The reaction: NO (g) + O 3 (g) --> NO 2 (g) + O 2 (g) was studied by performing

Question

The reaction: NO (g) + O3(g) --> NO2(g) + O2 (g) was studied by performing 2 experiments. In the first experiment, the rate of disappearance of NO was measured in the presence of a large excess of O3. The results were as follows (O3 remains effectively constant at 1.0 x 1014 molecules/ cm3).

In the second experiment [NO] was held constant at 2.00 x 1014 molecules/cm3. The data for the disappearance of O3 is in the table below.

[NO] and [O3] are both first order, and overall rate law is rate = k [NO]1[O3]1

a) What is the value of the rate constant for each experiment?

Rate= k' [NO]x   Rate= k'' [O3]y

b) What is the value of the rate constant for the overall rate law?

Rate= k [NO]x [O3]y

Time (sec x 10-3) [NO] (molecules/cm3) 0 6.0 x 108 100 5.0 x 108 500 2.4 x 108 700 1.7 x 108 1000 9.9 x 107

Explanation / Answer

since NO and O3 are mentioned 1st order we use 1st order kinetics

k = ( 1/t) ln( a/a-x)   where a = initial amount , a-x = final amount.

for NO , k 1 = ( 1/100) ln( 6x10^8 /5x10^8) = 0.001823

k2 = ( 1/500) ln (6x10^8 /2.4x10^8) = 0.001832

k3 = ( 1/700) ln( 6x10^8 /1.7x10^8) = 0.0018

k4 = ( 1/1000) ln( 6x10^8/9.9x10^7) = 0.0018

average k' = 0.0018 s-1 for NO

similarly for O3 , k1 = ( 1/50) ln( 10^10/8.4x10^9) = 0.00349

k2 = ( 1/100) ln( 10^10 /7x10^9) = 0.003567

k3 = ( 1/200) ln( 10^10/4.9x10^9) = 0.003566

k4 = ( 1/300) ln( 10^10 /3.4x10^9) = 0.003596

average k" = 0.00355 s-1    is rate constant for O3

b) overall rate constant k = k' x k" = 0.0018 x 0.00355 = 6.4 x 10^ -6 s-1

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