Unit 2 Kinetics
2.2 Differential Rate Laws
OpenStax
Section Learning Objectives
- Explain the form and function of a rate law.
- Use rate laws to calculate reaction rates.
- Use rate and concentration data to identify reaction orders and derive rate laws.
✓ SECTION 2.2 CHECKLIST
| Learning Activity | Graded? | Estimated Time |
|---|---|---|
| Make notes about this section’s reading portion. | No | 135 min |
| Optional Resource: Watch a video example that relates graphical reaction rates to the differential rate law. | No | 6 min |
| Work on the self-check question. | No | 15 min |
| Work on practice exercises. | No | 150 min |
📖 READING PORTION
The Form of the Differential Rate Law
As described in the previous section, the rate of a reaction may be often affected by the concentrations of reactants. Rate laws (sometimes called differential rate laws) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation
a A + b B ⟶ products
where a and b are respective stoichiometric coefficients of reactants A and B. The rate law for this reaction has the form
rate = k[A]x[B]y
in which [A] and [B] represent the molar concentrations of reactants, and k is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents x and y are the reaction orders specific to the reactant and are typically positive integers, though they can be fractions, negative, or zero. In CHEM 1523, the reaction orders will be positive integers or zero. The reaction orders in a rate law describe the mathematical dependence of the overall reaction rate on reactant concentrations. The sum of the individual orders, here x+y, is the overall order of the reaction. The rate constant k and the reaction orders x and y must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the reactant concentrations, but it does vary with temperature. Here are some examples of rate laws:
The rate law
rate = k[H2O2]1
is first-order in hydrogen peroxide, indicated by the exponent 1 on the concentration, and first-order overall. Note that an exponent of 1 is usually not written out but is shown here to be explicit about the reaction order.
The rate law
rate = k[C4H6]2
is second-order in C4H6 and second-order overall.
The rate law
rate = k[H+][OH−]
is first-order in H+, first-order in OH−, and second-order overall.
The rate law
rate = k
is independent of reactant concentration. An exponent of zero on a concentration, i.e. [reactant]0, is equal to 1. The reaction is zeroth-order overall. Zeroth-order reactions are only possible for gases reacting on a solid catalytic surface or reactions with enzymes.
Rate Constant
The exact numerical value of the rate constant can be determined experimentally, but the unit on the rate constant follows a pattern. Given that concentrations have the unit mol/L, the unit on the rate constant is such that multiplying the rate constant with the concentration(s) raised to the reaction order(s) (exponents) yields a reaction rate in mol/(L · s). Table 1 shows the unit of the rate constant for common overall reaction orders. Note that Table 1 uses mol/L for concentration and seconds for time, but any suitable unit for concentration and time may be used.
|
Overall Reaction Order (x) |
Unit of Rate Constant (Lx−1 mol1−x s−1) |
| 0 (Zero) | mol L−1 s−1 |
| 1 (First) | s−1 |
| 2 (Second) | L mol−1 s−1 |
| 3 (Third) | L2 mol−2 s−1 |
|
Table 1 Commonly used units of the rate constant k. |
|
Determining Rate Law from Initial Rate Data
Example 1
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. One such reaction is the combination of nitric oxide, NO, with ozone, O3:
NO(g) + O3(g) ⟶ NO2(g) + O2(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) | [latex]\frac{{\Delta}[\text{NO}_2]}{{\Delta}t}\;(\text{mol L}^{-1}\text{s}^{-1})[/latex] |
|---|---|---|---|
| 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:
We can determine the values of m, n, and k from the experimental data using the following three-part process:
- 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.
- 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:
[latex]\text{rate} = k[\text{NO}]^1[\text{O}_3]^1 = k[\text{NO}][\text{O}_3][/latex]
- Determine the value of k from one set of concentrations and the corresponding rate.
[latex]\begin{array}{r @{{}={}} l} k &= \frac{\text{rate}}{[\text{NO}][\text{O}_3]} \\[0.5em] & \frac{6.60\;\times\;10^{-5}\;\rule[0.5ex]{2.5em}{0.1ex}\hspace{-2.5em}\text{mol L}^{-1}\text{s}^{-1}}{(1.00\;\times\;10^{-6}\;\rule[0.5ex]{2.75em}{0.1ex}\hspace{-2.75em}\text{mol L}^{-1})(3.00\;\times\;10^{-6}\;\text{mol L}^{-1})} \\[0.5em] & 2.20\;\times\;10^{7}\;\text{L mol}^{-1}\text{s}^{-1} \end{array}[/latex]
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.
