Problem 69 - General Chemistry

Question

A sample of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), weighing \(2.84 \mathrm{~g}\) was burned in an excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rose from \(25.00^{\circ} \mathrm{C}\) to \(33.73^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter and contents was \(9.63 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), what is the value of \(q\) for burning \(1 \mathrm{~mol}\) of ethanol at constant volume and \(25.00^{\circ} \mathrm{C}\) ? The reaction is $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$

Step-by-Step Solution

Answer

The heat of combustion is \(-1,364.94 \, \text{kJ/mol}\).

Step 1: Calculate the Temperature Change

First, determine the change in temperature, \( \Delta T \), using the initial and final temperatures:\[\Delta T = T_{\text{final}} - T_{\text{initial}} \]Substitute the given values:\[\Delta T = 33.73^{\circ} \text{C} - 25.00^{\circ} \text{C} = 8.73^{\circ} \text{C}\]

Step 2: Calculate the Heat Absorbed by the Calorimeter

Use the formula to calculate the heat absorbed by the calorimeter, \( q_{\text{cal}} \):\[q_{\text{cal}} = C \times \Delta T\]where \( C \) is the heat capacity. Substituting the known values:\[q_{\text{cal}} = 9.63 \, \text{kJ}/{ }^{\circ} \text{C} \times 8.73^{\circ} \text{C} = 84.06 \, \text{kJ}\]

Step 3: Calculate Moles of Ethanol Burned

Find the molar mass of ethanol, \( \text{C}_2 \text{H}_5 \text{OH} \):\[\text{Molar mass} = 2 \times 12.01 \text{ g/mol} + 6 \times 1.01 \text{ g/mol} + 16.00 \text{ g/mol} = 46.08 \text{ g/mol}\]Calculate the moles of ethanol burned:\[\text{Moles} = \frac{2.84 \, \text{g}}{46.08 \, \text{g/mol}} = 0.0616 \, \text{mol}\]

Step 4: Calculate the Heat Released Per Mole of Ethanol

Use the total heat released, \( q_{\text{cal}} \), to find the heat released per mole, \( q_{\text{ethanol}} \):\[q_{\text{ethanol}} = \frac{q_{\text{cal}}}{\text{moles of ethanol}}\]Substitute the values:\[q_{\text{ethanol}} = \frac{84.06 \, \text{kJ}}{0.0616 \, \text{mol}} = 1,364.94 \, \text{kJ/mol}\]

Step 5: Adjust for the Reaction Conditions

Since the measurement was taken at constant volume under calorimeter conditions, \( q \) represents the heat of combustion under these conditions.Thus, the value of \( q \) for burning 1 mol of ethanol is \[-1,364.94 \, \text{kJ/mol}. \]

Key Concepts

Heat of Combustion

The heat of combustion is an essential concept in calorimetry, representing the thermal energy released when a substance undergoes complete combustion with oxygen. In this exercise, we are interested in calculating the heat of combustion for ethanol, a common laboratory and industrial chemical. The reaction provided shows that ethanol reacts with oxygen to release carbon dioxide and water as products. This process releases energy, which in this context is absorbed by the bomb calorimeter, a device designed to measure such changes.
Understanding the heat of combustion helps us determine the energy content of fuels and other substances, which is crucial for applications such as energy production. This energy is typically expressed in terms of kilojoules per mole (kJ/mol), providing a standardized measure to compare the energy potentials of different substances.
In our example, the calculated heat of combustion for ethanol was found to be \[-1,364.94 \, \text{kJ/mol}\]. The negative sign indicates the exothermic nature of the combustion process, meaning heat is released.

Bomb Calorimeter

A bomb calorimeter is a device used to measure the heat of combustion of a substance. It's a robust and sealed apparatus that allows the combustion reaction to occur under constant volume conditions, ensuring no gas leaks. The sample, in this case ethanol, is placed in a small container called a bomb, which is then filled with an excess of oxygen.
Once the ethanol is ignited, the heat released by the combustion reaction is absorbed by the surrounding water and calorimeter, causing a temperature rise. The calorimeter is equipped with a thermometer to accurately record these changes. By knowing the heat capacity of the calorimeter and the temperature change, the total heat absorbed by the system can be calculated. This captured heat reflects the energy released by the combustion process, allowing us to understand the substance's energy content efficiently.
Bomb calorimeters are extremely precise tools, providing accurate assessments of thermochemical properties which are crucial in fields such as chemistry and environmental science.

Temperature Change

Temperature change is a direct indicator of the heat exchange in a calorimetric process. In the bomb calorimeter exercise, the temperature of the system changed from 25.00°C to 33.73°C, resulting in a temperature difference (\(\Delta T\)). This change is central to calculating the total heat absorbed by the calorimeter.
The formula for calculating the temperature change is straightforward: \[\Delta T = T_{\text{final}} - T_{\text{initial}} \]Substituting the given values provides: \[\Delta T = 33.73^{\circ} \text{C} - 25.00^{\circ} \text{C} = 8.73^{\circ} \text{C} \]
This temperature change is then used along with the calorimeter's heat capacity in the equation \(q_{\text{cal}} = C \times \Delta T\) to find the heat absorbed, helping us further determine the combustion characteristics of ethanol.

Molar Mass Calculation

Calculating the molar mass of a chemical compound is a foundational step in stoichiometry which allows us to convert between grams and moles. In our exercise, the molar mass of ethanol (\(\text{C}_2 \text{H}_5 \text{OH}\)) was calculated by adding the atomic masses of its constituent elements:

• Carbon (C): 2 atoms \(\times\) 12.01 g/mol = 24.02 g/mol
• Hydrogen (H): 6 atoms \(\times\) 1.01 g/mol = 6.06 g/mol
• Oxygen (O): 1 atom \(\times\) 16.00 g/mol = 16.00 g/mol

Adding these together gives us a molar mass of \(46.08 \text{g/mol}\) for ethanol.
With the molar mass, we can find the number of moles present in a given mass of the compound. In this case, for 2.84 g of ethanol, the number of moles was calculated as follows:
\[\text{Moles} = \frac{2.84 \text{g}}{46.08 \text{g/mol}} = 0.0616 \text{mol}\]
This calculation is crucial for determining the heat released per mole of ethanol in the overall combustion process.