A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.
Determining a Rate Law from Initial Rates: Ozone formation
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 1 below). One such reaction is the combination of nitric oxide, NO, with ozone, O3:$$NO(g)+O_3(g)⟶NO_2(g)+O_2(g)$$
This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.
| Trial | [NO] (mol/L) | [O3] (mol/L) | $\frac{Δ[NO_2]}{Δt}\;(mol\;L^{-1})$ |
|---|---|---|---|
| 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:
Determine the values of m, n, and k from the experimental data using the following four-part process:
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Determine the value of m from the data in which [NO] varies and [O3] is constant.
In the last three experiments (3, 4, & 5), [NO] varies while [O3] remains constant. Any two of these three experiments can be compared to determine the order for n – you should get the same result. Below, trials 3 and 4 are used.
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.
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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. Any two of these three experiments can be compared to determine the order for n – you should get the same result. Below, trials 1 and 2 are used.
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] (n = 1).
- Write the rate law:
We have determined that m is 1 and n is also equal to 1.The rate law is thus: $$rate=k[NO]^1[O_3]^1=k[NO][O_3]$$ -
Determine the value of k from one set of concentrations and the corresponding rate. The data from trial 1 are used below:
$$k=\frac{\text{rate}}{[NO][O_3]}$$ $$k=\frac{6.60×10^{-5}\;\require{enclose}\enclose{horizontalstrike}{mol\;L^{-1}}\;s^{-1}}{(1.00×10^{-6}\;\enclose{horizontalstrike}{mol\;L^{-1}})(3.00×10^{-6}\;mol\;L^{-1})}$$ $$k=2.20×10^7\;mol\;L^{-1}s^{-1}$$ The complete rate law is: $$rate=\left( 2.20×10^7\;mol\;L^{-1}s^{-1}\right) [NO]^1[O_3]^1=k[NO][O_3]$$
Check Your Learning
Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:
Determine the rate law and the rate constant for the reaction from the following experimental data:
| Trial | [CH3CHO] (mol/L) | $-\frac{Δ[CH_3CHO]}{Δt}\;(mol\;L^{-1}s^{-1})$ |
|---|---|---|
| 1 | $1.75×10^{-3}$ | $2.06×10^{-11}$ |
| 2 | $3.50×10^{-3}$ | $8.24×10^{-11}$ |
| 3 | $7.00×10^{-3}$ | $3.30×10^{-10}$ |
Solution:
By comparing experiments 1&2 or 2&3, you will find that as the concentration of acetaldehyde doubles, the rate quadruples (x 4). Since doubling = (x 2), to get quadruple you need (2)m = 4, therefore m must be 2 for this reaction. The order of reaction for acetaldehyde is 2. $$rate=k[CH_3CHO]^2$$ $$k=6.73×10^{-6} L/mol/s$$ $$rate=\left( 6.73×10^{-6} L/mol/s \right) [CH_3CHO]^2$$
Determining Rate Laws from Initial Rates: Ratio Method
Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:
| Trial | [NO] (mol/L) | [Cl2] (mol/L) | $-\frac{Δ[NO]}{Δt}\;(mol\;L^{-1}s^{-1})$ |
|---|---|---|---|
| 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:
As in the ozone example above, approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to build the rate law and determine the value of k. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of m and n. This method can also be used in more complex examples where there is not two experiments with a concentration held constant.
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Determine the value of m from the data in which [NO] varies and [Cl2] is constant. Write the ratios with subscripts to indicate data from two different trials:
$$\frac{\text{rate}_{run 1}}{\text{rate}_{run 2}}=\frac{k[NO]_{run 1}^m[Cl_2]_{run 1}^n}{k[NO]_{run 2}^m[Cl_2]_{run 2}^n}$$Using the third trial and the first trial, in which [Cl2] does not vary, gives:
$$\frac{\text{rate 3}}{\text{rate 1}}=\frac{0.00675}{0.00300}=\frac{k(0.15)^m(0.10)^n}{k(0.10)^m(0.10)^n}$$Canceling equivalent terms in the numerator and denominator leaves:
$$\frac{0.00675}{0.00300}=\frac{(0.15)^m}{(0.10)^m}$$which simplifies to:
$$2.25=(1.5)^m$$Use logarithms to determine the value of the exponent m:
$$ln(2.25)=m\cdot ln(1.5)$$ $$\frac{ln(2.25)}{ln(1.5)}=m$$ $$2=m$$Confirm the result
$$1.5^2=2.25$$ -
Determine the value of n from data in which [Cl2] varies and [NO] is constant.
$$\frac{\text{rate 2}}{\text{rate 1}}=\frac{0.00450}{0.00300}=\frac{k(0.10)^m(0.15)^n}{k(0.10)^m(0.10)^n}$$Cancelation gives:
$$\frac{0.0045}{0.0030}=\frac{(0.15)^n}{(0.10)^n}$$which simplifies to:
$$1.5=(1.5)^n$$ Therefore n must be 1. -
Write the rate law:
We have determined that the order of reaction on NO is 2 and the order on Cl2 is 1, so the rate law is: $$\text{rate}=k[NO]^m[Cl_2]^n=k[NO]^2[Cl_2]$$ -
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:
$$0.00300\;mol\;L^{-1}\;s^{-1}=k(0.10\;mol\;L^{-1})^1$$ $$k=3.0\;mol^{-2}\;L^2\;s^{-1}$$ $$\text{rate}=\left( 3.0\;mol^{-2}\;L^2\;s^{-1} \right)[NO]^2[Cl_2]$$
Check Your Learning
Use the provided initial rate data to derive the rate law for the reaction whose equation is:
| 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.
Solution:
Using the ratio method as in the example above (with x for the order of reaction on OCl– and y for the order on I–): $$\frac{\text{rate 2}}{\text{rate 3}}=\frac{0.00092}{0.00046}=\frac{k(0.0020)^x(0.0040)^y}{k(0.0020)^x(0.0020)^y}$$ $$2.00=2.00^y$$ $$y=1$$ $$\frac{\text{rate 1}}{\text{rate 2}}=\frac{0.00184}{0.00092}=\frac{k(0.0040)^x(0.0020)^y}{k(0.0020)^x(0.0040)^y}$$ $$2.00=\frac{2^x}{2^y}$$ $$2.00=\frac{2^x}{2^1}$$ $$4.00=2^x$$ $$x=2$$ The rate law is therefore: $$\text{rate}=k[OCl^-]^2[I^-]^1$$ Substituting the concentration data from trial 1 and solving for k yields: $$0.00184=k(0.0040)^2(0.0020)^1$$ $$k=5.75×10^4\;mol^{-2}\;L^2\;s^{-1}$$ $$\text{rate}=\left( 5.75×10^4\;mol^{-2}\;L^2\;s^{-1} \right) [OCl^-]^2[I^-]$$