The equilibrium constant, K, for the following reaction is 1.54×10-2 at 506 K:

PCl5(g) PCl3(g) + Cl2(g)

An equilibrium mixture of the three gases in a 15.7 L container at 506 K contains 0.207 M PCl5, 5.64×10-2 M PCl3 and 5.64×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 6.47 L?

To solve this problem, we can use the concept of the equilibrium expression and the stoichiometry of the reaction.

First, let's write down the balanced equation for the reaction:
PCl5(g) ⇌ PCl3(g) + Cl2(g)

The equilibrium expression is given by:
K = [PCl3] * [Cl2] / [PCl5]

We are given the value of K, which is 1.54×10^-2 at 506 K.

Now, let's calculate the initial concentrations of each gas based on the given information:
[PCl5] = 0.207 M
[PCl3] = 5.64×10^-2 M
[Cl2] = 5.64×10^-2 M

Next, we need to find the final concentrations after the volume is compressed to 6.47 L.

To do this, we can use the concept of the ideal gas law:
PV = nRT

Since the temperature (T) is constant, we can rewrite the equation as:
PV = constant

With this, we can set up the initial and final conditions of the gases:
P1V1 = P2V2

The initial conditions:
P1 = ?
V1 = 15.7 L

The final conditions:
P2 = ?
V2 = 6.47 L

We need to find the ratio of pressures (P1/P2) to determine the change in concentration of each gas.

Now, let's calculate the ratio of pressures:
P1/P2 = V2/V1 = 6.47 L / 15.7 L

P1/P2 = 0.412 (approximately)

So, the ratio of pressures is 0.412.

To find the change in concentration for each gas, we need to multiply the initial concentration by the ratio of pressures.

For PCl5:
[PCl5] equilibrium = [PCl5] initial * (P1/P2)
[PCl5] equilibrium = 0.207 M * 0.412
[PCl5] equilibrium = 0.085 M (approximately)

For PCl3:
[PCl3] equilibrium = [PCl3] initial * (P1/P2)
[PCl3] equilibrium = 5.64×10^-2 M * 0.412
[PCl3] equilibrium = 2.32×10^-2 M (approximately)

For Cl2:
[Cl2] equilibrium = [Cl2] initial * (P1/P2)
[Cl2] equilibrium = 5.64×10^-2 M * 0.412
[Cl2] equilibrium = 2.32×10^-2 M (approximately)

So, the concentrations of the three gases once equilibrium has been reestablished in the compressed volume of 6.47 L are:
[PCl5] = 0.085 M
[PCl3] = 2.32×10^-2 M
[Cl2] = 2.32×10^-2 M