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) \(-3637 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short answer

(c) 9-3270 9kJ9mol^{-1}

Step-by-step solution

Understand the Problem

We need to find the heat of combustion of benzene at constant pressure, given the heat at constant volume. The given data is for the heat evolved when 7.8 g of benzene undergo combustion at constant volume, which is 327 kJ.

Relation between Constant Volume and Constant Pressure Heat

At constant volume, the heat is denoted as \( q_v \), and the relation to constant pressure heat \( q_p \) (also known as \( 9H \)) is given by the equation:\[ q_p = q_v + 1n 9RT \]where \( 1n \) is the change in the number of moles of gases during the reaction, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

Determine the Molar Mass and Moles of Benzene

Benzene, \( C_6H_6 \), has a molar mass of 78 g/mol. The moles of benzene combusted is given by:\[ 9n = \frac{97.8 9g}{78 9g/9mol} = 0.1 9mol \]

Calculate Change in Mole Number (\( 9n \))

The combustion reaction for benzene is:\[ 2C_6H_6 + 15O_2 9rightarrow 12CO_2 + 6H_2O \]The change in moles is: \( 9n = (12 + 6) - (2 + 15) = 18 - 17 = 1 \)

Convert Temperature to Kelvin

The temperature in Celsius is 27°C. Convert it to Kelvin:\[ T = 27 + 273 = 300 9K \]

Calculate \( nRT \)

Using \( R = 8.3 9J/9mol 9K \) and \( 1n = 1 \) (from Step 4), calculate:\[ nRT = 1 9mol 9× 8.3 9J/9mol 9K 9× 300 9K = 2490 9J = 2.49 9kJ \]

Find Heat of Combustion per Mole at Constant Pressure

Using step 2, apply:\[ q_p = q_v + 9nRT = -327 9kJ + 2.49 9kJ \]Now, convert the result to kJ per mole:Since 0.1 moles of benzene gives -324.51 kJ, 1 mole would give:\[ q_p = \frac{-324.51 9kJ}{0.1 9mol} = -3245.1 9kJ/9mol \]

Round to Closest Option

Round \( -3245.1 9kJ/9mol \) to the closest given option: \( -3270 9kJ/9mol \) is the closest choice.

Key concepts

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In this context, it's pivotal for understanding chemical reactions and combustion, as these processes are governed by the laws of thermodynamics. The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed.
In the case of benzene combustion, the chemical energy stored in molecular bonds is transformed. It changes into heat and work as benzene reacts with oxygen. Knowing how energy transitions from bonds to heat is crucial for calculating the heat of combustion, whether at constant pressure or volume.
Understanding the difference between constant pressure and constant volume processes helps us apply the right calculations according to the conditions of the experiment or industrial process.

Benzene Combustion

When benzene combusts, it reacts with oxygen in a burning reaction to form carbon dioxide and water. This type of reaction releases a significant amount of heat, which is known as the heat of combustion. The chemical equation for benzene combustion is:
\[ 2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O \]
This equation shows that two moles of benzene react with fifteen moles of oxygen to produce twelve moles of carbon dioxide and six moles of water. This reaction is exothermic, meaning it releases energy in the form of heat, due to the strong carbon-oxygen and hydrogen-oxygen bonds formed, which are more stable than the original bonds.
Understanding the stoichiometry of the reaction, meaning the balance of moles, is important. It helps us calculate not only the molecular transformations but also the quantitative aspects of energy changes involved.

Constant Pressure and Constant Volume Heat

In thermodynamics, heat measured either at constant pressure, often denoted as\( q_p \), or at constant volume, denoted as \( q_v \), reveals different aspects of energy transformations. The distinction between these two measurements is key in chemical thermodynamic calculations.
When a reaction occurs at constant volume, the system does no work, and the heat change is equal to the change in internal energy. This is the\( q_v \). Conversely, at constant pressure, where the system might do work by expanding, the heat change, \( q_p \), is greater by the work done in expansion. Hence, the relation \( q_p = q_v + \Delta nRT \) where \( \Delta n \) is the change in moles of gas, \( R \) is the gas constant, and \( T \) the temperature in Kelvin.
This difference arises because, at constant pressure, there is the potential for gas expansion or contraction, requiring more energy input or release beyond internal energy changes alone. Recognizing and calculating this is crucial for accurate thermodynamic analyses in chemical reactions, as seen in the benzene combustion problem.