Chemical Kinetics
Complete Revision Guide for CBSE Board Exams
1. Introduction to Chemical Kinetics
Thermodynamics tells us whether a chemical reaction is feasible (spontaneous) or not based on $\Delta G$. However, it doesn't say anything about the speed or rate of the reaction. Chemical Kinetics is the branch of chemistry that deals with the study of reaction rates, the factors affecting them, and the mechanism by which the reactions proceed.
Reactions can be classified on the basis of rates into three categories:
- Very Fast Reactions: Occur almost instantaneously (e.g., ionic reactions like precipitation of AgCl from $AgNO_3$ and NaCl).
- Very Slow Reactions: Take days, months, or years to show measurable change (e.g., rusting of iron).
- Moderate Reactions: Proceed at measurable rates at room temperature (e.g., hydrolysis of sucrose, decomposition of $N_2O_5$). Kinetics mainly focuses on these.
2. Master Formula Cheat Sheet
A. Rate Law and Rate Constant
- Rate of Reaction (for $aA + bB \rightarrow cC + dD$):
$$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$$
- Rate Law Expression: For a general reaction, $\text{Rate} = k[A]^x[B]^y$ where $x+y$ = Overall Order of Reaction ($n$).
- Units of Rate Constant ($k$): The unit depends on the order of the reaction ($n$).
$$\text{Unit of } k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$$For Zero Order (n=0): $\text{mol L}^{-1} \text{s}^{-1}$
For First Order (n=1): $\text{s}^{-1}$
For Second Order (n=2): $\text{L mol}^{-1} \text{s}^{-1}$
B. Integrated Rate Equations & Half-Life
- Zero Order Reaction ($R \rightarrow P$):
$$[R] = [R]_0 - kt \quad \implies \quad k = \frac{[R]_0 - [R]}{t}$$Half-Life ($t_{1/2}$): $$t_{1/2} = \frac{[R]_0}{2k}$$ (Half-life is directly proportional to initial concentration).
- First Order Reaction ($R \rightarrow P$):
$$k = \frac{2.303}{t} \log_{10} \frac{[R]_0}{[R]}$$Half-Life ($t_{1/2}$): $$t_{1/2} = \frac{0.693}{k}$$ (Half-life is strictly independent of initial concentration).
C. Temperature Dependence (Arrhenius Equation)
- Arrhenius Equation: $$k = A e^{-E_a/RT}$$ (Where $k$ = Rate constant, $A$ = Arrhenius frequency factor, $E_a$ = Activation energy, $R$ = Gas constant $8.314 \text{ J K}^{-1} \text{ mol}^{-1}$, $T$ = Temperature in Kelvin).
- Logarithmic Form:
$$\ln k = \ln A - \frac{E_a}{RT} \quad \implies \quad \log k = \log A - \frac{E_a}{2.303 RT}$$
- Comparison at Two Temperatures ($T_1$ and $T_2$):
$$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$$
3. Exhaustive Theoretical Concepts
3.1 Rate of a Chemical Reaction
The rate of a reaction is defined as the change in concentration of any one of the reactants or products per unit time.
- Average Rate: Rate calculated over a large, measurable time interval ($\Delta t$). $\text{Rate}_{avg} = -\frac{\Delta [R]}{\Delta t} = +\frac{\Delta [P]}{\Delta t}$
- Instantaneous Rate: Rate measured at any particular instant of time (when $\Delta t \rightarrow 0$). $\text{Rate}_{inst} = -\frac{d[R]}{dt} = +\frac{d[P]}{dt}$
The negative sign indicates a decrease in the concentration of reactants as time progresses.
3.2 Rate Law, Order, and Molecularity
Rate Law
Rate Law is the expression in which the reaction rate is given in terms of molar concentrations of reactants, with each term raised to some power, which may or may not be identical to the stoichiometric coefficient of the reacting species in a balanced chemical equation.
