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

The pressure of O2 over a sample of NiO at 25.00°C can be calculated using the Van't Hoff equation. The equilibrium constant is approximately 9.02 x 10^-12, and the partial pressure of O2 would be 0.12 atm.

Step-by-step explanation:

To calculate the pressure of O2 (in atm) over a sample of NiO at 25.00°C, we can use the Van't Hoff equation:

ln(K) = -ΔG°/RT

Where K is the equilibrium constant, ΔG° is the standard Gibbs free energy change, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the given ΔG° value from kJ/mol to J/mol:

ΔG° = 212 kJ/mol x 1000 J/kJ = 212,000 J/mol

Next, we need to convert the temperature from Celsius to Kelvin:

T = 25.00°C + 273.15 = 298.15 K

Substituting the values into the Van't Hoff equation:

ln(K) = -212,000 J/mol / (8.3144 J/K·mol x 298.15 K)

Solving for ln(K):

ln(K) ≈ -26.33

Finally, exponentiating both sides of the equation to solve for K:

K ≈ e-26.33 ≈ 9.02 x 10-12

Therefore, the equilibrium constant is approximately 9.02 x 10-12. To calculate the pressure of O2, we can assume that the pressure of NiO is negligible compared to O2. So, at equilibrium, the partial pressure of O2 would be equal to:

PO2 = 0.24 atm x (1/2) = 0.12 atm

Answer 2

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) +
`(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))^ (1)/(2)`So, (1.51 X
`10^(37)`)
`^ (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