Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. $$\begin{array}{lll} {\text { }} & {\text { Boiling Point (1 atm)}} & { \Delta H_{\mathrm{vap}} } \\ \hline {\text { Methane }} & \quad\quad\quad {112 \mathrm{K}} & {8.20 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Hexane }} & \quad\quad\quad {342 \mathrm{K}} & {28.9 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Compare the molar volume of gaseous methane at 112 \(\mathrm{K}\) with that of gaseous hexane at 342 \(\mathrm{K}\) . How do the differences in molar volume affect the values of \(\Delta S_{\mathrm{vap}}\) for these liquids?

To find the entropy change for both methane and hexane, we'll use the formula and the given values: For methane: \(\Delta S_{vap}^{CH_4} = \frac{\Delta H_{vap}^{CH_4}}{T^{CH_4}}\) \(\Delta S_{vap}^{CH_4} = \frac{8200 J/mol}{112 K}\) \(\Delta S_{vap}^{CH_4} = 73.21 J/molK\) For hexane: \(\Delta S_{vap}^{C_6H_{14}} = \frac{\Delta H_{vap}^{C_6H_{14}}}{T^{C_6H_{14}}}\) \(\Delta S_{vap}^{C_6H_{14}} = \frac{28900 J/mol}{342 K}\) \(\Delta S_{vap}^{C_6H_{14}} = 84.50 J/molK\) So, the entropy change for the vaporization of methane is 73.21 J/molK, and for hexane, it is 84.50 J/molK.

We'll use the Ideal Gas Law to calculate the molar volumes of gaseous methane and hexane. Since we are calculating the molar volumes, we'll rearrange the formula: \(\frac{V}{n} = \frac{RT}{P}\) For methane at 112 K: \(\frac{V}{n}^{CH_4} = \frac{8.314 J/molK \cdot 112 K}{101325 Pa}\) \(\frac{V}{n}^{CH_4} = 9.216 \times 10^{-3} m^3/mol\) For hexane at 342 K: \(\frac{V}{n}^{C_6H_{14}} = \frac{8.314 J/molK \cdot 342 K}{101325 Pa}\) \(\frac{V}{n}^{C_6H_{14}} = 2.817 \times 10^{-2} m^3/mol\) Thus, the molar volume of gaseous methane at 112 K is 9.216 x 10^{-3} m^3/mol, and of gaseous hexane at 342 K is 2.817 x 10^{-2} m^3/mol.

The entropy change for the vaporization of a substance depends on the molar volume of the substance. The larger the molar volume of the substance, the larger the entropy change during vaporization. This is due to the greater dispersion of particles upon vaporization, resulting in increased disorder. In this case, the molar volume of gaseous hexane is larger than that of gaseous methane, resulting in a larger entropy change in the vaporization of hexane compared to methane (84.50 J/molK for hexane and 73.21 J/molK for methane). As seen through the calculated values of molar volumes and entropy changes for vaporization, the differences in molar volumes affect the values of entropy change for liquids. The larger the molar volume, the larger the entropy change upon vaporization, due to the greater dispersion of particles, leading to increased disorder.