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 : In the reaction of NiO(s) ⇌ Ni(s) + 
 
We have the given data of Δ = 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 +RTlnQ

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

0 = Δ + RTlnK

lnK = Δ  / (RT)

ln K = 212 / (0.00831 X 298) = 85.6 
 therefore K =  = 1.51 X

Now, K = 
So, 1.51 X =  

So, the pressure of oxygen will be = 3.87 X Pa

On converting, Pa to atm it will be 3.756 X  atm

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer