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Chapter 7: Problem 44

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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Most popular questions from this chapter

An alternative approach to bomb calorimetry is to establish the heat capacity of the calorimeter, exclusive of the water it contains. The heat absorbed by the water and by the rest of the calorimeter must be calculated separately and then added together. A bomb calorimeter assembly containing \(983.5 \mathrm{g}\) water is calibrated by the combustion of \(1.354 \mathrm{g}\) anthracene. The temperature of the calorimeter rises from 24.87 to \(35.63^{\circ} \mathrm{C} .\) When \(1.053 \mathrm{g}\) citric acid is burned in the same assembly, but with 968.6 g water, the temperature increases from 25.01 to \(27.19^{\circ} \mathrm{C}\). The heat of combustion of anthracene, \(\mathrm{C}_{14} \mathrm{H}_{10}(\mathrm{s}),\) is \(-7067 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{C}_{14} \mathrm{H}_{10} \cdot\) What is the heat of combustion of citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) expressed in \(\mathrm{kJ} / \mathrm{mol} ?\)

Determine \(\Delta H^{\circ}\) for this reaction from the data below. \(\mathrm{N}_{2} \mathrm{H}_{4}(1)+2 \mathrm{H}_{2} \mathrm{O}_{2}(1) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(1)\) $$\begin{array}{r} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-622.2 \mathrm{kJ} \end{array}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(1) \quad \Delta H^{\circ}=-187.8 \mathrm{kJ}$$

What is the change in internal energy of a system if the system (a) absorbs \(58 \mathrm{J}\) of heat and does \(58 \mathrm{J}\) of work; (b) absorbs 125 J of heat and does 687 J of work; (c) evolves 280 cal of heat and has 1.25 kJ of work done on it?

The heat of solution of \(\mathrm{NaOH}(\mathrm{s})\) in water is \(-41.6 \mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} .\) When \(\mathrm{NaOH}(\mathrm{s})\) is dissolved in water the solution temperature (a) increases; (b) decreases; (c) remains constant; (d) either increases or decreases, depending on how much NaOH is dissolved.

In a student experiment to confirm Hess's law, the reaction $$\mathrm{NH}_{3}(\text { concd aq })+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})$$ was carried out in two different ways. First, \(8.00 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\text { aq })\) was added to \(100.0 \mathrm{mL}\) of 1.00 M HCl in a calorimeter. [The NH \(_{3}(\) aq) was slightly in excess.] The reactants were initially at \(23.8^{\circ} \mathrm{C},\) and the final temperature after neutralization was \(35.8^{\circ} \mathrm{C} .\) In the second experiment, air was bubbled through \(100.0 \mathrm{mL}\) of concentrated \(\mathrm{NH}_{3}(\mathrm{aq})\) sweeping out \(\mathrm{NH}_{3}(\mathrm{g})\) (see sketch). The \(\mathrm{NH}_{3}(\mathrm{g})\) was neutralized in \(100.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). The temperature of the concentrated \(\mathrm{NH}_{3}(\text { aq })\) fell from 19.3 to \(13.2^{\circ} \mathrm{C} .\) At the same time, the temperature of the 1.00 M HCl rose from 23.8 to 42.9 ^ C as it was neutralized by \(\mathrm{NH}_{3}(\mathrm{g}) .\) Assume that all solutions have densities of \(1.00 \mathrm{g} / \mathrm{mL}\) and specific heats of \(4.18 \mathrm{Jg}^{-1 \circ} \mathrm{C}^{-1}\) (a) Write the two equations and \(\Delta H\) values for the processes occurring in the second experiment. Show that the sum of these two equations is the same as the equation for the reaction in the first experiment. (b) Show that, within the limits of experimental error, \(\Delta H\) for the overall reaction is the same in the two experiments, thereby confirming Hess's law.
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