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(a) When a 0.47-g sample of benzoic acid is combusted in a bomb calorimeter (Figure 5.19), the temperature rises by \(3.284^{\circ} \mathrm{C}\). When a 0.53 -g sample of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2}\), is burned, the temperature rises by \(3.05^{\circ} \mathrm{C}\). Using the value of \(26.38 \mathrm{~kJ} / \mathrm{g}\) for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncertainty of \(0.002^{\circ} \mathrm{C}\) in each temperature reading and that the masses of samples are measured to \(0.001 \mathrm{~g},\) what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

Short Answer

Expert verified
The heat of combustion per mole of caffeine is calculated to be 13771 kJ/mol, and the estimated uncertainty in this value is approximately 16.66 kJ/mol.

Step by step solution

01

Calculate the heat released by benzoic acid

First we need to find out the heat released by the benzoic acid when it is combusted. We are given the temperature rise and the mass of the benzoic acid sample and we will use the heat of combustion of benzoic acid to find the heat released using the following equation: Heat released (Q) = Mass of sample × Heat of combustion × Temperature rise Q = 0.47 g × 26.38 kJ/g × 3.284 °C Q = 40.61 kJ
02

Calculate the heat released by caffeine

Now, using the temperature rise and the mass of the caffeine sample, we will find the heat released by the caffeine using the information from the benzoic acid combustion. Heat released by caffeine, \( Q_{caffeine} \) = Heat released by benzoic acid × (Temperature rise of caffeine / Temperature rise of benzoic acid) \( Q_{caffeine} \) = 40.61 kJ × (3.05 °C/3.284 °C) \( Q_{caffeine} \) = 37.64 kJ
03

Calculate the moles of caffeine

Now we need to find the number of moles of caffeine in the sample. We are given the mass of the caffeine sample and the molecular formula C8H10N4O2. Using the molar mass of each element, we can calculate the molar mass of caffeine: Molar mass of C = 12.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of N = 14.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of caffeine = 8 × 12.01 + 10 × 1.01 + 4 × 14.01 + 2 × 16.00 Molar mass of caffeine = 194.19 g/mol Now, we will find the moles of caffeine sample: Moles of caffeine = mass of caffeine / molar mass of caffeine Moles of caffeine = 0.53 g / 194.19 g/mol Moles of caffeine = 0.00273 mol
04

Calculate the heat of combustion per mole of caffeine

To find the heat of combustion per mole of caffeine, we will divide the heat released by the sample by the moles of caffeine: Heat of combustion = Heat released by caffeine / moles of caffeine Heat of combustion = 37.64 kJ / 0.00273 mol Heat of combustion = 13771 kJ/mol Therefore, the heat of combustion per mole of caffeine is 13771 kJ/mol.
05

