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 in a bomb calorimeter is: \(C_6H_5OH + 7.5O_2 → 6CO_2 + 3H_2O\). The heat of combustion for phenol is 32.42 kJ/g and 3055.49 kJ/mol.

Step-by-step solution

1. Write a balanced chemical equation for the combustion of phenol

Phenol has the chemical formula C6H5OH. The products of the combustion of an organic compound are carbon dioxide and water. Hence, the balanced chemical equation for the combustion of phenol in a bomb calorimeter is: \(C_6H_5OH + 7.5O_2 → 6CO_2 + 3H_2O\)

2. Calculate the heat released in the combustion process

The calorimeter's total heat capacity is given as 11.66 kJ/°C. We are given the initial and final temperatures of the calorimeter, from which we can find the temperature change. The temperature change can then be used to calculate the heat released by the combustion process (q). Initial temperature (T1) = 21.36°C Final temperature (T2) = 26.37°C Temperature change (ΔT) = T2 - T1 = 26.37 - 21.36 = 5.01°C Heat capacity of calorimeter (Ccal) = 11.66 kJ/°C Heat released (q) = Ccal × ΔT = 11.66 kJ/°C × 5.01°C = 58.36 kJ

3. Calculate the amount (in moles) of phenol in the given sample

The mass of the phenol sample is given (1.800 g). We need to find the molar mass of phenol to convert this mass into moles. Molar mass of phenol (M) = (6 × 12.01) + (5 × 1.01) + 16.00 = 94.11 g/mol Amount of phenol (n) = mass / molar mass = 1.800 g / 94.11 g/mol = 0.0191 mol

4. Calculate the heat of combustion per gram and per mole of phenol

Now we can use the heat released (q) to find the heat of combustion per gram and per mole of phenol. Heat of combustion per gram (qg) = q / mass = 58.36 kJ / 1.800 g = 32.42 kJ/g Heat of combustion per mole (qm) = q / moles = 58.36 kJ / 0.0191 mol = 3055.49 kJ/mol The heat of combustion for phenol is 32.42 kJ/g and 3055.49 kJ/mol.

Key concepts

Heat of Combustion

The heat of combustion is a measure of the energy released as heat when a compound combusts completely with oxygen under standard conditions. It is an essential aspect in understanding energetic processes in chemistry.
In the case of phenol, you calculate the heat of combustion using the bomb calorimeter, which captures the energy released during combustion. This energy can then be expressed per unit mass or per mole of the substance.
The calculation begins by determining the total heat released using the calorimeter's heat capacity and the observed temperature change. For phenol, when burned, the measured temperature increase was 5.01°C, and the calorimeter's heat capacity was 11.66 kJ/°C.
The heat released during the combustion (q) is found using:

• \( q = C_{cal} \times \Delta T = 11.66 \text{ kJ/°C} \times 5.01 °C = 58.36 \text{ kJ} \)

This q value represents the energy released by burning the sample of phenol under these conditions. By translating this to per gram and per mole values, you get a thorough understanding of the energy density of phenol.

Phenol Combustion

Phenol combustion involves the chemical reaction where phenol (C_6H_5OH) reacts with oxygen (O_2) to produce carbon dioxide (CO_2) and water (H_2O).
This type of reaction is typical for organic compounds, undergoing combustion in the presence of excess oxygen, leading to complete oxidation.
The products, CO_2 and H_2O, are indicative of complete combustion, meaning all carbon atoms in phenol convert to carbon dioxide and all hydrogen atoms to water.
The balanced chemical equation for the combustion of phenol is:

• \( C_6H_5OH + 7.5 O_2 \rightarrow 6 CO_2 + 3 H_2O \)

This equation must always be balanced to adhere to the law of conservation of mass, ensuring that the number of atoms for each element is equal on both sides of the equation. Understanding this balance helps you accurately assess the energetic and material transactions within the reaction.

Chemical Equation Balancing

Chemical equation balancing is a crucial skill in chemistry. It ensures that what you start with as reactants equals what you end with as products, respecting the law of conservation of mass.
Each side of the equation must have an identical number of each type of atom. For phenol combustion, start with the formula for phenol (C_6H_5OH) and add oxygen (O_2) to yield carbon dioxide (CO_2) and water (H_2O).
The initial equation may look something simple like this:

• \( C_6H_5OH + O_2 \rightarrow CO_2 + H_2O \)

However, balancing it requires adjusting the coefficients that indicate the number of molecules or atoms.

• For phenol: 6 carbon atoms, thus 6 molecules of CO_2 are needed.
• For hydrogen: 5 hydrogen atoms on one side and water molecules (H_2O) need to balance; hence add 3 waters.
• Finally, oxygen must be balanced. Initially placing 7.5 molecules of O_2 balances the complete equation.

Completing these steps successfully confirms that balancing is not only about numbers; it's about understanding the stoichiometric relationships in chemical reactions.