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In a solution with carbon tetrachloride as the solvent, the compound VCl_ undergoes dimerization:$$2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8}$$ When \(6.6834 \mathrm{g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(\left.1.696 \mathrm{g} / \mathrm{cm}^{3}, \text { and } K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol} \text { for } \mathrm{CCl}_{4} .\right)\).

Short Answer

Expert verified
The equilibrium constant for the dimerization of VCl₄ in a carbon tetrachloride solution, based on the given information, is approximately 0.806.

Step by step solution

01

Calculate the moles of VCl₄

First, we need to calculate the number of moles of VCl₄ that have been dissolved in the solution. We do this by converting the mass of VCl₄ into moles using its molar mass. The molar mass of VCl₄ is (50.94 + 4 * 35.45) g/mol. \[ \text{moles of } \mathrm{VCl_{4}} = \frac{6.6834 \;\mathrm{g}}{(50.94 + 4 \cdot 35.45) \;\mathrm{g/mol}} = 0.024855 \;\mathrm{mol} \]
02

Calculate the molality of the VCl₄ solution

Next, we will determine the molality of the VCl₄ solution by dividing the moles of VCl₄ by the mass of the solvent in kg. \[ \text{molality} = \frac{0.024855 \;\mathrm{mol}}{0.1000\;\mathrm{kg}} = 0.24855 \;\mathrm{mol/kg} \]
03

Calculate the molality of dissociated VCl₄

The molality of dissociated VCl₄ can be determined using the freezing point depression formula: \[ \Delta T_{f} = K_{f} \cdot m \cdot i \] Where \(\Delta T_{f}\) is the change in freezing point, \(K_{f}\) is the freezing point depression constant, \(m\) is the molality of the solution, and \(i\) is the van 't Hoff factor. In this case, i = 1 since dimerization of VCl₄ is a 1:1 reaction. Rearranging the formula to solve for the molality of dissociated VCl₄ gives: \[ m_{dissociated} = \frac{\Delta T_{f}}{K_{f}} = \frac{5.97^{\circ}\mathrm{C}}{29.8^{\circ}\mathrm{C \cdot kg/mol}} = 0.20033\;\mathrm{mol/kg} \]
04

Calculate the equilibrium constant

We can now determine the equilibrium constant for the dimerization of VCl₄ by dividing the molality of dissociated VCl₄ by the molality of the initial VCl₄ solution: \[ K_{eq} = \frac{m_{dissociated}}{m_{initial}} = \frac{0.20033\;\mathrm{mol/kg}}{0.24855\;\mathrm{mol/kg}} = 0.806 \] So, the equilibrium constant for the dimerization of VCl₄ is approximately 0.806.

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Key Concepts

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

Dimerization of VCl4
Understanding the dimerization of vanadium(IV) chloride, VCl4, involves exploring how individual VCl4 molecules associate to form V2Cl8 dimers. This reaction is represented by the equilibrium equation:

\[2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2}\mathrm{Cl}_{8}\]

In this chemical process, two VCl4 molecules reversibly combine into one V2Cl8 molecule. The dimerization is governed by an equilibrium constant, denoted as \(K_{eq}\), which quantifies the ratio of the concentration of products to that of reactants at equilibrium. A higher \(K_{eq}\) signifies a greater extent of dimerization. To calculate \(K_{eq}\), knowledge of molality and the mole fraction of VCl4 in the equilibrium mixture is essential. It's important to note that in a dimerization reaction, the van 't Hoff factor \(i\) is considered 1, as no increase in the number of particles occurs in the solution phase.
Freezing Point Depression
Freezing point depression is a colligative property of solutions that describes how the freezing point of a solvent decreases when a solute is dissolved in it. The extent of the freezing point depression depends on the number of solute particles in the solution, not on their identity.

This phenomenon is quantitatively expressed through the equation:
\[\Delta T_{f} = K_{f} \cdot m \cdot i\]

Where \(\Delta T_{f}\) is the change in freezing point, \(K_{f}\) is the freezing point depression constant specific to the solvent, \(m\) is the molality of the solution, and \(i\) is the van 't Hoff factor. For the dimerization of VCl4, the \(\Delta T_{f}\) observed provides a crucial link to finding the molality of dissociated VCl4, which in turn feeds into the calculation of the equilibrium constant for the reaction.
Molality Calculation
Molality is a measure of the concentration of a solute in a solution, defined by the moles of solute per kilogram of solvent.

Calculating the molality requires the following formula:
\[\text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kilograms}}\]

This concentration metric is pivotal when dealing with properties that depend on the amount of solute in the solution, such as freezing point depression. Since molality is based on the mass of the solvent, it remains unchanged with variations in temperature, unlike molarity, which is volume-based and therefore temperature-dependent. This characteristic makes molality particularly useful in thermodynamic calculations such as those required for determining equilibrium constants in reactions like VCl4 dimerization.
van 't Hoff Factor
The van 't Hoff factor, denoted by \(i\), is a dimensionless quantity that represents the number of particles a solute yields when dissolved in a solvent. It is crucial in the study of colligative properties of solutions, which include freezing point depression, boiling point elevation, and osmotic pressure.

For non-dissociating substances like the VCl4 dimer (V2Cl8), the van 't Hoff factor is essentially 1 since the solute does not increase the number of particles in solution. This value plays a direct role in the freezing point depression equation, indicating that the solute particles' effect on the solvent's freezing point is not magnified by dissociation or ionization. In the context of the VCl4 equilibrium exercise, correctly applying the van 't Hoff factor is key to accurately determining the molality of the dissociated VCl4 and ultimately the equilibrium constant of the dimerization reaction.

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Most popular questions from this chapter

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

An \(8.00-\mathrm{g}\) sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ At equilibrium the total pressure and the density of the gaseous mixtures were 1.80 atm and \(1.60 \mathrm{g} / \mathrm{L},\) respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

Consider the decomposition of the compound \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}\) as follows:$$\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+3 \mathrm{CO}(g)$$.When a 5.63 -g sample of pure \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}_{3}(g)\) was sealed into an otherwise empty 2.50 -L flask and heated to \(200 .^{\circ} \mathrm{C},\) the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate \(K\) for this reaction.

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