Class 12 Chemistry Unit 3: Chemical Kinetics | Notes + Important Questions

Contents

Score Target: 7/70 Marks

Topics: Rate Laws, Zero & First Order, Half-Life, Arrhenius Equation, and Collision Theory.

Welcome to the ultimate revision guide for CBSE Class 12 Chemistry. Chemical Kinetics is a physical chemistry unit that is highly scoring. Unlike Organic Chemistry, there are no exceptions here—just pure logic, formulas, and graphs. If you master the three major derivations and the Arrhenius calculations, the 7 marks are guaranteed.

1. Rate of Chemical Reaction

Chemical Kinetics deals with the speed (rate) of a reaction. The rate is defined as the change in concentration of a reactant or product per unit time.

$\text{Rate} = \frac{\Delta c}{\Delta t}$

A. Average vs. Instantaneous Rate

Average Rate ($r_{av}$): The rate measured over a measurable time interval.
For $R \to P$:
$r_{av} = -\frac{\Delta [R]}{\Delta t} = +\frac{\Delta [P]}{\Delta t}$
Instantaneous Rate ($r_{inst}$): The rate at a specific instant (when $\Delta t \to 0$).
$r_{inst} = -\frac{d[R]}{dt} = +\frac{d[P]}{dt}$

Why the Negative Sign?
The negative sign in $-\frac{d[R]}{dt}$ indicates that the concentration of Reactants decreases with time. Since rate cannot be negative, we multiply the negative slope by -1 to make the rate positive.

B. Relation with Stoichiometry

For a general reaction: $aA + bB \to cC + dD$

The rate of reaction is equated by dividing the rate of disappearance/appearance by the stoichiometric coefficient:

Rate $= -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}$

2. Factors Affecting Rate of Reaction

Remember the acronym S.C.T.C:

State (Surface Area): For solid reactants, greater surface area $\rightarrow$ faster reaction (e.g., powdered wood burns faster than logs).
Concentration: Higher concentration means more molecules per unit volume $\rightarrow$ more effective collisions $\rightarrow$ higher rate.
Temperature: Increasing temp increases kinetic energy. Roughly, for every 10° rise, the rate constant doubles.
Catalyst: It provides an alternate pathway with lower Activation Energy ($E_a$), increasing the rate without being consumed.

3. Rate Law, Order, and Molecularity

Rate Law

The representation of rate of reaction in terms of concentration of reactants is known as Rate Law. It is determined experimentally.

$\text{Rate} = k [A]^x [B]^y$

$k$: Specific Rate Constant (Rate when conc. is unity).
$x$ and $y$: Experimental powers (may or may not equal stoichiometry).

Order of Reaction ($n$)

The sum of powers of the concentration of the reactants in the rate law expression ($n = x + y$).

Molecularity

The number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction which must collide simultaneously.

Parameter	Order of Reaction	Molecularity
Definition	Sum of powers in rate law.	Number of colliding species.
Value	Can be 0, fractional, integer.	Always whole number (1, 2, 3). Never 0.
Determination	Experimental.	Theoretical.
Scope	Applicable to elementary & complex reactions.	Only for elementary steps.

Unit Trick for 'k':
Unit $= (\text{mol})^{1-n} (\text{L})^{n-1} s^{-1}$

Zero Order: mol L$^{-1}$ s$^{-1}$ (Same as rate)
First Order: s$^{-1}$ (Time inverse)
Second Order: mol$^{-1}$ L s$^{-1}$

Tip: In exam, look at the unit of $k$ to identify the order of reaction!

4. Integrated Rate Equations (High Priority)

The differential rate equations are integrated to give a relation between concentration and time.

A. Zero Order Reaction

Rate is independent of concentration. $R \to P$.

Differential form: $-\frac{d[R]}{dt} = k[R]^0 = k$

On integrating, we get:

$[R] = -kt + [R]_0$
Or
$k = \frac{[R]_0 - [R]}{t}$

Graph: Plot $[R]$ vs $t$. Straight line sloping down.
Slope: $-k$
Intercept: $[R]_0$

B. First Order Reaction

Rate is proportional to the first power of concentration.

