AP Chem 5.2: Introduction to Rate Law

Rate and concentration, quantitatively

Lesson 2.11.2 showed that higher concentration means a faster rate. AP goes further: if a concentration doubles, by what factor does the rate change? A rate law速率定律 answers that. For reactants A and B:

\text{rate} = k[\ce{A}]^{m}[\ce{B}]^{n}

m and n are the orders级数 in A and B, usually 0, 1, or 2 on the AP exam; m + n is the overall order总级数.
k is the rate constant速率常数. At a given temperature it is constant and doesn’t change with concentration; raising the temperature or adding a catalyst changes k.

Orders come only from experiment

Orders aren’t tied to the coefficients in the equation. OpenStax reports this experimental result at 100 °C:

\ce{NO2(g) + CO(g) -> NO(g) + CO2(g)} \qquad \text{rate} = k[\ce{NO2}]^2

The reaction is second order in NO₂ and zero order in CO, though both have a coefficient of 1. So you can’t write a rate law by looking at an overall equation (U5.4 explains when you can).

What the orders mean:

Order in A	When [A] doubles, the rate is multiplied by	When [A] triples, the rate is multiplied by
0	1 (no change)	1
1	2	3
2	4	9

Zero order doesn’t mean “A doesn’t react”: A is still used up, but the rate doesn’t depend on [A]. As U5.8 explains, a reaction is usually zero order in a reactant because that reactant isn’t involved in the slowest step (the rate-determining step) and is used up in a faster later step.

The method of initial rates

Run several trials with different starting concentrations, and measure each initial rate初始速率 (the rate right at the start, before much product forms). Compare two trials in which only one reactant’s concentration changes:

\left(\frac{[\ce{A}]_2}{[\ce{A}]_1}\right)^{m} = \frac{\text{rate}_2}{\text{rate}_1}

If [A] doubles and the rate quadruples, 2ᵐ = 4, so m = 2. If the factors aren’t neat, take logs: m = log(rate ratio) ÷ log(concentration ratio). Once you know every order, substitute any trial’s data to find k.

Check two trials in the table and see whether they give an order; then choose orders and test whether the rate law explains every trial. The first data set is real data from the textbook:

Try it: find a rate law by the method of initial rates

NO(g) + O₃(g) → NO₂(g) + O₂(g)

Trial	[NO] (M)	[O₃] (M)	Initial rate (M/s)
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

Data: OpenStax Chemistry 2e, Example 12.4 (25 °C); the rate is the rate of formation of NO₂.

2NO(g) + O₂(g) → 2NO₂(g)

Trial	[NO] (M)	[O₂] (M)	Initial rate (M/s)
1	0.0100	0.0100	7.10 × 10⁻³
2	0.0200	0.0100	2.84 × 10⁻²
3	0.0100	0.0200	1.42 × 10⁻²
4	0.0300	0.0200	1.28 × 10⁻¹

Hypothetical data (the form of the rate law matches the experimental result in the textbook).

NO₂(g) + CO(g) → NO(g) + CO₂(g) (100 °C)

Trial	[NO₂] (M)	[CO] (M)	Initial rate (M/s)
1	0.100	0.100	5.40 × 10⁻³
2	0.200	0.100	2.16 × 10⁻²
3	0.100	0.200	5.40 × 10⁻³
4	0.300	0.300	4.86 × 10⁻²

Hypothetical data (the form of the rate law matches the experimental result in the textbook).

2C₄H₆(g) → C₈H₁₂(g)

Trial	[C₄H₆] (M)	Initial rate (M/s)
1	0.0100	5.00 × 10⁻⁵
2	0.0300	4.50 × 10⁻⁴
3	0.0200	2.00 × 10⁻⁴

Hypothetical data (the form of the rate law matches the experimental result in the textbook).

Choose the order in each reactant and test the rate law:

Units of k

Rates are always in M/s, so the units of k depend on the overall order n:

Overall order	Example rate law	Units of k
0	rate = k	M/s
1	rate = k[A]	s⁻¹
2	rate = k[A]² or k[A][B]	M⁻¹ s⁻¹
3	rate = k[A]²[B]	M⁻² s⁻¹

In general, k has units of M¹⁻ⁿ s⁻¹. Working backward, the units of k tell you the overall order.