The complete rate law is
rate = 2.20×107 L mol−1 s−1 [NO][O3]
and is first-order in nitric oxide, first-order in ozone, and second-order overall.
Example 2
Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction
2 NO(g) + Cl2(g) ⟶ 2 NOCl(g)
| Trial | [NO] (mol/L) | [Cl2] (mol/L) | [latex]-\frac{{\Delta}[\text{NO}]}{{\Delta}t}\text({mol L}^{-1}\text{s}^{-1})[/latex] |
|---|---|---|---|
| 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 Example 1, 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 :
- 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:
[latex]\frac{\text{rate}_x}{\text{rate}_y} = \frac{k[\text{NO}]_x^m[\text{Cl}_2]_x^n}{k[\text{NO}]_y^m[\text{Cl}_2]_y^n}[/latex]
Using the third trial and the first trial, in which [Cl2] does not vary, gives:
[latex]\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}[/latex]After canceling equivalent terms in the numerator and denominator, we are left with:
[latex]\frac{0.00675}{0.00300} = \frac{(0.15)^m}{(0.10)^m}[/latex]which simplifies to:
[latex]2.25 = (1.5)^m[/latex]We can use natural logs to determine the value of the exponent m :
[latex]\begin{array}{r @{{}={}} l} \text{ln}(2.25) &= m\text{ln}(1.5) \\[0.5em] \frac{\text{ln}(2.25)}{\text{ln}(1.5)} &= m \\[0.5em] 2 &= m \end{array}[/latex]We can confirm the result easily, since:
[latex]1.5^2 = 2.25[/latex] - Determine the value of n from data in which [Cl2] varies and [NO] is constant.
[latex]\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}[/latex]
Cancelation gives:
[latex]\frac{0.0045}{0.0030} = \frac{(0.15)^n}{(0.10)^n}[/latex]which simplifies to:
[latex]1.5 = (1.5)^n[/latex]Thus n must be 1, and the form of the rate law is:
[latex]\text{Rate} = k[\text{NO}]^m[\text{Cl}_2]^n = k[\text{NO}]^2[\text{Cl}_2][/latex] - 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 :
[latex]\begin{array}{r @{{}={}} l} 0.00300\;\text{mol L}^{-1}\text{s}^{-1} &= k(0.10\;\text{mol L}^{-1})^2(0.10\;\text{mol L}^{-1})^1 \\[0.5em] k &= 3.0\;\text{mol}^{-2}\text{L}^2\text{s}^{-1} \end{array}[/latex]
The complete rate law is
rate = 3.0 mol−2 L2 s−1 [NO]2[Cl2]
and is second-order in nitric oxide, first-order in chlorine, and third-order overall.
In our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case. It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.
Pseudo First-Order Reactions
For a second-order reaction, A + B ⟶ products, that has the rate law
rate = k[A][B]
an excess of one reactant, say B, means that the concentration of B stays approximately constant as the reaction proceeds. The rate law can be approximated as
rate = k[A][B] ≈ k′[A]
where the rate constant k and the nearly constant [B] are grouped together into the pseudo rate constant k′. The rate law appears first-order in [A]. We can also have an excess of A and limiting quantities of B, which would give a rate law that appears first-order in [B]. By using an excess of one reactant, we make a second-order reaction behave as a first-order reaction.
Optional Resource
Watch a video example that relates graphical reaction rates to the differential rate law.
Video 1 Introduction to Chemical Kinetics (5 min 41 s).
Which of the following is true?
A) The rate constant k does not vary with temperature.
B) Zeroth-order reactions depend on concentration for the rate.
C) The reaction orders do not necessarily match the stoichiometry of the net reaction.
D) The reaction orders are derived from stoichiometry of the net reaction.
Click to see answer
C) The reaction orders do not necessarily match the stoichiometry of the net reaction.
