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Chapter 16: Problem 37

Ethanethiol \((\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\) also called ethyl mercaptan) is 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?

Short Answer

Expert verified
The entropy of vaporization for ethanethiol can be calculated using the formula ΔS_vap = ΔH_vap / bP. First, convert the boiling point to Kelvin: bP (K) = 35°C + 273.15 = 308.15K. Then, convert the heat of vaporization to J/mol: ΔH_vap (J/mol) = 27.5 kJ/mol * (1000 J/1 kJ) = 27500 J/mol. Finally, calculate the entropy of vaporization: ΔS_vap = 27500 J/mol / 308.15K ≈ 89.25 J/mol K.

Step by step solution

01

Convert the boiling point from Celsius to Kelvin

Formula to convert Celsius to Kelvin: Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15 Given boiling point, bP = 35°C Now, convert the boiling point to Kelvin. bP (K) = 35°C + 273.15 = 308.15K
02

Calculate entropy of vaporization

The entropy of vaporization can be calculated using the following formula: Entropy of vaporization (ΔS_vap) = Heat of vaporization (ΔH_vap) / Boiling point (bP) Given the heat of vaporization, ΔH_vap = 27.5 kJ/mol. Now, we need to convert the ΔH_vap from kJ/mol to J/mol: ΔH_vap (J/mol) = 27.5 kJ/mol * (1000 J/1 kJ) = 27500 J/mol Now, calculate the entropy of vaporization, ΔS_vap: ΔS_vap = ΔH_vap (J/mol) / bP (K) = 27500 J/mol / 308.15K ΔS_vap ≈ 89.25 J/mol K (rounded to two decimal places) The entropy of vaporization for ethanethiol is approximately 89.25 J/mol K.

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