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4.4: Determining Rate Laws from Initial Rates (Differential Rate Laws)

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    81958
  • It is sometimes helpful to use an explicit algebraic method, often referred to as the method of initial rates, to determine the orders in rate laws. To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. We then solve this equation for the coefficient.

    Example \(\PageIndex{1}\): Determining a Rate Law from Initial Rates

    Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure \(\PageIndex{1}\)). One such reaction is the combination of nitric oxide, \(NO\), with ozone, O3:

    \[\ce{NO}(g)+\ce{O3}(g)⟶\ce{NO2}(g)+\ce{O2}(g) \nonumber\]

    Figure \(\PageIndex{1}\): Over the past several years, the atmospheric ozone concentration over Antarctica has decreased during the winter. This map shows the decreased concentration as a purple area. The Antarctic hole is stabilizing and may be slowly recovering. (credit: modification of work by NASA)

    This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.

    Trial \([NO]\) (mol/L) \([O_3]\) (mol/L) \(\dfrac{Δ[\ce{NO2}]}{Δt}\:\mathrm{(mol\:L^{−1}\:s^{−1})}\)
    1 1.00 × 10−6 3.00 × 10−6 6.60 × 10−5
    2 1.00 × 10−6 6.00 × 10−6 1.32 × 10−4
    3 1.00 × 10−6 9.00 × 10−6 1.98 × 10−4
    4 2.00 × 10−6 9.00 × 10−6 3.96 × 10−4
    5 3.00 × 10−6 9.00 × 10−6 5.94 × 10−4

    Determine the rate law and the rate constant for the reaction at 25 °C.

    Solution

    The rate law will have the form:

    \[\ce{rate}=k[\ce{NO}]^m[\ce{O3}]^n\nonumber\]

    We can determine the values of m, n, and k from the experimental data using the following three-part process:

    1. Determine the value of m from the data in which [NO] varies and [O3] is constant. In the last three experiments, [NO] varies while [O3] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.

    2. Determine the value of n from data in which [O3] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O3] varies. The reaction rate changes in direct proportion to the change in [O3]. When [O3] doubles from trial 1 to 2, the rate doubles; when [O3] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O3], and n is equal to 1.The rate law is thus:

      \[\ce{rate}=k[\ce{NO}]^1[\ce{O3}]^1=k[\ce{NO}][\ce{O3}]\nonumber\]
    3. Determine the value of k from one set of concentrations and the corresponding rate.

      \[\begin{align}
      k&=\mathrm{\dfrac{rate}{[NO][O_3]}} \nonumber\\
      &=\mathrm{\dfrac{6.60×10^{−5}\cancel{mol\: L^{−1}}\:s^{−1}}{(1.00×10^{−6}\cancel{mol\: L^{−1}})(3.00×10^{−6}\:mol\:L^{−1})}} \nonumber\\
      &=\mathrm{2.20×10^7\:L\:mol^{−1}\:s^{−1}}\nonumber
      \end{align}\nonumber\]

      The large value of k tells us that this is a fast reaction that could play an important role in ozone depletion if [NO] is large enough.

    Exercise \(\PageIndex{1}\)

    Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:

    \[\ce{CH3CHO}(g)⟶\ce{CH4}(g)+\ce{CO}(g)\nonumber\]

    Determine the rate law and the rate constant for the reaction from the following experimental data:

    Trial [CH3CHO] (mol/L) \(−\dfrac{Δ[\ce{CH3CHO}]}{Δt}\mathrm{(mol\:L^{−1}\:s^{−1})}\)
    1 1.75 × 10−3 2.06 × 10−11
    2 3.50 × 10−3 8.24 × 10−11
    3 7.00 × 10−3 3.30 × 10−10
    Answer:

    \(\ce{rate}=k[\ce{CH3CHO}]^2\) with k = 6.73 × 10−6 L/mol/s

    Example \(\PageIndex{2}\): Determining Rate Laws from Initial Rates

    Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:

    \[\ce{2NO}(g)+\ce{Cl2}(g)⟶\ce{2NOCl}(g)\nonumber\]

    Trial [NO] (mol/L) [Cl2] (mol/L) \(−\dfrac{Δ[\ce{NO}]}{Δt}\mathrm{(mol\:L^{−1}\:s^{−1})}\)
    1 0.10 0.10 0.00300
    2 0.10 0.15 0.00450
    3 0.15 0.10 0.00675

    Solution

    The rate law for this reaction will have the form:

    \[\ce{rate}=k[\ce{NO}]^m[\ce{Cl2}]^n\nonumber\]

