A plot of ln k versus 1T has slope -9000 K. Calculate E_a in J mol^-1. [2 points]

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Why does temperature have such a large effect on reaction rate, and how does the collision model and the Arrhenius equation explain it?

Topic 5.5 Collision Model: use collision theory and the Arrhenius equation to explain how activation energy, temperature, orientation and collision frequency control the rate constant.

A focused answer to AP Chemistry Topic 5.5, covering collision theory, activation energy, the Maxwell-Boltzmann distribution, molecular orientation, and the Arrhenius equation linking rate constant to temperature, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

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

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Quick answer

Collision theory says a reaction happens only when reactant particles collide with at least the activation energy and with the correct orientation. The activation energy is the minimum energy needed to reach the transition state. Most collisions fail because they are too weak or wrongly aligned. Raising the temperature shifts the Maxwell-Boltzmann distribution of molecular energies to the right, so a much larger fraction of collisions clear the activation energy; this exponential effect, not the small rise in collision frequency, is why a modest temperature rise can double the rate. The Arrhenius equation $k = Ae^{-E_a/RT}$ packages all of this: $A$ is the frequency and orientation factor, and $e^{-E_a/RT}$ is the fraction of collisions energetic enough to react. A catalyst lowers $E_a$ and so raises $k$ .

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What this topic is asking
Collision theory
Temperature and the Maxwell-Boltzmann distribution
The Arrhenius equation
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What this topic is asking

The College Board (Topic 5.5) wants you to use collision theory and the Arrhenius equation to explain how activation energy, temperature, molecular orientation and collision frequency together set the rate constant. This topic explains why $k$ depends so strongly on temperature and why catalysts work, tying the macroscopic rate law back to the behavior of individual particles.

Collision theory

The vast majority of collisions are ineffective. Even at a high collision frequency, only those with enough energy and the right geometry produce product. This is why rates are far lower than the raw collision rate would predict, and why both an energy factor and an orientation factor appear in the rate constant.

Temperature and the Maxwell-Boltzmann distribution

A common exam graph shows two distribution curves, one for a higher temperature, with $E_a$ marked. The area to the right of $E_a$ is the fraction of effective collisions, and it grows dramatically between the two temperatures even though the total number of collisions rises only modestly. This is the molecular picture behind the rule of thumb that a small temperature increase can double a rate.

The Arrhenius equation

The Arrhenius equation ties the rate constant to temperature and activation energy:

k = A e^{-E_a/RT}

Here $A$ is the frequency factor (collision frequency and the orientation requirement), $E_a$ is the activation energy, $R$ is the gas constant and $T$ is the absolute temperature. The exponential term $e^{-E_a/RT}$ is the fraction of collisions with at least the activation energy. Raising $T$ makes the exponent less negative, so $k$ increases; lowering $E_a$ (by catalysis) does the same. Taking logarithms gives the linear form $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$ , so a plot of $\ln k$ against $\dfrac{1}{T}$ is a straight line of slope $-\dfrac{E_a}{R}$ .

Reasoning with the Arrhenius equation

A reaction has $E_a = 75\ \text{kJ mol}^{-1}$ . Explain qualitatively what happens to $k$ when the temperature rises from $300.$ K to $310.$ K.

step 1 Identify the temperature-dependent term

In $k = Ae^{-E_a/RT}$ , only the exponent $-\dfrac{E_a}{RT}$ depends on temperature.

step 2 Evaluate the exponent at each temperature

At $300.$ K: $-\dfrac{75000}{(8.314)(300.)} = -30.1$ . At $310.$ K: $-\dfrac{75000}{(8.314)(310.)} = -29.1$ .

step 3 Compare the exponentials

The exponent rose by about $1.0$ , so $k$ increases by a factor of about $e^{1.0} \approx 2.7$ .

step 4 Interpret

A 10 K rise nearly triples the rate constant, because the fraction of collisions exceeding $E_a$ climbs exponentially even though the average speed rises only slightly.

Try this

Q1. State the two requirements collision theory places on an effective collision. [2 points]

Cue. Energy at least equal to $E_a$ , and correct orientation of the colliding particles.

Q2. A plot of $\ln k$ versus $\dfrac{1}{T}$ has slope $-9000$ K. Calculate $E_a$ in $\text{J mol}^{-1}$ . [2 points]

Cue. Slope $= -\dfrac{E_a}{R}$ , so $E_a = 9000 \times 8.314 = 7.5 \times 10^{4}\ \text{J mol}^{-1}$ .

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). A reaction has activation energy

E_a = 50.\ \text{kJ mol}^{-1}

. (a) Using collision theory, identify the two conditions a collision must meet to lead to reaction. (b) Explain, in terms of the Maxwell-Boltzmann distribution, why raising the temperature increases the rate far more than the small rise in collision frequency alone would suggest. (c) The rate constant is

k = 1.0 \times 10^{-3}\ \text{s}^{-1}

300.

K. Using the Arrhenius equation, determine qualitatively whether

k

350.

K is larger or smaller, and justify. (d) Explain the effect of a catalyst on

E_a

and on

k

Show worked answer →

A 4-point conceptual FRQ on the collision model.

(a) Conditions (1 point): the colliding particles must have at least the activation energy, and they must collide with the correct orientation.
(b) Distribution (1 point): the Maxwell-Boltzmann curve shifts to higher energies and broadens at higher temperature, so the fraction of molecules with energy $\geq E_a$ (the area beyond $E_a$ ) rises sharply; this exponential increase in the effective-collision fraction dominates over the modest rise in total collision frequency.
(c) Arrhenius (1 point): $k = Ae^{-E_a/RT}$ ; raising $T$ makes the exponent less negative, so $e^{-E_a/RT}$ grows and $k$ at $350.$ K is larger.
(d) Catalyst (1 point): a catalyst lowers $E_a$ by providing an alternative pathway, which increases $e^{-E_a/RT}$ and so increases $k$ at a given temperature.

Markers reward the two collision conditions, the Maxwell-Boltzmann reasoning, the Arrhenius direction with justification, and the catalyst effect on both $E_a$ and $k$ .

AP 2021 (style)1 marksSection I (multiple choice). Increasing the temperature of a reaction increases the rate mainly because (A) the activation energy decreases (B) a larger fraction of collisions have energy at least equal to

E_a

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

Temperature does not change $E_a$ or the reaction enthalpy; it shifts the Maxwell-Boltzmann distribution so that a larger fraction of molecules have at least the activation energy, sharply raising the proportion of effective collisions. The trap is (A): only a catalyst lowers $E_a$ .

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Sources & how we know this

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