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Chemical Kinetics - Arrhenius equation

Grade 12ICSEChemistry

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

πŸ”‘Concepts

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The Arrhenius equation expresses the dependence of the rate constant kk of a chemical reaction on the absolute temperature TT.

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Activation Energy (EaE_a): The minimum amount of energy required by reactant molecules to result in a chemical reaction. It is measured in Jβ‹…molβˆ’1J \cdot mol^{-1} or kJβ‹…molβˆ’1kJ \cdot mol^{-1}.

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Pre-exponential factor (AA): Also known as the frequency factor or Arrhenius factor. It represents the frequency of collisions with the correct orientation for a reaction to occur.

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The term eβˆ’Ea/RTe^{-E_a/RT} represents the fraction of molecules that possess kinetic energy greater than or equal to the activation energy EaE_a at temperature TT.

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Temperature Coefficient (Ξ·\eta): Usually defined as the ratio of rate constants at temperatures differing by 10 K10\,K, typically Ξ·=kT+10kTβ‰ˆ2Β toΒ 3\eta = \frac{k_{T+10}}{k_T} \approx 2 \text{ to } 3.

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A plot of log⁑k\log k versus 1T\frac{1}{T} yields a straight line with a slope of βˆ’Ea2.303R-\frac{E_a}{2.303 R} and a y-intercept of log⁑A\log A.

πŸ“Formulae

k=Aeβˆ’EaRTk = A e^{-\frac{E_a}{RT}}

ln⁑k=ln⁑Aβˆ’EaRT\ln k = \ln A - \frac{E_a}{RT}

log⁑10k=log⁑10Aβˆ’Ea2.303RT\log_{10} k = \log_{10} A - \frac{E_a}{2.303 RT}

log⁑10(k2k1)=Ea2.303R(T2βˆ’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)

πŸ’‘Examples

Problem 1:

The rate constant of a reaction is 1.2Γ—10βˆ’3 sβˆ’11.2 \times 10^{-3} \, s^{-1} at 300 K300\,K and 2.4Γ—10βˆ’3 sβˆ’12.4 \times 10^{-3} \, s^{-1} at 310 K310\,K. Calculate the activation energy EaE_a for the reaction. (Given R=8.314 J Kβˆ’1 molβˆ’1R = 8.314 \, J \, K^{-1} \, mol^{-1})

Solution:

Given: k1=1.2Γ—10βˆ’3 sβˆ’1k_1 = 1.2 \times 10^{-3} \, s^{-1}, k2=2.4Γ—10βˆ’3 sβˆ’1k_2 = 2.4 \times 10^{-3} \, s^{-1}, T1=300 KT_1 = 300\,K, T2=310 KT_2 = 310\,K. Using the formula: log⁑k2k1=Ea2.303R(T2βˆ’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 values: log⁑2.4Γ—10βˆ’31.2Γ—10βˆ’3=Ea2.303Γ—8.314(310βˆ’300300Γ—310)\log \frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = \frac{E_a}{2.303 \times 8.314} \left( \frac{310 - 300}{300 \times 310} \right). log⁑2=Ea19.147(1093000)\log 2 = \frac{E_a}{19.147} \left( \frac{10}{93000} \right). 0.3010=EaΓ—1017806710.3010 = \frac{E_a \times 10}{1780671}. Ea=0.3010Γ—178067110=53598.2 J molβˆ’1β‰ˆ53.6 kJ molβˆ’1E_a = \frac{0.3010 \times 1780671}{10} = 53598.2 \, J \, mol^{-1} \approx 53.6 \, kJ \, mol^{-1}.

Explanation:

The problem uses the two-temperature form of the Arrhenius equation to find the activation energy. Note that when the rate constant doubles for a 10 K10\,K rise, EaE_a is typically around 5050-60 kJ molβˆ’160 \, kJ \, mol^{-1}.

Problem 2:

For a reaction, the slope of the plot of log⁑k\log k vs 1T\frac{1}{T} is βˆ’8000 K-8000\,K. Calculate the activation energy EaE_a.

Solution:

We know that the slope of the log⁑k\log k vs 1T\frac{1}{T} graph is given by: Slope=βˆ’Ea2.303R\text{Slope} = -\frac{E_a}{2.303 R}. Given Slope=βˆ’8000 K\text{Slope} = -8000\,K. Therefore, βˆ’8000=βˆ’Ea2.303Γ—8.314-8000 = -\frac{E_a}{2.303 \times 8.314}. Ea=8000Γ—2.303Γ—8.314E_a = 8000 \times 2.303 \times 8.314. Ea=153177.3 J molβˆ’1β‰ˆ153.2 kJ molβˆ’1E_a = 153177.3 \, J \, mol^{-1} \approx 153.2 \, kJ \, mol^{-1}.

Explanation:

This example demonstrates how to extract the activation energy from graphical data using the linear form of the Arrhenius equation.

Arrhenius equation - Revision Notes & Key Formulas | ICSE Class 12 Chemistry