Collision Theory of Reaction Rates

Q: What is activation energy?

Activation energy Ea is the minimum extra energy that reactants must acquire above their average energy to reach the transition state (activated complex) and form products. It is the energy barrier between reactants and products.

Q: What is the difference between activation energy and threshold energy?

Threshold energy Et is the absolute minimum total energy that colliding molecules must possess to react. Activation energy is the excess of threshold energy over the average energy of the reactants: Ea = Et - E(avg). In practice, activation energy is what appears in the Arrhenius equation.

Q: What is the Arrhenius equation?

The Arrhenius equation is k = A exp(-Ea/RT), where k is the rate constant, A is the pre-exponential (frequency) factor, Ea is the activation energy, R is the gas constant, and T is the absolute temperature. It quantitatively links rate constant to temperature and activation energy.

Q: What is the steric factor P?

The steric factor P (also called orientation factor) is the fraction of collisions with the geometrically correct orientation for reaction. It corrects the simple collision theory rate Z exp(-Ea/RT) to P Z exp(-Ea/RT). Typical values range from about 0.1 down to 10^-6 or lower for complex molecules.

Q: How is activation energy determined experimentally?

Measure the rate constant k at several temperatures. Plot log k (or ln k) against 1/T; the slope is -Ea/(2.303 R) (or -Ea/R for ln). Alternatively, measure k at two temperatures and use the two-point form log(k2/k1) = (Ea/2.303R)((T2 - T1)/(T1 T2)).

Q: What does the pre-exponential factor A physically represent?

The pre-exponential factor A (also called the frequency factor) equals the product of the collision frequency Z and the steric factor P. It has the same units as k and represents the value of k in the hypothetical limit Ea/RT to 0, i.e. when every collision is effective.

Q: Why do reactions with higher activation energy show larger temperature-dependence?

From the two-point Arrhenius form, log(k2/k1) is proportional to Ea for a fixed temperature change. A larger Ea therefore gives a larger fractional rise in k for the same delta T. Physically, when Ea is high, only a small fraction of molecules are effective at T1; raising T has a large multiplicative effect on that fraction.

ChemistryChemical KineticsFor JEE aspirants

The collision theory of reaction rates explains how chemical reactions actually happen at the molecular level. For a reaction to occur, molecules must collide, but not every collision leads to reaction. Only those collisions in which the molecules have energy at least equal to the activation energy $E_{a}$ and are correctly oriented lead to product formation. These are called effective collisions. From this simple picture, the collision theory gives the rate equation $rate = P \cdot Z \cdot e^{- E_{a} / R T}$ , and Arrhenius put it in the celebrated compact form $k = A e^{- E_{a} / R T}$ .

Key Formulas - Quick Reference

Rate from collision theory: $rate = Z \cdot f = Z e^{- E_{a} / R T}$
With steric (orientation) factor $P$ : $k = P \cdot Z \cdot e^{- E_{a} / R T}$
Arrhenius equation: $k = A e^{- E_{a} / R T}$ ; $ln k = ln A - \frac{E _{a}}{R T}$
Log form: $lo g k = lo g A - \frac{E _{a}}{2.303 R T}$
Two-point form: $lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) = \frac{E _{a}}{2.303 R} (\frac{T _{2} - T _{1}}{T _{1} T _{2}})$
Slope of $lo g k$ vs $1/ T$ : $- E_{a} / (2.303 R)$ ; intercept: $lo g A$ .
For a reversible reaction: $Δ H = E_{a, forward} - E_{a, backward}$

1. Basic Postulates of Collision Theory

The collision theory rests on three postulates:

Reactant molecules must collide with each other for a reaction to occur.
Only those collisions in which the colliding molecules have energy greater than or equal to a certain minimum (threshold energy) are effective.
Colliding molecules must have the correct orientation so that the atoms that need to bond can actually come into contact.

The number of collisions per unit volume per unit time is called the collision frequency and is denoted $Z$ . For a typical gas-phase reaction at ordinary temperature and pressure, $Z \sim 1 0^{28}$ collisions per litre per second - an astronomical number. Yet the reaction may take hours to complete. Only a tiny fraction of collisions are effective.

2. Energy Factor: Threshold Energy and Activation Energy

Threshold energy (

E_{T}

): the minimum total energy that colliding molecules must possess for a fruitful collision.

