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20 October, 12:41

The following reaction: NO2 (g) - -> NO (g) 1/2 O2 (g) is second-order in the reactant. The rate constant for this reaction is 3.40 L/mol*min. Determine the time needed for the concentration of NO2 to decrease from 2.00 M to 1.50 M.

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  1. 20 October, 15:37
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    t = 0.049 mins or 2.94 secs

    Explanation:

    For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;

    Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.

    Using the integrated rate law:

    I/[NO2]t - 1/[NO2]o = Kt

    Where K = 3.40 L/mol/min

    [NO2]t = 1.5 mol/L

    [N02]o = 2.0 mol/L

    t = ?

    Making t subject of formula;

    t = (1/[NO2]t - 1/[NO2]o) / K

    t = (1/1.5 - 1/2.0) / 3.40

    t = 0.049 mins or 2.94 secs
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