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Chapter 10: Problem 98

The molar heat of fusion of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(9.92 \mathrm{~kJ} / \mathrm{mol}\). Its molar heat of vaporization is \(30.7 \mathrm{~kJ} / \mathrm{mol}\). Calculate the heat required to melt \(8.25 \mathrm{~g}\) benzene at its normal melting point. Calculate the heat required to vaporize \(8.25 \mathrm{~g}\) benzene at its normal boiling point. Why is the heat of vaporization more than three times the heat of fusion?

Short Answer

Expert verified
In short, for \(8.25 \mathrm{~g}\) of benzene, the heat required to melt it at its normal melting point is approximately \(1.047 \mathrm{~kJ}\) and the heat required to vaporize it at its normal boiling point is approximately \(3.241 \mathrm{~kJ}\). The heat of vaporization is more than three times the heat of fusion because vaporizing requires overcoming all intermolecular forces holding the molecules together in the liquid state, which takes significantly more energy than the partial overcoming of these forces during melting.

Step by step solution

01

Calculate the number of moles for the given mass of benzene.

To determine the number of moles, we can use the formula: n = \( \frac{mass}{molar~mass} \) The molecular formula for benzene is C6H6, and the molar mass is: \(6 \times 12.01 \mathrm{~g/mol} + 6 \times 1.01 \mathrm{~g/mol} = 78.12 \mathrm{~g/mol} \) Now, let's calculate the number of moles of benzene: n = \( \frac{8.25 \mathrm{~g}}{78.12 \mathrm{~g/mol}} \approx 0.1056 \mathrm{~mol} \)
02

Calculate the heat required to melt the benzene.

To calculate the heat required to melt the benzene, we can use the formula: q = n × ΔHf where q represents the heat required, n denotes the number of moles, and ΔHf stands for the molar heat of fusion (9.92 kJ/mol) Now, let's calculate the heat required to melt the benzene: q = 0.1056 mol × 9.92 kJ/mol ≈ 1.047 kJ
03

Calculate the heat required to vaporize the benzene.

To calculate the heat required to vaporize the benzene, we can use the formula: q = n × ΔHv where q represents the heat required, n denotes the number of moles, and ΔHv stands for the molar heat of vaporization (30.7 kJ/mol) Now, let's calculate the heat required to vaporize the benzene: q = 0.1056 mol × 30.7 kJ/mol ≈ 3.241 kJ
04

Explain why the heat of vaporization is more than three times the heat of fusion.

The heat of vaporization is more than three times the heat of fusion because when a substance vaporizes, its molecules must completely overcome the intermolecular forces holding them together in the liquid state. This requires significantly more energy than simply overcoming a portion of these forces during the process of melting (fusion). As a result, it takes more heat energy to vaporize a substance than to melt it.

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Most popular questions from this chapter

\(X\) rays of wavelength \(2.63 \mathrm{~A}\) were used to analyze a crystal. The angle of first-order diffraction \((n=1\) in the Bragg equation) was \(15.55\) degrees. What is the spacing between crystal planes, and what would be the angle for second-order diffraction \((n=2) ?\)

What quantity of energy does it take to convert \(0.500 \mathrm{~kg}\) ice at \(-20 .{ }^{\circ} \mathrm{C}\) to steam at \(250 .{ }^{\circ} \mathrm{C}\) ? Specific heat capacities: ice, \(2.03 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ;\) liquid, \(4.2 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ;\) steam, \(2.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} ; \Delta H_{\text {vep }}=\) 40.7 \(\mathrm{kJ} / \mathrm{mol} ; \Delta H_{\mathrm{fec}}=6.02 \mathrm{~kJ} / \mathrm{mol}\).

What type of solid (network, metallic, Group \(8 \mathrm{~A}\), ionic, or molecular) will each of the following substances form? a. \(\mathrm{Kr}\) d. \(S i \mathrm{O}_{2}\) b. \(\mathrm{SO}_{2}\) e. \(\mathrm{NH}_{3}\) c. \(\mathrm{N} i\) f. \(\mathrm{Pt}\)

Why is a burn from steam typically much more severe than a burn from boiling water?

Predict which substance in each of the following pairs would have the greater intermolecular forces. a. \(\mathrm{CO}_{2}\) or \(\mathrm{OCS}\) b. \(\mathrm{SeO}_{2}\) or \(\mathrm{SO}_{2}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) d. \(\mathrm{CH}_{3} \mathrm{CH}_{3}\) or \(\mathrm{H}_{2} \mathrm{CO}\) e. \(\mathrm{CH}_{3} \mathrm{OH}\) or \(\mathrm{H}_{2} \mathrm{CO}\)
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