Question

19. For the reaction 2NO + O2 + 2NO2 the following reaction mechanism have been suggested; step1: NO + NO NZOZ Fast step 2: N2O2 + O2 k2, 2NO2 Slow and the apparent rate law was found to be R = kes [NO]? What would be the estimated effective activation energy, Eel, value, if the activation energies for forward and backward reactions in step 1 are 82 kJ mol- and 205 kJ mol', respectively, and the activation energy of step 2 is 82 kJ mol'? -41 kJ mol-1 205 kJ mol! 82 kJ mol 365 kJ mol-1 E) 41 kJ mol-

Answer 1

Rate of the reaction is the speed at which reactants are converted into products
Reaction 2N0 +0, —> 2NO, Mechanism Step 1 Notno kao NaO fast Stef2 2,0 +0ka > 2NO, Slow Rate = keff [N07² Rate of Reaction :- Rate of reaction is the speed at which reactants are converted into products, Rate in terms of concentration of produel is written as y =Rate = 4 d[NO] -O 3) Naw applying haul of mass action in Step2 Show Step ) d[NO] = 2 ky [N202] [6] - dt

3 Here N20, is an intermediate so, Now applying sabid equillibrium for fost Kfawaid = (conc. of Product) Stiochiometry kbackward ( Cone. Of Reactant) Stochiometry, Steb | R, = (No.1 Ki [no]² dal dolu LL LLL LLL LLL Kri [N90s] = kı. [No]{ -W Now from equation & ① and @ we get Rate = d Encin] Rate = (2K2 CN20,](0,7) Rute k, ki [No7°CQ7 Hesie Keft = kikz - KI kal K-1 a using Arrhenius Equation K= A é Ea/RT

Applying Arrhenius. Equation in - Keff = ki kz Aeffe Bartlet - A, EU/R1. A, e6z/RT An, e-E/RT lemparing Activation Energy DOBLE Eeft = 67 +6,-64) Eett = 82 KJ mal" +82 kJ mob"' –205 kJ mol- Eelt = -41 kJ mol-') Ang