Crucial Point: The rate law cannot be predicted merely by looking at the balanced equation; it must be determined experimentally.
| Order of Reaction | Molecularity of Reaction |
|---|---|
| The sum of the powers of the concentration of the reactants in the rate law expression. | The number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction, which must collide simultaneously to bring about a chemical reaction. |
| It is an experimental quantity. | It is a theoretical concept. |
| It can be zero, a fraction, or an integer. | It is always a whole number. It can never be zero or a fractional number. |
| Applicable to both elementary and complex reactions. | Applicable only to elementary reactions. For complex reactions, molecularity has no meaning. |
3.3 Pseudo First Order Reactions
A reaction that is not strictly of the first order but behaves as a first-order reaction under certain specific conditions (usually when one reactant is present in large excess) is called a pseudo first-order reaction.
Example: Inversion of Cane Sugar (Hydrolysis of Sucrose)
$$C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 (\text{Glucose}) + C_6H_{12}O_6 (\text{Fructose})$$
The theoretical rate law would be $\text{Rate} = k[C_{12}H_{22}O_{11}][H_2O]$. However, water is present in such a large excess that its concentration remains practically constant during the reaction. Thus, the effective rate law becomes $\text{Rate} = k'[C_{12}H_{22}O_{11}]$. The reaction is artificially reduced to first-order kinetics.
3.4 Temperature Dependence of Reaction Rate
For a typical chemical reaction, the rate constant is observed to nearly double (or triple) for every $10^\circ C$ rise in temperature. This is defined by the Temperature Coefficient.
Activation Energy ($E_a$)
According to Arrhenius, a chemical reaction occurs through collisions between reactant molecules. However, only those collisions result in a chemical reaction where the colliding molecules possess a certain minimum energy. This minimum extra energy required by reacting molecules to cross the energy barrier and form the transition state (activated complex) is called Activation Energy.
Effect of a Catalyst
A catalyst is a substance that alters the rate of a reaction without itself undergoing any permanent chemical change. It operates by providing an alternate pathway or reaction mechanism with a lower Activation Energy between reactants and products, thus increasing the rate of reaction.
- A catalyst does not alter the Gibbs Free Energy ($\Delta G$) of the reaction.
- It does not change the equilibrium constant ($K_c$); it only helps to attain equilibrium faster.
- It catalyzes both the forward and reverse reactions to the exact same extent.
3.5 Collision Theory of Chemical Kinetics
This theory provides a mechanical picture of reaction rates. For a reaction to occur:
- Energy Criterion: Colliding molecules must possess energy equal to or greater than the Threshold Energy ($E_{threshold} = E_{activation} + E_{reactants}$).
- Orientation Criterion: The molecules must collide with the proper spatial orientation to facilitate the breaking of old bonds and formation of new ones.
Thus, the rate of reaction is given by $\text{Rate} = Z_{AB} e^{-E_a/RT} \times P$
Where $Z_{AB}$ is the collision frequency, $e^{-E_a/RT}$ is the fraction of molecules with energy $\ge E_a$, and $P$ is the probability or steric factor (which accounts for the orientation constraint).
4. Most Important CBSE Board Questions
Curated strictly according to the latest CBSE sample papers and marking schemes.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. The unit of rate constant for a zero order reaction is: 1 Mark
- $\text{s}^{-1}$
- $\text{mol L}^{-1} \text{s}^{-1}$
- $\text{L mol}^{-1} \text{s}^{-1}$
- $\text{L}^2 \text{mol}^{-2} \text{s}^{-1}$
Explanation: For zero order, Rate = $k[A]^0$. Therefore, the unit of $k$ is exactly the same as the unit of Rate, which is concentration/time ($\text{mol L}^{-1} \text{s}^{-1}$).
Q2. For a chemical reaction, a graph plotted between $\ln k$ versus $1/T$ is a straight line. What is the slope of this line? 1 Mark
- $-E_a/R$
- $E_a/R$
- $-E_a/2.303R$
- $R/E_a$
Explanation: From the Arrhenius equation, $\ln k = \ln A - \frac{E_a}{RT}$. This is in the form of a straight line equation $y = mx + c$, where $y = \ln k$ and $x = 1/T$. The slope $m = -E_a/R$. Note: If the plot was $\log k$ vs $1/T$, the slope would be $-E_a/2.303R$.
Assertion-Reasoning
Q3. Assertion (A): Order of a reaction can be zero or fractional.
