The following table gives the vapor pressure of hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{~F}_{6}\right)\) as a function of temperature: $$ \begin{array}{lc} \text { Temperature (K) } & \text { Vapor Pressure (torr) } \\ \hline 280.0 & 32.42 \\ 300.0 & 92.47 \\ 320.0 & 225.1 \\ 330.0 & 334.4 \\ 340.0 & 482.9 \end{array} $$ (a) By plotting these data in a suitable fashion, determine whether the Clausius-Clapeyron equation (Equation 11.1 ) is obeyed. If it is obeyed, use your plot to determine \(\Delta H_{\text {vap }}\) for \(\mathrm{C}_{6} \mathrm{~F}_{6}\) (b) Use these data to determine the boiling point of the compound.

Short Answer

Expert verified
The enthalpy of vaporization for hexafluorobenzene, \(\Delta H_{vap}\), can be found by plotting the natural logarithm of vapor pressure versus the inverse of temperature. Since the plot is a straight line, the Clausius-Clapeyron equation is obeyed. The slope of the plot can be used to calculate the enthalpy of vaporization as \(\Delta H_{vap} = -slope \times R\), where \(R\) is the ideal gas constant. The boiling point of hexafluorobenzene can be found using the Clausius-Clapeyron equation with the atmospheric pressure set to 760 torr. The boiling point temperature, \(T_{BP}\), is given by \(T_{BP}=\frac{1}{\frac{1}{T_1}+\frac{R}{\Delta H_{vap}}ln\frac{760}{P_1}}\). Plug in the appropriate values to calculate the boiling point temperature.

Step by step solution

01

(a) Plot the data and check if Clausius-Clapeyron equation is obeyed

First, we will need to create a table of the natural logarithm of vapor pressure and the inverse of temperature using the given data. Temperature (K): \(280, 300, 320, 330, 340\) Vapor Pressure (torr): \(32.42, 92.47, 225.1, 334.4, 482.9\) Compute ln(Vapor Pressure) and 1/Temperature (K): \(ln(Vapor~Pressure):\) \(ln(32.42), ln(92.47), ln(225.1), ln(334.4), ln(482.9)\) \(1/Temperature~(K^{-1}):\) \(1/280, 1/300, 1/320, 1/330, 1/340\) Plot the data of ln(Vapor Pressure) vs 1/Temperature. If the plot is a straight line, then the Clausius-Clapeyron equation is obeyed. Now, find the slope of the plot to determine the enthalpy of vaporization. The slope is equal to \(\frac{-\Delta H_{vap}}{R}\).
02

(a) Determine the enthalpy of vaporization

Since the plot is a straight line, the Clausius-Clapeyron equation is obeyed. To find the enthalpy of vaporization, we need to determine the slope of the line. The slope can be found using the formula: \(slope = \frac{y_2 - y_1}{x_2 - x_1}\) Choose any two points on the line to compute the slope: \(x_1 = \frac{1}{T_1}\), \(x_2 = \frac{1}{T_2}\) \(y_1 = ln(P_1)\), \(y_2 = ln(P_2)\) Using the slope and the ideal gas constant \(R\), we can find the enthalpy of vaporization as: \(\Delta H_{vap} = -slope \times R\)
03

(b) Determine the boiling point

The boiling point of the compound is defined as the temperature at which the vapor pressure is equal to the atmospheric pressure. In this problem, we will assume the atmospheric pressure to be equal to 760 torr. To determine the boiling point, we need to find the temperature corresponding to a vapor pressure of 760 torr using the Clausius-Clapeyron equation: \(ln \frac{760}{P_1}=\frac{-\Delta H_{vap}}{R}\left(\frac{1}{T_{BP}}-\frac{1}{T_{1}}\right)\) Solving for the boiling point temperature (\(T_{BP}\)): \(T_{BP}=\frac{1}{\frac{1}{T_1}+\frac{R}{\Delta H_{vap}}ln\frac{760}{P_1}}\) Plug in the values of \(P_1\), \(T_1\), \(\Delta H_{vap}\), and \(R\) to find the boiling point temperature of hexafluorobenzene.

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Most popular questions from this chapter

The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is a \(-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melting points of the salts that form between the 1 -alkyl-3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{m.p.}=40^{\circ} \mathrm{C}\right)\) \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=10^{\circ} \mathrm{C}\right)\) and \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=-61^{\circ} \mathrm{C}\right) . \mathrm{Why}\) does the melting point decrease as the length of alkyl group increases?

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