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14.7: Reaction Kinetics: A Summary

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    Learning Objectives
    • To use graphs to analyze the kinetics of a reaction.

    You learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order.

    We will illustrate the use of these graphs by considering the thermal decomposition of NO2 gas at elevated temperatures, which occurs according to the following reaction:

    \[(\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \label{14.26}\]

    Experimental data for this reaction at 330°C are listed in Table \(\PageIndex{1}\); they are provided as [NO2], ln[NO2], and 1/[NO2] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO2 are plotted versus time in part (a) in Figure \(\PageIndex{1}\).

    Figure \(\PageIndex{1}\): The Decomposition of NO2. These plots show the decomposition of a sample of NO2 at 330°C as (a) the concentration of NO2 versus t, (b) the natural logarithm of [NO2] versus t, and (c) 1/[NO2] versus t.

    Because the plot of [NO2] versus t is not a straight line, we know the reaction is not zeroth order in NO2. A plot of ln[NO2] versus t (part (b) in Figure \(\PageIndex{1}\)) shows us that the reaction is not first order in NO2 because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO2] versus t (part (c) in Figure \(\PageIndex{1}\)). This plot is a straight line, indicating that the reaction is second order in NO2.

    Table \(\PageIndex{1}\): Concentration of \(NO_2\) as a Function of Time at 330°C
    Time (s) [NO2] (M) ln[NO2] 1/[NO2] (M−1)
    0 1.00 × 10−2 −4.605 100
    60 6.83 × 10−3 −4.986 146
    120 5.18 × 10−3 −5.263 193
    180 4.18 × 10−3 −5.477 239
    240 3.50 × 10−3 −5.655 286
    300 3.01 × 10−3 −5.806 332
    360 2.64 × 10−3 −5.937 379

    We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure \(\PageIndex{2}\), the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates required multiple experiments at different NO2 concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.

    Figure \(\PageIndex{2}\): Properties of Reactions That Obey Zeroth-, First-, and Second-Order Rate Laws
    Example \(\PageIndex{1}\)

    Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction:

    2N2O5(soln) → 4NO2(soln) + O2(g)

    This reaction is carried out in a CCl4 solution at 45°C. The concentrations of N2O5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N2O5 concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant.

    Time (s) [N2O5] (M) ln[N2O5] 1/[N2O5] (M−1)
    0 0.0365 −3.310 27.4
    600 0.0274 −3.597 36.5
    1200 0.0206 −3.882 48.5
    1800 0.0157 −4.154 63.7
    2400 0.0117 −4.448 85.5
    3000 0.00860 −4.756 116
    3600 0.00640 −5.051 156

    Given: balanced chemical equation, reaction times, and concentrations

    Asked for: graph of data, rate law, and rate constant


    A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure \(\PageIndex{1}\) to determine the reaction order.

    B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction.


    A Here are plots of [N2O5] versus t, ln[N2O5] versus t, and 1/[N2O5] versus t:

    Ex 9 A.jpg

    The plot of ln[N2O5] versus t gives a straight line, whereas the plots of [N2O5] versus t and 1/[N2O5] versus t do not. This means that the decomposition of N2O5 is first order in [N2O5].

    B The rate law for the reaction is therefore

    rate = k[N2O5]

    Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s,

    \[\textrm{slope}=\dfrac{\ln[\mathrm{N_2O_5}]_{3000}-\ln[\mathrm{N_2O_5}]_0}{3000\textrm{ s}-0\textrm{ s}}=\dfrac{(-4.756)-(-3.310)}{3000\textrm{ s}}=-4.820\times10^{-4}\textrm{ s}^{-1}\]

    Thus k = 4.820 × 10−4 s−1.

    Exercise \(\PageIndex{1}\)

    1,3-Butadiene (CH2=CH—CH=CH2; C4H6) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C4H6 as a function of time at 326°C are listed in the following table along with ln[C4H6] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C4H6, the rate law, and the rate constant for the reaction.

    Time (s) [C4H6] (M) ln[C4H6] 1/[C4H6] (M−1)
    0 1.72 × 10−2 −4.063 58.1
    900 1.43 × 10−2 −4.247 69.9
    1800 1.23 × 10−2 −4.398 81.3
    3600 9.52 × 10−3 −4.654 105
    6000 7.30 × 10−3 −4.920 137


    Ex 9 Ans.jpg

    second order in C4H6; rate = k[C4H6]2; k = 1.3 × 10−2 M−1·s−1


    For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.

    Key Takeaway

    • Plotting the concentration of a reactant as a function of time produces a graph with a characteristic shape that can be used to identify the reaction order in that reactant.

    Contributors and Attributions

    • Anonymous

    14.7: Reaction Kinetics: A Summary is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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