The reaction NO(g) + O 3 -> NO 2 (g) + O 2 (g) was studied in 2 experiments unde

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 O3?

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?

Explanation / Answer

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]yinitial / [NO]yfinal

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]xinitial / [NO]xfinal

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

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