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1 March, 14:33

For the reaction 2N2O5 (g) 4NO2 (g) + O2 (g), the following data were colected:

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

1) The order of this reaction in N2O5 is:

a) 2

b) 3

c) 1

d) 0

2. The half-life of this reaction is approximately:

a) 18 minutes

b) 23 minutes

c) 36 minutes

d) 45 minutes

e) 15 minutes

3. The initial rate of production of NO2 for this reaction is approximately

A) 1.6

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Answers (1)
  1. 1 March, 14:35
    0
    a) The reaction is first order, that is, order 1. Option C is correct.

    b) The half life of the reaction is 23 minutes. Option B is correct

    c) The initial rate of production of NO2 for this reaction is approximately = (3.7 * 10⁻⁴) M/min. Option has been cut off.

    Explanation:

    First of, we try to obtain the order of the reaction from the data provided.

    t (minutes) [N2O5] (mol/L)

    0 1.24x10-2

    10 0.92x10-2

    20 0.68x10-2

    30 0.50x10-2

    40 0.37x10-2

    50 0.28x10-2

    70 0.15x10-2

    Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

    C₀ = Initial concentration of the reactant

    C = concentration of the reactant at any time.

    k = rate constant

    t = time since the reaction started

    T (1/2) = half life

    We Start from the first guess of zero order.

    For a zero order reaction, the general equation is

    C₀ - C = kt

    k = (C₀ - C) / t

    If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

    C₀ = 0.0124 M

    At t = 10 minutes, C = 0.0092 M

    k = (0.0124 - 0.0092) / 10 = 0.00032 M/min

    At t = 20 minutes, C = 0.0068 M

    k = (0.0124 - 0.0068) / 20 = 0.00028 M/min

    At t = 30 minutes, C = 0.0050 M

    k = (0.0124 - 0.005) / 30 = 0.00024 M/min

    It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

    We try first order next, for first order reaction

    In (C₀/C) = kt

    k = [In (C₀/C) ]/t

    C₀ = 0.0124 M

    At t = 10 minutes, C = 0.0092 M

    k = [In (0.0124/0.0092) ]/10 = 0.0298 / min

    At t = 20 minutes, C = 0.0068 M

    k = 0.030 / min

    At t = 30 minutes, C = 0.0050 M

    k = 0.0303

    At t = 40 minutes

    k = 0.0302 / min

    At t = 50 minutes,

    k = 0.0298 / min

    At t = 60 minutes,

    k = 0.031 / min

    This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

    The rate constant to be taken will be the average of them all.

    Average k = 0.0302 / min.

    b) The half life of a first order reaction is related to the rate constant through this relation

    T (1/2) = (In 2) / k

    T (1/2) = (In 2) / 0.0302

    T (1/2) = 22.95 minutes = 23 minutes.

    c) The initial rate of production of the product at the start of the reaction

    Rate = kC (first order)

    At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 / min

    Rate = 0.0302 * 0.0124 = 0.000374 M/min = (3.74 * 10⁻⁴) M/min
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