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 : Given data ;
Δ G° = 212 KJ/molTemperature is = 25+273 = 298 KAnd gas constant R = 0.008314 KJ/mol

The reaction is NiO(s) ⇌ Ni(s) + $\frac{1}{2} O_{2 _{(g)}$ We can find out the pressure on oxygen by using the equation of gibb's free energy with remainder quotient which is,                                      ΔG = ΔG° +RT lnQ
when at equilibrium Q = K, here K is equilibrim constant, and ΔG becomes 0;
so we get, 0 = ΔG° + RT ln K
on rearranging we get, ln K = ΔG° / (RT)
ln K = 212 / (0.00831 X 298) = 85.6  
Here, now K =  = 1.51 X $10^{37}$
When we get K = 1.51 X $10^{37}$

But, K = $(PO_{2})^ \frac{1}{2}$So, (1.51 X $10^{37}$) $^ \frac{1}{2}$ = 3.87 X $10^{18}$
Hence, the pressure of oxygen will be = 3.87 X $10^{18}$ Pa

Now, Pa to atm conversion will be 3.756 X $10^{13}$ atm