Question

Consider the equiLiBrium system described by the chemical reaction below. At equiLiBrium, a sample of gas from the system is collected into a 4.00 L flask at 600 K. The flask is found to contain 3.86 g of PCl₅, 12.7 g of PCl₃, and 13.3 g of Cl₂. What are the values of Kc and Kp for this reaction? PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Answer 1

The equilibrium constant expression for the given reaction is Kc = ([PCl₃][Cl₂])/[PCl₅]. We can determine the concentrations of the gases in the equilibrium mixture by calculating the moles of PCl₅, PCl₃, and Cl₂ and then converting them to molar concentrations using the volume of the flask. The values of Kc and Kp for this reaction are 0.231 and 11.4, respectively.

Step-by-step explanation:

The equilibrium constant expression for the given reaction is Kc = ([PCl₃][Cl₂])/[PCl₅]. To calculate the values of Kc and Kp, we need to determine the concentrations of the gases in the equilibrium mixture. To do this, we can use the given masses of PCl₅, PCl₃, and Cl₂ to calculate their moles and then convert them to molar concentrations using the volume of the flask. Once we have the concentrations, we can substitute them into the equilibrium constant expression to determine the values of Kc and Kp.

To calculate the molar concentrations of the gases, we will use the formula:

Concentration (M) = Moles/Volume (L)

Let's start by calculating the moles of PCl₅, PCl₃, and Cl₂:

Moles = Mass (g)/Molar mass (g/mol)

Molar mass of PCl₅ = 208.25 g/mol

Moles of PCl₅ = 3.86 g / 208.25 g/mol = 0.0186 mol

Molar mass of PCl₃ = 137.33 g/mol

Moles of PCl₃ = 12.7 g / 137.33 g/mol = 0.0925 mol

Molar mass of Cl₂ = 70.90 g/mol

Moles of Cl₂ = 13.3 g / 70.90 g/mol = 0.1874 mol

Next, we need to calculate the molar concentrations:

Concentration of PCl₅ = 0.0186 mol / 4.00 L = 0.00465 M

Concentration of PCl₃ = 0.0925 mol / 4.00 L = 0.0231 M

Concentration of Cl₂ = 0.1874 mol / 4.00 L = 0.0469 M

Now we can substitute these concentrations into the equilibrium constant expression:

Kc = ([0.0231][0.0469])/0.00465 = 0.231

Since the reaction involves gases, we can also calculate the value of Kp using the relationship:

Kp = Kc(RT)^(Δn)

Where R is the gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin (600 K), and Δn is the difference in the number of moles of gases on the product side and the reactant side. In this case, Δn = (1 + 1) - 1 = 1.

Substituting the values into the equation:

Kp = 0.231 (0.0821 L·atm/(mol·K))(600 K)^1 = 11.4

Therefore, the values of Kc and Kp for the given equilibrium system are 0.231 and 11.4, respectively.