Question

High levels of ozone (O3) cause rubber to deteriorate, green plants to turn brown, and many people to have difficulty breathing.
(a) Is the formation of O3 from O2 favored at all T, no T,highT,orlowT?

(b) Calculate ∆G ° for this reaction at 298 K.
(c) Calculate ∆G at 298 K for this reaction in urban smog where [O2] = 0.21 atm and [O3] = 5 x 10-7 atm.

Answer 1

The Chemical reaction for the formation of ozone is 3 O₂ (g) 2 O₃ (g).

(a) The Entropy change or this reaction will be -ive because the no of gas molecules are decresing (S = -ive).

It is an endothermic reaction because, O₂ molecules needs energy to convert into O₃ molecules so (H = +ive).

As we know the free energy equation G = H - TS

For a reaction to be favourable, G should be -ive.

Here for ozone formation, we have H = +ive and S = -ive

So G = H - TS , would be +ive (unfavorable reaction) at No Temp and high temp (+ive values).

Only in the case of low temp when G would attain a -ive value (due to -ive T) ozone formation will be favored.

(b) G =H- T S

In std conditions G^o =H^o- T S^o (T = 298 K)

Ozone's Enthalpy of Formation (H^o)= 142.7 kJ / mol

Calculation of S^o : standard entropy of O₂₍g): 205.03 J mol
                                    standard entropy of O₃(g): 238.82 J mol

S^o := products - reactants = 2O3(g)- 3O2(g) = 2(238.82) - 3(205.03) = -137.45

S^o (per mole of O3) = -137.45/2 = -68.725 J/mol = -68.725 * 10^-3 KJ/mol

G^o = 142.7 - (298)*(-68.725 * 10^-3) = 142.7+20.48 = 163.18 KJ/mol of O3

Since the reaction produces 2 mols of O³, so G^o = 2* 163.18 KJ/mol = 326.36 KJ/mol

(c) G = G° + RT log Q; here Q = [O3]² / [O2]³

G = G° + RT log Q; here Q = [O3]² / [O2]³

G= 326.36 + (8.314×10⁻³) (298) log[(5 x 10^-7)^² / (0.21)³]

G= 326.36 + (2.477) log[(5 x 10^-7 )^² / (0.21)³]

G= 326.36 + (2.477) * (-9.56)

G = 326.36 - 23.70
G = 302.66 kJ/mol