18.10: Determining The Rate Law From Experimental Data

Determining the Rate Law from Experimental Data

In order to experimentally determine a rate law, a series of experiments must be performed with various starting concentrations of reactants. The initial rate law is then measured for each of the reactions. Consider the reaction between nitrogen monoxide gas and hydrogen gas to form nitrogen gas and water vapor:

\[2 \ce{NO} \left( g \right) + 2 \ce{H_2} \left( g \right) \rightarrow \ce{N_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)\nonumber \]

The following data were collected for this reaction at \(1280^\text{o} \text{C}\) (see table below).

Reaction between nitrogen monoxide gas and hydrogen gas to form nitrogen gas and water vapor

Table \(\PageIndex{1}\)
Experiment	\(\left[ \ce{NO} \right]\)	\(\left[ \ce{H_2} \right]\)	Initial Rate \(\left( \text{M/s} \right)\)
1	0.0050	0.0020	\(1.25 \times 10^{-5}\)
2	0.010	0.0020	\(5.00 \times 10^{-5}\)
3	0.010	0.0040	\(1.00 \times 10^{-4}\)

Notice that the starting concentrations of \(\ce{NO}\) and \(\ce{H_2}\) were varied in a specific way. In order to compare the rates of reaction and determine the order with respect to each reactant, the initial concentration of each reactant must be changed while the other is held constant.

Comparing experiments 1 and 2: The concentration of \(\ce{NO}\) was doubled, while the concentration of \(\ce{H_2}\) was held constant. The initial rate of the reaction quadrupled, since \(\frac{5.00 \times 10^{-5}}{1.25 \times 10^{-5}} = 4\). Therefore, the order of the reaction with respect to \(\ce{NO}\) is 2. In other words, \(\text{rate} \propto \left[ \ce{NO} \right]^2\). Because \(2^2 = 4\), the doubling of \(\left[ \ce{NO} \right]\) results in a rate that is four times greater.

Comparing experiments 2 and 3: The concentration of \(\ce{H_2}\) was doubled while the concentration of \(\ce{NO}\) was held constant. The initial rate of the reaction doubled, since \(\frac{1.00 \times 10^{-4}}{5.00 \times 10^{-5}} = 2\). Therefore, the order of the reaction with respect to \(\ce{H_2}\) is 1, or \(\text{rate} \propto \left[ \ce{H_2} \right]^1\). Because \(2^1 = 2\), the doubling of \(\ce{H_2}\) results in a rate that is twice as great. The overall rate law then includes both of these results.

\[\text{rate} = k \left[ \ce{NO} \right]^2 \left[ \ce{H_2} \right]\nonumber \]

The sum of the exponents is \(2 + 1 = 3\), making the reaction third-order overall. Once the rate law for a reaction is determined, the specific rate constant can be found by substituting the data for any of the experiments into the rate law and solving for \(k\).

\[k = \frac{\text{rate}}{\left[ \ce{NO} \right]^2 \left[ \ce{H_2} \right]} = \frac{1.25 \times 10^{-5} \: \text{M/s}}{\left( 0.0050 \: \text{M} \right)^2 \left( 0.0020 \: \text{M} \right)} = 250 \: \text{M}^{-2} \text{s}^{-1}\nonumber \]

Notice that the rate law for the reaction does not relate to the balanced equation for the overall reaction. The coefficients of \(\ce{NO}\) and \(\ce{H_2}\) are both 2, while the order of the reaction with respect to the \(\ce{H_2}\) is only one. The units for the specific rate constant vary with the order of the reaction. So far, we have seen reactions that are first or second order with respect to a given reactant. Occasionally, the rate of a reaction may not depend on the concentration of one of the reactants at all. In this case, the reaction is said to be zero-order with respect to that reactant.