Problem 127 - Chemistry

Question

In a solution with carbon tetrachloride as the solvent, the compound \(\mathrm{VCl}_{4}\) undergoes dimerization: $$2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8}$$ When 6.6834 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 \(1.696 \mathrm{g} / \mathrm{cm}^{3},\) and \(K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\) for \(\mathrm{CCl}_{4} . )\)

Step-by-Step Solution

Answer

The equilibrium constant for the dimerization of VCl4 at this temperature is \(5.62 \times 10^{-2}\).

Step 1: Calculate the molality of VCl4 in the solution

We know that the freezing point depression is given by the formula: $$ΔT_f = K_f × \text{molality}$$ Where ΔT_f is the freezing point depression, K_f is the cryoscopic constant for the solvent, and the molality is the concentration of the solute. We are given ΔT_f = 5.97°C and K_f = 29.8°C kg/mol, so we can solve for the molality: $$\text{molality} = \dfrac{ΔT_f}{K_f} = \dfrac{5.97\,^{\circ}\text{C}}{29.8\,^{\circ}\text{C·kg/mol}} = 0.2 \text{mol/kg (carbon tetrachloride)}$$

Step 2: Calculate the moles of VCl4 and carbon tetrachloride

First, find the moles of VCl4 in the solution: We are given 6.6834 g of VCl4, and the molar mass of VCl4 = 152.88 g/mol $$\text{moles}_{VCl4} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{6.6834\,\text{g}}{152.88\,\text{g/mol}} = 0.0437\, \text{mol}$$ Next, find the moles of carbon tetrachloride solvent: Given 100 g of solvent and the molar mass of carbon tetrachloride = 153.82 g/mol $$\text{moles}_{CCl4} = \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{100\,\text{g}}{153.82\,\text{g/mol}} = 0.65\, \text{mol}$$

Step 3: Calculate the initial concentration of VCl4 and V2Cl8

Now, let us find the initial concentration of the two compounds in the equilibrium equation using the volume of the solution. The volume can be calculated using the density of the solution. $$\text{mass}_{total} = \text{mass}_{VCl4} + \text{mass}_{CCl4} = 6.6834\,\text{g} + 100\,\text{g} = 106.6834\,\text{g}$$ Given, density \(=1.696\, \text{g/cm}^{3}\) \(\implies\) Volume = \(\dfrac{\text{mass}_{total}}{\text{density}} = \dfrac{106.6834\,\text{g}}{1.696\, \text{g/cm}^{3}} = 62.92\,\text{cm}^{3} = 62.92 \times 10^{-3}\, \text{L}\) Now, we can find the initial concentration of VCl4 and V2Cl8: $$[\text{VCl4}]_{initial} = \dfrac{\text{moles}_{VCl4}}{\text{volume}} = \dfrac{0.0437\,\text{mol}}{62.92 \times 10^{-3}\, \text{L}} = 0.694\,\text{M}$$ Since there is no dimer to start, \([\text{V2Cl8}]_{initial} = 0\,\text{M}\). Partial equilibrium equation: $$2 \,\text{VCl4} \rightleftharpoons \text{V2Cl8}$$ $$(0.694 - 2x_\text{eq})\quad \text{--->}\quad \quad x_\text{eq}$$

Step 4: Calculate the mole ratio of VCl4 and V2Cl8 at equilibrium

Now using the initial concentrations and the molality we calculated in step 1, we can calculate the moles of VCl4 and V2Cl8 at equilibrium based on their mole fraction in the solution $$\text{mole fraction}_{VCl4} = \dfrac{\text{moles}_{VCl4}}{\text{moles}_{VCl4} + \dfrac{1}{2}\text{moles}_{V2Cl8}}$$ Using the molality and moles of carbon tetrachloride, we can calculate the mole fraction $$\text{x}_{VCl4} = \dfrac{\text{molality} \times \text{moles}_{CCl4}}{\text{moles}_{VCl4} + \dfrac{1}{2}\text{moles}_{V2Cl8}}$$ $$0.2\, \text{mol/kg} × 0.65\, \text{mol} = \dfrac{0.0437\, \text{mol}-2x_\text{eq}+x_\text{eq}}{0.0437\, \text{mol}+0.5x_\text{eq}}$$ By solving this equation, we get: $$x_\text{eq} = 0.0235\, \text{M}$$ Now, we can find the \([\text{VCl4}]_{eq}\) and \([\text{V2Cl8}]_{eq}\). $$[\text{VCl4}]_{eq} = 0.694\, \text{M} - 2x_\text{eq} = 0.694\, \text{M} - 2(0.0235\, \text{M}) = 0.647\, \text{M}$$ $$[\text{V2Cl8}]_{eq} = x_\text{eq} = 0.0235\, \text{M}$$

