For 2HI(g) → H_2(g) + I_2(g), HI disappears at 0.040\ M s^-1. Calculate the rate of appearance of H_2. [2 points]

What is not dividing by the coefficient?

The single reaction rate requires dividing each species rate by its stoichiometric coefficient; species rates alone differ.

College Board AP 2023 (style)-style practice: Section II (long FRQ, part). For the reaction 2N_2O_5(g) → 4NO_2(g) + O_2(g), the concentration of N_2O_5 falls from 0.0800 M to 0.0500 M over 120. s. (a) Calculate the average rate of disappearance of N_2O_5. (b) Calculate the average rate of appearance of NO_2. (c) Calculate the rate of the reaction as defined by AP. (d) Justify why the rate of appearance of NO_2 is not equal to the rate of disappearance of N_2O_5.

A 4-point quantitative FRQ on rate definitions. (a) Disappearance of N_2O_5 (1 point): -[N_2O_5] t = -0.0500 - 0.0800120. = 2.50 × 10^-4\ M s^-1. (b) Appearance of NO_2 (1 point): the mole ratio N_2O_5 : NO_2 is 2:4 = 1:2, so NO_2 appears twice as fast: 2 × 2.50 × 10^-4 = 5.00 × 10^-4\ M s^-1. (c) Reaction rate (1 point): dividing each by its coefficient, rate = 12(2.50 × 10^-4) = 1.25 × 10^-4\ M s^-1. (d) Justify (1 point): the stoichiometry makes four NO_2 for every two N_2O_5 consumed, so NO_2 is produced at twice the rate that N_2O_5 disappears; the rates differ by the ratio of coefficients. Markers reward each correct rate, dividing by coefficients for the reaction rate, and explaining the difference through stoichiometry.

College Board AP 2021 (style)-style practice: Section I (multiple choice). For A + 3B → 2C, if B disappears at 0.30\ M s^-1, the rate of appearance of C is (A) 0.10\ M s^-1 (B) 0.20\ M s^-1 (C) 0.30\ M s^-1 (D) 0.45\ M s^-1. Justify your reasoning.

A 1-point conceptual MCQ. The answer is (B). The reaction rate is -13[B] t = 0.303 = 0.10\ M s^-1. C appears at 2 × 0.10 = 0.20\ M s^-1 because its coefficient is 2. The trap is forgetting to scale by the coefficients on both B and C.

United StatesChemistrySyllabus dot point

How do we define and measure the rate of a chemical reaction, and what controls how fast it goes?

Topic 5.1 Reaction Rates: express the rate of a reaction in terms of the change in concentration of a reactant or product over time, relate rates through the stoichiometric coefficients, and identify the factors that influence rate.

A focused answer to AP Chemistry Topic 5.1, covering the definition of reaction rate, average versus instantaneous rate, relating rates through stoichiometric coefficients, and the factors that change the rate of a reaction, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The rate of a reaction is the change in concentration of a species per unit time. For a reactant the concentration falls, so we put a minus sign in front to keep the rate positive; for a product it rises. To get a single rate for the whole reaction, divide each species rate by its stoichiometric coefficient. For $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$ , the reaction rate is $-\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta[\text{B}]}{\Delta t} = \frac{1}{c}\frac{\Delta[\text{C}]}{\Delta t} = \frac{1}{d}\frac{\Delta[\text{D}]}{\Delta t}$ . The average rate is measured over an interval; the instantaneous rate is the slope of the concentration-time curve at one moment. Rate is raised by higher concentration, higher temperature, greater surface area and a catalyst.

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What this topic is asking
Defining the rate
Average and instantaneous rate
What changes the rate
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What this topic is asking

The College Board (Topic 5.1) wants you to define the rate of a reaction as the change in concentration of a reactant or product per unit time, distinguish average from instantaneous rate, relate the rates of different species through the stoichiometric coefficients, and name the factors that change how fast a reaction goes. Kinetics is about how fast, not how far: a thermodynamically favorable reaction can still be immeasurably slow.

Defining the rate

Because the species in a reaction are consumed and made in fixed ratios, their individual rates differ. To give the reaction one unambiguous rate, the College Board divides each species rate by its coefficient. For $a\text{A} + b\text{B} \rightarrow c\text{C} + d\text{D}$ :

\text{rate} = -\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta[\text{B}]}{\Delta t} = \frac{1}{c}\frac{\Delta[\text{C}]}{\Delta t} = \frac{1}{d}\frac{\Delta[\text{D}]}{\Delta t}

This is why a product with a large coefficient appears faster than a reactant with a small one disappears.

Average and instantaneous rate

Reaction rates almost always decrease as a reaction proceeds, because the reactant concentrations fall. The concentration-time graph for a reactant is therefore a curve that levels off, and its slope (the instantaneous rate) gets shallower with time. Measuring the initial rate, before products build up and reverse reactions matter, isolates the dependence on the starting concentrations.

