Problem 57 - Chemistry The Central Science

Question

A 2.200 -g sample of quinone \(\left(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}\right)\) is burned in a bomb calorimeter whose total heat capacity is \(7.854 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\). The temperature of the calorimeter increases from \(23.44^{\circ} \mathrm{C}\) to \(30.57^{\circ} \mathrm{C}\). What is the heat of combustion per gram of quinone? Per mole of quinone?

Step-by-Step Solution

Answer

The heat of combustion per gram of quinone is \( 25.46 \frac{kJ}{g} \) and per mole of quinone is \( 2751 \, \frac{kJ}{mol} \).

Step 1: Calculate the heat released by the combustion of quinone

To do this, we will use the formula: \( q = C_{cal} \times \Delta T \) where q is the heat released, \( C_{cal} \) is the heat capacity of the calorimeter, and \( \Delta T \) is the change in temperature. Given, \( C_{cal} = 7.854 \frac{kJ}{^\circ C} \), the initial temperature \( T_i = 23.44 ^\circ C \) and the final temperature \( T_f = 30.57 ^\circ C \). First, let's find the change in temperature: \( \Delta T = T_f - T_i \)

Step 2: Calculate the change in temperature

\( \Delta T = 30.57 ^\circ C - 23.44 ^\circ C = 7.13 ^\circ C \) Now that we have the change in temperature, let's calculate the heat released:

Step 3: Calculate the heat released (q)

\(q = C_{cal} \times \Delta T = 7.854 \frac{kJ}{^\circ C} \times 7.13 ^\circ C \) \(q = 56.01 \, kJ\)

Step 4: Calculate the heat of combustion per gram of quinone

Given that we have 2.200 g sample of quinone, we can find the heat of combustion per gram by dividing the heat released by the mass: \( q_{per\,gram} = \frac{q}{mass} = \frac{56.01 \, kJ}{2.200 \, g} \) \( q_{per\,gram} = 25.46 \frac{kJ}{g} \)

Step 5: Calculate the molar mass of quinone

Quinone has the chemical formula \( \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2} \). We can calculate its molar mass by adding up the molar mass of each element: Molar mass of \(C = 12.01 \, g/mol\), of \(H = 1.01 \, g/mol\) and of \(O = 16.00 \, g/mol\). \( Molar\,mass\,of\,quinone = 6 \times 12.01 + 4 \times 1.01 + 2 \times 16.00 \) \( Molar\,mass\,of\,quinone = 108.10 \, g/mol \)

Step 6: Calculate the heat of combustion per mole of quinone

Now that we have the molar mass of quinone, we can calculate the heat of combustion per mole: \( q_{per\,mole} = q_{per\,gram} \times Molar\,mass\,of\,quinone = 25.46 \frac{kJ}{g} \times 108.10 \, g/mol \) \( q_{per\,mole} = 2751 \, \frac{kJ}{mol} \) Therefore, the heat of combustion per gram of quinone is \( 25.46 \frac{kJ}{g} \) and per mole of quinone is \( 2751 \, \frac{kJ}{mol} \).

Key Concepts

Heat of Combustion

The heat of combustion refers to the total energy released when a substance burns completely in the presence of oxygen. In the context of quinone, we first calculate this based on the heat change in the surroundings. Using the formula \( q = C_{cal} \times \Delta T \), where \( q \) represents the heat released, \( C_{cal} \) is the calorimeter heat capacity, and \( \Delta T \) is the temperature change, we determine the energy released as 56.01 kJ for the given sample. This value signifies how much heat is produced when the 2.200 g of quinone combusts. By dividing this by the mass, we obtain the heat of combustion per gram, which is 25.46 kJ/g. This figure tells us the energy released for every gram of quinone burned.
To find the heat of combustion per mole, we multiply by the molar mass, calculated later, providing a more comprehensive understanding of this energetic transformation.

Calorimeter Heat Capacity

Calorimeter heat capacity, noted as \( C_{cal} \), represents the amount of heat required to raise the temperature of the calorimeter by one degree Celsius. In this exercise, the calorimeter's heat capacity is given as 7.854 kJ/°C.
This figure is crucial because it allows us to determine the total heat absorbed by the calorimeter when quinone is combusted. By multiplying this heat capacity by the temperature change, we effectively calculate the total heat generated. Understanding the calorimeter's parameters ensures accurate readings since any miscalibration or error in the heat capacity would lead to incorrect conclusions about the energy changes in the chemical process.

• Key formula: \( q = C_{cal} \times \Delta T \)
• Understanding \( C_{cal} \) allows us to account for how much of the heat is retained by the calorimeter itself.

Temperature Change

Temperature change, denoted as \( \Delta T \), is simply the difference between the final and initial temperatures during the reaction. In our example, the initial temperature was 23.44°C, and the final was 30.57°C, leading to a \( \Delta T \) of 7.13°C.
This change in temperature is directly linked to the heat generated by burning the sample. A larger temperature difference indicates more heat production. Using \( \Delta T \), we calculate the total energy change in the system, which is pivotal in finding the heat of combustion.

• Formula: \( \Delta T = T_f - T_i \)
• Important for understanding how the reaction affected the calorimeter's state.

Recognizing this change helps quantify the experiment's impact and allows for precise calculations of energy transformations.

Molar Mass Calculation

Molar mass is the mass of one mole of a given substance, calculated by summing up the atomic masses of all atoms in its chemical formula. Quinone, with the formula \( \mathrm{C}_6\mathrm{H}_4\mathrm{O}_2 \), comprises carbon (C), hydrogen (H), and oxygen (O).
The calculation involves:

• Carbon: \( 6 \times 12.01 \) g/mol
• Hydrogen: \( 4 \times 1.01 \) g/mol
• Oxygen: \( 2 \times 16.00 \) g/mol

Summing these yields a molar mass of 108.10 g/mol for quinone. Knowing the molar mass is essential for relating the gram-based heat of combustion to a per mole basis, thereby allowing us to express the energy change in terms of standard chemical amounts, making it easier to compare with other substances.