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Chemical Kinetics - Factors affecting rate of reaction

Grade 12ICSEChemistry

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

πŸ”‘Concepts

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The rate of a chemical reaction is influenced by several factors including the concentration of reactants, temperature, surface area, and the presence of a catalyst.

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Nature of Reactants: Ionic reactions are generally faster than molecular reactions because they involve simple ion combinations rather than the breaking and forming of covalent bonds. For example, the precipitation of AgClAgCl from AgNO3AgNO_3 and NaClNaCl is almost instantaneous.

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Concentration of Reactants: According to the Collision Theory, an increase in concentration leads to an increase in the frequency of effective collisions per unit volume, thereby increasing the rate. This is expressed by the Rate Law: Rate=k[A]x[B]yRate = k[A]^x[B]^y.

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Surface Area: In heterogeneous reactions involving solids, the rate increases with an increase in surface area. Powdered reactants react faster than large chunks (e.g., powdered CaCO3CaCO_3 reacts faster with HClHCl than marble chips).

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Temperature: Generally, the rate of reaction increases with temperature. For most reactions, the rate constant doubles or triples for every 10∘C10^{\circ}C rise in temperature. This is quantified by the Temperature Coefficient: η=kT+10kT\eta = \frac{k_{T+10}}{k_T}.

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Activation Energy (EaE_a): It is the minimum extra energy required by reactant molecules to undergo a successful collision. A catalyst increases the reaction rate by providing an alternative pathway with a lower EaE_a.

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Effect of Radiation: Some reactions, known as photochemical reactions, occur only in the presence of light (photons). For example, the reaction between H2H_2 and Cl2Cl_2 to form HClHCl is significantly accelerated by hΞ½h\nu (ultraviolet light).

πŸ“Formulae

Rate=k[A]x[B]yRate = k[A]^x[B]^y

TemperatureΒ Coefficient(Ξ·)=kT+10kTβ‰ˆ2Β toΒ 3\text{Temperature Coefficient} (\eta) = \frac{k_{T+10}}{k_T} \approx 2 \text{ to } 3

k=Aeβˆ’Ea/RTk = A e^{-E_a/RT}

ln⁑k=ln⁑Aβˆ’EaRT\ln k = \ln A - \frac{E_a}{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’3sβˆ’11.2 \times 10^{-3} s^{-1} at 300K300 K. When the temperature is increased to 310K310 K, the rate constant doubles. Calculate the activation energy (EaE_a) of the reaction. (Given: R=8.314JKβˆ’1molβˆ’1R = 8.314 J K^{-1} mol^{-1}, log⁑2=0.3010\log 2 = 0.3010)

Solution:

Given T1=300KT_1 = 300 K, T2=310KT_2 = 310 K, k2=2k1k_2 = 2k_1. Using the Arrhenius equation: 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) log⁑2=Ea2.303Γ—8.314(310βˆ’300310Γ—300)\log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{310 - 300}{310 \times 300} \right) 0.3010=Ea19.147Γ—10930000.3010 = \frac{E_a}{19.147} \times \frac{10}{93000} Ea=0.3010Γ—19.147Γ—9300010β‰ˆ53598J/mol=53.6kJ/molE_a = \frac{0.3010 \times 19.147 \times 93000}{10} \approx 53598 J/mol = 53.6 kJ/mol

Explanation:

The problem applies the integrated form of the Arrhenius equation to find the energy barrier molecules must overcome. Since the rate doubles with a 10K10 K rise, the calculation determines the EaE_a responsible for this sensitivity.

Problem 2:

How does the addition of a catalyst affect the Gibbs free energy (Ξ”G\Delta G) and the enthalpy (Ξ”H\Delta H) of a reaction?

Solution:

A catalyst has no effect on the thermodynamic parameters like Ξ”G\Delta G or Ξ”H\Delta H. It only lowers the activation energy (EaE_a) for both the forward and backward reactions.

Explanation:

Catalysts change the kinetics (speed) of the reaction by providing a different mechanism, but they do not change the initial and final energy states of the reactants and products, meaning the equilibrium constant and enthalpy remain unchanged.

Factors affecting rate of reaction Revision - Class 12 Chemistry ICSE