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Unit 3: Chemical Kinetics

3.1 Rate of a Chemical Reaction

Chemical kinetics is the branch of chemistry that deals with the study of reaction rates and their mechanisms.

The rate of a reaction is the change in the concentration of any one of the reactants or products per unit time. For a reaction \( R \rightarrow P \):

  • Average rate of reaction \( = \frac{-\Delta [R]}{\Delta t} = \frac{+\Delta [P]}{\Delta t} \)
  • Instantaneous rate of reaction \( = \frac{-d[R]}{dt} = \frac{+d[P]}{dt} \)

Factors influencing rate of a reaction:

  • Concentration of reactants
  • Temperature
  • Nature of reactants and products
  • Presence of a catalyst
  • Surface area of reactants (for heterogeneous reactions)
  • Exposure to radiation (for photochemical reactions)

3.2 Rate Expression and Rate Constant

The experimental expression of the rate of reaction in terms of the concentration of reactants is known as the rate law or rate expression. For a general reaction: \( aA + bB \rightarrow cC + dD \) \[ \text{Rate} = k [A]^x [B]^y \] where \( x \) and \( y \) are determined experimentally. \( k \) is the rate constant or specific reaction rate.

3.3 Order and Molecularity of a Reaction

Order of a Reaction: The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. Let the rate law be Rate = \( k[A]^x[B]^y \). Then, Order = \( x + y \). Order can be 0, 1, 2, 3 and even a fraction.

Molecularity: The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. Molecularity is always a whole number (1, 2, 3…).

3.4 Integrated Rate Equations

Zero Order Reactions

The rate of the reaction is independent of the concentration of reactants. \[ \text{Rate} = \frac{-d[R]}{dt} = k [R]^0 = k \] Integrated form: \[ [R] = -kt + [R]_0 \] where \( [R]_0 \) is the initial concentration and \( [R] \) is the concentration at time \( t \).

Time (t) [R] Slope = -k Zero Order Reaction [R]₀

First Order Reactions

The rate of the reaction is proportional to the first power of the concentration of the reactant \( R \). \[ \text{Rate} = \frac{-d[R]}{dt} = k [R] \] Integrated form: \[ \ln [R] = -kt + \ln [R]_0 \] Or using base 10 logarithm: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \]

Time (t) ln [R] Slope = -k First Order Reaction

3.5 Half-Life of a Reaction (\( t_{1/2} \))

The time in which the concentration of a reactant is reduced to one half of its initial concentration.

  • For a zero order reaction: \( t_{1/2} = \frac{[R]_0}{2k} \)
  • For a first order reaction: \( t_{1/2} = \frac{0.693}{k} \) (Independent of initial concentration)

3.6 Temperature Dependence of the Rate of a Reaction

Arrhenius Equation describes the effect of temperature on the rate constant (\( k \)) of a reaction: \[ k = A e^{-E_a/RT} \] Taking natural logarithm on both sides: \[ \ln k = -\frac{E_a}{RT} + \ln A \] Taking log base 10: \[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]

where:

  • \( A \) is the Arrhenius factor or frequency factor or pre-exponential factor.
  • \( R \) is gas constant.
  • \( E_a \) is activation energy in joules/mole.

Competency-Based Questions (CBQs)

Q1. (CBSE 2023) The decomposition of \( N_2O_5 \) in \( CCl_4 \) at 318 K has been studied by monitoring the concentration of \( N_2O_5 \) in the solution. Initially, the concentration of \( N_2O_5 \) is 2.33 mol/L and after 184 minutes, it is reduced to 2.08 mol/L. Give the sequence of steps to determine the average rate of this reaction in terms of hours, minutes and seconds.


Answer: Average rate \( = -\frac{\Delta [N_2O_5]}{\Delta t} \) \[ \Delta [N_2O_5] = [\text{Final}] - [\text{Initial}] = 2.08 - 2.33 = -0.25 , \text{mol/L} \] Rate in minutes: \[ \text{Time } \Delta t = 184 \text{ min} \] \[ \text{Rate} = - \frac{-0.25}{184} = 1.36 \times 10^{-3} , \text{mol L}^{-1} \text{min}^{-1} \]

Rate in hours: \[ \Delta t = \frac{184}{60} = 3.067 \text{ hrs} \] \[ \text{Rate} = - \frac{-0.25}{3.067} = 8.15 \times 10^{-2} , \text{mol L}^{-1} \text{h}^{-1} \]

Rate in seconds: \[ \Delta t = 184 \times 60 = 11040 \text{ sec} \] \[ \text{Rate} = - \frac{-0.25}{11040} = 2.26 \times 10^{-5} , \text{mol L}^{-1} \text{s}^{-1} \]

Q2. (Sample Paper 2024) A first order reaction takes 40 min for 30% decomposition. Calculate \( t_{1/2} \).


Answer: For a first order reaction, \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \] Here, let initial concentration \( [R]_0 = 100 \). Since 30% decomposes, the remaining concentration \( [R] = 100 - 30 = 70 \). Time \( t = 40 \) min. \[ k = \frac{2.303}{40} \log \frac{100}{70} \] \[ k = \frac{2.303}{40} \log 1.428 \] \[ k = \frac{2.303}{40} \times 0.1548 \] \[ k = 8.91 \times 10^{-3} , \text{min}^{-1} \] Now, calculate the half-life \( t_{1/2} \): \[ t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.91 \times 10^{-3}} = 77.7 , \text{min} \]

Q3. (CBSE 2021) The rate constant of a reaction increases by 5% when its temperature is raised from 27°C to 28°C. Calculate the activation energy of the reaction. (Given: \( R = 8.314 \) J K\(^{-1}\) mol\(^{-1}\))


Answer: Given:

  • \( T_1 = 27^\circ \text{C} = 300 , \text{K} \)
  • \( T_2 = 28^\circ \text{C} = 301 , \text{K} \)
  • Rate constant \( k_2 \) is 5% more than \( k_1 \), i.e., \( k_2 = k_1 + 0.05 k_1 = 1.05 k_1 \).
  • This means \( \frac{k_2}{k_1} = 1.05 \).

Using Arrhenius Equation: \[ \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] \] \[ \log (1.05) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{301 - 300}{300 \times 301} \right] \] \[ 0.0212 = \frac{E_a}{19.147} \left[ \frac{1}{90300} \right] \] \[ E_a = 0.0212 \times 19.147 \times 90300 \] \[ E_a = 36653 , \text{J/mol} \] \[ E_a = 36.65 , \text{kJ/mol} \]