Step 5: Calculate the equilibrium constant

Finally, we can calculate the equilibrium constant using the concentrations of VCl4 and V2Cl8 at equilibrium. $$K_{eq} = \dfrac{[\text{V2Cl8}]_{eq}}{[\text{VCl4}]_{eq}^2} = \dfrac{0.0235\, \text{M}}{(0.647\, \text{M})^2} = 5.62 \times 10^{-2}$$ Hence, the equilibrium constant for the dimerization of VCl4 at this temperature is \(5.62 \times 10^{-2}\).

Key Concepts

Vanadium(IV) chloride

Vanadium(IV) chloride, commonly written as \( \mathrm{VCl}_{4} \), is a chemical compound that can behave uniquely in a solution. It is a strong Lewis acid due to vanadium being in a high oxidation state. In solution, especially those like carbon tetrachloride, \( \mathrm{VCl}_{4} \) tends to form dimers. This is a process where two molecules combine to form a single unit, in this case forming \( \mathrm{V}_{2} \mathrm{Cl}_{8} \). Dimerization occurs because it can lead to a more stable structure. Remember, stability in compounds can often be achieved by decreasing their energy through stronger bonds or the sharing of electrons more effectively.

Understanding \( \mathrm{VCl}_{4} \) is important because it helps clarify why certain reactions, like dimerization, occur naturally. This specific reaction is crucial to point because it changes the properties of the compound, influencing how it behaves when dissolved.

Freezing point depression

Freezing point depression is a colligative property observed in solutions. Colligative properties depend on the number of solute particles in a solution, not on their identities. When a solute like \( \mathrm{VCl}_{4} \) is dissolved in a solvent such as carbon tetrachloride, it causes the freezing point of the solution to decrease. This is because solute particles disrupt the formation of solid structures necessary for the solvent to solidify, thus requiring a lower temperature to freeze.

This concept is quantified using the formula \( \Delta T_f = K_f \times \text{molality} \), where \( \Delta T_f \) is the change in freezing point, \( K_f \) is the cryoscopic constant (a property of the solvent), and molality is the concentration of the solute. The ability to calculate \( \Delta T_f \) allows scientists and students to determine important properties of solutions, such as the extent of dissolution or interactions between particles.

Equilibrium constant

The equilibrium constant, denoted as \( K_{eq} \), is a vital concept in chemistry that describes the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction. For the dimerization of \( \mathrm{VCl}_{4} \), the equilibrium constant can be expressed as:

• \( K_{eq} = \frac{[\text{V}_2\text{Cl}_8]_{eq}}{[\text{VCl}_4]^2_{eq}} \)

Knowing \( K_{eq} \) helps us understand the position of equilibrium and predict how changes in concentration, pressure, or temperature might shift the equilibrium. A small \( K_{eq} \) value, like in our case of \( 5.62 \times 10^{-2} \), indicates that the reactants are favored, meaning not much of \( \mathrm{VCl}_{4} \) converts to \( \mathrm{V}_{2} \mathrm{Cl}_{8} \) under the given conditions.

Mastering equilibrium constants equips you with the ability to analyze chemical systems comprehensively, crucial for both educational and laboratory settings.

Mole fraction

Mole fraction is another way to express concentrations in a mixture. It is defined as the ratio of the number of moles of a component to the total number of moles of all components present in the mixture. In the context of \( \mathrm{VCl}_{4} \) dimerization, mole fraction allows us to examine the relative proportions of \( \mathrm{VCl}_{4} \) and \( \mathrm{V}_{2} \mathrm{Cl}_{8} \) at equilibrium.

• \( \text{Mole fraction}_{VCl4} = \frac{\text{moles of } \mathrm{VCl}_{4}}{\text{moles of } \mathrm{VCl}_{4} + \text{moles of } \mathrm{V}_{2} \mathrm{Cl}_{8}} \)

Understanding and using mole fraction is fundamental when dealing with solutions because it provides a dimensionless figure that makes it easy to compare the different components, which can be immensely helpful in predicting the behavior of a solution during chemical reactions.