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) + [tex] \frac{1}{2} O_{2 _{(g)} [/tex] 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 [tex] 10^{37} [/tex]
When we get K = 1.51 X [tex] 10^{37} [/tex]
But, K = [tex](PO_{2})^ \frac{1}{2} [/tex]So, (1.51 X [tex] 10^{37} [/tex]) [tex]^ \frac{1}{2}[/tex] = 3.87 X [tex] 10^{18} [/tex]
Hence, the pressure of oxygen will be = 3.87 X [tex] 10^{18} [/tex] Pa
Now, Pa to atm conversion will be 3.756 X [tex] 10^{13} [/tex] atm
