The following data were collected for the rate of disappearance of NO in the rea

Question

The following data were collected for the rate of disappearance of NO in the reaction 2 NO(g) + O2(g) ? 2 NO2(g).
Experiment [NO] (M) [O2] (M) Initial Rate (M/s)
1 0.0126 0.0125 1.41 10-2
2 0.0252 0.0125 5.64 10-2
3 0.0252 0.0250 1.13 10-1
(a) What is the rate law for the reaction?
(b) what is the average value of the rate constant calculated from the three data sets?
(c) what is the rate of disaperance of NO when [NO] =0.0750 M and [O2]= 0.0100M
(d) what is the rate of disappearance of 02 at the concentrations given in part (d)?

Explanation / Answer

let rate law be
r = k[NO]a[O2]b

(a)

take ratio of rates of expt no. 1 and 2, you will get

2a = 4

=> a = 2

similarly, take ratios of rates of trial 2 and 3, you will get

2b = 2

=> b = 1

hence the rate law is

r = k[NO]2[O2]1

insert values of rates and concentrations of trial 1

1.41x10-2 = k[0.0126]2[0.0125]1

so, k1 = 7105.064 M-2s-1

similarly, insert values of rates and concentrations of trial 2

k2 = 7105.064 M-2s-1

similarly, insert values of rates and concentrations of trial 2

k3 = 7117.662 M-2s-1

so, average k = (k1+k2+k3)/3 = 7109.263 M-2s-1

complete rate law

r = 7109.263[NO]2[O2]1

(b)

as calculated above, kavg = 7109.263 M-2s-1

(c)

when [NO] =0.0750 M and [O2]= 0.0100M

rate = 0.4 M/s

rate of dissapperance of NO = 0.5*rate = 0.2 M/s

(d)

when [NO] =0.0750 M and [O2]= 0.0100M

rate = 0.4 M/s

rate of dissappearance of O2 = 0.4 M/s

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