Chemical Kinetics · Lesson 2 of 10
Rate Laws, Reaction Order and Rate Constants
“Infer concentration dependence from experiments and calculate rate constants with units appropriate to the observed order.”
• Determine partial and overall orders from a measured rate law. • Infer orders by comparing controlled initial-rate experiments. • Calculate the rate constant and derive its units. • Use stoichiometry to update concentrations during reaction without guessing a new rate law.
A rate law describes an experimental relationship
A balanced equation tells us amounts consumed and formed, but usually does not reveal how the rate changes when one concentration is varied. To find that dependence, measure initial rates in separate mixtures at the same temperature while controlling solvent, catalyst and other conditions. For a useful range of conditions, a measured power-law expression may describe the observations.
The rate constant k is the proportionality coefficient for that chosen rate law under specified conditions. Changing concentration changes the rate according to the powers, without changing k if temperature and other kinetic conditions remain fixed. Changing temperature or the catalytic pathway generally changes k. A numerical value of k alone has little meaning without its units, conditions and rate definition.
The exponent describing concentration dependence in an experimentally supported rate law. Overall order is the sum of the exponents; it need not equal the sum of stoichiometric coefficients.
Order can be zero, a positive integer, fractional or even negative. A zero exponent means that varying that reactant concentration within the studied range leaves the measured rate unchanged. A negative exponent means the observed rate decreases as that concentration rises, often because of inhibition or a more complex mechanism. These are empirical behaviours, not claims that negative numbers of molecules collide.
Problem
Determine overall orders for r=k[A]¹ᐟ²[B]³ᐟ² and r=k[A]³ᐟ²[B]⁻¹.
- 1.For the first law, add the partial orders: 1/2+3/2=2. The overall order is two, although neither partial order is an integer.
- 2.For the second, 3/2+(−1)=1/2. The inverse dependence on B reduces the overall order.
- 3.Increasing B alone lowers the second rate; increasing A alone raises it. Each effect follows its own exponent.
- 4.Do not replace these exponents by coefficients from an overall balanced equation.
Examples: balance and empirical powers answer different questions
| Balanced overall conversion | Reported concentration dependence | Partial-order interpretation |
|---|---|---|
| 3NO → N₂O + NO₂ | k[NO]² | Second order in NO; the repaired equation conserves N and O. |
| H₂O₂ + 3I⁻ + 2H⁺ → 2H₂O + I₃⁻ | k[H₂O₂][I⁻] in the stated regime | First order in each displayed concentration; acidity and other conditions are part of the observation. |
| CH₃CHO → CH₄ + CO | k[CH₃CHO]³ᐟ² | Fractional order does not count reacting molecular entities. |
| C₂H₅Cl → C₂H₄ + HCl | k[C₂H₅Cl] | A first-order law for the specified conditions. |
| CH₃OCH₃ → CH₄ + H₂ + CO | k[CH₃OCH₃]³ᐟ² | Pressure-based and concentration-based coefficients need distinct units. |
Use controlled rate ratios to infer exponents
| Experiment | Initial [NO] / M | Initial [O₂] / M | NO₂ appearance rate / M s⁻¹ |
|---|---|---|---|
| 1 | 0.30 | 0.30 | 0.096 |
| 2 | 0.60 | 0.30 | 0.384 |
| 3 | 0.30 | 0.60 | 0.192 |
| 4 | 0.60 | 0.60 | 0.768 |
Problem
For 2NO+O₂→2NO₂, infer the law from the table and calculate k for the stated NO₂ appearance rate.
- 1.Compare experiments 1 and 2: O₂ stays fixed while NO doubles. The rate ratio is 0.384/0.096=4=2², so the NO partial order is two.
- 2.Compare 1 and 3: NO stays fixed while O₂ doubles. The rate doubles, so the O₂ order is one. Overall order is three.
- 3.For NO₂ appearance v(NO₂)=k_app[NO]²[O₂], k_app=0.096/(0.30²×0.30)=3.55556 L² mol⁻² s⁻¹.
- 4.If r is instead the normalised reaction rate, r=v(NO₂)/2, the corresponding k is half as large, 1.77778 L² mol⁻² s⁻¹. The orders remain unchanged.
- 5.The match of these exponents with the overall coefficients in this example does not establish a single three-body mechanism.
Problem
Three experiments have ([A],[B],r)=(0.20,0.30,5.07×10⁻⁵), (0.20,0.10,5.07×10⁻⁵), (0.40,0.05,1.43×10⁻⁴), using M and M s⁻¹. Infer the orders.
- 1.The first two keep A fixed while B changes by a factor of three; rate is unchanged. Thus the observed B order is zero in this range.
- 2.Compare the first and third: A doubles. Because B has zero order, rate ratio=(1.43×10⁻⁴)/(5.07×10⁻⁵). x=ln(rate ratio)/ln2=1.49596≈1.50.
