Answer:
[tex]\Delta G=133053.681574\ J[/tex]
Explanation:
T = Temperature = 298 K
[tex]n_{O_3}=1\ mole[/tex]
[tex][O_2]=0.21\ atm[/tex]
[tex][O_3]=5\times 10^{-7}\ atm[/tex]
Reaction quotient is given by
[tex]Q=\dfrac{P(O_3)}{P(O_2)^{1.5}}\\\Rightarrow Q=\dfrac{5\times 10^{-7}}{(0.21)^{1.5}}\\\Rightarrow Q=0.00000519566405324[/tex]
[tex]\Delta G^{\circ}=1\times\Delta G_f^{\circ}O_3-1.5\Delta G_f^{\circ}O_2\\\Rightarrow \Delta G^{\circ}=1\times 163.2-1.5\times 0\\\Rightarrow \Delta G^{\circ}=163.2\ kJ[/tex]
We know that
[tex]\Delta G=\Delta G^{\circ}+RTlnQ\\\Rightarrow \Delta G=163200+(8.314\times 298)\times ln(0.00000519566405324)\\\Rightarrow \Delta G=133053.681574\ J[/tex]
[tex]\Delta G=133053.681574\ J[/tex]