Question

The reaction

                                    NO(g) + O₃ -> NO₂(g) + O₂(g)

was studied in 2 experiments under pseudo-first order conditions.

a) [O₃] = 1x10¹⁴ molecules/cc in excess

the [NO] varied as follows   {Note, time is in msec (1 msec = 1x10^-3 s)!}

            time (msec)                 NO (molecules/cc)

                        0                      6x10⁸  

                        100                  5x10⁸  

                        500                  2.4x10⁸

                        700                  1.7x10⁸

                        1000                9.9x10⁷

b) [NO] = 2x10¹⁴ molecules/cc in excess     

            time (msec)                 O₃ (molecules/cc)

                        0                      1x10¹⁰

                        50                    8.4x10⁹           

                        100                  7x10⁹

                        200                  4.9x10⁹

                        300                  3.4x10⁹

a) what is the order of reaction with respect to NO?

b) what is the order of reaction with respect to O_3?

c) what is the overall rate law?

d) what is the pseudo-first order rate from experiment a); what overall rate coefficient does this give? (overall rate should be in cm³ molecule^-1 s^-1.)

e) what is the pseudo-first order rate from experiment b); what overall rate coefficient does this give?

Answer 1

rate law

rate = K [O₃]^x[NO]^y

If the concentration of one relative reactant remains constant because it is supplied in great excess, its concentration can be absorbed at the expressed constant rate, obtaining the pseudo first order reaction constant, because in fact, it depends on the same concentration of only one of the two reactants.

in 1st case O3 is in excess

so the rate now becomes

rate = K' [NO]^y ; where K' = K/[O₃]

rate = ([NO]_final -[NO]_initial)/ t

rate for 100 sec = ([6 X 10⁸] -[5 X 10⁸]) / 100 , = 10⁶ M L^-1 s^-1

similarly for 500 sec = ([6 X 10⁸] -[2.4 X 10⁸]) / 500 , = 0.72 X 10⁶ M L^-1 s^-1

for 700 sec = 0.614 X 10⁶ M L^-1 s^-1

for 1000 = 0.501 X 10⁶ M L^-1 s^-1

thus rate = K' [NO]^y

since rate constant is going to be nearly same

so (rate 1 /rate 2 )= [NO]^y_initial / [NO]^y_final

taking log on both sides

log (rate 1 /rate 2 ) = y log([NO]_initial / [NO]_final)

or y = log  (rate 1 /rate 2 ) / log([NO]_initial / [NO]_final)

y =  log  (10⁶ /0.72 X 10⁶ ) / log(6 X 10⁸ / 5 X 10⁸)

=0.42

thus calculating same way for other rates.. we have average y = 0.42

[O3]

rate = K' [O₃]^x ; where K' = K/[NO]

rate = ([O3]_final -[O3]_initial)/ t

rate for 100 sec = ([1 X 10¹⁰] -[0.82 X 10¹⁰]) / 50 , = 3.2 X10⁷ M L^-1 s^-1

similarly for 500 sec = ([1 X 10¹⁰] -[0.7 X 10¹⁰]) / 100 , = 3 X 10⁷ M L^-1 s^-1

for 200 sec = 2.55 X 10⁷ M L^-1 s^-1

for 300 s = 0.501 X 10⁷ M L^-1 s^-1

thus rate = K' [O3]^x

since rate constant is going to be nearly same

so (rate 1 /rate 2 )= [O3]^x_initial / [NO]^x_final

taking log on both sides

log  (rate 1 /rate 2 ) = y log([O3]_initial / [O3]_final)

or x = log  (rate 1 /rate 2 ) / log([NO]_initial / [NO]_final)

x =  log  (0.72 X 10⁷ / 0.614X 10⁷ ) / log(8.4 X 10⁹ / 7 X 10⁹)

x= 0.356

thus calculating same way for other rates.. we have average x = 0.40

thus rate law

rate = K [NO]^0.42[O₃]^0.40