One reactant in large excess: pseudo-first order

If B’s concentration is far larger than A’s, say [B]₀ = 1.00 M and [A]₀ = 0.010 M (reacting 1 : 1), then even when all of A has reacted, [B] has fallen by only 1%, so it can be treated as constant. Then

\text{rate} = k[\ce{A}][\ce{B}] \approx k'[\ce{A}], \qquad k' = k[\ce{B}]_0

The reaction behaves as if it were first order in A, which is called a pseudo-first-order reaction准一级反应. A graph of [B] vs. t is then almost flat, while [A] falls the way a first-order reactant does (U5.3). Putting one reactant in large excess lets you study the order in the other on its own; once you know k′, divide by [B]₀ to get k.

Example 1: A rate law from initial rates

Hypothetical data for $\ce{2NO(g) + Cl2(g) -> 2NOCl(g)}$ are below. Find the rate law and k.

Trial	[NO] (M)	[Cl₂] (M)	Initial rate (M/s)
1	0.10	0.10	1.8 × 10⁻²
2	0.20	0.10	7.2 × 10⁻²
3	0.20	0.20	1.44 × 10⁻¹

Step-by-step solution

Write the general form. Orders can only be found by experiment, so let:
$\text{rate} = k[\ce{NO}]^m[\ce{Cl2}]^n$
Order in NO, m. Pick two trials where only [NO] changes: trials 1 and 2. [NO] doubles, and the rate ratio is:
$\begin{aligned} \frac{\text{rate}_2}{\text{rate}_1} &= \frac{7.2 \times 10^{-2}}{1.8 \times 10^{-2}} = 4 \\ 2^m &= 4 \Rightarrow m = 2 \end{aligned}$
Order in Cl₂, n. Trials 2 and 3 keep [NO] constant while [Cl₂] doubles:
$\begin{aligned} \frac{\text{rate}_3}{\text{rate}_2} &= \frac{1.44 \times 10^{-1}}{7.2 \times 10^{-2}} = 2 \\ 2^n &= 2 \Rightarrow n = 1 \end{aligned}$
Write the rate law.
$\boxed{\text{rate} = k[\ce{NO}]^2[\ce{Cl2}]}$
It is 2 + 1 = 3, third order overall.
Find k. Use trial 1 (any trial works):
$\begin{aligned} k &= \frac{\text{rate}}{[\ce{NO}]^2[\ce{Cl2}]} \\ &= \frac{1.8 \times 10^{-2}}{(0.10)^2(0.10)} \\ &= \boxed{18\ \text{M}^{-2}\,\text{s}^{-1}} \end{aligned}$
Check. With trial 3: 18 × (0.20)² × 0.20 = 0.144 = 1.44 × 10⁻¹ M/s, matching the data.

Example 2: Predicting with a rate law

A reaction has the rate law rate = k[A][B]². If [A] is tripled while [B] is halved, how does the rate change? How does k change?

Step-by-step solution

Write the two changes. [A] → 3[A] and [B] → ½[B], at the same temperature.
Effect of A. First order in A, so the rate is multiplied by the concentration factor to the first power:
$3^1 = 3$
Effect of B. Second order in B, so the rate is multiplied by the factor squared:
$\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$
Combine them. Multiply the factors:
$\begin{aligned} \frac{\text{rate}_\text{new}}{\text{rate}_\text{old}} &= 3 \times \frac{1}{4} \\ &= \boxed{0.75} \end{aligned}$
The rate becomes 0.75 times as large.
What happens to k. k changes only with temperature (or a catalyst), not with concentration; at the same temperature, k is unchanged. The change in rate comes entirely from the concentrations.

Practice: after this change, the rate is multiplied by what factor?

rate = k[A][B]; [A] doubles
rate = k[A]²; [A] doubles
rate = k[A]²[B]; [A] and [B] both double
rate = k[B]; [A] doubles
rate = k[A]²; [A] is halved
rate = k[A][B]; [A] doubles and [B] is halved
rate = k[A]²[B]; [B] doubles
rate = k[A][B]²; [B] doubles
rate = k[A]² (gases); the container volume doubles
rate = k[A][B] (gases); the container volume is halved
rate = k (zero order); [A] is multiplied by 4
rate = k[A][B]²; [A] is multiplied by 4 and [B] is halved

Common mistakes

Mistake 1: “The exponents in the rate law are the coefficients in the equation.” For an overall reaction, orders come only from experiment. NO₂ + CO at 100 °C is zero order in CO even though CO’s coefficient is 1. Only for an elementary step (U5.4) do orders equal coefficients.

Mistake 2: “Raising the concentration raises k.” Changing concentration changes the rate, not k. At a given temperature k is constant; only temperature or a catalyst changes it.

