Question

For the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ at \(600 . \mathrm{K}\), the equilibrium constant, \(K_{\mathrm{p}}\), is \(11.5 .\) Suppose that \(2.450 \mathrm{~g} \mathrm{PCl}_{5}\) is placed in an evacuated \(500 .-\mathrm{mL}\) bulb, which is then heated to \(600 . \mathrm{K}\). a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the degree of dissociation of \(\mathrm{PCl}_{5}\) at equilibrium?

Short answer

a. The pressure of PCl5 if it did not dissociate is 11.5 atm. b. The partial pressure of PCl5 at equilibrium is 4.34 atm. c. The total pressure in the bulb at equilibrium is 11.5 atm. d. The degree of dissociation of PCl5 at equilibrium is 0.3927.

Step-by-step solution

Calculate the initial pressure of PCl5 if it did not dissociate

Given the mass of PCl5 and the volume of the bulb, we can calculate the initial pressure using the ideal gas equation: \[ PV = nRT \] We need to first convert the mass of PCl5 to moles. Molar mass of PCl5 = 1 × (30.97 g/mol) + 5 × (35.45 g/mol) = 208.25 g/mol Therefore, moles of PCl5 = 2.450 g / 208.25 g/mol = 0.01176 mol Now we can solve for the pressure: \[ P = \frac{nRT}{V} \] Given: R = 0.0821 L atm / (K mol), T = 600 K, and V = 500 mL = 0.5 L a. \( P_{\mathrm{PCl}_5} = \frac{(0.01176 \, \text{mol})(0.0821 \, \frac{\text{L atm}}{\text{K mol}})(600 \, \text{K})}{0.5 \, \text{L}} = 11.5 \, \text{atm} \)

Use the expression for equilibrium constant and stoichiometry to write the equations for partial pressures

Let x be the change in the moles of PCl5 as it dissociates. Then the change in moles of PCl3 and Cl2 will be x (as per the reaction stoichiometry). At equilibrium: PCl5 = 0.01176 - x moles PCl3 = x moles Cl2 = x moles Now, we can express their partial pressures, \(P_{\mathrm{PCl}_5}\), \(P_{\mathrm{PCl}_3}\), and \(P_{\mathrm{Cl}_2}\) in terms of x. \[P_{\mathrm{PCl}_5} = \frac{(0.01176 - x) \cdot RT}{V} = \frac{(0.01176 - x) \cdot 0.0821 \cdot 600}{0.5}\] \[P_{\mathrm{PCl}_3} = \frac{x \cdot 0.0821 \cdot 600}{0.5}\] \[P_{\mathrm{Cl}_2} = \frac{x \cdot 0.0821 \cdot 600}{0.5}\] Next, we can write the equation for the equilibrium constant: \[ K_{\mathrm{p}} = \frac{P_{\mathrm{PCl}_3} \cdot P_{\mathrm{Cl}_2}}{P_{\mathrm{PCl}_5}} \]

Solve for x and partial pressures at equilibrium

Given Kp = 11.5, substitute the expressions for the partial pressures in the equilibrium constant equation and solve for x: \[ 11.5 = \frac{\frac{x \cdot 0.0821 \cdot 600}{0.5} \cdot \frac{x \cdot 0.0821 \cdot 600}{0.5}}{\frac{(0.01176 - x) \cdot 0.0821 \cdot 600}{0.5}} \] \[ x = 0.00462 \, \text{mol} \] Now, we can find the partial pressures at equilibrium: b. \( P_{\mathrm{PCl}_5} = \frac{(0.01176 - 0.00462) \cdot 0.0821 \cdot 600}{0.5} = 4.34 \, \text{atm} \) \[ P_{\mathrm{PCl}_3} = \frac{0.00462 \cdot 0.0821 \cdot 600}{0.5} = 3.58 \, \text{atm} \] \[ P_{\mathrm{Cl}_2} = \frac{0.00462 \cdot 0.0821 \cdot 600}{0.5} = 3.58 \, \text{atm} \] c. The total pressure in the bulb at equilibrium = \( P_{\mathrm{PCl}_5} + P_{\mathrm{PCl}_3} + P_{\mathrm{Cl}_2} = 4.34 + 3.58 + 3.58 = 11.5 \, \text{atm} \)

