Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

The heat liberated on complete combustion of \(7.8 \mathrm{~g}\) benzene is \(327 \mathrm{~kJ}\). This heat was measured at constant volume and at \(27^{\circ} \mathrm{C}\). Calculate the heat of combustion of benzene at constant pressure \(\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\). (a) \(-3274 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-1637 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-3270 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-3637 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The heat of combustion of benzene at constant pressure is (a) \(-3274 \mathrm{~kJ} \mathrm{~mol}^{-1}\).

Step 1: Determine the molar mass of benzene

The molar mass of benzene (C₆H₆) is calculated by summing the atomic masses: \(C=12\times6=72\) and \(H=1\times6=6\). Thus, the molar mass of benzene is \(78 \; \text{g/mol}\).

Step 2: Calculate the moles of benzene combusted

Using the mass of benzene \( ( 7.8 \; \text{g} ) \) and its molar mass \((78 \; \text{g/mol})\), calculate the moles of benzene: \[\text{moles} = \frac{7.8 \; \text{g}}{78 \; \text{g/mol}} = 0.1 \; \text{mol}\]

Step 3: Calculate heat of combustion at constant volume per mole

Given that \(7.8 \; \text{g} \) of benzene releases \(327 \; \text{kJ}\), calculate the heat released per mole of benzene: \[327 \; \text{kJ} \text{ per } 0.1 \; \text{mol} = 3270 \; \text{kJ/mol}\] This is the heat of combustion at constant volume \((q_v) = -3270 \; \text{kJ/mol}\).

Step 4: Calculate the heat of combustion at constant pressure

Use the relationship between the heat at constant volume \((q_v)\) and constant pressure \((q_p)\) given by the formula: \[q_p = q_v + \Delta nRT\]In this combustion, \(\Delta n = (7-(6+3)) = -2\). Convert temperature to Kelvin: \(T = 27 + 273 = 300 \; \text{K}\). Thus,\[q_p = -3270 \; \text{kJ/mol} + (-2)(8.3 \times 10^{-3} \; \text{kJ/mol/K})(300 \; \text{K})\]\[q_p = -3270 \; \text{kJ/mol} - 4.98 \; \text{kJ/mol} \]\[q_p = -3274.98 \; \text{kJ/mol}\]

Step 5: Determine the heat of combustion value

Round the calculated value \(-3274.98 \; \text{kJ/mol}\) to match given answer choices and determine which option corresponds to this value. The closest option is (a) \(-3274 \; \text{kJ/mol}\).

Key Concepts

Thermochemistry

Thermochemistry is a branch of chemistry focused on the study of energy and heat associated with chemical reactions and changes in state. It is primarily concerned with measuring and understanding how and why heat is transferred during transformations. In reactions like combustion, thermochemical concepts help us quantify the energy released or absorbed.
In this exercise, we calculate the heat of combustion of benzene both at constant volume and constant pressure. This process involves understanding how heat changes under different conditions. Constant volume measurements, often called bomb calorimetry, involve no change in volume, while constant pressure processes, like those open to the atmosphere, focus on changes under atmospheric conditions. Both measurements provide insightful data on energy changes and are foundational in energy-related studies.

• Heat of combustion is often measured in kilojoules per mole (kJ/mol), representing the amount of energy released when one mole of a substance is combusted.
• The transition from constant volume to constant pressure requires accounting for work done on the system, which is calculated using the expression involving gas laws.

Molar Mass Calculation

Molar mass is the mass of one mole of a given particle—atoms, molecules, or ions—and is expressed in grams per mole (g/mol). It is crucial for converting between grams and moles, making it a fundamental step in stoichiometry.
In the exercise, determining the molar mass of benzene involves summing the atomic masses of carbon and hydrogen atoms in its formula, C₆H₆. Each carbon atom has a molar mass of 12 g/mol, and with six carbon atoms in benzene, we calculate 6 x 12 = 72 g/mol for carbon. Similarly, hydrogen, with a molar mass of 1 g/mol, contributes 6 g/mol for the six hydrogen atoms. Adding these, we find benzene’s molar mass is 78 g/mol.

• Correct molar mass calculation allows for accurate conversion between mass and moles, enabling precise stoichiometric calculations.
• Errors in calculating molar mass can lead to significant miscalculations in chemical reactions and energy evaluations.

Ideal Gas Law

The Ideal Gas Law is an essential concept in understanding gas behaviors and is often expressed as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles of gas, \(R\) is the ideal gas constant, and \(T\) is temperature.
In calculating the heat at constant pressure from the constant volume heat, we use a concept derived from the Ideal Gas Law, specifically \( \Delta nRT \). \( \Delta n \) refers to the change in moles of gas. For combustion reactions, this involves calculating the total moles of gaseous products and reactants to find \( \Delta n \).

• The Ideal Gas Law helps determine the work done by or against the atmosphere during reactions, which is key to understanding energy changes in processes involving gases.
• In our exercise, we converted the temperature to Kelvin (as required in gas law calculations) and used this in conjunction with the known values for R to determine the additional energy component added to the heat at constant volume to find the constant pressure difference.