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Chemical Kinetics - Factors Influencing Rate of a Reaction

Grade 12CBSEChemistry

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

🔑Concepts

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Concentration of Reactants: Increasing the concentration of reactants increases the frequency of collisions, thereby increasing the reaction rate. This is quantitatively expressed by the Rate Law: r=k[A]x[B]yr = k[A]^x[B]^y.

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Temperature: For most reactions, the rate increases with an increase in temperature. A general rule of thumb is that for every 10∘C10^{\circ}C rise in temperature, the rate constant kk nearly doubles.

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Nature of Reactants: The physical state (solid, liquid, gas), surface area, and chemical identity of reactants influence the rate. For solids, a larger surface area (finely divided powder) leads to a faster reaction.

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Catalyst: A catalyst increases the rate of reaction by providing an alternative pathway with a lower Activation Energy (EaE_a). It does not change the equilibrium constant (ΔG\Delta G) or the enthalpy of the reaction (ΔH\Delta H).

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Activation Energy (EaE_a): This is the minimum extra energy required by a reactant molecule to get converted into a product. According to collision theory, only collisions with energy ≥Ea\ge E_a and proper orientation result in a reaction.

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Arrhenius Equation: It quantifies the effect of temperature on the rate constant: k=Ae−Ea/RTk = A e^{-E_a / RT}, where AA is the frequency factor and RR is the gas constant (8.314 J mol−1 K−18.314 \, J \, mol^{-1} \, K^{-1}).

📐Formulae

r=k[A]x[B]yr = k[A]^x [B]^y

k=Ae−EaRTk = A e^{-\frac{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]

k=PZABe−EaRTk = P Z_{AB} e^{-\frac{E_a}{RT}}

💡Examples

Problem 1:

The rate constant of a reaction is 1.2×10−3 s−11.2 \times 10^{-3} \, s^{-1} at 300 K300 \, K and 2.4×10−3 s−12.4 \times 10^{-3} \, s^{-1} at 310 K310 \, K. Calculate the activation energy (EaE_a) for the reaction. (Use R=8.314 J K−1 mol−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⁡2.4×10−31.2×10−3=Ea2.303×8.314[310−300310×300]\log \frac{2.4 \times 10^{-3}}{1.2 \times 10^{-3}} = \frac{E_a}{2.303 \times 8.314} [\frac{310 - 300}{310 \times 300}] log⁡2=Ea19.147[1093000]\log 2 = \frac{E_a}{19.147} [\frac{10}{93000}] 0.3010=Ea×5.61×10−60.3010 = E_a \times 5.61 \times 10^{-6} Ea=0.30105.61×10−6≈53597 J mol−1=53.6 kJ mol−1E_a = \frac{0.3010}{5.61 \times 10^{-6}} \approx 53597 \, J \, mol^{-1} = 53.6 \, kJ \, mol^{-1}

Explanation:

The Arrhenius equation in its logarithmic form is used to relate the change in rate constant with temperature to the activation energy.

Problem 2:

What is the effect of adding a catalyst on (i) Activation energy and (ii) Gibbs free energy (ΔG\Delta G) of a reaction?

Solution:

(i) Activation energy (EaE_a) decreases. (ii) Gibbs free energy (ΔG\Delta G) remains unchanged.

Explanation:

A catalyst provides an alternative reaction mechanism with a lower EaE_a barrier. Since ΔG\Delta G is a state function depending only on the initial and final states of the reactants and products, it is not affected by the path taken or the catalyst.