Use the following data to estimate the standard molar entropy of gaseous benzene at \(298.15 \mathrm{K} ;\) that is, \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \mathrm{atm})\right] .\) For \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{s}, 1 \mathrm{atm})\) at its melting point of \(5.53^{\circ} \mathrm{C}, S^{\circ}\) is \(128.82 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}\). The enthalpy of fusion is \(9.866 \mathrm{kJ} \mathrm{mol}^{-1} .\) From the melting point to 298.15 K, the average heat capacity of liquid benzene is \(134.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1} .\) The enthalpy of vaporization of \(\mathrm{C}_{6} \mathrm{H}_{6}(1)\) at \(298.15 \mathrm{K}\) is \(33.85 \mathrm{kJ} \mathrm{mol}^{-1},\) and in the vapor- ization, \(\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g})\) is produced at a pressure of 95.13 Torr. Imagine that this vapor could be compressed to 1 atm pressure without condensing and while behaving as an ideal gas. Calculate \(S^{\circ}\left[\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{g}, 1 \text { atm) }] .[ \text { Hint: Refer to }\right.\) Exercise \(88,\) and note the following: For infinitesimal quantities, \(d S=d q / d T ;\) for the compression of an ideal gas, \(d q=-d w ;\) and for pressure-volume work, \(d w=-P d V\).

Step by step solution

01

Calculate entropy change during fusion

First, entropy change associated with the process when benzene changes from solid to liquid at its melting point can be calculated using the formula: \[ \Delta S_{\text{fusion}} = \dfrac{\Delta H_{\text{fusion}}}{T_{\text{m}}} \] where \(\Delta H_{\text{fusion}} = 9.866 \, \text{kJ/mol}\) (enthalpy of fusion), and \(T_{\text{m}} = 5.53^{\circ}C = 278.68 \, K\). Substituting these into the formula yields \(\Delta S_{\text{fusion}}\).

02

Calculate entropy change during heating of liquid

Then, entropy change during heating the liquid benzene from melting point to 298.15 K is calculated using the formula: \[ \Delta S_{\text{heating}} = \int_{T_{\text{m}}}^{T_{\text{b}}} \dfrac{C_{p}}{T} \, dT \] where \(C_{p}\) is the heat capacity at constant pressure which is \(134.0 \, \text{JK}^{-1} \text{mol}^{-1}\), \(T_{\text{m}}\) is the melting temperature (278.68 K), and \(T_{\text{b}}\) is the given temperature (298.15 K). On simplification the integral yields an expression for \(\Delta S_{\text{heating}}\).

03

Calculate entropy change during vaporization

Next, entropy change during vaporization of benzene at 298.15 K is calculated by the formula: \[ \Delta S_{\text{vaporization}} = \dfrac{\Delta H_{\text{vaporization}}}{T_{\text{b}}} \] where \(\Delta H_{\text{vaporization}} = 33.85 \, \text{kJ/mol}\) and \(T_{\text{b}} = 298.15 \, K\). Substituting these into the formula will yield \(\Delta S_{\text{vaporization}}\).

04

Calculate entropy change during pressure change of gas

Entropy change during compression process of an ideal gas is calculated by the formula: \[ \Delta S_{\text{compression}} = R \ln \left( \dfrac{P_{2}}{P_{1}} \right) \] where \(R\) is universal gas constant (8.314 Jmol-1K-1), \(P_{1}\) is initial pressure (95.13 Torr = 0.125 atm) and \(P_{2}\) is final pressure (1 atm). Substituting these into the formula yields \(\Delta S_{\text{compression}}\).

05

Calculate total entropy change

Total entropy change is given by the sum of all individual entropy changes: \[ S^{\circ}[\text{C}_{6} \text{H}_{6}(\text{g}, 1 \text{atm})] = S^{\circ}[\text{C}_{6} \text{H}_{6}(\text{s}, 1 \text{atm})] + \Delta S_{\text{fusion}} + \Delta S_{\text{heating}} + \Delta S_{\text{vaporization}} + \Delta S_{\text{compression}} \] by substituting known values, the value for \( S^{\circ}[\text{C}_{6} \text{H}_{6}(\text{g}, 1 \text{atm})] \) is calculated.

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