Reason (R): We cannot determine order from balanced chemical equation. 1 Mark
Explanation: Order is an experimental quantity and can indeed be zero or fractional (A is true). Order cannot be predicted from a balanced chemical equation; it must be found via experimental rate laws (R is true). However, the reason why it can be fractional is tied to the mechanism of complex reactions, not merely because it can't be read from the equation.
Section B: 2 Mark Questions (Very Short Answer)
Q4. A first-order reaction is found to have a rate constant, $k = 5.5 \times 10^{-14} \text{ s}^{-1}$. Find the half-life of the reaction. 2 Marks
$$t_{1/2} = \frac{0.693}{k}$$
Given $k = 5.5 \times 10^{-14} \text{ s}^{-1}$.
$$t_{1/2} = \frac{0.693}{5.5 \times 10^{-14} \text{ s}^{-1}}$$ $$t_{1/2} = 0.126 \times 10^{14} \text{ s} = \mathbf{1.26 \times 10^{13} \text{ s}}$$
Q5. Define pseudo first-order reaction with an example. 2 Marks
Example: Acid-catalyzed hydrolysis of ethyl acetate.
$CH_3COOC_2H_5 + H_2O \xrightarrow{H^+} CH_3COOH + C_2H_5OH$
Rate = $k' [CH_3COOC_2H_5]$. Water is the solvent and is present in immense excess, so its concentration is omitted from the rate law.
Section C: 3 Mark Questions (Short Answer)
Q6. Show that in a first-order reaction, the time required for completion of 99.9% of the reaction is 10 times the half-life ($t_{1/2}$) of the reaction. 3 Marks
Step 1: Calculate $t_{99.9\%}$
Amount reacted = 99.9. Amount left, $[R] = 100 - 99.9 = 0.1$.
For a 1st order reaction: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$
$t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log 1000$
$t_{99.9\%} = \frac{2.303}{k} \times 3 \quad \text{--- (Equation 1)}$
Step 2: Express $t_{1/2}$
We know that $t_{1/2} = \frac{0.693}{k}$. Note that $0.693 = 2.303 \times 0.3010$ (which is $2.303 \times \log 2$).
So, $t_{1/2} = \frac{2.303 \times 0.3010}{k} \quad \text{--- (Equation 2)}$
Step 3: Compare both
Divide Eq 1 by Eq 2:
$$\frac{t_{99.9\%}}{t_{1/2}} = \frac{3}{0.3010} \approx 9.96 \approx 10$$ Therefore, $t_{99.9\%} \approx 10 \times t_{1/2}$. (Proved).
Q7. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. ($R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$) 3 Marks
$T_1 = 293\text{ K}$, $T_2 = 313\text{ K}$
Rate quadruples, meaning $k_2 = 4k_1$, therefore $\frac{k_2}{k_1} = 4$.
$R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$
Using the Arrhenius Equation:
$$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$$
$$\log(4) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{313 - 293}{293 \times 313} \right]$$
$$0.6021 = \frac{E_a}{19.147} \left[ \frac{20}{91709} \right]$$
$$E_a = \frac{0.6021 \times 19.147 \times 91709}{20}$$
$$E_a = \frac{1057303.4}{20}$$
$$E_a = 52865.17\text{ J mol}^{-1} = \mathbf{52.86\text{ kJ mol}^{-1}}$$.
Section D: 4 Mark Case-Based Questions
Q8. Read the passage given below and answer the following questions:
(i) How does the addition of a catalyst affect the activation energy and $\Delta G$ of a reaction? (1 Mark)
(ii) What does the term $e^{-E_a/RT}$ in the Arrhenius equation represent? (1 Mark)
(iii) The activation energy for a reaction is zero. If the rate constant at 280 K is $1.6 \times 10^6 \text{ s}^{-1}$, what will be the rate constant at 300 K? Show working. (2 Marks)
(ii) The term $e^{-E_a/RT}$ represents the fraction of total molecules that possess kinetic energy equal to or greater than the activation energy ($E_a$). These are the molecules capable of undergoing effective collisions.
