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A 1.397 g sample of thymol, \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O}(\mathrm{s})\) (a preservative and a mold and mildew preventative), is burned in a bomb calorimeter assembly. The temperature increase is \(11.23^{\circ} \mathrm{C},\) and the heat capacity of the bomb calorimeter is \(4.68 \mathrm{kJ} /^{\circ} \mathrm{C}\). What is the heat of combustion of thymol, expressed in kilojoules per mole of \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O} ?\)

Short Answer

Expert verified
The heat of combustion of thymol is \(-5656.99 \mathrm{kJ/mol}\)

Step by step solution

01

Calculate Heat Change

Using the formula for heat change (Q), which is given by Q=mcΔT, where m is the heat capacity, c is the increase in temperature and ΔT is the change in temperature. So, we have \[Q = (4.68 \, \mathrm{kJ/^{\circ}C})(11.23\, ^{\circ}C) = 52.59 \, \mathrm{kJ}\] The heat change is positive since heat is gained (combustion process is exothermic).
02

Convert gaseous weight to mol

Convert the weight of thymol from grams to moles using the molar mass of thymol (C10H14O), which is 150.22 g/mol. Thus, \[\mathrm{moles\, of\, Thymol} = \frac{1.397\, \mathrm{g}}{150.22\, \mathrm{g/mol}} = 0.0093\, \mathrm{mol}\]
03

Calculate heat of combustion

Finally, calculate the heat of combustion (ΔH) in kJ per mole of thymol using the formula ΔH = -Q/n, where Q is the heat change calculated in step 1 and n is the amount of thymol in moles calculated in step 2. Thus, \[\Delta H = \frac{-52.59\, \mathrm{kJ}}{0.0093\, \mathrm{mol}} = -5656.99\, \mathrm{kJ/mol}\] The negative sign indicates that the process is exothermic as heat is released during the combustion process.

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Key Concepts

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

Bomb Calorimeter
A bomb calorimeter is an essential tool in thermodynamics, specifically for measuring the heat of combustion of a substance. It is a sealed container designed to withstand high pressure, allowing reactions to occur while capturing the heat released. When a substance such as thymol is burned inside this calorimeter, the heat produced leads to a noticeable temperature increase in the surrounding water bath or assembly.
This change is crucial for calculating energy changes.The key components of a bomb calorimeter include:
  • A strong, sealed container called a "bomb," which contains the sample and oxygen.
  • A surrounding water bath to absorb the heat from the reaction.
  • A temperature sensor to measure the change in water temperature.
  • An ignition source to start the combustion reaction inside the bomb.
The bomb calorimeter is calibrated to know exactly how much energy a specific temperature change represents. Thus, when a sample like thymol is combusted, it is possible to calculate the heat released by using the formula for heat change, which is given by \( Q = mc\Delta T \). Here, \( m \) refers to the calorimeter's heat capacity, \( c \) represents the increase in temperature, and \( \Delta T \) is the temperature change as measured by the sensor.
Molar Mass
Understanding molar mass is crucial when working with chemical reactions and stoichiometry. Molar mass refers to the mass of one mole of a substance and is expressed in grams per mole (g/mol). It is the weight of all the atoms in a molecule combined.For our compound, thymol (\( \mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O} \)), the molar mass is calculated by summing the atomic masses of all its constituent elements:
  • Carbon (C): 10 atoms, each with an atomic mass of approximately 12.01 g/mol.
  • Hydrogen (H): 14 atoms, with an atomic mass of approximately 1.01 g/mol.
  • Oxygen (O): 1 atom, with an atomic mass of approximately 16.00 g/mol.
Adding these together gives the molar mass of thymol as \( 150.22 \text{ g/mol} \).In our exercise, the molar mass is used to convert the mass of thymol in grams to moles. This conversion is essential for calculating other properties, such as the heat of combustion, on a per mole basis. This step is vital to expressing the reaction energetics in a standardized form, helping scientists compare data efficiently.
Exothermic Process
An exothermic process is a chemical reaction that releases energy to its surroundings, usually in the form of heat. Combustion reactions, like burning thymol, are classic examples of exothermic processes. This is reflected in the rise in temperature observed when a material is combusted in a bomb calorimeter.The essential characteristics of an exothermic process include:
  • Heat is released, causing the temperature of the surroundings to increase.
  • The products have lower energy than the reactants, leading to the release of excess energy.
  • The enthalpy change \( (\Delta H) \) of the process is negative, indicating that energy is flowing out of the system.
In our exercise, the negative value of \( \Delta H = -5656.99 \text{ kJ/mol} \) for thymol indicates it's an exothermic reaction. This negative sign is a direct representation of the energy release experienced during the combustion, supporting the usage of bomb calorimeters to measure such reactions.Understanding exothermic and endothermic processes helps in predicting and controlling chemical reactions. It also aids in applying these reactions to practical purposes, such as harnessing the released energy in fuels or other industrial applications.

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Most popular questions from this chapter

The standard molar enthalpy of formation of \(\mathrm{CO}_{2}(\mathrm{g})\) is equal to (a) \(0 ;\) (b) the standard molar heat of combustion of graphite; (c) the sum of the standard molar enthalpies of formation of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) (d) the standard molar heat of combustion of \(\mathrm{CO}(\mathrm{g})\)

Some of the butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) in a \(200.0 \mathrm{L}\) cylinder at \(26.0^{\circ} \mathrm{C}\) is withdrawn and burned at a constant pressure in an excess of air. As a result, the pressure of the gas in the cylinder falls from 2.35 atm to 1.10 atm. The liberated heat is used to raise the temperature of 132.5 L of water in a heater from 26.0 to 62.2 ^ C. Assume that the combustion products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) exclusively, and determine the efficiency of the water heater. (That is, what percent of the heat of combustion was absorbed by the water?)

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Is it possible for a chemical reaction to have \(\Delta U<0\) and \(\Delta H>0 ?\) Explain.

A clay pot containing water at \(25^{\circ} \mathrm{C}\) is placed in the shade on a day in which the temperature is \(30^{\circ} \mathrm{C} .\) The outside of the clay pot is kept moist. Will the temperature of the water inside the clay pot (a) increase; (b) decrease; (c) remain the same?

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