}$, which perfectly matches option (a). The negative sign confirms that the combustion is highly exothermic, releasing a massive amount of energy into the surroundings.