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Chapter 17: Problem 112

Carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) form ideal solutions. Consider an equimolar solution of \(\mathrm{CCl}_{4}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) at \(25^{\circ} \mathrm{C}\). The vapor above the solution is collected and condensed. Using the following data, determine the composition in mole fraction of the condensed vapor.

Short Answer

Expert verified
Hence, the composition (mole fraction) of the condensed vapor for Carbon tetrachloride (CCl4) and Benzene (C6H6) is: \[y_{CCl_4} = 0.691\] \[y_{C_6H_6} = 0.309\]

Step by step solution

01

(Step 1: Understand Raoult's Law for ideal solutions)

Raoult's Law states that the partial pressure of a component in an ideal solution is the product of the mole fraction of that component in the solution and its vapor pressure. Mathematically, it can be expressed as: \[P_i = x_i P_i^*\] Where: - \(P_i\): Partial pressure of component i in the vapor phase - \(x_i\): Mole fraction of component i in the solution - \(P_i^*\): Vapor pressure of component i in its pure state
02

(Step 2: Calculate the partial pressures for CCl4 and C6H6 using Raoult's Law)

Since the solution is equimolar, we know that both CCl4 and C6H6 have the same mole fraction in the solution: \[x_{CCl_4} = x_{C_6H_6} = 0.5\] Now, we are given the vapor pressures of CCl4 and C6H6 at 25°C: \[P_{CCl_4}^* = 213 \, \mathrm{torr}\] \[P_{C_6H_6}^* = 95.1 \, \mathrm{torr}\] Using Raoult's Law, we can calculate the partial pressures of both components in the vapor phase: \[P_{CCl_4} = x_{CCl_4} P_{CCl_4}^* = 0.5 \times 213 \, \mathrm{torr} = 106.5 \, \mathrm{torr}\] \[P_{C_6H_6} = x_{C_6H_6} P_{C_6H_6}^* = 0.5 \times 95.1 \, \mathrm{torr} = 47.55 \, \mathrm{torr}\]
03

(Step 3: Calculate the total pressure in the vapor phase)

Now that we have the partial pressures of CCl4 and C6H6, we can calculate the total pressure in the vapor phase: \[P_{total} = P_{CCl_4} + P_{C_6H_6} = 106.5 \, \mathrm{torr} + 47.55 \, \mathrm{torr} = 154.05 \, \mathrm{torr}\]
04

(Step 4: Calculate the mole fractions of CCl4 and C6H6 in the vapor phase)

Using the partial pressures and total pressure, we can determine the mole fractions in the vapor phase. For CCl4: \[y_{CCl_4} = \frac{P_{CCl_4}}{P_{total}} = \frac{106.5 \, \mathrm{torr}}{154.05 \, \mathrm{torr}} = 0.691\] And for C6H6: \[y_{C_6H_6} = \frac{P_{C_6H_6}}{P_{total}} = \frac{47.55 \, \mathrm{torr}}{154.05 \, \mathrm{torr}} = 0.309\]
05

(Conclusion)

Hence, the composition (mole fraction) of the condensed vapor for Carbon tetrachloride (CCl4) and Benzene (C6H6) is: \[y_{CCl_4} = 0.691\] \[y_{C_6H_6} = 0.309\]

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Most popular questions from this chapter

At \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}, \Delta H^{\circ}=40.6 \mathrm{~kJ} / \mathrm{mol}\) for the vaporiza- tion of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .{ }^{\circ} \mathrm{C}\) and \(110 .{ }^{\circ} \mathrm{C}\). Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) do not depend on temperature.

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 temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). 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 \(\mathrm{Ni}(\mathrm{CO})_{4}\) 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 $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\). a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process \(\mathrm{S}_{\text {rhombic }}(s) \longrightarrow \mathrm{S}_{\text {monoclinic }}(s)\) b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Consider the following reaction: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}\), \(P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

In the text, the equation $$ \Delta G=\Delta G^{\circ}+R T \ln (Q) $$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of \(\mathrm{mol} / \mathrm{L}\) for the quantities in \(Q\), specifically for aqueous reactions. With this in mind, consider the reaction $$ \mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) $$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C}\). a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{~F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} M\) Based on the calculated \(\Delta G\) values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?
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