Activation energy (

E_{a}

): the extra energy above the average energy of reactants that is needed to reach

E_{T}

E_{a} = E_{T} - E_{avg reactants}

When two high-energy molecules collide, they overcome mutual repulsion and form an unstable transient species called the activated complex or transition state. The activated complex sits at the top of the energy barrier; from there it can either fall forward to give products or fall back to reactants.

The activation energy of the forward reaction ( $E_{a, f}$ ) is the energy needed to go from reactants to the top of the barrier. The activation energy of the backward reaction ( $E_{a, b}$ ) is the energy needed from products to reach the same barrier. The two are related to the enthalpy of reaction:

$Δ H = E_{a, f} - E_{a, b}$

An exothermic reaction has $E_{a, b} > E_{a, f}$ (products lie below reactants). An endothermic reaction has $E_{a, b} < E_{a, f}$ .

3. Orientation Factor (Steric Factor)

Even when colliding molecules have enough energy, the collision may still fail if the molecules approach with the wrong orientation. Consider the reaction of a diatomic $A B$ with $C$ : if the $A$ end of $A B$ hits $C$ , no reaction occurs; only if the $B$ end faces $C$ can the new $B - C$ bond form.

Figure: Only collisions with correct orientation lead to product. Fraction of correctly oriented collisions is the steric factor

P

To account for this, we introduce a steric (or orientation) factor $P$ , defined as the fraction of collisions that have the correct orientation. Typically $P$ lies between $1 0^{- 1}$ and $1 0^{- 6}$ ; for very complex molecules with strict geometric requirements it can be smaller still.

4. The Full Rate Expression

Combining energy and orientation factors, the collision-theory rate is

$rate = Z \times f \times P$

where $Z$ is collision frequency, $f = e^{- E_{a} / R T}$ is the fraction of collisions with sufficient energy (from the Maxwell-Boltzmann distribution), and $P$ is the steric factor. Since rate is proportional to $k$ , we can write

$k = P \cdot Z \cdot e^{- E_{a} / R T}$

The product $P \cdot Z$ is a constant (for a given reaction at a given $T$ ). Arrhenius denoted it $A$ and called it the pre-exponential factor or frequency factor. This gives the celebrated form of the Arrhenius equation:

$k = A e^{- E_{a} / R T}$

5. The Arrhenius Equation

Historically, Arrhenius arrived at this equation from the temperature dependence of the equilibrium constant. Starting from the Van't Hoff equation

$\frac{d ln K _{p}}{d T} = \frac{Δ H}{R T ^{2}}$

and writing $K_{p} = k_{f} / k_{b}$ (forward and backward rate constants), with $Δ H = E_{a, f} - E_{a, b}$ , one can split the equation into two parts, one for each direction. Setting a constant to zero, Arrhenius wrote

$\frac{d ln k}{d T} = \frac{E _{a}}{R T ^{2}}$

Integrating (assuming $E_{a}$ is roughly constant over the relevant temperature range):

$ln k = - \frac{E _{a}}{R T} + ln A ⟹ k = A e^{- E_{a} / R T}$

5.1 Interpretation

The exponential factor $e^{- E_{a} / R T}$ measures the fraction of molecules that have energy $\geq E_{a}$ . If $n_{E_{a}}$ is the number of such molecules out of a total $n$ , then

$\frac{n _{E_{a}}}{n} = e^{- E_{a} / R T}$

The pre-exponential factor $A$ has the same units as $k$ and represents the rate constant that would be observed if there were no activation-energy barrier (that is, in the hypothetical limit $T \to \infty$ ). It contains the collision frequency and orientation factor.

6. The $lo g k$ vs $1/ T$ Plot

Taking $lo g_{10}$ of the Arrhenius equation:

$lo g k = lo g A - \frac{E _{a}}{2.303 R} \cdot \frac{1}{T}$

A plot of $lo g k$ against $1/ T$ is a straight line with slope $- E_{a} / (2.303 R)$ and intercept $lo g A$ . This is the standard experimental method for determining $E_{a}$ : measure $k$ at several temperatures, plot $lo g k$ vs $1/ T$ , and read off the slope.