What changes the rate

Four factors raise the rate, each explained by the collision model developed later in this unit: higher reactant concentration (more collisions per second), higher temperature (more collisions, and a far greater fraction with enough energy), greater surface area of a solid (more exposed particles), and adding a catalyst (a lower-energy pathway). Pressure raises the rate of gas reactions for the same reason as concentration.

Relating rates through coefficients

For $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$ , hydrogen is consumed at $0.090\ \text{M s}^{-1}$ . Find the reaction rate and the rate of appearance of ammonia.

step 1 Write the reaction rate from $\text{H}_2$

$\text{rate} = -\dfrac{1}{3}\dfrac{\Delta[\text{H}_2]}{\Delta t} = \dfrac{1}{3}(0.090) = 0.030\ \text{M s}^{-1}$ .

step 2 Use the rate to find $\text{NH}_3$

$\dfrac{1}{2}\dfrac{\Delta[\text{NH}_3]}{\Delta t} = 0.030$ , so $\dfrac{\Delta[\text{NH}_3]}{\Delta t} = 2 \times 0.030 = 0.060\ \text{M s}^{-1}$ .

step 3 Check against stoichiometry

Three $\text{H}_2$ disappear for every two $\text{NH}_3$ formed, so ammonia appears at $\tfrac{2}{3}$ the rate hydrogen disappears: $\tfrac{2}{3}(0.090) = 0.060\ \text{M s}^{-1}$ , which matches.

Try this

Q1. For $2\text{HI}(g) \rightarrow \text{H}_2(g) + \text{I}_2(g)$ , HI disappears at $0.040\ \text{M s}^{-1}$ . Calculate the rate of appearance of $\text{H}_2$ . [2 points]

Cue. Reaction rate $= \tfrac{1}{2}(0.040) = 0.020\ \text{M s}^{-1}$ ; $\text{H}_2$ appears at $0.020\ \text{M s}^{-1}$ (coefficient 1).

Q2. Explain why the instantaneous rate of a reaction is largest at the start. [2 points]

Cue. Reactant concentrations are highest at the start, giving the most frequent collisions; as reactants are consumed the rate falls.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). For the reaction

2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g) + \text{O}_2(g)

, the concentration of

\text{N}_2\text{O}_5

falls from

0.0800

M to

0.0500

M over

120.

s. (a) Calculate the average rate of disappearance of

\text{N}_2\text{O}_5

. (b) Calculate the average rate of appearance of

\text{NO}_2

. (c) Calculate the rate of the reaction as defined by AP. (d) Justify why the rate of appearance of

\text{NO}_2

is not equal to the rate of disappearance of

\text{N}_2\text{O}_5

Show worked answer →

A 4-point quantitative FRQ on rate definitions.

(a) Disappearance of $\text{N}_2\text{O}_5$ (1 point): $-\dfrac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = -\dfrac{0.0500 - 0.0800}{120.} = 2.50 \times 10^{-4}\ \text{M s}^{-1}$ .
(b) Appearance of $\text{NO}_2$ (1 point): the mole ratio $\text{N}_2\text{O}_5 : \text{NO}_2$ is $2:4 = 1:2$ , so $\text{NO}_2$ appears twice as fast: $2 \times 2.50 \times 10^{-4} = 5.00 \times 10^{-4}\ \text{M s}^{-1}$ .
(c) Reaction rate (1 point): dividing each by its coefficient, rate $= \dfrac{1}{2}(2.50 \times 10^{-4}) = 1.25 \times 10^{-4}\ \text{M s}^{-1}$ .
(d) Justify (1 point): the stoichiometry makes four $\text{NO}_2$ for every two $\text{N}_2\text{O}_5$ consumed, so $\text{NO}_2$ is produced at twice the rate that $\text{N}_2\text{O}_5$ disappears; the rates differ by the ratio of coefficients.

Markers reward each correct rate, dividing by coefficients for the reaction rate, and explaining the difference through stoichiometry.

AP 2021 (style)1 marksSection I (multiple choice). For

\text{A} + 3\text{B} \rightarrow 2\text{C}

, if B disappears at

0.30\ \text{M s}^{-1}

, the rate of appearance of C is (A)

0.10\ \text{M s}^{-1}

(B)

0.20\ \text{M s}^{-1}

(C)

0.30\ \text{M s}^{-1}

(D)

0.45\ \text{M s}^{-1}

. Justify your reasoning.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

The reaction rate is $-\dfrac{1}{3}\dfrac{\Delta[\text{B}]}{\Delta t} = \dfrac{0.30}{3} = 0.10\ \text{M s}^{-1}$ . C appears at $2 \times 0.10 = 0.20\ \text{M s}^{-1}$ because its coefficient is 2. The trap is forgetting to scale by the coefficients on both B and C.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)