Question

For the reaction NO(g) + O3(g) ↔ NO2(g) + O2(g) at a temperature of 298 K, the standard Gibbs free energy change is ΔGº = -198 kJ/mol. Calculate the non-standard Gibbs free energy ΔG in kJ/mol when the partial pressures of these gases are those shown below.

Species Partial pressure in atm

NO(g) 1.0 x 10-8

O3(g) 1.0 x 10-9

NO2(g) 1.0 x 10-6

O2(g) 0.20


I am most interested in what form of R I should use, 0.008314 kj or 8.314 J

And what Q is??? Should I leave out pure elements when calculation Q?


Thanks!

Answer 1

ΔG = -139kJ/mol

Step-by-step explanation:

The formula to find ΔG in a non-standard state is:

ΔG = ΔG° + RT ln Q

Where ΔG° is ΔG value for standard state, R is gas constant (0.008314 kJ/molK); T is absolute temperature (298K); And Q is reaction quotient.

For the reaction:

NO(g) + O₃(g) ⇄ NO₂(g) + O₂(g)

Reaction quotient, Q, is:

`Q = (P_(O_2)P_(NO_2))/(P_(NO)P_(O_3))`

Where P represents the pressures in the non-standard state

Replacing:

`Q = (0.20atm*1.0x10^(-6)atm)/(1.0x10^(-8)atm*1.0x10^(-9)atm)`

Q = 2.0x10¹⁰

ΔG = -198kJ/mol + 0.008314kJ/molK×298K ln 2.0x10¹⁰

-You can use also 8.314kJ/molK as constant but it is necessary the consistency in units-

ΔG = -139kJ/mol