- 3.The rounded data support a 3/2 order in A and overall 3/2, rather than requiring the fitted number 1.49596 to be an exact physical exponent.
- 4.Evidence of zero order over a finite range does not prove that B never participates chemically.
Derive the units instead of memorising a single unit
| Overall order n | Units of k for M and s |
|---|---|
| 0 | mol L⁻¹ s⁻¹ |
| 1 | s⁻¹ |
| 2 | L mol⁻¹ s⁻¹ |
| 3 | L² mol⁻² s⁻¹ |
| 3/2 | L¹ᐟ² mol⁻¹ᐟ² s⁻¹ |
| 1/2 | mol¹ᐟ² L⁻¹ᐟ² s⁻¹ |
Problem
Identify the order of k=2.3×10⁻⁵ L mol⁻¹ s⁻¹ and k=3.0×10⁻⁴ s⁻¹.
- 1.L mol⁻¹ is inverse concentration, so 1−n=−1 and n=2 for the first constant.
- 2.The second constant has no concentration dimension: 1−n=0, so n=1.
- 3.This inference assumes that rate has the usual concentration/time units and the law is expressed using concentration, rather than a different measured observable.
Use stoichiometry to update a later rate
Problem
The normalised law is r=k[A][B]² with k=2.0×10⁻⁶ L² mol⁻² s⁻¹. Initially [A]=0.100 M and [B]=0.200 M. Find r initially and when A reaches 0.060 M at constant volume.
- 1.Initially r=2.0×10⁻⁶×0.100×0.200²=8e-09 M s⁻¹.
- 2.A has decreased by 0.040 M. Because two A are consumed per B, B decreases by 0.020 M to 0.180 M.
- 3.Later r=2.0×10⁻⁶×0.060×0.180²=3.888e-09 M s⁻¹. Holding B at its initial value would ignore its chemical consumption.
- 4.The coefficients update composition; the experimentally specified exponents determine how that composition affects the rate.
Examples with coefficients different from empirical powers include CHCl₃+Cl₂→CCl₄+HCl with a reported dependence [CHCl₃][Cl₂]¹ᐟ² under appropriate conditions, and hydrolysis measured with excess water that appears independent of further water changes. The precise medium and concentration range belong to the meaning of a measured law. A single overall equation is insufficient to assign a universal rate expression.
Quick Quiz
Quiz
What is the overall order of r=k[A]¹ᐟ²[B]²?
Explanation: Add the measured exponents: 0.5+2=2.5.
If a second-order dependence on A applies, tripling A at unchanged other conditions multiplies rate by:
Explanation: The factor is 3²=9.
Which units indicate first-order k when rate is M s⁻¹?
Explanation: For n=1, concentration^(1−n) disappears from the rate-constant unit.
Why must a typical overall rate law be measured?
Explanation: Stoichiometry alone does not specify the sequence and concentration dependence of the controlling elementary events.
For r=k[A][B]², doubling both concentrations changes rate by:
Explanation: The multipliers are 2¹ and 2², whose product is eight.
Practice Problems
- Determine the order and k units for laws k[NO]², k[H₂O₂][I⁻], k[CH₃CHO]³ᐟ² and k[C₂H₅Cl], with rate in M s⁻¹.
- A pressure-based disappearance law is v_p=k_p p³ᐟ², with pressure in bar and time in min. Give units of v_p and k_p.
- For r=k[A][B]², state the effect of tripling B alone and doubling both A and B.
- Experiments for 2A+B→C+D give ([A],[B],v_D): (0.1,0.1,0.006), (0.3,0.2,0.072), (0.3,0.4,0.288), (0.4,0.1,0.024), with time in min. Find the law and k.
- For a law first order in A and zero order in B, v=0.020 M min⁻¹ at [A]=0.10 M. Find k, the rate at [A]=0.40 M and the A concentration giving v=0.040 M min⁻¹.
- List factors that may change observed reaction rate and explain why increasing concentration does not guarantee a faster rate for every possible law.
Add exponents in each law: orders are 2,2,3/2 and 1. Using [k]=M^(1−n)s⁻¹ gives M⁻¹s⁻¹, M⁻¹s⁻¹, M⁻¹ᐟ²s⁻¹ and s⁻¹. A balanced possible conversion for the first example is 3NO→N₂O+NO₂; 3NO→N₂O alone would not conserve atoms. The exponent still comes from the supplied law, not from that repaired stoichiometry.
Key Takeaways
• A rate law is an experimentally supported concentration dependence under specified conditions. • Partial orders are powers in that law; overall order is their sum. • Orders may be zero, fractional or negative and are distinct from molecular counts. • Controlled initial-rate comparisons isolate each concentration effect. • Rate-constant units follow from concentration^(1−n) divided by time. • Normalising a species rate can rescale k while leaving orders unchanged. • Use balanced coefficients to update concentrations and empirical exponents to update rates.