Figure: Arrhenius plot. Slope =

- E_{a} / (2.303 R)

; intercept =

lo g A

7. Two-Point Form of the Arrhenius Equation

If $k_{1}$ and $k_{2}$ are the rate constants at temperatures $T_{1}$ and $T_{2}$ , subtracting the two Arrhenius equations gives

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) = \frac{E _{a}}{2.303 R} \cdot \frac{T _{2} - T _{1}}{T _{1} T _{2}}$

This is the workhorse formula for finding $E_{a}$ from rate-constant data at two temperatures, or for predicting $k$ at a new temperature.

Solved Example 1

The rate constant of a first-order reaction increases from

2 \times 1 0^{- 3} s^{- 1}

300 K

8 \times 1 0^{- 3} s^{- 1}

320 K

. Find the activation energy.

Solution:

$lo g \frac{k _{2}}{k _{1}} = lo g 4 = 0.602$ .

$\frac{T _{2} - T _{1}}{T _{1} T _{2}} = \frac{20}{300 \times 320} = 2.083 \times 1 0^{- 4} K^{- 1}$ .

$E_{a} = \frac{2.303 \times 8.314 \times 0.602}{2.083 \times 1 0 ^{- 4}} = 55.3 kJ/mol$ .

Solved Example 2

A catalyst lowers the activation energy of the forward reaction by

20 kJ/mol

. By how much does it change the activation energy of the backward reaction?

Solution:

The catalyst offers a lower barrier on the same energy profile, so the peak comes down by the same amount whichever direction we approach it from. Since $Δ H = E_{a, f} - E_{a, b}$ is unchanged (a catalyst does not alter thermodynamics), if $E_{a, f}$ drops by $20 kJ/mol$ then $E_{a, b}$ must drop by exactly the same $20 kJ/mol$ .

Answer: the activation energy of the backward reaction is also lowered by $20 kJ/mol$ .

Solved Example 3

For a certain reaction, the rate constant at

700 K

0.011 s^{- 1}

and at

800 K

it is

0.105 s^{- 1}

. Estimate the activation energy.

Solution:

$lo g (k_{2} / k_{1}) = lo g (0.105/0.011) = lo g (9.545) = 0.980$ .

$\frac{T _{2} - T _{1}}{T _{1} T _{2}} = \frac{100}{700 \times 800} = 1.786 \times 1 0^{- 4}$ .

$E_{a} = \frac{2.303 \times 8.314 \times 0.980}{1.786 \times 1 0 ^{- 4}} \approx 105 kJ/mol$ .

Solved Example 4

Two reactions have activation energies

E_{a 1}

and

E_{a 2}

with

E_{a 1} > E_{a 2}

. If the temperature is increased from

T_{1}

T_{2}

, which reaction shows a greater proportional increase in rate constant?

Solution:

From the two-point form, $lo g (k_{2} / k_{1}) = \frac{E _{a}}{2.303 R} \cdot \frac{T _{2} - T _{1}}{T _{1} T _{2}}$ . For a given temperature interval, $lo g (k_{2} / k_{1})$ is directly proportional to $E_{a}$ .

Therefore the reaction with larger $E_{a}$ (i.e. $E_{a 1}$ ) shows a greater proportional increase in $k$ when the temperature is raised. Reactions with higher activation energies are more temperature-sensitive.

Common Mistakes to Avoid

Watch out

Confusing threshold energy $E_{T}$ with activation energy $E_{a}$ . Threshold is the absolute minimum energy of the colliding molecules; activation energy is the extra above the average energy of reactants: $E_{a} = E_{T} - E_{a v g}$ .
Forgetting the orientation (steric) factor $P$ . Simple collision theory (without $P$ ) often overestimates the rate constant of complex reactions by orders of magnitude.
Assuming a catalyst affects the forward and backward activation energies differently. It lowers both by the same amount, because $Δ H = E_{a, f} - E_{a, b}$ is unchanged.
Reading the slope of the Arrhenius plot as $- E_{a} / R$ instead of $- E_{a} / (2.303 R)$ . The extra factor of $2.303$ applies when you use $lo g_{10}$ rather than $ln$ .
Getting the sign convention wrong on the two-point form. The formula $lo g (k_{2} / k_{1}) = (E_{a} /2.303 R) (1/ T_{1} - 1/ T_{2})$ gives a positive result when $T_{2} > T_{1}$ (so $k_{2} > k_{1}$ ), because $1/ T_{1} > 1/ T_{2}$ .
Using the Arrhenius equation for reactions whose mechanism changes with temperature. The equation assumes $E_{a}$ is constant; for reactions with multiple pathways at different temperatures, an Arrhenius plot may curve.