Calculate the degree of dissociation at equilibrium

The degree of dissociation (α) is the ratio of the change in moles of PCl5 at equilibrium to the initial moles of PCl5: \[ \alpha = \frac{x}{0.01176} \] d. Degree of dissociation, α = \( \frac{0.00462}{0.01176} = 0.3927 \) In conclusion, the answers for the given problem are: a. The pressure of PCl5 if it did not dissociate is 11.5 atm. b. The partial pressure of PCl5 at equilibrium is 4.34 atm. c. The total pressure in the bulb at equilibrium is 11.5 atm. d. The degree of dissociation of PCl5 at equilibrium is 0.3927.

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, often denoted as \(K\), is crucial in predicting how far a reaction will proceed before reaching equilibrium. For gas-phase reactions, we specifically use \(K_p\), which is framed in terms of the partial pressures of the gases involved. This dimensionless number derives from the concentrations (or pressures) of the reactants and products at equilibrium.
The general expression for a reaction \(aA + bB \rightleftharpoons cC + dD\) is:

• \( K_p = \left(\frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}\right)\)

Here, \(P_{X}\) represents the partial pressure of component X. The equilibrium constant is a fundamental way to determine the position of equilibrium and can indicate whether the reactants or products are favored at a given temperature.
In our exercise, the given \(K_p\) value of 11.5 suggests that at 600 K, the formation of products \(\text{PCl}_3\) and \(\text{Cl}_2\) is favored over the reactant \(\text{PCl}_5\). Understanding \(K_p\) allows chemists to predict behavior and optimize reactions accordingly.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and amount (in moles) of a gas. Mathematically, it is expressed as \(PV = nRT\), where:

• \(P\) is the pressure of the gas
• \(V\) is the volume it occupies
• \(n\) is the number of moles
• \(R\) is the ideal gas constant (0.0821 L atm K⁻¹ mol⁻¹)
• \(T\) is the temperature in Kelvin

This law is predicated on the behavior of an "ideal" gas, where the molecules have no volume and experience no intermolecular forces. Though real gases deviate from this model to some extent, especially under high pressure or low temperature, it provides an efficient approximation for many practical purposes.
In the exercise you provided, the Ideal Gas Law was used to calculate the initial pressure of \(\text{PCl}_5\) if it did not dissociate by using the mass to find moles and then applying the other known values. This step is critical to establishing a baseline for further calculations involving equilibration.

Partial Pressure

In gas mixtures, each individual gas exerts pressure independently of the others, known as its partial pressure. The total pressure exerted by the gas mixture is the sum of the partial pressures of all gases present. This is a key concept in chemistry, especially when dealing with reactions involving gases, as seen in our exercise.
For a given gas in a mixture, its partial pressure can be found through the relation:

• \(P_i = \chi_i \cdot P_{\text{total}}\)

where \(\chi_i\) is the mole fraction of the gas in the mixture, and \(P_{\text{total}}\) is the total pressure of the mixture.
In chemical equilibria, knowing the partial pressures is essential for calculating the equilibrium constant \(K_p\). In our exercise, the partial pressures of \(\text{PCl}_5\), \(\text{PCl}_3\), and \(\text{Cl}_2\) were found at equilibrium, showing how the initial assumption of ideal behavior helps predict the distribution of species in a system.

Degree of Dissociation

The degree of dissociation is a measure of the extent to which a compound dissociates into its components. It is expressed as a fraction or percentage, which shows the number of moles dissociated over the total initial moles.
Mathematically, it's given as:

• \(\alpha = \frac{x}{n_0}\)

where \(x\) is the change in moles at equilibrium, and \(n_0\) is the initial moles.
Understanding the degree of dissociation is vital because it informs the extent of a reaction and how the reactants and products are distributed at equilibrium. It can be influenced by several factors, including temperature and pressure.
In our example, the degree of dissociation of \(\text{PCl}_5\) was calculated to be approximately 39.27%. This suggests that a significant fraction of the initial \(\text{PCl}_5\) molecules have transformed into products, helping us understand the dynamics and directionality of the reaction under given conditions.