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 : 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) +  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 
When we get K = 1.51 X 

But, K = So, (1.51 X ) = 3.87 X
Hence, the pressure of oxygen will be = 3.87 X Pa

Now, Pa to atm conversion will be 3.756 X  atm

What do you mean by that? like the contents? the thc level?