Skip to main content
Chemistry LibreTexts

4.8: Rates and Concentrations

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Learning Objectives
    • Explain chemical reaction rates.
    • Explain the concentration effects on reaction rate.
    • Define the order of a reaction with respect to a reactant.
    • Define the overall order of a chemical reaction.
    • Define the rate constant of a chemical reaction.
    • Determine the order of a reaction by experiment.

    Concentration and Chemical Reaction Rate

    reaction.gifIn the introduction to chemical kinetics, we have already defined chemical reaction rates. Rates of chemical reactions depend on the nature of the reactants, the temperature, the presence of a catalyst, and concentration. This page discusses how the concentration affects the chemical reaction rates. Concentration effect is important because chemical reactions are usually carried out in solutions.

    Chemical Reaction Rates

    The reaction rates of chemical reactions are the amounts of a reactant reacted or the amount of a product formed per unit time (moles per second). Often, the amount can be expressed in terms of concentrations.

    \(Rate = \dfrac{amount\: reacted\: or\: produced}{time\: interval}\: \textrm{units: }\mathrm{g/s,\: mol/s,\: or\: \%/s}\)

    At certain conditions, the rates are functions of concentrations. Depending on the time interval between measurements, the rates are called

    • average rate: rate measured between long time interval
    • instantaneous rate: rate measured between very short interval
    • initial rate: instantaneous rate at the beginning of an experiment

    However, a more realistic representation for a reaction rate is the change in concentration per unit time, either the decrease of concentration per unit time of a reactant or the increase of concentration per unit time of a product. In this case, the rate is expressed in mol/(L sec).

    \(Rate = \dfrac{concentration\: change\: of\: a\: reactant\: or\: product}{time\: interval}\: \textrm{units: g/(M s), M/s, ppm/s etc.}\)

    Measuring Reaction Rate

    To measure a reaction rate, we usually monitor either a product or a reactant for its change. Any physical characteristic related to the quantity or concentration of a product or reactant can be monitored. Some of the characteristics to be monitored are:

    • change in pressure,
    • change in color (spectroscopic measurement),
    • temperature for exothermic or endothermic reaction, and
    • presence of certain key substance

    The change can be plotted on a graph, and from the graph, we can get the average rate or the instantaneous rate by either graphical methods or using a computer for the data analysis.

    Rate Constants and the Orders

    Usually, the rate of a reaction is a function of the concentrations of reactants. For example, the rate of the reaction

    \(\ce{2 NO + O2 \rightarrow 2 NO2}\)

    has the form:

    \(Rate = k \ce{[O2] [NO]^2}\)

    The rate is proportional to the concentration of \(\ce{O2}\), usually written as \(\ce{[O2]}\) and is proportional to the square of \(\ce{[NO]}\), or \(\ce{[NO]^2}\). The orders of 1 and 2 for \(\ce{[O2]}\) and \(\ce{[NO]}\) respectively have been determined by experiment, NOT from the chemical equation. The total order of this reaction is 3 (=2+1).

    Note the rates and order in the following example reactions:

    \(\ce{H2 + I2 \rightarrow 2 HI}\),
    \(Rate = k \ce{[H2] [I2]}\),

    Total order 2.

    \(\ce{H2 + Br2 \rightarrow 2 HBr}\),
    \(Rate = k \ce{[H2] [Br2]}^{1/2}\),

    Total order 1.5.

    In particular, note that orders are NOT determined from the stoichiometry of the reaction equation.

    Rates as Functions of Reactant Concentrations

    The order with respect to (wrt) a reactant is determined experimentally by keeping the concentration of other reactants constant, but varying the concentration of one of the reactants, say \(\ce{A}\) in a general reaction

    \(a \,\textrm A + b \,\textrm B + c \,\textrm C = products\)

    If concentrations of \(\ce{B}\) and \(\ce{C}\) are kept constant, you can measure the reaction rate of \(\ce{A}\) at various concentrations. You can then plot the rate as a function of \(\ce{[A]}\). For a zeroth order reaction, you will get a horizontal line, because

    \(rate = k\) (a horizontal line)

         |      /
         |     /rate = k [A]
         |    /
         |- -/- - - - - rate = k
         |  /
         | /
         0   1   2   3   4  [A]

    For a first order reaction, the plot is a straight line (linear), as shown above, because

    \(rate = k \ce{[A]}\) (a straight line)

    Note that \(rate = k\) when \(\mathrm{[A] = 1}\).

    For a second order reaction, the plot is a branch of a parabola, because

    \(rate = k \ce{[A]}^2\)

         |      .
         |       rate = k [A]2 
         |     .   (a branch of 
         |          a parabola)
         |    .
         | - . - - - - - -
         |  .
         0   1   2   3   4  [A]

    For a reaction with an infinite order, the plot is a step function. The rate is small, almost zero, when \(\ce{[A]}\) is less than 1. When \(\ce{[A]}\) is greater than or equal to 1, then the reaction rate is very large. This model applies to nuclear explosion, except that \(\mathrm{[A] = 1}\) is actually the critical mass of the fission material.

