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Chemical Kinetics - Collision Theory of Chemical Reactions

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Collision Theory, proposed by Max Trautz and William Lewis, provides a qualitative and quantitative insight into the energetic and mechanistic aspects of chemical reactions.

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It assumes reactant molecules are hard spheres and reactions occur only when molecules collide with each other.

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The number of collisions per second per unit volume of the reaction mixture is called collision frequency (ZZ).

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Not all collisions result in a chemical reaction. A collision must be an 'effective collision' to produce products.

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Two criteria for an effective collision: (1) Activation Energy Barrier: Molecules must possess a minimum amount of energy called threshold energy. (2) Orientation Barrier: Molecules must collide in a specific spatial arrangement.

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The fraction of molecules having energy equal to or greater than the activation energy (EaE_a) is represented by the Boltzmann factor e−Ea/RTe^{-E_a / RT}.

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The Steric Factor (PP), or probability factor, is introduced to account for the requirement of proper orientation during collision.

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In collision theory, the rate constant is expressed as k=PZABe−Ea/RTk = P Z_{AB} e^{-E_a / RT}.

📐Formulae

Rate=ZABe−EaRTRate = Z_{AB} e^{-\frac{E_a}{RT}}

k=PZABe−EaRTk = P Z_{AB} e^{-\frac{E_a}{RT}}

Threshold Energy=Activation Energy(Ea)+Average Kinetic EnergyThreshold\ Energy = Activation\ Energy (E_a) + Average\ Kinetic\ Energy

f=e−EaRTf = e^{-\frac{E_a}{RT}}

💡Examples

Problem 1:

For a reaction involving the collision of AA and BB, the collision frequency is ZAB=1030 s−1m−3Z_{AB} = 10^{30} \text{ s}^{-1}\text{m}^{-3} and the activation energy is 100 kJ mol−1100 \text{ kJ mol}^{-1}. If the steric factor PP is 0.010.01, calculate the rate constant at 300 K300 \text{ K}.

Solution:

Using the formula k=PZABe−Ea/RTk = P Z_{AB} e^{-E_a / RT}:

  1. Convert EaE_a to J: Ea=100×103 J mol−1E_a = 100 \times 10^3 \text{ J mol}^{-1}.
  2. Calculate the exponential factor: f=e−1000008.314×300≈e−40.09≈3.8×10−18f = e^{-\frac{100000}{8.314 \times 300}} \approx e^{-40.09} \approx 3.8 \times 10^{-18}.
  3. Substitute values: k=0.01×1030×3.8×10−18=3.8×1010 unitsk = 0.01 \times 10^{30} \times 3.8 \times 10^{-18} = 3.8 \times 10^{10} \text{ units}.

Explanation:

This demonstrates how the rate constant kk is reduced significantly by both the activation energy barrier (the exponential term) and the orientation requirement (the steric factor PP).

Problem 2:

Explain the significance of the orientation factor using the reaction between CH3BrCH_3Br and OH−OH^-.

Solution:

The reaction is CH3Br+OH−→CH3OH+Br−CH_3Br + OH^- \rightarrow CH_3OH + Br^-. For a successful reaction, the OH−OH^- ion must attack the carbon atom from the side opposite to the bulky bromine atom (back-side attack).

Explanation:

If the OH−OH^- ion collides with the bromine atom side, the molecules simply bounce apart due to repulsion and improper alignment. This requirement for specific alignment is quantified by the steric factor (PP) in collision theory.

Collision Theory of Chemical Reactions Revision - Class 12 Chemistry CBSE