In a study of the decomposition of nitrous oxide at 565 °C the following data we

Question

In a study of the decomposition of nitrous oxide at 565 °C the following data were obtained [N2O], M 1.27 0.635 0.318 0.159 seconds 870 60x103 6.08x103 Hint: It is not necessary to graph these data. The observed half life for this reaction when the starting concentration is 1.27 M is s and when the starting concentration is 0.635 M is M' s. The average ?( 1/[N20]) / ?? from t-0 s to t-870 s is The average ?(1/[N20]) / ? t from t-870 s to t-2.60x10's is M1 s Based on these data, the rate constant for this order reaction is M1 s-l

Explanation / Answer

ans)

1. observed half life for starting concentration of 1.27 M = 870 s

observed half life for starting concentration of 0.635 M = 2.60×103 s - 870s = 1.73×103 s

2. change in 1/ [N2O] for t =0 to t= 870 s = 1/0.635 - 1/1.27 = 0.7874M-1

now, 1/[N2O] per second will be = ( 0.787 M-1 ) /(870 s ) = 0.905 ×10-3 M-1s-1

change in1/ [N2O] for t= 870 s to t = 2.60 ×103 = 1/0.318 - 1/0.635 = 1.57 M-1

now, 1/[N2O] per second will be = (1.57 M-1 ) /(1.73×103 s) = 0.907×10-3 M-1s-1

3. it is a second order reaction. because with change in time per second the change in 1/[N2O] is two times [ 1.57 / 0.787 =1.99 aproximately 2]

rate constant is 0.907×10-3 M-1s-1

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