Question

3. To reduce the amount of nitric oxide emitted from cars, we use catalytic converters, where NO + CO react on the catalyst surface to form CO, and N, via a first order reaction with a rate constant of 0.005 m/s (more information is given in the diagram below). The NO diffuses to the catalyst surface (Doco 10* m/s) Determine: a. The molar concentration of CO at the catalyst surface. b. The rate of removal of CO at the surface. c. The rate of removal if the process is diffusion limited. SCHEMATIC: Exhaust gases .kbar XAL 015 Mixture, DAB:104m2/s As.MA,s NO Catalytic surfaceX

Answer 1

We use catalytic converts, where NO + CO react on the surface to form CO_2 and 1/2 N_2. Calculate the mole fraction of No. at the surface X_Delta 1/X_delta 1 = 1/1 + (l k''_1/DAB) Here, X delta, s is the mole fraction of 'NO' at the surface, x_delta l is mole fraction of 'NO', k, is the reaction rate constant, DAB is the diffusivity of 'NO' and l is the length. x_delta, s/0.15 = 1/1 + [10^-3 times 0.05 m62ls/106-4 m^2/s] x_delta, s = 0.15 times [1/1 + 0.15] = 0.1 \r\n Calculate the rate of consumption of 'NO' n''_A, s = M_A n_A, s Here, NA''s is molar flex of 'NO', n''_Delta, x is molar flex rate of 'NO' at surface and M_A is molecular weight of 'NO'. Calculate the molar flex at the surface N_A, s'' = k'', Cx_A, l/1 + (1k''_1/DAB) Here, 'C' is the molar concentration C = P/RT = 1.2 bar times (1 atm/1.0133 bar)/8.314 times 10^-2