(iii) According to the Arrhenius relation for two temperatures:
$$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$$
Given $E_a = 0$.
$$\log \frac{k_2}{k_1} = \frac{0}{2.303 R} \times \left[ \frac{300 - 280}{300 \times 280} \right] = 0$$
If $\log \frac{k_2}{k_1} = 0$, then taking antilog gives $\frac{k_2}{k_1} = 10^0 = 1$.
Therefore, $k_2 = k_1$.
The rate constant at 300 K will remain unchanged at $\mathbf{1.6 \times 10^6 \text{ s}^{-1}}$. (A reaction with zero activation energy is independent of temperature).
Section E: 5 Mark Long Answer Questions
Q9. (a) For the reaction $2A + B \rightarrow A_2B$, the rate = $k[A][B]^2$ with $k = 2.0 \times 10^{-6}\text{ mol}^{-2}\text{L}^2\text{s}^{-1}$. Calculate the initial rate of the reaction when $[A] = 0.1\text{ mol L}^{-1}$, $[B] = 0.2\text{ mol L}^{-1}$. Calculate the rate of reaction after $[A]$ is reduced to $0.06\text{ mol L}^{-1}$. (3 Marks)
(b) State the two main postulates of the collision theory of chemical kinetics. (2 Marks)
Step 1: Initial Rate Calculation
Rate = $k[A][B]^2$
Rate $= (2.0 \times 10^{-6}) \times (0.1) \times (0.2)^2$
Rate $= (2.0 \times 10^{-6}) \times 0.1 \times 0.04 = \mathbf{8.0 \times 10^{-9}\text{ mol L}^{-1}\text{ s}^{-1}}$.
Step 2: Rate after [A] is reduced to 0.06 mol L⁻¹
Initial moles of A = 0.1 M. Final moles of A = 0.06 M.
Amount of A reacted = $0.1 - 0.06 = 0.04\text{ M}$.
From stoichiometry, 2 moles of A react with 1 mole of B.
Therefore, amount of B reacted = $\frac{1}{2} \times (\text{Amount of A reacted}) = \frac{1}{2} \times 0.04 = 0.02\text{ M}$.
Amount of B remaining in the solution = Initial B - Reacted B = $0.2 - 0.02 = \mathbf{0.18\text{ M}}$.
New rate = $k[\text{New A}][\text{New B}]^2$
New rate = $(2.0 \times 10^{-6}) \times (0.06) \times (0.18)^2$
New rate = $(2.0 \times 10^{-6}) \times 0.06 \times 0.0324$
New rate = $\mathbf{3.89 \times 10^{-9}\text{ mol L}^{-1}\text{ s}^{-1}}$.
(b) Collision Theory Postulates:
1. Energy Factor (Threshold Energy): For a collision to be effective (i.e., result in a chemical reaction), the colliding molecules must possess a minimum amount of energy called Threshold Energy ($E_a + E_R$). If they collide with less energy, they simply bounce off each other.
2. Orientation Factor (Steric Factor): Even if molecules possess sufficient energy, a reaction will only happen if they collide with the proper spatial orientation so that old bonds can break and new bonds can form efficiently.
Mastering 'Chemical Kinetics'
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Spot the Unit Trap: Often in MCQ questions, you are asked to identify the order of the reaction just by looking at the unit of the rate constant '$k$'. Memorize: $s^{-1}$ is 1st order, $mol\ L^{-1} s^{-1}$ is 0th order, and $L\ mol^{-1} s^{-1}$ is 2nd order.
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The $t_{99.9\%}$ Proof: The 3-mark question proving $t_{99.9\%} = 10 \times t_{1/2}$ is one of the most frequently repeated questions in CBSE history. Practice deriving it exactly as shown above.
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Graphs are Critical: Understand the $ln(k)$ vs $1/T$ (Arrhenius) graph, and the Concentration vs Time graphs for Zero and First-order reactions. You might be asked to identify the order from the slope.
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Stoichiometry in Rates: Remember that rate of disappearance/appearance must be divided by the stoichiometric coefficient (e.g., $-\frac{1}{2}\frac{d[N_2O_5]}{dt}$). Students often forget the fraction!
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