What you'll be able to do: Determine reaction orders and the rate constant from initial-rate data, and state the correct units of k.
Best after: Reaction Rates and How They Are Measured
A rate law is an experimental fact, not something you can read off a balanced equation. The method of initial rates finds it by changing one concentration at a time and watching what the rate does.
These are recommended, not required. You can start this lesson at any time.
A rate law states how the instantaneous rate depends on concentrations at that moment. The exponents are the reaction orders, and they are found by experiment. They are frequently whole numbers, but fractional and negative orders exist.
rate = k[A]ᵐ[B]ⁿ
For a multistep reaction, the experimentally measured orders reflect the mechanism as a whole. A slow step can help explain the rate, but its reactant coefficients alone do not always give the observed overall rate law, especially when earlier equilibria or intermediates are involved.
Only for a single elementary step can its reactant coefficients be used directly as exponents in that step’s rate expression.
Find two experiments in which one concentration changes and all others are held constant. Take the ratio of the rates and the ratio of the concentrations. The order is the exponent that connects them: if the concentration ratio is 2 and the rate ratio is 4, then 2ᵐ = 4 and m = 2.
| Trial | [A] / M | [B] / M | Initial rate / (M/s) |
|---|---|---|---|
| 1 | 0.100 | 0.100 | 0.00200 |
| 2 | 0.200 | 0.100 | 0.00400 |
| 3 | 0.100 | 0.200 | 0.00800 |
Trials 1 and 2 isolate A, giving 2 = 2ᵐ and m = 1. Trials 1 and 3 isolate B, giving 4 = 2ⁿ and n = 2. Thus rate = k[A][B]². The orders describe how the rate responds to concentrations; the overall equation A + B → C need not display them.
If both concentrations change, use rate₂/rate₁ = ([A]₂/[A]₁)ᵐ([B]₂/[B]₁)ⁿ. Once m is known, divide out its contribution to isolate n. If neither order is known, enough independent trials can supply simultaneous equations; taking logarithms makes those equations linear in the unknown orders.
rate2/rate1 = ([A]2/[A]1)ᵐ
Doubling a concentration and finding no rate change means zero order in that species. Double the rate means first order. Four times the rate means second order. Tripling a concentration and getting nine times the rate is also second order.
Substitute one full set of experimental data into the rate law and solve for k. The units follow from the overall order: zero order M/s, first order 1/s, second order 1/(M s), third order 1/(M² s). Quoting k without units is an incomplete answer.
For the table, k = 0.00200/[(0.100)(0.100)²] = 2.00 M⁻² s⁻¹. Recalculate from the other trials: all should agree within experimental uncertainty. If they do not, inspect temperature, units, trial selection and whether a simple power law fits the whole concentration range. The general unit result follows directly from (M/s)/Mm+n, rather than from memorizing a separate list.
units of k = M1 - overall order s⁻¹
In rate = k[A]ᵐ[B]ⁿ, changing A by a factor f multiplies the rate by fᵐ; changing B by g multiplies it by gⁿ. A zero-order reactant still participates in the overall reaction but its concentration does not affect the measured rate in the observed regime. Combine independent factors only after applying each exponent.
A concentration held in very large excess may barely change during one run. For rate = k[A][B], approximately constant [B] allows k′ = k[B] and rate ≈ k′[A], called pseudo-first-order behavior. This does not prove the true order in B is zero: repeating the experiment at a different excess [B] changes k′. True zero order is established by changing that concentration without changing the measured rate in the studied regime. State which variables were experimentally varied and which were effectively held fixed.
An experimentally determined equation giving rate as a function of concentrations.
The exponent on a concentration term in the rate law.
The sum of the individual orders, which sets the units of the rate constant.
Compare experiments in which only one concentration differs, otherwise two effects are mixed together.
k is treated as constant for a specified mechanism at fixed temperature, medium and catalyst conditions; ordinary positive-Ea Arrhenius behavior gives increasing k on heating.
rate = k[A]ᵐ[B]ⁿ
m = ln(rate2/rate1) / ln([A]2/[A]1)
For A + B → C: Exp 1, [A] = 0.10, [B] = 0.10, rate = 2.0 × 10⁻³ M/s. Exp 2, [A] = 0.20, [B] = 0.10, rate = 4.0 × 10⁻³ M/s. Exp 3, [A] = 0.10, [B] = 0.20, rate = 8.0 × 10⁻³ M/s. Find the rate law and k.
Try it first: Pick the pair of experiments where only [A] changes, then the pair where only [B] changes.
0 of 4 steps revealed.
Doubling [X] leaves the rate unchanged. What is the order in X, and what does that mean physically?
0 of 2 steps revealed.
A reaction has rate law rate = k[A]². The rate is 0.024 M/s when [A] = 0.20 M. Find k and predict the rate at [A] = 0.50 M.
0 of 2 steps revealed.
For rate = k[A]²[B]⁰, [A] is halved while [B] is doubled. What happens to rate?
Try it first: Calculate a factor for each concentration separately.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Reading the orders off the coefficients of the balanced equation. | The overall equation gives net stoichiometry, not the sequence of elementary steps. For a multistep mechanism, measured orders can depend on several steps and intermediates. | Determine orders from initial-rate data unless the prompt explicitly identifies a single elementary step; then compare the proposed mechanism with the measured rate law. |
| Quoting k without units. | The units of k depend on the overall order, so a bare number is ambiguous. | Use M1 - overall order s⁻¹ and state it every time. |
| Comparing two experiments in which both concentrations changed. | The two effects are combined, so neither order can be isolated. | Choose a pair with exactly one concentration different, or handle the algebra with logarithms. |
| Believing the rate constant changes as the reaction proceeds. | At fixed temperature and reaction medium, k does not change just because reactant concentrations fall. Changing temperature, solvent or medium, or catalyst can change the observed rate constant. | Hold temperature, solvent or medium, and catalyst conditions fixed when treating k as constant. |
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After the lesson · Question bank
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The rate law rate = k[A]ᵐ[B]ⁿ relates rate to concentrations through orders established experimentally. Comparing experiments where only one concentration changes isolates that order. For multistep reactions, the observed orders reflect the mechanism as a whole; only an elementary step lets you read exponents directly from its reactant coefficients. Overall order sets the units of k. At fixed reaction conditions, k stays constant as concentrations change.
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