    As in Example \(\PageIndex{21}\), we can approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k. In this example, however, we will use a different approach to determine the values of m and n:

    1. Determine the value of m from the data in which [NO] varies and [Cl2] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:

      \[\dfrac{\ce{rate}_x}{\ce{rate}_y}=\dfrac{k[\ce{NO}]^m_x[\ce{Cl2}]^n_x}{k[\ce{NO}]^m_y[\ce{Cl2}]^n_y}\nonumber\]

      Using the third trial and the first trial, in which [Cl2] does not vary, gives:

      \[\mathrm{\dfrac{rate\: 3}{rate\: 1}}=\dfrac{0.00675}{0.00300}=\dfrac{k(0.15)^m(0.10)^n}{k(0.10)^m(0.10)^n}\nonumber\]

      After canceling equivalent terms in the numerator and denominator, we are left with:

      \[\dfrac{0.00675}{0.00300}=\dfrac{(0.15)^m}{(0.10)^m}\nonumber\]

      which simplifies to:

      \[2.25=(1.5)^m\nonumber\]

      We can use natural logs to determine the value of the exponent m:

      \[\begin{align}
      \ln(2.25)&=m\ln(1.5)
      \dfrac{\ln(2.25)}{\ln(1.5)}&=m
      2&=m
      \end{align}\nonumber\]

      We can confirm the result easily, since:

      \[1.5^2=2.25\nonumber\]
    2. Determine the value of n from data in which [Cl2] varies and [NO] is constant.

      \[\mathrm{\dfrac{rate\: 2}{rate\: 1}}=\dfrac{0.00450}{0.00300}=\dfrac{k(0.10)^m(0.15)^n}{k(0.10)^m(0.10)^n}\nonumber\]

      Cancelation gives:

      \[\dfrac{0.0045}{0.0030}=\dfrac{(0.15)^n}{(0.10)^n}\nonumber\]

      which simplifies to:

      \[1.5=(1.5)^n\nonumber\]

      Thus n must be 1, and the form of the rate law is:

      \[\ce{Rate}=k[\ce{NO}]^m[\ce{Cl2}]^n=k[\ce{NO}]^2[\ce{Cl2}]\nonumber\]
    3. Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol3/L3. The units for k should be mol−2 L2/s so that the rate is in terms of mol/L/s.

      To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k:

      \[\begin{align}
      \mathrm{0.00300\:mol\:L^{−1}\:s^{−1}}&=k\mathrm{(0.10\:mol\:L^{−1})^2(0.10\:mol\:L^{−1})^1}\nonumber\\
      k&=\mathrm{3.0\:mol^{−2}\:L^2\:s^{−1}} \nonumber
      \end{align}\nonumber\]

    Exercise \(\PageIndex{2}\)

    Use the provided initial rate data to derive the rate law for the reaction whose equation is:

    \[\ce{OCl-}(aq)+\ce{I-}(aq)⟶\ce{OI-}(aq)+\ce{Cl-}(aq)\]

    Trial [OCl] (mol/L) [I] (mol/L) Initial Rate (mol/L/s)
    1 0.0040 0.0020 0.00184
    2 0.0020 0.0040 0.00092
    3 0.0020 0.0020 0.00046

    Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.

    Answer:

    \[\mathrm{\dfrac{rate\: 2}{rate\: 3}}=\dfrac{0.00092}{0.00046}=\dfrac{k(0.0020)^x(0.0040)^y}{k(0.0020)^x(0.0020)^y}\nonumber\]

    \[2.00 = 2.00^y\nonumber\]

    so

    \[y = 1\nonumber\]

    Now the other reactant

    \[\mathrm{\dfrac{rate\: 1}{rate\: 2}}=\dfrac{0.00184}{0.00092}=\dfrac{k(0.0040)^x(0.0020)^y}{k(0.0020)^x(0.0040)^y}\nonumber\]

    \[\begin{align}
    2.00&=\dfrac{2^x}{2^y}\nonumber\\
    2.00&=\dfrac{2^x}{2^1}\nonumber\\
    4.00&=2^x\nonumber\\
    x&=2 \nonumber
    \end{align}\nonumber\]

    Substituting the concentration data from trial 1 and solving for k yields:

    \[\begin{align}
    \ce{rate}&=k[\ce{OCl-}]^2[\ce{I-}]^1\nonumber\\
    0.00184&=k(0.0040)^2(0.0020)^1\nonumber\\
    k&=\mathrm{5.75×10^4\:mol^{−2}\:L^2\:s^{−1}}\nonumber
    \end{align} \nonumber\]

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