Frequently Asked Questions

Q1. What is collision theory of reaction rates?

Collision theory states that a chemical reaction occurs only when reactant molecules collide with (i) sufficient energy (at least the activation energy $E_{a}$ ) and (ii) the correct orientation. The overall rate is proportional to the collision frequency, the fraction of collisions with sufficient energy, and the fraction with correct orientation.

Q2. What is activation energy?

Activation energy $E_{a}$ is the minimum extra energy that reactants must acquire above their average energy to reach the transition state (activated complex) and form products. It is the energy barrier between reactants and products.

Q3. What is the difference between activation energy and threshold energy?

Threshold energy $E_{T}$ is the absolute minimum total energy that colliding molecules must possess to react. Activation energy is the excess of threshold energy over the average energy of the reactants: $E_{a} = E_{T} - E_{a v g}$ . In practice, activation energy is what appears in the Arrhenius equation.

Q4. What is the Arrhenius equation?

The Arrhenius equation is $k = A e^{- E_{a} / R T}$ , where $k$ is the rate constant, $A$ is the pre-exponential (frequency) factor, $E_{a}$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. It quantitatively links rate constant to temperature and activation energy.

Q5. What is the steric factor $P$ ?

The steric factor $P$ (also called orientation factor) is the fraction of collisions with the geometrically correct orientation for reaction. It corrects the simple collision theory rate $Z e^{- E_{a} / R T}$ to $P Z e^{- E_{a} / R T}$ . Typical values range from about $1 0^{- 1}$ down to $1 0^{- 6}$ or lower for complex molecules.

Q6. What is an activated complex?

The activated complex (or transition state) is the highest-energy configuration of atoms that appears momentarily as reactants convert into products. It sits at the top of the energy barrier and can either fall forward to give products or roll back to reactants.

Q7. How is activation energy determined experimentally?

Measure the rate constant $k$ at several temperatures. Plot $lo g k$ (or $ln k$ ) against $1/ T$ ; the slope is $- E_{a} / (2.303 R)$ (or $- E_{a} / R$ for $ln$ ). Alternatively, measure $k$ at two temperatures and use the two-point form $lo g (k_{2} / k_{1}) = (E_{a} /2.303 R) ((T_{2} - T_{1}) / T_{1} T_{2})$ .

Q8. What does the pre-exponential factor $A$ physically represent?

The pre-exponential factor $A$ (also called the frequency factor) equals the product of the collision frequency $Z$ and the steric factor $P$ . It has the same units as $k$ and represents the value of $k$ in the hypothetical limit $E_{a} / R T \to 0$ , i.e. when every collision is effective.

Q9. Why do reactions with higher activation energy show larger temperature-dependence?

From the two-point Arrhenius form, $lo g (k_{2} / k_{1})$ is proportional to $E_{a}$ for a fixed temperature change. A larger $E_{a}$ therefore gives a larger fractional rise in $k$ for the same $Δ T$ . Physically, when $E_{a}$ is high, only a small fraction of molecules are effective at $T_{1}$ ; raising $T$ has a large multiplicative effect on that fraction.

Previous year questions on Collision Theory of Reaction Rates

8 questions from past papers, each with a step-by-step solution.

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1. Basic Postulates of Collision Theory

2. Energy Factor: Threshold Energy and Activation Energy

3. Orientation Factor (Steric Factor)

4. The Full Rate Expression

5. The Arrhenius Equation

5.1 Interpretation

6. The logk vs 1/T Plot

7. Two-Point Form of the Arrhenius Equation

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is collision theory of reaction rates?

Q2. What is activation energy?

Q3. What is the difference between activation energy and threshold energy?

Q4. What is the Arrhenius equation?

Q5. What is the steric factor P?

Q6. What is an activated complex?

Q7. How is activation energy determined experimentally?

Q8. What does the pre-exponential factor A physically represent?

Q9. Why do reactions with higher activation energy show larger temperature-dependence?

Previous year questions on Collision Theory of Reaction Rates

Ready to master Chemical Kinetics?

6. The $lo g k$ vs $1/ T$ Plot

Q5. What is the steric factor $P$ ?

Q8. What does the pre-exponential factor $A$ physically represent?