    \(rate = k \ce{[A]}^{\infty}\)

         |       (order = infinity)
         |   |   rate = k [A]00
         |   |   (a vertical line)
         |   |
         |   |
         |   |
         |   |
         0   1   2   3   4  [A]

    Is there a chemical process like this? Well, we all know that one of the key conditions in an atomic bomb is to have a critical mass of the fission material, \(\ce{^235U}\) or \(\ce{^239Pu}\). When such a mass is put together, the reaction rate increases dramatically, leading to an explosion. Thus, this model seems to apply; however, the mechanism for the fission reaction is not described by the order of the fission material.

    Variation of Rate, Rate Constant, and Order of a Reaction

    If only \(\ce{[A]}\) is varied in experiments, and the order wrt \(\ce{[A]}\) is n, then the rate has the general expression,

    \(rate = k \ce{[A]}^n\)

    In this expression, k is the specific rate constant, or the rate when \(\mathrm{[A] = 1.00}\). Again, the order n is not necessarily an integer, but its most common values are 0.5 (1/2), 1, 2, or 3. Cases in which n is a negative number are rare.

    Mathematical models for the effect of concentration on rates are interesting. In general, the rate of a reaction of order n with respect to \(\ce{A}\) can be represented by the equation:

    \(y = k x^n\), (n = various values including 0.5, 1, 2, 3, ...)

    Plots of equations for various values of n illustrate the dependence of rate on concentration for various orders.

    Evaluation of Order by Experiments

    For a chemical reaction, we often determine the order with respect to a reagent by determining the initial rate. When more than one reactants are involved, we vary the concentrations in a systematic way so that the effect of concentration of one of the reactants can be measured.

    For example, in a reaction involving three reactants, \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\), we vary \(\ce{[A]}\) from experiment 1 to experiment 2 and find out how the rate varies. Similarly, we vary concentrations of \(\ce{B}\) or \(\ce{C}\) in other experiments, keeping others constant, and investigate its effect. The example below illustrates the strategy for such an approach.


    Derive the rate law for the reaction \(\mathrm{A + B + C \rightarrow products}\) from the following data, where rate is measured as soon as the reactants are mixed.

    Experiment 1 2 3 4
    [A]o 0.100 0.200 0.200 0.100
    [B]o 0.100 0.100 0.300 0.100
    [C]o 0.100 0.100 0.100 0.400
    rate 0.100 0.800 7.200 0.400

    Assuming the orders to be x, y, and z respectively for \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\), we have

    \(rate = k \ce{[A]}^{\Large x} \ce{[B]}^{\Large y} \ce{[C]}^{\Large z}\)

    From experiment 1 and 2, we have:

    \(\dfrac{0.800}{0.100}=\dfrac{k\: 0.2^{\Large x}\: 0.1^{\Large y}\: 0.1^{\Large z}}{k\: 0.1^{\Large x}\: 0.1^{\Large y}\: 0.1^{\Large z}}\)

    Thus, \(8 = 2^{\Large x}\); and \(x = \dfrac{\ln8}{\ln2} = 3\)

    By similar procedures, we get \(y = 2\) and \(z = 1\). Thus, the rate law is:

    \(rate = k \ce{[A]}^3 \ce{[B]}^2 \ce{[C]}\)

    Note the following relationships:

    \(x = y^{\Large z}\)

    \(\ln x = z \ln y\)

    \(z = \dfrac{\ln x}{\ln y}\)


    The variation of reaction rates as functions of order and concentrations are summarized in the form of a Table below.

    rate law
    Plot of rate vs
    - d[A]/dt = k horizontal
    - d[A]/dt = k[A] straight line with
    slope = k
    - d[A]/dt = [A]2 a branch of
    = infinity
    - d[A]/dt = k [A]oo rate = 0 when [A] < 1
    rate = infinite when [A] > 1
    a vertical line at [A] = 1

    Confidence Building Questions

    1. Hint: First order wrt \(\ce{A}\)

      Skill -
      Recognize the order when rate is linear dependent on \(\ce{[A]}\). Only when n = 1 does the rate depend linearly on the concentration.

    2. Hint: Zeroth order wrt \(\ce{A}\).

      Skill -
      Recognize the order when \(rate = k\) when \(n = 0\).

    3. For a reaction that is second order with respect to a reactant \(\ce{A}\), how many times does the rate increase as \(\ce{[A]}\) increases by a factor of 2?

      Hint: The rate increases four times.

      Skill -
      Predict rate increases as the concentration doubles for reactions of various orders.

    4. What is the reaction rate when the concentrations of all reactants are unity (\(\mathrm{[A] = [B] = 1}\)) for a second order reaction?

      Hint: Rate = rate constant k.

      Discussion -
      \(Rate = k\), when \(\mathrm{[A] = [B] = 1}\) regardless of the order of the reaction. Note the difference between rate and rate constant, k.

    5. For a reaction that is 2nd order with respect to \(\ce{A}\), you can keep concentrations of other reactants constant, and vary the concentration of \(\ce{A}\). You can measure the initial rate in the experiments, and then plot rate as a function of \(\ce{[A]}\). What type of curve or line is such a plot?

      Hint: A branch of a parabola.

    6. In general, the rates increase as the concentrations increase for any reaction with a positive order, true or false?

      Hint: True

      Skill -
      What if n is a negative value?

    4.8: Rates and Concentrations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Chung (Peter) Chieh.

    • Was this article helpful?