Mistake 3: “Zero order in B means B doesn’t react (or that B is a catalyst).” Zero order only means the rate doesn’t depend on [B]; B is still used up, usually in a fast step after the rate-determining step (U5.8). Zero order also has nothing to do with the size of B’s coefficient. When [B] changes between trials and the rate doesn’t, the reaction is zero order in B.

Quick check

With [B] held constant, tripling [A] multiplies the initial rate by 9. What is the order in A?

For rate = k[A][B], with concentrations in M and time in s, what are the units of k?

The rate law for H₂ + 2NO → N₂O + H₂O is rate = k[NO]²[H₂]. What is the overall order?

A reaction is zero order in B. If [B] is tripled, how does the rate change?

A reaction has k = 0.25 s⁻¹. What is its overall order?

Which change alters the rate constant k?

rate = k[A]². When [A] = 0.20 M, the rate is 0.012 M/s. What is k?

A reaction is first order in A. A student says, “Doubling [A] doubles k.” Which evaluation is correct?

The reaction 2A + B → C is zero order in B. Which explanation is most reasonable?

rate = k[A][B]. In an experiment, [B]₀ = 2.0 M and [A]₀ = 0.020 M. Which statement is correct?

rate = k[X]²[Y]. With [Y] constant, [X] is tripled. By what factor does the rate change, and why?

AP-style free response

Three trials were run for $\ce{H2(g) + 2NO(g) -> N2O(g) + H2O(g)}$ (hypothetical data):

Trial	[NO] (M)	[H₂] (M)	Initial rate of formation of N₂O (M/s)
1	0.0050	0.0020	1.25 × 10⁻⁵
2	0.0100	0.0020	5.00 × 10⁻⁵
3	0.0100	0.0040	1.00 × 10⁻⁴

(a) Determine the order in NO and in H₂, stating which trials you used.

(b) Write the rate law, and calculate k with units.

(c) Predict the initial rate of formation of N₂O when [NO] = 0.0150 M and [H₂] = 0.0030 M.

(d) In trial 1, what is the initial rate of disappearance of NO?

(e) A student says, “NO has a coefficient of 2, so the reaction must be second order in NO.” Evaluate the claim.

Model answer and scoring

(a) Trials 1 and 2: [H₂] is constant and [NO] doubles; the rate quadruples, so the reaction is second order in NO. Trials 2 and 3: [NO] is constant and [H₂] doubles; the rate doubles, so it is first order in H₂.

(b) rate = k[NO]²[H₂]. From trial 1: k = (1.25 × 10⁻⁵ M/s) ÷ ((0.0050 M)² × 0.0020 M) = 250 M⁻² s⁻¹.

(c) rate = 250 M⁻² s⁻¹ × (0.0150 M)² × 0.0030 M = 1.7 × 10⁻⁴ M/s.

(d) NO’s coefficient is twice N₂O’s: 2 × 1.25 × 10⁻⁵ M/s = 2.50 × 10⁻⁵ M/s.

(e) The conclusion happens to be right, but the reasoning isn’t valid. Orders for an overall reaction come only from experiment; here, second order in NO follows from trials 1 and 2, not from the coefficient.

Scoring: (a) both orders, each from a pair of trials in which only one concentration changes; (b) rate law, value of k, and units; (c) correct substitution; (d) uses the 2 : 1 coefficient ratio; (e) states that orders must be determined by experiment.

Summary

Rate law: rate = k[A]ᵐ[B]ⁿ. m and n are orders, and m + n is the overall order; orders come only from experiment, not from an overall equation’s coefficients.
Method of initial rates: compare two trials in which only one concentration changes, use (concentration ratio)ᵐ = rate ratio to find the order, then substitute one trial to find k.
k has units of M¹⁻ⁿ s⁻¹: M/s for zero order, s⁻¹ for first order, M⁻¹ s⁻¹ for second order.
k changes only with temperature and catalysts; when concentrations change, the rate changes according to the rate law and k stays the same.
When one reactant is in large excess, its concentration barely changes and the reaction is pseudo-first order: rate ≈ k′[A], with k′ = k[B]₀.

AP tip: In your justification, name the two trials you used and state which concentration stays constant, what factor the other changes by, and what factor the rate changes by. When you calculate k, include units, which depend on the overall order.

Introduction to Rate Law速率定律入门

Learning goals

Before you start

Rate and concentration, quantitatively

Orders come only from experiment

The method of initial rates

Units of k

One reactant in large excess: pseudo-first order

Example 1: A rate law from initial rates

Example 2: Predicting with a rate law

Common mistakes

Quick check

AP-style free response

Summary

Feedback / Report a problem