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

I wonder if the delta G you wrote is delta Go? Delta G at equilibrium is zero, then

delta Go = -RTlnK

To calculate the value of the equilibrium constant (Kp) for the given reaction, you need to use the Gibbs free energy change (ΔG) and the temperature (T = 298.0 K). There is a relationship between these values given by the equation:

ΔG = -RTln(Kp)

Where:
ΔG is the Gibbs free energy change
R is the gas constant (8.314 J/(K·mol))
T is the temperature in Kelvin
Kp is the equilibrium constant

To calculate Kp, we need to rearrange this equation to solve for Kp:

Kp = e^(-ΔG/RT)

Let's substitute the given values into the equation and calculate Kp:

ΔG = 140.0 kJ/mol = 140.0 × 10^3 J/mol
R = 8.314 J/(K·mol)
T = 298.0 K

Kp = e^(-ΔG/RT)
= e^(-((140.0 × 10^3 J/mol)/(8.314 J/(K·mol) × 298.0 K))

Now, let's calculate the value of Kp using a scientific calculator or software with exponential functions:

Kp ≈ e^(-((140.0 × 10^3)/(8.314 × 298.0)))

Kp ≈ e^(-59.64) (rounded to two decimal places)

Kp ≈ 1.344 (rounded to three decimal places)

Therefore, the value of the equilibrium constant (Kp) for the given reaction at 298.0 Kelvin is approximately 1.344.