Question

The second-order rate constant for the decomposition of NO2 (to NO and O2) at 573 K is 0.54 M-1 s-1. Calculate the time for an initial NO2 concentration of 0.20 mol/L to decrease to a) one-half; b) one-sixteenth; c) one-ninth of its initial concentration.

Answer 1

The rate of second order reaction depends on the change in the concentration of the reactants.

Second order reaction:

A second order reaction is one in which the rate of reaction depends on the concentration of reactants.

For a second order reaction;

1/[A] = kt + 1/[A]o

When concentration deceresases to one-half;

1/0.1 = 0.54t + 1/0.20

1/0.1 -  1/0.20 = 0.54t

10 - 5/0.54 = t

t = 9.3 s

When concentration deceresases to one-sixteenth;

1/0.0125 = 0.54t + 1/0.20

1/0.0125 -   1/0.20 = 0.54t

80 - 5 =  0.54t

t = 80 - 5/0.54

t = 149 s

When concentration deceresases to one-ninth;

1/0.022 =  0.54t + 1/0.20

1/0.022  -   1/0.20 = 0.54t

45 - 5 =  0.54t

t = 45 - 5/0.54

t = 74 s

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