Calculate the uncertainty of the heat of combustion

We are given the uncertainties in the temperature readings and the mass of the samples. We can use these uncertainties to estimate the uncertainty in the heat of combustion per mole of caffeine. In general, the uncertainty in the heat of combustion, δQ = Q × (δT / T + δM / M) Where δT and δM are the uncertainties in the temperature readings and mass of the samples, respectively, and T and M are the actual temperature readings and mass of the samples. For benzoic acid, δQ = Q × (δT / T + δM / M) δQ = 40.61 kJ × (0.002 °C / 3.284 °C + 0.001 g / 0.47 g) δQ = 40.61 kJ × 0.00123 δQ = 0.0499 kJ For caffeine, δQ = Q × (δT / T + δM / M) δQ = 37.64 kJ × (0.002 °C / 3.05 °C + 0.001 g / 0.53 g) δQ = 37.64 kJ × 0.00121 δQ = 0.0455 kJ Now, we will propagate the uncertainties to find the uncertainty in the heat of combustion per mole: δ(Heat of combustion) = Heat of combustion × (δQ / Q) δ(Heat of combustion) = 13771 kJ/mol × (0.0455 kJ / 37.64 kJ) δ(Heat of combustion) = 13771 kJ/mol × 0.00121 δ(Heat of combustion) = 16.66 kJ/mol Therefore, the estimated uncertainty in the heat of combustion per mole of caffeine is approximately 16.66 kJ/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat of Combustion
The heat of combustion refers to the energy released when a substance undergoes complete combustion in the presence of an oxidizer, typically oxygen, at constant volume. This process is essential for understanding how much energy can be obtained from burning fuels or specific substances. In our case, we calculate the heat of combustion for both benzoic acid and caffeine using a bomb calorimeter. The device measures how much the temperature rises when the sample burns, providing a basis to calculate the energy released.
It's important to note the heat of combustion is expressed per gram or per mole, allowing us to compare the energy yields from different compounds. In this exercise, we used the heat of combustion per gram for benzoic acid to find the heat released, which assisted in calculating caffeine's heat of combustion per mole.
Benzoic Acid
Benzoic acid plays a crucial role as a calibration substance in bomb calorimetry. It has a known and consistent heat of combustion, making it an ideal standard for calibrating the calorimeter. When we combust 0.47 g of benzoic acid, the energy released causes a measurable temperature rise, used to calculate the calorimeter's heat capacity.
This step is vital because it sets the baseline for future experiments by creating a relationship between the temperature change and the heat released. In this specific exercise, the heat liberated by burning benzoic acid allowed us to determine the heat capacity of the bomb calorimeter, which was subsequently used to calculate the energy released by caffeine.
Caffeine Combustion
Caffeine is an organic compound whose heat of combustion we aim to determine. After determining the calorimeter's heat capacity using benzoic acid, the next step is to combust caffeine and note the temperature rise, which was 3.05°C. By comparing this temperature increase with that from benzoic acid, we calculate how much energy caffeine releases.
The formula for calculating this takes into account the mass of caffeine and applies the heat capacity derived from benzoic acid combustion. From there, dividing the total energy by the number of moles of caffeine gives us the heat of combustion per mole, a critical measure for understanding its energy potential. The final value calculated was 13771 kJ/mol for caffeine.
Uncertainty Calculation
Understanding uncertainty is crucial for generating reliable experimentation results. It involves quantifying the possible deviations in measurement from their exact value. In this exercise, we consider uncertainties in temperature readings and mass measurements.
For each component—benzoic acid and caffeine—we calculate their respective uncertainties using the given margin of error. These are then combined to find the overall uncertainty in the heat of combustion calculation. For caffeine, the estimated uncertainty in the measurement was about 16.66 kJ/mol.
  • Temperature uncertainty affects how accurately we can link the temperature rise to energy released.
  • Mass uncertainty influences precise determination of how much sample was burned.
Calculating these helps assess the reliability of the heat of combustion value obtained.

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Most popular questions from this chapter

Calculate \(\Delta E\) and determine whether the process is endothermic or exothermic for the following cases: \((\mathbf{a}) q=0.763 \mathrm{~kJ}\) and \(w=-840 \mathrm{~J}\). (b) A system releases \(66.1 \mathrm{~kJ}\) of heat to its surroundings while the surroundings do \(44.0 \mathrm{~kJ}\) of work on the system.

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The air bags that provide protection in automobiles in the event of an accident expand because of a rapid chemical reaction. From the viewpoint of the chemical reactants as the system, what do you expect for the signs of \(q\) and \(w\) in this process?

At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat \(\mathrm{KClO}_{3}\) : $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H=-89.4 \mathrm{~kJ} $$ For this reaction, calculate \(\Delta H\) for the formation of (a) \(1.36 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) and \((\mathbf{b}) 10.4 \mathrm{~g}\) of \(\mathrm{KCl} .(\mathbf{c})\) The decomposition of \(\mathrm{KClO}_{3}\) proceeds spontaneously when it is heated. Do you think that the reverse reaction, the formation of \(\mathrm{KClO}_{3}\) from \(\mathrm{KCl}\) and \(\mathrm{O}_{2},\) is likely to be feasible under ordinary conditions? Explain your answer.

During a deep breath, our lungs expand about \(2.0 \mathrm{~L}\) against an external pressure of \(101.3 \mathrm{kPa}\). How much work is involved in this process (in J)?

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