The method of initial rates is used to determine how the rate of a reaction depends on the concentrations of its reactants. The reaction is performed several times using different initial concentrations, and the initial reaction rate is measured for each run.
For a reaction of the general form
\[ A+B\rightarrow\text{products} \]the rate law may be written as
\[ \text{Rate}=k[A]^{\alpha}[B]^{\beta} \]where \(k\) is the rate constant and the exponents \(\alpha\) and \(\beta\) must be determined experimentally.
To determine one exponent, two experiments are compared in which the concentration of only one reactant changes. For example, if \([B]\) is held constant while \([A]\) changes, the ratio of the two measured rates is
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[A]_2^{\alpha}[B]_2^{\beta}} {k[A]_1^{\alpha}[B]_1^{\beta}} \]Because \(k\) is the same in both experiments and \([B]_2=[B]_1\), these factors cancel:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^{\alpha} \]The exponent \(\alpha\) can then be found by examining how the rate changes when the concentration of \(A\) changes.
| Change in concentration | Change in rate | Reaction order |
|---|---|---|
| Doubled | No change | Zeroth order |
| Doubled | Doubled | First order |
| Doubled | Quadrupled | Second order |
| Doubled | Increased eightfold | Third order |
These simple patterns can often be recognized by inspection. For example, if doubling \([A]\) causes the reaction rate to increase by a factor of four, then
\[ 4=2^{\alpha} \]and therefore
\[ \alpha=2 \]so the reaction is second order with respect to \(A\).
If the concentrations do not change by a simple factor, or if the rate change is not easily recognized, the reaction order can be calculated using logarithms:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^{\alpha} \] \[ \ln\left(\frac{\text{Rate}_2}{\text{Rate}_1}\right) = \alpha \ln\left(\frac{[A]_2}{[A]_1}\right) \] \[ \alpha = \frac{ \ln\left(\text{Rate}_2/\text{Rate}_1\right) }{ \ln\left([A]_2/[A]_1\right) } \]The same procedure is used to determine the order with respect to each additional reactant. Once all of the exponents are known, the complete rate law can be written.
After the reaction orders have been determined, the rate constant can be calculated by rearranging the rate law:
\[ k = \frac{\text{Rate}} {[A]^{\alpha}[B]^{\beta}} \]The data from any experimental run may be used. In an ideal data set, each run should give approximately the same value of \(k\); small differences may result from experimental uncertainty.
Big picture: The method of initial rates determines a rate law by comparing experiments in which reactant concentrations are varied systematically. By holding all but one concentration constant, chemists can isolate the effect of that reactant, determine its reaction order, and then use the complete rate law to calculate the rate constant.
The following initial-rate data were collected for the reaction
\[ A+B\rightarrow\text{products} \]| Run | \([A]\) (M) | \([B]\) (M) | Initial rate (\(\mathrm{M\,s^{-1}}\)) |
|---|---|---|---|
| 1 | 0.100 | 0.100 | \(2.00\times10^{-3}\) |
| 2 | 0.200 | 0.100 | \(4.00\times10^{-3}\) |
| 3 | 0.200 | 0.200 | \(1.60\times10^{-2}\) |
Determine the order with respect to \(A\) and \(B\), write the rate law, and calculate the rate constant.
Begin with the general form of the rate law:
\[ \text{Rate}=k[A]^{\alpha}[B]^{\beta} \]Compare Runs 1 and 2. The concentration of \(B\) remains constant, while the concentration of \(A\) doubles:
\[ \frac{[A]_2}{[A]_1} = \frac{0.200}{0.100} = 2 \]The reaction rate also doubles:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{4.00\times10^{-3}}{2.00\times10^{-3}} = 2 \]Therefore,
\[ 2=2^{\alpha} \] \[ \alpha=1 \]The reaction is first order in \(A\).
Compare Runs 2 and 3. The concentration of \(A\) remains constant, while the concentration of \(B\) doubles:
\[ \frac{[B]_3}{[B]_2} = \frac{0.200}{0.100} = 2 \]The reaction rate increases by a factor of four:
\[ \frac{\text{Rate}_3}{\text{Rate}_2} = \frac{1.60\times10^{-2}}{4.00\times10^{-3}} = 4 \]Therefore,
\[ 4=2^{\beta} \] \[ \beta=2 \]The reaction is second order in \(B\).
