Question

Calculate the pressure of O2 (in atm) over a sample of NiO at 25.00°C if ΔG o= 212 kJ/mol for the reaction. For this calculation, use the value R = 8.3144 J/K·mol.
NiO(s) ⇌ Ni(s) + 1/2 O2(g)

Answer 1

Answer : In the reaction of NiO(s) ⇌ Ni(s) + $\frac{1}{2} O_{2} _{(g)}$
 
We have the given data of Δ$G^{0}$ = 212 KJ/mol
Temperature is = 25+273 = 298 K
And gas constant R = 0.008314 KJ/mol

we can use the relatinoship of gibb's free energy and remainder quotient (Q).
ΔG = Δ$G^{0}$ +RTlnQ

at equilibrium Q = K where K is equilibrim constant, and ΔG = 0;

0 = Δ$G^{0}$ + RTlnK

lnK = Δ$G^{0}$  / (RT)

ln K = 212 / (0.00831 X 298) = 85.6 
 therefore K = $e^{85.6}$ = 1.51 X$10^{37}$

Now, K = $(PO_{2}) ^{\frac{1}{2}}$
So, 1.51 X$10^{37}$ = $(PO_{2}) ^{\frac{1}{2}}$ 

So, the pressure of oxygen will be = 3.87 X $10^{18}$ Pa


On converting, Pa to atm it will be 3.756 X $10^{13}$ atm