Calculate the value of the equilibrium constant, Kp , for the following reaction at 298.0 Kelvin.

(Use the reaction free energy given below.)
2SO3 = 2SO2 + O2 in the gaseous state
ΔG = 140.0 kJ/mol

To calculate the value of the equilibrium constant, Kp, we need to make use of the reaction free energy (ΔG) at a given temperature. The equation relating ΔG and Kp is as follows:

ΔG = -RT ln Kp

Where:
ΔG - reaction free energy
R - ideal gas constant (8.314 J/(mol·K))
T - temperature in Kelvin
Kp - equilibrium constant

In this case, we are given ΔG as 140.0 kJ/mol, and the temperature (T) is 298.0 Kelvin.

First, we need to convert ΔG to joules/mol:
ΔG = 140.0 kJ/mol = 140000 J/mol

Next, we substitute the values into the equation and solve for Kp:

ΔG = -RT ln Kp

140000 J/mol = - (8.314 J/(mol·K)) * 298.0 K * ln Kp

Now, let's solve for ln Kp:

ln Kp = - (140000 J/mol) / [(8.314 J/(mol·K)) * 298.0 K]

ln Kp ≈ -59.19

To find Kp, we need to take the exponential (e) of both sides of the equation:

Kp ≈ e^(-59.19)

Calculating this in most calculators, we get:

Kp ≈ 7.47 x 10^(-26)

Therefore, the value of the equilibrium constant, Kp, for the given reaction at 298.0 Kelvin is approximately 7.47 x 10^(-26).