The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at c

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

The volume occupied by liquids and solids is negligible compared to gases. Therefore, the work done by expansion or compression ($P\Delta V$) is almost entirely due to the change in the number of moles of gas.

Q: Why is water considered a liquid and not a gas in this reaction?

The standard state of water at $25^\circ \text{C}$ (298 K) is liquid. Unless specified otherwise or if the reaction occurs at a high temperature (above $100^\circ \text{C}$), water produced in combustion is taken as a liquid.

Q: What does the negative sign in $\Delta U$ and $\Delta H$ indicate?

The negative sign indicates that the reaction is exothermic, meaning heat is released from the system into the surroundings.

Q: Why did we divide $R$ by 1000?

The value of $\Delta U$ is given in $\text{kJ mol}^{-1}$, while the standard value of $R$ is $8.314 \text{ J K}^{-1} \text{mol}^{-1}$. To add the $\Delta n_g RT$ term to $\Delta U$, both must be in the same units. Dividing $R$ by 1000 converts it to $\text{kJ K}^{-1} \text{mol}^{-1}$.

Q: Is heat of combustion at constant volume always $\Delta U$?

Yes! By the first law of thermodynamics, $q = \Delta U - W$. At constant volume, work done ($W = -P\Delta V$) is zero. Therefore, the heat exchanged at constant volume ($q_v$) is exactly equal to the change in internal energy ($\Delta U$).

Question

The combustion of benzene ($l$) gives $CO_2(g)$ and $H_2O(l)$. Given that heat of combustion of benzene at constant volume is $-3263.9 	ext{ kJ mol}^{-1}$ at $25^\circ 	ext{C}$; heat of combustion (in $	ext{kJ mol}^{-1}$) of benzene at constant pressure will be ($R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$)

Options:

(A) $4152.6$

(B) $-452.46$

(C) $3260$

(D) $-3267.6$

Sigma Percentile · Accepted Answer

### The Thermodynamics of Burning Benzene: Constant Volume vs. Constant Pressure

Imagine you are a chemist tasked with measuring the energy released when benzene burns. You have two choices: you can burn it inside a sealed, rigid steel container called a bomb calorimeter, or you can burn it in an open beaker exposed to the atmosphere.

Will the heat released be the same? The answer is a resounding no, and this classic JEE problem explores exactly why.

### Analyzing the Setup

When we burn benzene in a bomb calorimeter, the volume is strictly constant. By the First Law of Thermodynamics, any heat exchanged at constant volume ($q_v$) is exactly equal to the change in internal energy, $\Delta U$. The problem states that $\Delta U = -3263.9 	ext{ kJ mol}^{-1}$.

However, if we burn it in an open container, the pressure is constant (atmospheric pressure). The heat exchanged at constant pressure ($q_p$) is equal to the change in enthalpy, $\Delta H$.

To find $\Delta H$, we must first write the perfectly balanced chemical equation for the combustion of liquid benzene:

$$C_6H_6(l) + \frac{15}{2}O_2(g) \longrightarrow 6CO_2(g) + 3H_2O(l)$$

**Notice the physical states carefully!** At $25^\circ 	ext{C}$ (298 K), benzene is a liquid, oxygen and carbon dioxide are gases, and water is a liquid. This distinction is the beating heart of the problem.

### The Master Equation

The relationship between enthalpy and internal energy for a chemical reaction involving gases is given by:

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

Here, $\Delta n_g$ represents the change in the number of moles of *gaseous* species only. Why? Because the volume occupied by liquids and solids is negligible compared to gases. The term $\Delta n_g RT$ essentially represents the pressure-volume work ($P\Delta V$) done by or on the system.

Let's calculate $\Delta n_g$:

$$\Delta n_g = n_{g(	ext{products})} - n_{g(	ext{reactants})}$$
$$\Delta n_g = 6 - \frac{15}{2} = -1.5$$

Because $\Delta n_g$ is negative, the volume of the system shrinks during the reaction.

### The Unit Trap and Final Calculation

Before we substitute our values, we must navigate a classic examiner's trap. The internal energy $\Delta U$ is given in $	ext{kJ mol}^{-1}$, but the universal gas constant $R$ is given as $8.314 	ext{ J K}^{-1} 	ext{mol}^{-1}$. We must convert $R$ to kiloJoules by multiplying by $10^{-3}$.

Let's plug everything into our master equation:

$$\Delta H = -3263.9 + (-1.5) 	imes (8.314 	imes 10^{-3}) 	imes 298$$
$$\Delta H = -3263.9 - 3.716$$
$$\Delta H = -3267.6 	ext{ kJ mol}^{-1}$$

### The Physical Insight

Did you get the feel of it? Why is $\Delta H$ more negative than $\Delta U$?

Because the number of gaseous moles decreased, the system contracted. The surrounding atmosphere pushed down on the system, doing work *on* it. This added work energy is subsequently released as extra heat. Therefore, the heat released at constant pressure ($\Delta H$) is slightly greater in magnitude than the heat released at constant volume ($\Delta U$).

Select Answer:

Visualized Solution

\text{Balanced Chemical Equation}

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

\text{Calculating } \Delta n_g

\text{Unit Conversion and Substitution}

\text{Raw Setup}

\text{Final Calculation}

\text{Physical Significance}

The Sigma Insight: First Law of Thermodynamics

The Thermodynamics of Burning Benzene

Analyzing the Setup

The Master Equation

The Unit Trap and Final Calculation

The Physical Insight

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