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Chapter 5: Problem 107

(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 \(0.53-\mathrm{g}\) 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 approximately 4218 kJ/mol, with an uncertainty of about 3.88 kJ/mol.

Step by step solution

01

Calculate Total Heat Released by Benzoic Acid

First, calculate the total heat released during the combustion of the benzoic acid sample using the formula: \[ q = m \cdot \Delta H \] where \( m = 0.47 \, \text{g} \) and \( \Delta H = 26.38 \, \text{kJ/g} \). Substituting the values, we get: \[ q = 0.47 \, \text{g} \times 26.38 \, \text{kJ/g} = 12.3976 \, \text{kJ} \].
02

Calculate Calorimeter Constant

The calorimeter constant \( C \) is calculated using the heat absorbed by the calorimeter and the temperature change:\[ C = \frac{q}{\Delta T} \]where \( \Delta T = 3.284 \, ^\circ \text{C} \). Substituting the values calculated,\[ C = \frac{12.3976 \, \text{kJ}}{3.284 \, ^\circ \text{C}} \approx 3.776 \, \text{kJ/} ^\circ \text{C} \].
03

Calculate Heat Released by Caffeine

Using the calorimeter constant, calculate the heat released by the caffeine combustion:\[ q_{caffeine} = C \cdot \Delta T \]where \( \Delta T = 3.05 \, ^\circ \text{C} \).Thus,\[ q_{caffeine} = 3.776 \, \text{kJ/} ^\circ \text{C} \times 3.05 \, ^\circ \text{C} \approx 11.52 \, \text{kJ} \].
04

Calculate Moles of Caffeine

Calculate the moles of caffeine using its molecular weight:\[ \text{Molar mass of } \text{C}_8\text{H}_{10}\text{N}_4\text{O}_2 = 194.19 \, \text{g/mol} \].Thus,\[ n = \frac{0.53 \, \text{g}}{194.19 \, \text{g/mol}} \approx 0.00273 \, \text{mol} \].
05

Calculate Heat of Combustion per Mole of Caffeine

The heat of combustion per mole is calculated as:\[ \Delta H_{comb} = \frac{q_{caffeine}}{n} \]Substituting values,\[ \Delta H_{comb} = \frac{11.52 \, \text{kJ}}{0.00273 \, \text{mol}} \approx 4217.95 \, \text{kJ/mol} \].
06

Estimate Uncertainty

First, calculate the uncertainty in temperature change, \(\Delta T\), using the given uncertainty in temperature:\[ \text{Uncertainty in } \Delta T = \sqrt{(0.002)^2 + (0.002)^2} = 0.0028 \]Then, calculate the uncertainty in heat:\[ \text{Uncertainty in } q = C \cdot \text{Uncertainty in } \Delta T = 3.776 \, \text{kJ/} ^\circ \text{C} \times 0.0028 \, \text{C} \approx 0.0106 \, \text{kJ} \].Finally, calculate the uncertainty in heat of combustion per mole:\[ \text{Uncertainty in } \Delta H_{comb} = \frac{0.0106 \, \text{kJ}}{0.00273 \, \text{mol}} \approx 3.88 \, \text{kJ/mol} \].

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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 is an important concept in thermochemistry. It refers to the amount of heat energy released when a substance is burned completely in oxygen. This is often measured in kilojoules per mole (kJ/mol) for pure substances. When dealing with combustion reactions, especially in bomb calorimetry, knowing the heat of combustion helps us understand how much energy a substance can provide.
To determine the heat of combustion of a sample, such as when using benzoic acid in a calorimeter, we calculate the total heat released. This is done using the formula:
  • \[ q = m \cdot \Delta H \]
  • where \( q \) represents the heat released, \( m \) is the mass of the substance, and \( \Delta H \) is the heat of combustion per gram.
From this calculation, we can derive how much energy is produced per gram of the substance, giving us insight into its energetic value. In practical applications, such as designing engines, understanding the heat of combustion allows for optimization of fuel use.
Calorimeter Constant
The calorimeter constant, denoted as \( C \), is key in determining how much heat the calorimeter itself absorbs during a reaction. This value, often given in kilojoules per degree Celsius (kJ/掳C), allows us to quantify the relationship between heat absorbed and temperature change in the calorimeter.
To find the calorimeter constant, the equation used is:
  • \[ C = \frac{q}{\Delta T} \]
  • where \( q \) denotes the total heat transferred to the calorimeter, and \( \Delta T \) is the temperature change experienced.
Accurate determination of the calorimeter constant is crucial for any calorimetric measurement. It ensures that subsequent experiments provide reliable results. This constant allows us to correct for the heat that is not transferred to the surrounding but is instead absorbed by the calorimeter material itself. This adjustment is crucial for finding true enthalpy changes of reactions, especially for reactions conducted under fixed volume conditions, such as those in a bomb calorimeter.
Molar Mass
Molar mass is a fundamental concept in chemistry that describes the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Understanding molar mass is essential when converting between mass and moles, a common step in chemical calculations.
For a compound like caffeine, \( \text{C}_8\text{H}_{10}\text{N}_4\text{O}_2 \), calculating the molar mass involves summing the atomic masses of each atom present in the molecular formula. The calculation is straightforward:
  • Carbon (C): 12.01 g/mol, 8 atoms => \( 8 \times 12.01 \) g/mol
  • Hydrogen (H): 1.01 g/mol, 10 atoms => \( 10 \times 1.01 \) g/mol
  • Nitrogen (N): 14.01 g/mol, 4 atoms => \( 4 \times 14.01 \) g/mol
  • Oxygen (O): 16.00 g/mol, 2 atoms => \( 2 \times 16.00 \) g/mol
Adding these together gives the molar mass of caffeine as approximately 194.19 g/mol. This value is pivotal when translating the mass of a sample into the amount of substance, defined as moles. With moles, more complex calculations, such as finding out energy release per mole of substance (like in heat of combustion problems), become possible.

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Most popular questions from this chapter

A \(100-\mathrm{kg}\) man decides to add to his exercise routine by walking up six flights of stairs \((30 \mathrm{~m}) 10\) times per day. He figures that the work required to increase his potential energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption?

An aluminum can of a soft drink is placed in a freezer. Later, you find that the can is split open and its contents have frozen. Work was done on the can in splitting it open. Where did the energy for this work come from?

(a) What is meant by the term state function? (b) Give an example of a quantity that is a state function and one that is not. \((\mathbf{c})\) Is the volume of a system a state function? Why or why not?

The heat of combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(t),\) is -1367 \(\mathrm{kJ} / \mathrm{mol}\). A bottle of stout (dark beer) contains up to \(6.0 \%\) ethanol by mass. Assuming the density of the beer to be \(1.0 \mathrm{~g} / \mathrm{mL},\) what is the caloric content due to the alcohol (ethanol) in a bottle of beer (500 mL)?

Under constant-volume conditions, the heat of combustion of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) is \(40.18 \mathrm{~kJ} / \mathrm{g}\). A \(2.50-\mathrm{g}\) sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to \(28.83^{\circ} \mathrm{C}\). (a) What is the total heat capacity of the calorimeter? (b) A 1.50-g sample of a new organic substance is combusted in the same calorimeter. The temperature of the calorimeter increases from 21.14 to \(25.08^{\circ} \mathrm{C}\). What is the heat of combustion per gram of the new substance? (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?
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