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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.