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Chemical Kinetics - Concept of collision theory

Grade 12ICSEChemistry

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

🔑Concepts

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The Collision Theory, proposed by Max Trautz and William Lewis, provides a qualitative and quantitative explanation for the rates of chemical reactions, assuming reactant molecules to be hard spheres.

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A chemical reaction occurs only when reactant molecules collide with each other. However, the rate of reaction is generally much lower than the collision frequency because not every collision is effective.

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For a collision to be 'effective' (leading to product formation), it must overcome two barriers: (1) The Energy Barrier and (2) The Orientation Barrier.

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Energy Barrier: Reacting molecules must possess a minimum amount of energy called Threshold Energy (ETE_T). The additional energy required by reactant molecules to reach this threshold is the Activation Energy (EaE_a), where Ea=ET−EaverageE_a = E_T - E_{average}.

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Orientation Barrier: Even if molecules have sufficient energy, they must collide in the correct spatial orientation to facilitate the breaking of old bonds and the formation of new ones.

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The fraction of molecules with energy equal to or greater than EaE_a is given by the Boltzmann factor: f=e−EaRTf = e^{-\frac{E_a}{RT}}.

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The Steric Factor (PP), also known as the probability factor, is introduced to account for the requirement of proper orientation during collisions.

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Collision Frequency (ZZ): The number of collisions per second per unit volume of the reacting mixture. For a bimolecular reaction A+B→ProductsA + B \rightarrow \text{Products}, it is denoted as ZABZ_{AB}.

📐Formulae

Rate=ZAB⋅e−EaRT\text{Rate} = Z_{AB} \cdot e^{-\frac{E_a}{RT}}

Rate=P⋅ZAB⋅e−EaRT\text{Rate} = P \cdot Z_{AB} \cdot e^{-\frac{E_a}{RT}}

Ea=Ethreshold−EreactantsE_a = E_{\text{threshold}} - E_{\text{reactants}}

k=A⋅e−EaRTk = A \cdot e^{-\frac{E_a}{RT}}

ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}

log⁡10k2k1=Ea2.303R[T2−T1T1T2]\log_{10} \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]

💡Examples

Problem 1:

For a reaction, the rate constant kk is 3.2×10−5 s−13.2 \times 10^{-5} \, \text{s}^{-1} at 300 K300 \, \text{K} and 1.6×10−4 s−11.6 \times 10^{-4} \, \text{s}^{-1} at 310 K310 \, \text{K}. Calculate the Activation Energy (EaE_a). (Given R=8.314 J K−1mol−1R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1})

Solution:

We use the logarithmic form of the Arrhenius equation: log⁡k2k1=Ea2.303R[T2−T1T1T2]\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} [\frac{T_2 - T_1}{T_1 T_2}]. Substituting the values: log⁡1.6×10−43.2×10−5=Ea2.303×8.314[310−300310×300]\log \frac{1.6 \times 10^{-4}}{3.2 \times 10^{-5}} = \frac{E_a}{2.303 \times 8.314} [\frac{310 - 300}{310 \times 300}]. This simplifies to log⁡5=Ea19.147[1093000]\log 5 = \frac{E_a}{19.147} [\frac{10}{93000}]. Given log⁡5≈0.699\log 5 \approx 0.699, we get 0.699=Ea×1017806710.699 = \frac{E_a \times 10}{1780671}, resulting in Ea≈124468 J/molE_a \approx 124468 \, \text{J/mol} or 124.47 kJ/mol124.47 \, \text{kJ/mol}.

Explanation:

This example demonstrates how temperature changes affect the rate constant according to collision theory principles, specifically quantifying the energy barrier (EaE_a).

Problem 2:

Explain why the reaction between CH3BrCH_3Br and OH−OH^- to form CH3OHCH_3OH depends on the orientation of the molecules.

Solution:

In the reaction CH3Br+OH−→CH3OH+Br−CH_3Br + OH^- \rightarrow CH_3OH + Br^-, the OH−OH^- ion must collide with the carbon atom from the side opposite to the bromine atom. If the OH−OH^- hits the bromine side, the negatively charged bromine repels it, and no reaction occurs. This is represented by the steric factor P<1P < 1.

Explanation:

This illustrates the 'Orientation Barrier' of Collision Theory. Even if the kinetic energy is high, improper alignment prevents bond formation.

Concept of collision theory - Revision Notes & Key Formulas | ICSE Class 12 Chemistry