Question

A 1.800 -g sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is \(11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter plus contents increased from \(21.36^{\circ} \mathrm{C}\) to \(26.37{ }^{\circ} \mathrm{C}\). (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

Short answer

The balanced chemical equation for the combustion of phenol (C6H5OH) is: \(C_6H_5OH_{(l)} + 7.5O_2_{(g)} \rightarrow 6CO_2_{(g)} + 3H_2O_{(l)}\) The total heat released during the combustion process is \(58.29\, \mathrm{kJ}\). The heat of combustion per gram of phenol is \(32.39\, \mathrm{kJ\, g^{-1}}\). The heat of combustion per mole of phenol is \(3,050.26\, \mathrm{kJ\, mol^{-1}}\).

Step-by-step solution

(1. Write the balanced chemical equation)

The balanced chemical equation for the combustion of phenol (C6H5OH) is: \(C_6H_5OH_{(l)} + 7.5O_2_{(g)} \rightarrow 6CO_2_{(g)} + 3H_2O_{(l)}\)

(2. Calculate the total heat released)

To find the total heat released (q) during the combustion process, we'll use the formula: \(q = C_p \Delta T\), where \(C_p\) is the total heat capacity of the calorimeter, and \(\Delta T\) is the change in temperature. Given that \(C_p = 11.66 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), and the change in temperature is \(\Delta T = 26.37^{\circ} \mathrm{C} - 21.36^{\circ} \mathrm{C} = 5.01^{\circ} \mathrm{C}\). Now we can calculate q: \(q = (11.66\, \mathrm{kJ\,} { }^{\circ} \mathrm{C^{-1}})(5.01^{\circ} \mathrm{C}) \) \(q = 58.29\, \mathrm{kJ}\)

(3. Calculate the heat of combustion per gram of phenol)

We are given the mass of phenol as \(1.800\, \mathrm{g}\). To find the heat of combustion per gram, we'll divide the total heat released (q) by the mass of the phenol sample: Heat of combustion per gram = \(\frac{58.29\, \mathrm{kJ}}{1.800\, \mathrm{g}}\) Heat of combustion per gram = \(32.39\, \mathrm{kJ\, g^{-1}}\)

(4. Calculate the heat of combustion per mole of phenol)

To find the heat of combustion per mole of phenol, we'll first calculate the molar mass of phenol. The molar mass of phenol (C6H5OH): \((6 \times 12.01) + (6 \times 1.01) + (1 \times 16.00) = 94.11\, \mathrm{g}\, \mathrm{mol^{-1}}\) Now, we can calculate the heat of combustion per mole by dividing the total heat released (q) by the number of moles of phenol burned: Number of moles of phenol = \(\frac{1.800\, \mathrm{g}}{94.11\, \mathrm{g\, mol^{-1}}}\) Number of moles of phenol = \(0.0191\, \mathrm{mol}\) Heat of combustion per mole of phenol = \(\frac{58.29\, \mathrm{kJ}}{0.0191\, \mathrm{mol}}\) Heat of combustion per mole of phenol = \(3,050.26\, \mathrm{kJ\, mol^{-1}}\) So, the heat of combustion of phenol is \(32.39\, \mathrm{kJ\, g^{-1}}\) and \(3,050.26\, \mathrm{kJ\, mol^{-1}}\).

Key concepts

Heat of Combustion

The heat of combustion is an essential concept in thermochemistry. It is defined as the amount of energy released as heat when a substance undergoes complete combustion with oxygen. This energy is often measured in kJ per gram or per mole. Understanding heat of combustion helps in determining the energy efficiency of fuels and other substances.
The calculation of heat of combustion involves several steps. First, the change in temperature (∆T) resulting from the combustion is measured. This is crucial because it directly affects the total heat released, which can be calculated using the formula:

• \[q = C_p \Delta T\]

where \(C_p\) is the heat capacity of the calorimeter, and \(\Delta T\) is the temperature change.
In our given example, phenol was burned, resulting in a temperature increase. Knowing the heat capacity of the calorimeter and the temperature change allows the calculation of total heat released.
Finally, by knowing the mass or moles of the substance burned, the heat of combustion can be expressed either per gram or per mole. This helps to standardize the measurement and simplify comparisons between different substances.

Phenol Combustion

Phenol combustion is a specific chemical reaction where phenol (C6H5OH) reacts with oxygen in a process that produces carbon dioxide and water. This type of reaction is known as an exothermic reaction because it releases energy in the form of heat.
The balanced chemical equation for the combustion of phenol is:

• \[C_{6}H_{5}OH_{(l)} + 7.5O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_{2}O_{(l)}\]

This equation is crucial as it directly shows the reactants and products involved. It also highlights the stoichiometry of the reaction, indicating that each mole of phenol will react with 7.5 moles of oxygen to produce 6 moles of carbon dioxide and 3 moles of water.
Understanding this equation helps predict the amount of heat released during the combustion process. This prediction is based on the breaking and forming of bonds, with the ultimate goal of calculating how much energy is liberated.
For students, grasping the stoichiometry of this reaction and the concept of exothermic reactions forms the foundational knowledge required for more advanced thermodynamic calculations.

Thermodynamic Calculations

Thermodynamic calculations involve various techniques and equations used to determine the energetics of chemical reactions, particularly focusing on energy exchange.
In bomb calorimetry, a key thermodynamic calculation involves determining the total heat released during a reaction. For phenol combustion, students start by calculating how much temperature change occurred in the calorimeter. This change is then used along with the calorimeter's known heat capacity to find the total heat released.

• \[q = C_p \Delta T = (11.66\, \mathrm{kJ}\, { }^{\circ} \mathrm{C^{-1}})(5.01^{\circ} \mathrm{C})\]

Moreover, one is required to convert this energy release into applicable metrics like kJ per gram or kJ per mole. This involves additional steps of calculating moles of phenol and using molar mass to refine the results.
These calculations are not just mathematical exercises; they also provide insight into the efficiency of fuels and other substances undergoing thermal processes. Understanding these concepts is crucial for developing and selecting materials that deliver optimal energy productivity.