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Chapter 17: Problem 37

Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\right.\) 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 given heat of vaporization and boiling point, and the equation ΔS = ΔH/T. First, convert the boiling point to Kelvin: T(K) = 35°C + 273.15 = 308.15 K. Next, convert the heat of vaporization to J/mol: ΔH = 27.5 kJ/mol × 1000 J/kJ = 27500 J/mol. Finally, calculate the entropy of vaporization: ΔS = ΔH/T = 27500 J/mol / 308.15 K ≈ 89.2 J/mol·K. Thus, the entropy of vaporization for ethanethiol is approximately 89.2 J/mol·K.

Step by step solution

01

Convert the boiling point to Kelvin

In order to calculate the entropy of vaporization, we need to have the temperature in Kelvin, not in Celsius. To convert the boiling point from Celsius to Kelvin, we use the following equation: T(K) = T(°C) + 273.15 Let's apply this equation to the given boiling point: T(K) = 35°C + 273.15
02

Calculate the temperature in Kelvin

Add the value in Celsius to the constant value to obtain the temperature in Kelvin: T(K) = 35 + 273.15 T(K) = 308.15 K Now we have the required temperature in Kelvin.
03

Calculate the entropy of vaporization

Recall that the entropy of vaporization can be found using the equation: ΔS = ΔH/T, where ΔH is the heat of vaporization and T is the temperature in Kelvin. We are given the heat of vaporization ΔH as 27.5 kJ/mol. To perform the calculation using the temperature in Kelvin, we must convert ΔH to J/mol. To do so, we multiply by 1000: ΔH = 27.5 kJ/mol × 1000 J/kJ ΔH = 27500 J/mol Now we can calculate the entropy of vaporization by dividing the heat of vaporization by the temperature in Kelvin: ΔS = ΔH/T ΔS = 27500 J/mol / 308.15 K
04

Compute the entropy of vaporization

Divide the heat of vaporization by the temperature in Kelvin to get the entropy of vaporization: ΔS = 27500 J/mol / 308.15 K ΔS ≈ 89.2 J/mol·K The entropy of vaporization for ethanethiol is approximately 89.2 J/mol·K.

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Most popular questions from this chapter

Consider the system $$ \mathrm{A}(g) \longrightarrow \mathrm{B}(g) $$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mole}\) of \(\mathrm{A}(g)\) at \(1.00\) atm and \(1.00\) mole of \(\mathrm{B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{~g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent \(\left(\right.\) density \(\left.=0.879 \mathrm{~g} / \mathrm{cm}^{3}\right)\). The elevated boiling point of the solution was \(355.4 \mathrm{~K}\). What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{~kJ} / \mathrm{mol}\), the entropy of vaporization is \(95.95 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\), and the boiling-point elevation constant is \(2.5 \mathrm{~K} \cdot \mathrm{kg} / \mathrm{mol}\).

The equilibrium constant for a certain reaction decreases from \(8.84\) to \(3.25 \times 10^{-2}\) when the temperature increases from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\). Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise \(79 .\) )

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(\mathrm{g})\), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\). b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.
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