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Chemical Kinetics - Activation energy

Grade 12ICSEChemistry

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

🔑Concepts

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Activation Energy (EaE_a): The minimum amount of extra energy required by a reactant molecule to get converted into a product. It is the difference between the threshold energy and the average kinetic energy of the reactant molecules.

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Threshold Energy (ETE_T): The minimum energy which the colliding molecules must possess for the collision to be effective. ET=Ereactants+EaE_T = E_{reactants} + E_a.

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Arrhenius Equation: It expresses the quantitative relationship between the rate constant (kk) and the temperature (TT) in Kelvin: k=Ae−Ea/RTk = A e^{-E_a/RT}, where AA is the pre-exponential factor (frequency factor).

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Temperature Coefficient: For most chemical reactions, the rate constant nearly doubles or triples for every 10∘C10^{\circ}C rise in temperature. This is because the fraction of molecules with energy ≥Ea\ge E_a increases significantly.

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Effect of Catalyst: A catalyst increases the rate of reaction by providing an alternative pathway with a lower activation energy (EaE_a). It does not affect the enthalpy change (ΔH\Delta H) of the reaction.

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Activated Complex: An intermediate, unstable configuration of atoms formed during a collision at the peak of the energy barrier before products are formed.

📐Formulae

k=Ae−EaRTk = A e^{-\frac{E_a}{RT}}

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

log⁡10k=log⁡10A−Ea2.303RT\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}

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

💡Examples

Problem 1:

The rate constant of a reaction at 500 K500 \text{ K} and 700 K700 \text{ K} are 0.02 s−10.02 \text{ s}^{-1} and 0.07 s−10.07 \text{ s}^{-1} respectively. Calculate the value of activation energy (EaE_a). (Given: R=8.314 J mol−1 K−1R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1})

Solution:

Using the Arrhenius equation: log⁡k2k1=Ea2.303R[T2−T1T1T2]\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]. Substituting the values: log⁡0.070.02=Ea2.303×8.314[700−500700×500]\log \frac{0.07}{0.02} = \frac{E_a}{2.303 \times 8.314} \left[ \frac{700 - 500}{700 \times 500} \right]. log⁡3.5=Ea19.147[200350000]\log 3.5 = \frac{E_a}{19.147} \left[ \frac{200}{350000} \right]. 0.544=Ea×2.97×10−50.544 = E_a \times 2.97 \times 10^{-5}. Ea=0.5442.97×10−5≈18316 J/molE_a = \frac{0.544}{2.97 \times 10^{-5}} \approx 18316 \text{ J/mol} or 18.32 kJ/mol18.32 \text{ kJ/mol}.

Explanation:

The ratio of rate constants at two different temperatures is used to calculate the activation energy via the integrated form of the Arrhenius equation. Ensure TT is in Kelvin and RR matches the units of EaE_a.

Problem 2:

What happens to the rate of reaction when the activation energy is decreased by 10 kJ/mol10 \text{ kJ/mol} at 300 K300 \text{ K}?

Solution:

Let the original rate be k1=Ae−Ea1/RTk_1 = A e^{-E_{a1}/RT} and new rate be k2=Ae−(Ea1−10000)/RTk_2 = A e^{-(E_{a1} - 10000)/RT}. The ratio k2k1=e10000/(8.314×300)=e4.008≈55\frac{k_2}{k_1} = e^{10000 / (8.314 \times 300)} = e^{4.008} \approx 55.

Explanation:

Decreasing the activation energy increases the rate exponentially. A small decrease in EaE_a leads to a significantly higher fraction of molecules possessing the required energy to react.

Activation energy - Revision Notes & Key Formulas | ICSE Class 12 Chemistry