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Chapter 7: Problem 97

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 \(\mathrm{kJ} / \mathrm{mol} ?\)

Short Answer

Expert verified
The heat of combustion of citric acid is \(-2335.70 kJ/mol\).

Step by step solution

01

Calculate the heat capacity of the calorimeter

For anthracene combustion, the total heat calculated as how much kcal/mole (or kJ/mole) it emits, multiplied by how many moles of anthracene is combusted, obtaining \(-7067 \mathrm{kJ/mol} \times \frac{1.354g}{178.23g/mol} = -53.73 kJ\), where \(178.23 g/mol\) is the molecular weight of anthracene. The heat capacity (\(C\)) of the calorimeter is calculated as follows: \(C = -\frac{Q}{\Delta T}\) where \(Q\) is the total heat produced and \(\Delta T\) is the change in temperature. Plugging in the values: \(C = -\frac{-53.73 kJ}{35.63^{\circ}\mathrm{C} - 24.87^{\circ}\mathrm{C}} = 5.66 kJ/^{\circ}\mathrm{C}\).
02

Calculate the heat produced from the combustion of citric acid

Using the heat capacity of the calorimeter, calculate the heat produced from the combustion of citric acid \(\Delta H = C \Delta T = (5.66 kJ/^{\circ}\mathrm{C})(27.19^{\circ}\mathrm{C} - 25.01^{\circ}\mathrm{C}) = -12.35 kJ\). \( \Delta H \) is the heat absorbed by the calorimeter and the water it contains.
03

Calculate the heat of combustion of citric acid (Q)

Now, we know the total heat produced and the mass of citric acid that was burned. To find the heat of combustion of citric acid per mole, we first calculate the amount of citric acid in moles, which is given by \(\frac{1.053g}{M}\), where M is the molar mass of citric acid \( = 210.14 g/mol \). Finally, to obtain the heat of combustion, the heat produced is divided by the mole of citric acid burned. \(Q = -\frac{-12.35 kJ}{\frac{1.053g}{210.14 g/mol}} = -2335.70 kJ/mol\).

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Most popular questions from this chapter

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

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.

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$

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 \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O} ?\)

A plausible final temperature when \(75.0 \mathrm{mL}\) of water at \(80.0^{\circ} \mathrm{C}\) is added to \(100.0 \mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\) is (a) \(28^{\circ} \mathrm{C} ;\) (b) \(40^{\circ} \mathrm{C} ;\) (c) \(46^{\circ} \mathrm{C} ;\) (d) \(50^{\circ} \mathrm{C}\)
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