The equation of gibbs free energy is given by
ΔG = ΔG° - RT ln(Qc)
We need ΔG° and Qc.
Qc = Qp x (RT)Δng
Δng = change in gaseous moles = 1-2 =-1
R = 0.0821 atm L / K mol is used as pressure is in atm
T = 298K
Given pO3 = 5 x 10^ (-7) atm
pO2 = 0.210 atm
pO = 1 x 10^(-5) atm
Qp = reaction quotient in terms of pressure
= pO3 / (pO2 x pO) = 0.238 atm-1
from above equation,
Now Qc = reaction quotient in terms of concentration = 0.238 /( 0.0821 x 298) = 9.7 x 10^(-3)
As Δng was -1 the RT factor comes in denominator.
And
ΔG° = ΔG° of product - ΔG° of reactant
= 142.7 - 249.17 = -106.47 kJ /mol
Now, as we have aquired all the missing quantities,
ΔG = ΔG° - RTlnQc
= -106.47 - [8.314 x 298 x ln( 9.7 x 10^(-3))÷ 1000]
= -106.47 + 11.48 = -94.99kJ/mol
R= 8.314 J /K mol is taken here as free energy is in joules and is divided by 1000 for conversion in kJ
I hope you understand the numerical. Kindly provide us with a feedback.
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