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).