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Chemical Kinetics - Temperature Dependence of the Rate of a Reaction (Arrhenius Equation)

Grade 12CBSEChemistry

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

🔑Concepts

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Most chemical reactions accelerate as temperature increases. For many reactions, the rate constant (kk) nearly doubles with a 10 K10\ K rise in temperature.

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The Arrhenius Equation provides a mathematical relationship between the rate constant (kk), activation energy (EaE_a), and absolute temperature (TT): k=Ae−Ea/RTk = A e^{-E_a/RT}.

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The term AA is the Arrhenius factor or frequency factor, representing the frequency of collisions. It is constant for a specific reaction.

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The term e−Ea/RTe^{-E_a/RT} represents the fraction of molecules that have kinetic energy greater than or equal to the activation energy (EaE_a).

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Activation Energy (EaE_a) is the minimum energy required by reactant molecules to form an activated complex and subsequently lead to product formation.

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The transition state or activated complex is an unstable intermediate state formed during the conversion of reactants to products, where bonds are in the process of breaking and forming.

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A plot of ln⁡k\ln k against 1/T1/T yields a straight line with a slope of −EaR-\frac{E_a}{R} and an intercept of ln⁡A\ln A.

📐Formulae

k=Ae−Ea/RTk = A e^{-E_a/RT}

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

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

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)

Slope=−Ea2.303R (for log⁡k vs 1/T graph)Slope = -\frac{E_a}{2.303 R} \text{ (for } \log k \text{ vs } 1/T \text{ graph)}

💡Examples

Problem 1:

The rate constants of a reaction at 500 K500\ K and 700 K700\ K are 0.02 s−10.02\ s^{-1} and 0.07 s−10.07\ s^{-1} respectively. Calculate the values of EaE_a and AA. (R=8.314 J K−1mol−1R = 8.314\ J\ K^{-1} mol^{-1})

Solution:

Using the formula: 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⁡0.070.02=Ea2.303×8.314[700−500700×500]\log \frac{0.07}{0.02} = \frac{E_a}{2.303 \times 8.314} [\frac{700 - 500}{700 \times 500}] log⁡3.5=Ea19.147[200350000]\log 3.5 = \frac{E_a}{19.147} [\frac{200}{350000}] 0.544=Ea19.147×0.0005710.544 = \frac{E_a}{19.147} \times 0.000571 Ea=0.544×19.1470.000571≈18230.8 J mol−1=18.23 kJ mol−1E_a = \frac{0.544 \times 19.147}{0.000571} \approx 18230.8\ J\ mol^{-1} = 18.23\ kJ\ mol^{-1}. To find AA, use log⁡k=log⁡A−Ea2.303RT\log k = \log A - \frac{E_a}{2.303 RT} at 500 K500\ K: log⁡0.02=log⁡A−18230.82.303×8.314×500\log 0.02 = \log A - \frac{18230.8}{2.303 \times 8.314 \times 500} −1.699=log⁡A−1.904-1.699 = \log A - 1.904 log⁡A=0.205  ⟹  A=100.205≈1.60 s−1\log A = 0.205 \implies A = 10^{0.205} \approx 1.60\ s^{-1}.

Explanation:

We use the two-temperature form of the Arrhenius equation to solve for the unknown activation energy first, and then substitute back into the logarithmic form to find the frequency factor AA.

Temperature Dependence of the Rate of a Reaction (Arrhenius Equation) Revision - Class 12…