Substituting the experimentally determined exponents gives
\[ \boxed{\text{Rate}=k[A][B]^2} \]The reaction is first order in \(A\), second order in \(B\), and third order overall.
Any of the experimental runs may be used. Using Run 1:
\[ k = \frac{\text{Rate}}{[A][B]^2} \] \[ k = \frac{2.00\times10^{-3}\ \mathrm{M\,s^{-1}}} {(0.100\ \mathrm{M})(0.100\ \mathrm{M})^2} \] \[ k = 2.00\ \mathrm{M^{-2}\,s^{-1}} \]Therefore,
\[ \boxed{k=2.00\ \mathrm{M^{-2}\,s^{-1}}} \]Check: Doubling \([A]\) doubles the rate because the reaction is first order in \(A\). Doubling \([B]\) quadruples the rate because the reaction is second order in \(B\).
The following initial-rate data were collected for the reaction
\[ A+B\rightarrow\text{products} \]| Run | \([A]\) (M) | \([B]\) (M) | Initial rate (\(\mathrm{M\,s^{-1}}\)) |
|---|---|---|---|
| 1 | 0.120 | 0.200 | \(8.64\times10^{-4}\) |
| 2 | 0.300 | 0.200 | \(5.40\times10^{-3}\) |
| 3 | 0.300 | 0.350 | \(1.654\times10^{-2}\) |
Determine the order with respect to \(A\) and \(B\), write the rate law, and calculate the rate constant.
Begin with the general form of the rate law:
\[ \text{Rate}=k[A]^{\alpha}[B]^{\beta} \]Compare Runs 1 and 2. The concentration of \(B\) remains constant, so the effect of changing \([A]\) can be isolated:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \left(\frac{[A]_2}{[A]_1}\right)^{\alpha} \]Substitute the experimental values:
\[ \frac{5.40\times10^{-3}} {8.64\times10^{-4}} = \left(\frac{0.300}{0.120}\right)^{\alpha} \] \[ 6.25=(2.50)^{\alpha} \]Because this relationship may not be immediately obvious, take the natural logarithm of both sides:
\[ \ln(6.25)=\alpha\ln(2.50) \] \[ \alpha = \frac{\ln(6.25)}{\ln(2.50)} = 2.00 \]Therefore, the reaction is second order in \(A\).
Compare Runs 2 and 3. The concentration of \(A\) remains constant, so the effect of changing \([B]\) can be isolated:
\[ \frac{\text{Rate}_3}{\text{Rate}_2} = \left(\frac{[B]_3}{[B]_2}\right)^{\beta} \]Substitute the experimental values:
\[ \frac{1.654\times10^{-2}} {5.40\times10^{-3}} = \left(\frac{0.350}{0.200}\right)^{\beta} \] \[ 3.063=(1.75)^{\beta} \]Take the natural logarithm of both sides:
\[ \ln(3.063)=\beta\ln(1.75) \] \[ \beta = \frac{\ln(3.063)}{\ln(1.75)} = 2.00 \]Therefore, the reaction is second order in \(B\).
Substituting the experimentally determined exponents gives
\[ \boxed{\text{Rate}=k[A]^2[B]^2} \]The reaction is second order in \(A\), second order in \(B\), and fourth order overall.
Rearrange the rate law:
\[ k= \frac{\text{Rate}}{[A]^2[B]^2} \]Using the data from Run 1:
\[ k = \frac{8.64\times10^{-4}\ \mathrm{M\,s^{-1}}} {(0.120\ \mathrm{M})^2(0.200\ \mathrm{M})^2} \] \[ k = 1.50\ \mathrm{M^{-3}\,s^{-1}} \]Therefore,
\[ \boxed{k=1.50\ \mathrm{M^{-3}\,s^{-1}}} \]Check: Increasing \([A]\) by a factor of 2.50 increases the rate by a factor of \((2.50)^2=6.25\). Increasing \([B]\) by a factor of 1.75 increases the rate by a factor of \((1.75)^2=3.063\), consistent with second-order dependence on each reactant.
Use the initial-rate data to determine the order with respect to \(A\), the order with respect to \(B\), and the rate constant \(k\).
Big picture: The method of initial rates uses carefully designed experiments to determine how the reaction rate depends on reactant concentrations. Once the reaction orders are known, the complete rate law and the rate constant can be determined, providing a mathematical description of the reaction kinetics.