Calculating a Heat of Vaporization Using the Clausius–Clapeyron Equation Ether has Pvap = 400 mm Hg at 17.9 °C and a normal boiling point of 34.6 °C. What is the heat of vaporization, ΔHvap, for ether in kJ/mol?

Question

Calculating a Heat of Vaporization Using the Clausius–Clapeyron Equation

Ether has [latex]P_{vap}[/latex] = 400 mm Hg at 17.9 °C and a normal boiling point of 34.6 °C. What is the heat of vaporization, [latex]ΔH_{vap}[/latex], for ether in kJ/mol?

STRATEGY

The heat of vaporization, [latex]ΔH_{vap}[/latex], of a liquid can be obtained either graphically from the slope of a plot of ln [latex]P_{vap}[/latex] versus 1/T or algebraically from the Clausius–Clapeyron equation. As derived in Worked Example 10.5,

[latex]\ln P_2=\ln P_1+\left(\frac{\Delta H_{\mathrm{vap}}}{R} ight)\left(\frac{1}{T_1}-\frac{1}{T_2} ight)[/latex]

which can be solved for [latex]ΔH_{vap}[/latex]:

[latex]\Delta H_{ ext {vap }}=\frac{\left(\ln P_2-\ln P_1 ight)(R)}{\left(\frac{1}{T_1}-\frac{1}{T_2} ight)}[/latex]

where [latex]P_1[/latex] = 400 mm Hg and ln [latex]P_1[/latex] = 5.991, [latex]P_2[/latex] = 760 mm Hg at the normal boiling point and ln [latex]P_2[/latex] = 6.633, R = 8.3145 J/(K . mol), [latex]T_1[/latex] = 291.1 K (17.9 °C), and [latex]T_2[/latex] = 307.8 K (34.6 °C).

Accepted Answer

[latex]\Delta H_{ ext {vap }}=\frac{(6.633-5.991)\left(8.3145 \frac{\mathrm{J}}{\mathrm{K} \cdot \mathrm{mol}} ight)}{\frac{1}{291.1 \mathrm{~K}}-\frac{1}{307.8 \mathrm{~K}}}=28,600 \mathrm{~J} / \mathrm{mol}=28.6 \mathrm{~kJ} / \mathrm{mol}[/latex]

Question 10.6: Calculating a Heat of Vaporization Using the Clausius–Clapey......

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