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Chapter 10: Problem 95

Diethyl ether $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\right)$ was one of the first chemicals used as an anesthetic. At \(34.6^{\circ} \mathrm{C}\) , diethyl ether has a vapor pressure of 760 . torr, and at \(17.9^{\circ} \mathrm{C},\) it has a vapor pressure of 400 . torr. What is the \(\Delta H\) of vaporization for diethyl ether?

Short Answer

Expert verified
The enthalpy of vaporization for diethyl ether is approximately \(26,089\,\mathrm{J/mol}\).

Step by step solution

01

Write down the Clausius-Clapeyron equation

The Clausius-Clapeyron equation is given by: \[\ln{\frac{P_2}{P_1}} = \frac{\Delta H_\text{vap}}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\] where \(P_1\) and \(P_2\) are the vapor pressures at temperatures \(T_1\) and \(T_2\), respectively, and \(R\) is the gas constant.
02

Convert the given temperature values to kelvin

To use the Clausius-Clapeyron equation, we need to convert the given temperature values from Celsius to Kelvin. \[T_1 = 17.9^{\circ}C + 273.15 = 291.05 K\] \[T_2 = 34.6^{\circ}C + 273.15 = 307.75 K\]
03

Plug in the given values and solve for \(\Delta H_\text{vap}\)

Using the given vapor pressures and converted temperature values, we can plug them into the Clausius-Clapeyron equation: \[\ln{\frac{760\,\mathrm{torr}}{400\,\mathrm{torr}}} = \frac{\Delta H_\text{vap}}{8.3145\, \mathrm{J/mol\cdot K}} \left(\frac{1}{291.05\,\mathrm{K}} - \frac{1}{307.75\,\mathrm{K}}\right)\] Now, solve for \(\Delta H_\text{vap}\): \[\Delta H_\text{vap} = 8.3145\, \mathrm{J/mol\cdot K} \cdot \left(\frac{1}{291.05\,\mathrm{K}} - \frac{1}{307.75\,\mathrm{K}}\right)^{-1} \cdot \ln{\frac{760\,\mathrm{torr}}{400\,\mathrm{torr}}}\] \[\Delta H_\text{vap} \approx 26,089\,\mathrm{J/mol}\] So, the enthalpy of vaporization for diethyl ether is approximately \(26,089\,\mathrm{J/mol}\).

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