Problem 17 - General Chemistry: Principles and Modern Applications

Question

\(1.00 \times 10^{-3} \mathrm{mol} \mathrm{PCl}_{5}\) is introduced into a \(250.0 \mathrm{mL}\) flask, and equilibrium is established at \(284^{\circ} \mathrm{C}\) : \(\mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) .\) The quantity of \(\mathrm{Cl}_{2}(\mathrm{g})\) present at equilibrium is found to be \(9.65 \times 10^{-4} \mathrm{mol}\) What is the value of \(K_{c}\) for the dissociation reaction at \(284^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The value of \(K_c\) for the dissociation reaction at \(284^{\circ}\mathrm{C}\) is approximately 107.04.

Step 1: Identify Initial Moles of Substances

From the problem, the initial moles of \(PCl_5\) is \(1.00 \times 10^{-3} \) mol. And it is given that the moles of \(Cl_2\) at equilibrium is \(9.65 \times 10^{-4}\) mol. Because the reaction has not started, there is initially no \(PCl_3\) and \(Cl_2\).

Step 2: Determine the Moles at Equilibrium

At equilibrium, the moles of \(PCl_5\) used will be equal to the moles of \(Cl_2\) formed since they follow a 1:1 ratio based on the balanced chemical reaction. So the moles of \(PCl_5\) at equilibrium will be \(1.00 \times 10^{-3} - 9.65 \times 10^{-4} = 0.35 \times 10^{-4}\) mol.

Step 3: Calculate the Concentrations at Equilibrium

To calculate the concentration, we need to divide the moles by the volume of the flask. The volume of the flask is 250.0 mL which is equivalent to 0.25 L. So, the concentration of \(PCl_5\) is \(0.35 \times 10^{-4} / 0.25 = 1.4 \times 10^{-4}\) M, the concentration of \(Cl_2\) is \(9.65 \times 10^{-4} / 0.25 = 3.86 \times 10^{-3}\) M, and the concentration of \(PCl_3\) is equal to the concentration of \(Cl_2\) which is \(3.86 \times 10^{-3}\) M.

Step 4: Calculate the Equilibrium Constant (\(K_c\))

Using the equilibrium constant expression for the balanced chemical reaction \(K_c = [PCl_3][Cl_2]/[PCl_5]\), we substitute the concentrations into the expression, then \(K_c = (3.86 \times 10^{-3})(3.86 \times 10^{-3})/(1.4 \times 10^{-4})\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state of a reversible chemical reaction where the rates of the forward and reverse reactions are equal. This balance results in no net change in the concentrations of the reactants or products over time. When equilibrium is reached, the system remains stable if undisturbed. In the case of the reaction \[ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g}) \] considered in the exercise, the equilibrium position is maintained by the constant formation and decomposition of \( \text{PCl}_5 \). At equilibrium, the concentration of \( \text{Cl}_2 \) remains constant, as does that of \( \text{PCl}_3 \) and the leftover \( \text{PCl}_5 \). Importantly, the relationship between the concentrations of reactants and products at equilibrium is used to calculate the equilibrium constant \( K_c \). This equilibrium constant helps predict the extent to which a reaction will proceed.

Mole Calculation

Mole calculation is a fundamental concept in chemistry that allows us to quantify substances in reactions. A mole is a unit that describes a quantity of substance containing as many elementary entities as there are atoms in exactly 12 grams of pure carbon-12, which is approximately Avogadro's number \( 6.022 \times 10^{23} \). In the exercise, we start with a given amount of moles of \( \text{PCl}_5 \), specifically \( 1.00 \times 10^{-3} \) mol, introduced into the reaction vessel. At equilibrium, not all initial \( \text{PCl}_5 \) remains, as some has decomposed to form \( \text{Cl}_2 \) and \( \text{PCl}_3 \). The problem specifies that \( 9.65 \times 10^{-4} \) mol of \( \text{Cl}_2 \) is present at equilibrium. By stoichiometry, this indicates an equivalent amount of \( \text{PCl}_3 \) is formed, and the corresponding moles of \( \text{PCl}_5 \) that reacted. Calculating the remaining \( \text{PCl}_5 \) by subtracting \( 9.65 \times 10^{-4} \) mol from the initial \( 1.00 \times 10^{-3} \) mol gives us \( 0.35 \times 10^{-4} \) mol remaining.

Reaction Concentration

Understanding reaction concentration is crucial for calculating equilibrium constants like \( K_c \). The concentration (usually denoted in molarity \( M \), moles per liter) of each substance in the reaction determines the system's behavior at equilibrium.In the example provided, concentrations are calculated by dividing the number of moles by the volume of the solution, where the volume of the flask is \( 250.0 \) mL (or \( 0.25 \) L). The concentration for each species at equilibrium is:

• \( \text{PCl}_5 \): \( (0.35 \times 10^{-4} \, \text{mol}) / 0.25 \, \text{L} = 1.4 \times 10^{-4} \, \text{M} \)
• \( \text{Cl}_2 \): \( (9.65 \times 10^{-4} \, \text{mol}) / 0.25 \, \text{L} = 3.86 \times 10^{-3} \, \text{M} \)
• \( \text{PCl}_3 \): Same as \( \text{Cl}_2 \), \( 3.86 \times 10^{-3} \, \text{M} \)

The equilibrium constant \( K_c \) is calculated using these concentrations, providing insights into how the reaction progresses and to what extent the forward reaction has taken place.