In the upper atmosphere, ozone is produced from oxygen gas in the following reaction. 3O2(g) → 2O3(g)

Question

ΔH^o for this reaction= 285 kJ/mol

ΔS^o for ozone formation= -141 kJ/mol*K

ΔG^o for ozone formation= 326 kJ/mol

Assume an atmosphere where p(O₂) = 0.150 atm, and where T = 298 K. Below what pressure of O₃ will ozone production be spontaneous? (Enter your answer in atm).

Answer 1

The first thing to understand is that the Gibb's free energy (ΔG) of a spontaneous reaction must be negative. The standard Gibb's free energy (ΔG°) is 326 KJ/mol. So the reaction is not spontaneous under standard conditions.

The equation for finding the spontaneity of a reaction not at equilibrium is

ΔG = ΔG° + RTLn(Q).

Q = P²_ozone / P³_oxygen

R = 8.314 J/K·mol

T = 298 K

ΔG° = 326 KJ/mol

Since we need ΔG to be less than 0, if we solve the equation as if it were at equilibrium, that will tell us what pressure the reaction must be below. Therefore,

ΔG = 0

Now we plug everything in and solve for Q.

0 = 326(KJ/mol) + (8.314*10^-3 KJ/K·mol)(298K)Ln(Q)

-326 KJ/mol = 2.478 KJ/mol ·Ln(Q)

-131.58 = Ln(Q)

e^-131.58 = Q

Q = 7.167*10^-58

Substituting in the equation for the reaction quotient, we can now solve for the pressure of ozone.

Q = P²_ozone / P³_oxygen

7.167*10^-58 = P²_ozone / (0.150 atm)³

2.418*10^-60 = P²_ozone

√2.418*10^-60 = √P²_ozone

P_ozone = 1.55*10^-30atm

The reaction will be spontaneous when the pressure of ozone is less than 1.55*10^-30atm.