Flammability is a material’s ability to burn in the presence of
1 answer:
You might be interested in
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
Δ G° = 212 KJ/molTemperature is = 25+273 = 298 KAnd gas constant R = 0.008314 KJ/mol
The reaction is NiO(s) ⇌ Ni(s) +
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 =
Hence, the pressure of oxygen will be = 3.87 X
Now, Pa to atm conversion will be 3.756 X