Ethanethiol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ; \text { also called ethyl mercaptan) is }\right.$ commonly added to natural gas to provide the “rotten egg” smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is 27.5 $\mathrm{kJ} / \mathrm{mol}$ . What is the entropy of vaporization for this substance?

We will use the equation for the entropy of vaporization, which is given by: \[\Delta S_{vap} = \frac{\Delta H_{vap}}{T}\] Where: - \(\Delta S_{vap}\) is the entropy of vaporization - \(\Delta H_{vap}\) is the heat of vaporization (given in the problem) - \(T\) is the temperature in Kelvin at which the vaporization occurs (boiling point)

We are given the boiling point of ethanethiol as \(35^{\circ} \mathrm{C}\). To convert it to Kelvin, we will add 273.15 to the Celsius temperature: \[T = 35 + 273.15 = 308.15 \, \mathrm{K}\]

Now, we can substitute the given heat of vaporization and the calculated temperature in Kelvin into the formula to find the entropy of vaporization: \[\Delta S_{vap}=\frac{27.5 \, \mathrm{kJ/mol}}{308.15 \, \mathrm{K}}\] First, we need to convert the heat of vaporization to J/mol. Since 1 kJ = 1000 J: \[\Delta H_{vap} = 27.5 \, \mathrm{kJ/mol} \times \frac{1000 \, \mathrm{J}}{1 \, \mathrm{kJ}} = 27500 \, \mathrm{J/mol}\] Now substitute the value of \(\Delta H_{vap}\) in J/mol and \(T\) in the entropy of vaporization formula: \[\Delta S_{vap} = \frac{27500 \, \mathrm{J/mol}}{308.15 \, \mathrm{K}} = 89.2 \, \frac{\mathrm{J}}{\mathrm{mol} \cdot \mathrm{K}}\] The entropy of vaporization for ethanethiol is approximately 89.2 J/mol K.