Question

A 10.7 g sample of PCl5 is added to a sealed 1.75 L flask and the reaction is allowed to come to equilibrium at a constant temperature. At equilibrium, 47.4% of the PCl5 remains. What is the equilibrium constant, c , for the reaction?

Answer 1

To find the equilibrium constant, we first calculated the moles of PCl5 that reacted and the moles remaining at equilibrium. We then found the concentrations of PCl5, PCl3, and Cl2 at equilibrium by dividing the moles by the flask volume. Finally, we plugged these values into the equilibrium expression to calculate the Kc value of 0.0211.

Step-by-step explanation:

To calculate the equilibrium constant for the decomposition of PCl5 into PCl3 and Cl2, we need to use the information given about the percentage of PCl5 remaining at equilibrium and the initial amount of PCl5. Since 47.4% of PCl5 remains at equilibrium, this means that 52.6% of PCl5 has reacted to form PCl3 and Cl2.

We start by calculating the moles of PCl5 in the 10.7 g sample using its molar mass (208.24 g/mol for PCl5):

Moles PCl5 = 10.7 g / 208.24 g/mol = 0.0514 mol

At equilibrium, 47.4% remains, which is 0.0514 mol * 47.4/100 = 0.0244 mol

The change in the mole amount for PCl5 is therefore 0.0514 mol - 0.0244 mol = 0.027 mol. Since the stoichiometry of the reaction is 1:1:1, this is also the amount of PCl3 and Cl2 formed.

To find the concentrations at equilibrium, we divide each mole amount by the volume of the flask, 1.75 L:

[PCl5] = 0.0244 mol / 1.75 L = 0.01394 M

[PCl3] = [Cl2] = 0.027 mol / 1.75 L = 0.01543 M

The equilibrium expression for the reaction PCl5 (g) → PCl3 (g) + Cl2 (g) is Kc = [PCl3][Cl2]/[PCl5]. Plugging in the equilibrium concentrations gives us:

Kc = (0.01543 M)(0.01543 M) / (0.01394 M) = 0.0211