How to find the rate law from experimental data

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. It has the form extRate=k[A]m[B]n ext{Rate} = k[A]^m[B]^n, where kk is the rate constant, and mm and nn are the reaction orders with respect to reactants AA and BB.

To find the rate law from experimental data, use the method of initial rates. This applies when you have a table of initial reactant concentrations and the corresponding initial reaction rates for several trials.

The setup

You need experimental data consisting of at least two trials for each reactant, where the concentration of one reactant changes while all others remain constant. The general rate law is written as extRate=k[A]m[B]n ext{Rate} = k[A]^m[B]^n\dots, where the orders m,n,m, n, \dots must be determined experimentally.

The steps

  1. Select two trials where the concentration of one reactant changes while the concentrations of all other reactants are held constant.
  2. Set up a ratio of the rate laws for these two trials: extRate2extRate1=k[A]2m[B]2nk[A]1m[B]1n\frac{ ext{Rate}_2}{ ext{Rate}_1} = \frac{k[A]_2^m[B]_2^n}{k[A]_1^m[B]_1^n}.
  3. Cancel the rate constant kk and the constant concentration terms.
  4. Solve the resulting equation for the exponent (the reaction order) of the varied reactant. Often this can be done by inspection; otherwise, use logarithms.
  5. Repeat steps 1-4 for each remaining reactant.
  6. Write the final rate law using the determined orders.
  7. To find the rate constant kk, substitute the concentrations and rate from any single trial into the complete rate law and solve for kk.

Checking the result

To verify the rate law and the value of kk, plug the reactant concentrations from a different trial into the calculated rate law. The computed rate should match the experimental rate for that trial within experimental error.

Common errors

  • Assuming reaction orders match stoichiometric coefficients from the balanced equation. Orders must be found experimentally.
  • Forgetting to include units for the rate constant kk. The units of kk depend on the overall reaction order.
  • Selecting two trials where multiple concentrations change simultaneously, making it impossible to isolate the effect of a single reactant.

Worked example

Determine the rate law and the value of the rate constant kk for the reaction A+BCA + B \rightarrow C using the following initial rate data: Trial 1: [A]=0.10extM[A] = 0.10 ext{ M}, [B]=0.10extM[B] = 0.10 ext{ M}, Rate =2.0imes103extM/s= 2.0 imes 10^{-3} ext{ M/s} Trial 2: [A]=0.20extM[A] = 0.20 ext{ M}, [B]=0.10extM[B] = 0.10 ext{ M}, Rate =4.0imes103extM/s= 4.0 imes 10^{-3} ext{ M/s} Trial 3: [A]=0.10extM[A] = 0.10 ext{ M}, [B]=0.20extM[B] = 0.20 ext{ M}, Rate =8.0imes103extM/s= 8.0 imes 10^{-3} ext{ M/s}

  1. Write the general rate law: extRate=k[A]m[B]n ext{Rate} = k[A]^m[B]^n.
  2. Find order mm with respect to AA using Trial 1 and Trial 2 (where [B][B] is constant): extRate2extRate1=k[0.20]m[0.10]nk[0.10]m[0.10]n\frac{ ext{Rate}_2}{ ext{Rate}_1} = \frac{k[0.20]^m[0.10]^n}{k[0.10]^m[0.10]^n} 4.0imes1032.0imes103=(0.200.10)m\frac{4.0 imes 10^{-3}}{2.0 imes 10^{-3}} = \left(\frac{0.20}{0.10}\right)^m 2=2m    m=12 = 2^m \implies m = 1
  3. Find order nn with respect to BB using Trial 1 and Trial 3 (where [A][A] is constant): extRate3extRate1=k[0.10]1[0.20]nk[0.10]1[0.10]n\frac{ ext{Rate}_3}{ ext{Rate}_1} = \frac{k[0.10]^1[0.20]^n}{k[0.10]^1[0.10]^n} 8.0imes1032.0imes103=(0.200.10)n\frac{8.0 imes 10^{-3}}{2.0 imes 10^{-3}} = \left(\frac{0.20}{0.10}\right)^n 4=2n    n=24 = 2^n \implies n = 2
  4. Write the specific rate law: extRate=k[A][B]2 ext{Rate} = k[A][B]^2
  5. Calculate kk using data from Trial 1: 2.0imes103extM/s=k(0.10extM)(0.10extM)22.0 imes 10^{-3} ext{ M/s} = k(0.10 ext{ M})(0.10 ext{ M})^2 2.0imes103=k(0.0010extM3)2.0 imes 10^{-3} = k(0.0010 ext{ M}^3) k=2.0imes1030.0010=2.0extM2exts1k = \frac{2.0 imes 10^{-3}}{0.0010} = 2.0 ext{ M}^{-2} ext{s}^{-1}

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References: Chemistry: The Central Science (Brown, LeMay, Bursten) · OpenStax Chemistry 2e, Chapter 12: Kinetics · Khan Academy: Kinetics unit

See also