Question

PLEASE please answer the following question asap, I would really appreciate it :)

6 The excessive production of ozone (0s) gas in the lower atmosphere causes rubber to deteriorate, green plants to turn brown, and persons with respiratory disease to have diffmculty breathing The reaction is: 3 02(N)20 Oxygen S 0/mol K) 205.1 161.1 238.9 249.2 142.7 ) Is the formation of Os from O2 favor at all temperature, at no temperature, at high temperature, or low temperature? ii) Calculate Δ(i" for this reaction at 298 K and 1 atm. iii) Calculate ΔG, at 298 K for this reaction urban smog where [O2] = 0.21 M and and [03]s 5x107 M [03]-5x10-7 M

Answer 1

i) At low temperature Ozone formation is favored as the O3 molecule loses O-atom at high temperature and forms O2 molecules. O+O3$\rightarrow$2O2

O3 is highly unstable so deteriorates at high temp

ii) we have to calculate the $\Delta$G value first for the reaction     3O2$\rightleftharpoons$2O3

step of ozone formation- O2 + O$\rightarrow$O3

$\Delta$Gof=$\Delta$Hof-T$\Delta$Sof

$\Delta$Gof(O2)=0-298K*205.1 J/K mol=-61119.8J/mol

$\Delta$Gof(O)=249.2 kj/mol-298k*161.1j/K mol=24920-48007.8 =201192.1J/mol

$\Delta$Gof(O3)=142.7 KJ/mol-298K* 238.9 J/K mol=71507.8 J/mol

$\Delta$G0rxn=$\Delta$G0_fproduct-$\Delta$G0_f reactant=2*71507.8 J/mol-3*(-61119.8J/mol)=326375J =326.375 KJ

iii)$\Delta$Grxn=$\Delta$Gorxn-RTln Qp=$\Delta$G0-8.314J/K mol * 298K ln [O3]^2/[O2]^3=326.375 KJ- 2477.5 J/mol ln (5*10^-7)^2/(0.21)^3=

                                                     =326.375 KJ-2477.5 J/mol ln 25*10^14/9.261* 10^-3

                                                      =326.375 KJ-2477.5 J/mol ln 2.70*10^-11

                                                       =326.375 KJ-(-60269.51J/mol)=386.644 KJ