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Chapter 6: Problem 92

I he enthaipy of combustion benzo1c acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be \(-3226.7 \mathrm{~kJ} / \mathrm{mol}\). When \(1.9862 \mathrm{~g}\) of benzoic acid are burned in a calorimeter, the temperature rises from \(21.84^{\circ} \mathrm{C}\) to \(25.67^{\circ} \mathrm{C}\). What is the heat capacity of the bomb? (Assume that the quantity of water surrounding the bomb is exactly \(2000 \mathrm{~g} .)\)

Short Answer

Expert verified
The heat capacity of the bomb is approximately \(13.74 \, \mathrm{kJ/ ^{\circ}C}\).

Step by step solution

01

Analyze the given data

We are given: \n\n1. The enthalpy of combustion for benzoic acid, which is \(-3226.7 \, \mathrm{kJ/mol}\). \n2. Mass of benzoic acid burned, which is \(1.9862 \, \mathrm{g}\). \n3. Initial temperature of the water, which is \(21.84^{\circ}C\). \n4. Final temperature of the water after the benzoic acid has burned, which is \(25.67^{\circ}C\). \n5. Mass of the water in the calorimeter, which is \(2000 \, \mathrm{g}\).
02

Calculate the amount of moles of benzoic acid burned

The molar mass of benzoic acid is \(122.12 \, \mathrm{g/mol}\). Thus, the amount of benzoic acid burned, in moles, can be calculated using the formula: \n\nAmount of moles \(= \frac{mass}{molar \, mass}\) \n\nSubstituting the given values into the formula you get: \n\nAmount of moles \(= \frac{1.9862 \, \mathrm{g}}{122.12 \, \mathrm{g/mol}} = 0.0163 \, \mathrm{mol}\).
03

Compute the heat change

The overall heat of combustion \(Q\) is calculated as the product of the number of moles and the molar heat of combustion: \n\n\(-Q = \Delta H_{combustion} \times n\) \n\nSubstitute the moles and heat of combustion in the formula: \n\n\(-Q = -3226.7 \, \mathrm{kJ/mol} \times 0.0163 \, \mathrm{mol} = 52.60 \, \mathrm{kJ}\). The negative sign is used because an exothermic process releases heat.
04

Calculate the Heat capacity

Finally, the heat capacity of the bomb, \(C\), can be calculated by dividing the heat change by the change in temperature: \n\n \(C = \frac{Q}{\Delta T}\) \n\nThe change in temperature, \(\Delta T\), is the final temperature minus the initial temperature: \( \Delta T = 25.67^{\circ}C - 21.84^{\circ}C = 3.83^{\circ}C\). \n\nSubstitute the values of \(Q\) and \( \Delta T\) in the formula you get: \n\n \(C = \frac{52.60 \, \mathrm{kJ}}{3.83^{\circ}C} = 13.74 \, \mathrm{kJ/ ^{\circ}C}\).

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Most popular questions from this chapter

The convention of arbitrarily assigning a zero enthalpy value for the most stable form of each element in the standard state at \(25^{\circ} \mathrm{C}\) is a convenient way of dealing with enthalpies of reactions. Explain why this convention cannot be applied to nuclear reactions.

Describe the interconversions of forms of energy occurring in these processes: (a) You throw a softball up into the air and catch it. (b) You switch on a flashlight. (c) You ride the ski lift to the top of the hill and then ski down. (d) You strike a match and let it burn down.

Explain the meaning of this thermochemical equation: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) &+6 \mathrm{H}_{2} \mathrm{O}(g) \\ \Delta H=&-904 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol: $$ \begin{aligned} 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-1452.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$

Define these terms: enthalpy, enthalpy of reaction. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction?
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