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Chapter 17: Problem 94

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?

Short Answer

Expert verified
The entropy change for the vaporization of methane is 73.21 J/molK, and for hexane, it is 84.50 J/molK. 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 larger molar volume of gaseous hexane leads to a larger entropy change during vaporization compared to methane due to the greater dispersion of particles upon vaporization, resulting in increased disorder.

Step by step solution

01

Calculate the entropy change for the vaporization of methane and hexane

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.
02

Calculate the molar volume of gaseous methane and gaseous hexane

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.
03

Analyze the effect of differences in molar volumes on entropy change values

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vaporization
Vaporization is the process where a liquid turns into a gas. This transition involves converting molecules in the liquid state to the gas phase. It's essential to overcome intermolecular forces to change from liquid to gas. During vaporization, energy, usually in the form of heat, is absorbed by the substance. This energy separates the molecules, allowing them to move freely as a gas.

The heat required for this process is known as the enthalpy of vaporization \(\Delta H_{\text{vap}}\). Each substance has a specific enthalpy of vaporization, which depends on the strength of intermolecular forces within the liquid. Understanding this is crucial when calculating related changes, like entropy, during vaporization.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation used in chemistry to describe the behavior of gases. It relates the pressure \(P\), volume \(V\), temperature \(T\), and the amount of gas in moles \(n\) using the equation \(PV = nRT\), where \(R\) is the gas constant.
  • Pressure (P) measures the force the gas exerts on the walls of its container.
  • Volume (V) is the space in which gas molecules move.
  • Absolute temperature (T) is measured in Kelvin.
  • The gas constant (R) is a constant, 8.314 J/molK.
In the context of vaporization, the Ideal Gas Law helps calculate the molar volume of the vaporized substance, which further aids in determining its entropy change.
Entropy of Vaporization
Entropy of vaporization \(\Delta S_{\text{vap}}\) refers to the change in entropy when a liquid becomes a gas. Entropy is the measure of disorder in a system, and when vaporization occurs, the disorder increases significantly as molecules spread apart in the gaseous state.

To calculate the entropy of vaporization, use the formula \(\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T}\), where \(\Delta H_{\text{vap}}\) is the enthalpy of vaporization and \(T\) is the temperature in Kelvin.
  • The larger the entropy change, the greater the increase in disorder.
  • Differences in entropy values are influenced by how much the molar volume changes during the transition.
Entropy of vaporization is essential for understanding energy changes and molecular dynamics during phase transitions.
Molar Volume
Molar volume is the volume occupied by one mole of a substance. For gases, it is calculated using the Ideal Gas Law. In vaporization studies, molar volume shows how much space the gas takes up compared to its liquid form.

To find the molar volume, rearrange the Ideal Gas Law to \(\frac{V}{n} = \frac{RT}{P}\). This equation allows us to calculate the space taken by each mole of gas at a given temperature and pressure.
  • An increase in molar volume usually means a larger entropy change, as more space leads to greater molecular disorder.
  • Larger molar volumes imply that molecules are much further apart in the gaseous state than in the liquid state, reflecting significant expansion during vaporization.
Understanding molar volume is key to analyzing the differences in entropy for different substances undergoing vaporization.

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Most popular questions from this chapter

When most biological enzymes are heated, they lose their catalytic activity. This process is called denaturing. The change original enzyme \(\rightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

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 85.)

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 \(\mathrm{K}\) . Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\) . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are \(\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}\) greater than the \(K\) value. \(]\)

Consider the dissociation of a weak acid \(\mathrm{HA}\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)\) in water: $$\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)$$ Calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\)

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) \(686^{\circ} \mathrm{C},\) its boiling point. Calculate \(\Delta S^{\circ}, q, w,\) and \(\Delta E\) for the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure.
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