A 0.239 mol sample of PCl5(g) is injected into an empty 2.85 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

(PCl5) = 0.239 mol/2.85L = approx 0.08 but you need a better answer than that estimate.

.........PCl5 ==> PCl3 + Cl2
I.......0.08.......0......0
C........-x........x......x
E......0.08-x......x......x

Substitute the E line into the Kc expression for PCl5 and solve for x and evaluate 0.08-x. By the way you can't work this problem without the Kc NUMBER. I assume you just didn't type this into the post.

To calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium, we need to use the ideal gas law and the balanced chemical equation for the reaction.

The balanced chemical equation for the decomposition of PCl5(g) is:
PCl5(g) ⇌ PCl3(g) + Cl2(g)

Let's start by calculating the initial moles of PCl5(g) in the reaction vessel. We are given that the sample has 0.239 mol of PCl5(g).

Now, we need to determine the total number of moles of gas in the reaction vessel. Since PCl3(g) and Cl2(g) are the products of the reaction, their moles will be zero initially.

Next, we can calculate the total moles of gas at equilibrium using the ideal gas law. The ideal gas law equation is given by:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = moles of gas
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

First, we convert the temperature from Celsius to Kelvin by adding 273.15:
T = 250 °C + 273.15 = 523.15 K

Since the reaction vessel is empty initially, the pressure inside is atmospheric pressure, which is approximately 1 atm.

Using the ideal gas law equation, we can rearrange it to calculate the moles of gas:
n = (PV) / (RT)

Plugging the values into the equation, we get:
n = (1 atm * 2.85 L) / (0.0821 L·atm/mol·K * 523.15 K)

Now, we substitute the values and calculate the total moles of gas at equilibrium.

n = (2.85 L) / (0.0821 L·atm/mol·K * 523.15 K)

n ≈ 0.0659 mol

Now, let's calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

The concentration is expressed in terms of moles per liter (mol/L). To find the concentration, divide the moles of substance by the volume of the reaction vessel.

Concentration of PCl5(g) at equilibrium:
[PCl5] = moles of PCl5 / volume of reaction vessel
[PCl5] = 0.239 mol / 2.85 L

Concentration of PCl3(g) at equilibrium:
[PCl3] = moles of PCl3 / volume of reaction vessel
[PCl3] = moles of PCl3 / 2.85 L

Substituting the values, we can calculate the concentrations of both substances at equilibrium.

Please note that to obtain the exact concentrations at equilibrium, we need to know the equilibrium constant (Kc) of the reaction. Without that information, we can only calculate the ratio of the concentrations.