Decomposition of ozone to oxygen proceeds in two steps : Setp 1 (fast) `O_(3) hArr O_(2) + (O) ` and Step 2 (slow) , `O_(3) + (O) hArr 2O_(2)`. The rate law is , rate =` K [O_(3)]^(2) [O_(2)]^(-1)` . Explain.

The formation of O_(3) from O_(2) is

A+ H_(2)O_(2) to B+ O_(2) . In this reaction, what is the role of H_(2)O_(2) ? Explain.

Depletion of ozone occurs as : 2O_3 to 3O_2 Step 1: O_3 overset(K_c)(iff) O_2 (O) (fast) Step 2: O_3 + (O) overset(K_c)(iff) 2O_2 (slow) What is the order of the reaction

The chemical recation 2O_(3) to 3O_(2) proceeds as follows : O_(3)overset ("Fast")(to)O_(2)+O & O+O_(3) overset ("Slow")(to)2O_(2) the rate law expression should be

Elementary unimolecular reaction have first order rate laws, elementary bimolecular reaction have second order rate laws. A rate law is often derived from a proposed mechanism by imposiing the steady state approximation by assuming that there in pre equilibrium A proposal mechanism must be constant with the experimental rate law. The decomposition of O_(3) obeys the mechanism given below, Step 1: O_(3)hArrO_(2)+(O)- fast Step 2: O_(3)+(O)toO_(2)+O_(2)- slow H_(2)+Br_(2)to2HBr Mechaism : Br_(2)hArr2Br (fast) H_(2)+brtoHBr+H (slow) H+BrtoHBR (Fast) Order of the reaction is

(A) Conversin of PbS to PbSO _(4) consumes four moles of ozone (R ) O_(3) to O _(2) + (O)

(A) Ozone is used for bleaching oils, ivory, flour, starch ect. (R ) O _(3) to O _(2) + (O)

N_(2) O_(2) hArr 2NO, K_(1) , (1)/(2) N_(2) + (1)/(2) O_(2) hArr NO. K_(2) , 2NO hArr N_(2) + O_(2). K_(3) , NO hArr (1)/(2) N_(2) + (1)/(2) O_(2). K_(4)

The decomposition of nitrogen pentoxide is given as 2N_(2)O_(5) to 4NO_(2) + O_(2) . The rates of reaction are ([N_(2)O_(5)])/(Delta t ) =k_(1)[N_(2)O_(5)],(Delta[NO_(2)])/(Deltat) = k_(2) [N_(2)O_(5)] and (Delta [O_(2)])/(Delta t) = k_(3) [N_(2)O_(5)] Relate the rate constants k_(1) , k_(2) and k_(3) .