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Chapter 16: Problem 114

Carbon tetrachloride (CCl\(_4\)) and benzene (C \(_{6} \mathrm{H}_{6}\) ) 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
The composition of the condensed vapor is: - Mole fraction of Benzene: \(y_{C_{6}H_{6}} = 0.4536\) - Mole fraction of Carbon Tetrachloride: \(y_{CCl_{4}} = 0.5464\)

Step by step solution

01

Write down the given information and variables

Temperature: \(25^{\circ} C\) Benzene: \(C_{6}H_{6}\) Carbon Tetrachloride: \(CCl_{4}\) \[x_{C_{6}H_{6}} = x_{CCl_4} = 0.5\] Vapor Pressures: - Pure Benzene: \(P_{C_{6}H_{6}}^{sat}= 12.7\ kPa\) - Pure Carbon Tetrachloride: \(P_{CCl_{4}}^{sat}= 15.3\ kPa\)
02

Apply Raoult's Law to both components

We'll use Raoult's Law for each component to find the partial pressures. Raoult's Law for Benzene: \(P_{C_{6}H_{6}} = x_{C_{6}H_{6}}P_{C_{6}H_{6}}^{sat}\) Raoult's Law for Carbon Tetrachloride: \(P_{CCl_{4}} = x_{CCl_4}P_{CCl_{4}}^{sat}\)
03

Calculate partial pressures for both components

For Benzene: \(P_{C_{6}H_{6}} = 0.5 \times 12.7\ kPa = 6.35\ kPa\) For Carbon Tetrachloride: \(P_{CCl_{4}} = 0.5 \times 15.3\ kPa = 7.65\ kPa\)
04

Calculate total pressure of the mixture

The total pressure of the mixture can be calculated by adding the partial pressures. \(P_{total} = P_{C_{6}H_{6}} + P_{CCl_{4}}\) \(P_{total} = 6.35\ kPa + 7.65\ kPa = 14\ kPa\)
05

Calculate the mole fractions of Benzene and Carbon Tetrachloride in the vapor

Mole fraction of Benzene in the vapor: \(y_{C_{6}H_{6}} = \frac{P_{C_{6}H_{6}}}{P_{total}} = \frac{6.35\ kPa}{14\ kPa} = 0.4536\) Mole fraction of Carbon Tetrachloride in the vapor: \(y_{CCl_{4}} = \frac{P_{CCl_{4}}}{P_{total}} = \frac{7.65\ kPa}{14\ kPa} = 0.5464\)
06

Express the results

The composition of the condensed vapor: - Mole fraction of Benzene: \(y_{C_{6}H_{6}} = 0.4536\) - Mole fraction of Carbon Tetrachloride: \(y_{CCl_{4}} = 0.5464\)

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Most popular questions from this chapter

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Some water is placed in a coffee-cup calorimeter. When \(1.0 \mathrm{g}\) of an ionic solid is added, the temperature of the solution increases from \(21.5^{\circ} \mathrm{C}\) to \(24.2^{\circ} \mathrm{C}\) as the solid dissolves. For the dissolving process, what are the signs for \(\Delta S_{\mathrm{sys}}, \Delta S_{\text {surr, and }}\) \(\Delta S_{\text {univ }} ?\)

Consider the reaction $$2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S .\) b. Would the reaction be more spontaneous at high or low temperatures?

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(g),\) 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.

The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from \(300.0 \mathrm{K}\) to \(350.0 \mathrm{K}\). Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).
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