The heat of combustion of ethanol into carbon dioxides and water is -327 kcal at constant pressure.

Q: Why do we only consider gaseous moles for $\Delta n_g$?

The term $\Delta n_g RT$ comes from the pressure-volume work done by the system ($P\Delta V$). For solids and liquids, the change in volume during a reaction is negligible compared to gases. Therefore, we assume only gases contribute to the expansion or compression work.

Q: Why is the heat of combustion given as a negative value?

Combustion is an exothermic process, meaning it releases heat into the surroundings. In thermodynamics, by convention, energy leaving the system is assigned a negative sign.

Q: What is the fundamental difference between $q_p$ and $q_v$?

$q_p$ is the heat exchanged at constant pressure, which allows the system to do work by expanding. It is equal to the enthalpy change ($\Delta H$). $q_v$ is the heat exchanged at constant volume, where no expansion work can be done. It is equal to the internal energy change ($\Delta U$).

Q: Why did we have to convert kcal to cal?

The universal gas constant $R$ was given as $2 \text{ cal mol}^{-1} \text{ K}^{-1}$. To add or subtract terms in an equation, they must have the exact same units. Since $R$ is in calories, we must convert $\Delta H$ from kilocalories to calories to ensure mathematical consistency.

Q: What if the temperature was given in Celsius, can we use it directly in the formula?

No, you must always convert temperature to Kelvin when using the ideal gas law or related thermodynamic equations like this one. The Kelvin scale is an absolute temperature scale, which is required for the math to work correctly.

Question

The heat of combustion of ethanol into carbon dioxides and water is $-327 	ext{ kcal}$ at constant pressure. The heat evolved (in cal) at constant volume and $27^\circ	ext{C}$ (if all gases behave ideally) is ($R = 2 	ext{ cal mol}^{-1} 	ext{ K}^{-1}$) .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to a fascinating exploration of chemical thermodynamics! Imagine you are in a laboratory, and you've just ignited a sample of ethanol. When this combustion occurs in an open beaker—meaning it happens at a **constant pressure**—the heat released is given to us as $-327 	ext{ kcal}$. In thermodynamics, the heat exchanged at constant pressure is exactly equal to the change in enthalpy, denoted as $\Delta H$.

But what if we performed this exact same combustion inside a sealed, rigid steel container, like a bomb calorimeter? In this scenario, the volume cannot change. The heat evolved at **constant volume** corresponds to the change in internal energy, denoted as $\Delta U$. Our mission is to find this value.

### The Chemical Reality

To connect these two thermodynamic quantities, we must first understand the chemical reality of the reaction. Let's write down the balanced chemical equation for the combustion of liquid ethanol:

$$C_2H_5OH(l) + 3O_2(g) \longrightarrow 2CO_2(g) + 3H_2O(l)$$

Look closely at the physical states of our reactants and products. This is where many students make a critical error! To relate $\Delta H$ and $\Delta U$, we need to determine the change in the number of **gaseous moles**, denoted as $\Delta n_g$. We only count the gases because solids and liquids do not change their volume significantly enough to perform appreciable pressure-volume work.

On the product side, we have $2$ moles of carbon dioxide gas. On the reactant side, we have $3$ moles of oxygen gas. Therefore, we calculate $\Delta n_g$ as:

$$\Delta n_g = 2 - 3 = -1$$

### The Master Equation

From the First Law of Thermodynamics, we derive the master equation that bridges enthalpy and internal energy:

$$\Delta H = \Delta U + \Delta n_g RT$$

This equation is a cornerstone of chemical thermodynamics. Since our goal is to find the heat at constant volume ($\Delta U$), we rearrange the equation:

$$\Delta U = \Delta H - \Delta n_g RT$$

### Final Calculation and Unit Traps

Before we substitute our values, we must navigate a classic trap: **inconsistent units**. The enthalpy change is given in kilocalories ($-327 	ext{ kcal}$), but the universal gas constant $R$ is given in calories ($2 	ext{ cal mol}^{-1} 	ext{ K}^{-1}$). We must convert $\Delta H$ to calories by multiplying by $1000$, giving us $-327000 	ext{ cal}$.

Now, let's substitute all our values into the rearranged equation. The temperature $T$ is $27^\circ	ext{C}$, which is $300 	ext{ K}$.

$$\Delta U = -327000 - (-1)(2)(300)$$

The term $\Delta n_g RT$ evaluates to $-600$. Subtracting this negative value is mathematically identical to adding $600$:

$$\Delta U = -327000 + 600 = -326400 	ext{ cal}$$

Since the question specifically asks for the "heat evolved", we provide the magnitude of this energy release. The final answer is **$326400 	ext{ cal}$**.

Enter Numerical Value:

Visualized Solution

\text{Heat of Reaction: } q_p \text{ vs } q_v

\text{Balanced Combustion Equation}

\text{Change in Gaseous Moles } (\Delta n_g)

\text{Relation between } \Delta H \text{ and } \Delta U

\text{Substitution of Values}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Enthalpy and Hess's Law

Analyzing the Setup

The Chemical Reality

The Master Equation

Final Calculation and Unit Traps

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