Differential form: $-\frac{d[R]}{dt} = k[R]$

Integrated form (Natural Log): $\ln[R] = -kt + \ln[R]_0$

Converted to Log base 10 (Exam Formula):

$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$

Graph: Plot $\log[R]$ vs $t$. Straight line sloping down.
Slope: $\frac{-k}{2.303}$

C. Half-Life Period ($t_{1/2}$)

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

For Zero Order: $t_{1/2} = \frac{[R]_0}{2k}$ (Directly proportional to initial conc.)
For First Order: $t_{1/2} = \frac{0.693}{k}$ (Independent of initial conc.)

D. Pseudo First Order Reaction

Reactions which are not truly first order but strictly follow first order kinetics under certain conditions (usually when one reactant is in excess).

Example: Hydrolysis of Ethyl Acetate.
$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O (excess)} \to \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}$
Rate $= k' [\text{Ester}][\text{Water}] \approx k [\text{Ester}]$ (Since water conc. is constant).

5. Arrhenius Equation & Activation Energy

This section explains the effect of temperature on rate constant $k$.

Activation Energy ($E_a$): The minimum extra energy that reacting molecules must absorb to cross the energy barrier and form products.

Arrhenius Equation: $k = A e^{-E_a/RT}$

$A$: Arrhenius factor / Frequency factor.
$R$: Gas constant ($8.314 \, \text{J K}^{-1} \text{mol}^{-1}$).
$T$: Temperature in Kelvin.

Logarithmic Form (For Calculation):
taking log on both sides:

$\log k = \log A - \frac{E_a}{2.303 RT}$

Graph: Plot $\log k$ vs $1/T$. Slope $= \frac{-E_a}{2.303 R}$.

Two Temperature Form (Most Important for Numericals):

$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$

6. Collision Theory (Qualitative)

According to this theory, for a reaction to occur, reactant molecules must collide. However, not all collisions lead to products. Only Effective Collisions yield products.

Two barriers determine effectiveness:

Energy Barrier: Colliding molecules must have kinetic energy $\ge$ Threshold Energy.
Orientation Barrier: Molecules must collide in a specific orientation to break old bonds and form new ones.

Rate $= P Z_{AB} e^{-E_a/RT}$ (where $P$ is probability/steric factor, $Z$ is collision frequency).

7. Question & Answer Bank (Master These)

Q1: First Order Calculation (3 Marks)

Question: A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life ($t_{1/2}$).

Solution:
Step 1 Identify parameters.
$[R]_0 = 100$. For 30% decomp, 30 is gone, so $[R] = 100 - 30 = 70$.
Time $t = 40$ min.

Step 2 Calculate $k$.
$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
$k = \frac{2.303}{40} \log \frac{100}{70} = \frac{2.303}{40} (\log 10 - \log 7)$
$k = \frac{2.303}{40} (1 - 0.8451) = \frac{2.303}{40} (0.1549)$
$k = 0.00891 \, \text{min}^{-1}$

Step 3 Calculate $t_{1/2}$.
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.00891}$
$t_{1/2} \approx 77.78 \, \text{min}$.

Q2: Arrhenius & Activation Energy (3 Marks)

Question: The rate constant of a reaction increases by 5% when the temperature is raised from 300K to 301K. Calculate the activation energy ($E_a$).

Solution:
Given $T_1 = 300K, T_2 = 301K$.
Let $k_1 = k$. Then $k_2 = k + 5\% \text{ of } k = 1.05k$.
$\frac{k_2}{k_1} = 1.05$.

Formula
$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
$\log (1.05) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{301 - 300}{300 \times 301} \right]$
$0.0211 = \frac{E_a}{19.147} \left[ \frac{1}{90300} \right]$
$E_a = 0.0211 \times 19.147 \times 90300$
$E_a = 36,483 \, \text{J/mol} \approx 36.5 \, \text{kJ/mol}$.

Q3: Gas Phase Kinetics (Tricky - 3 Marks)

Question: For the decomposition $A(g) \to B(g) + C(g)$, initial pressure is $P_i$. After time $t$, total pressure is $P_t$. Derive the expression for $k$.

Solution:
Reaction: $A \to B + C$
At $t=0$: $P_i \quad 0 \quad 0$
At time $t$: $P_i - x \quad x \quad x$

Total Pressure $P_t = (P_i - x) + x + x = P_i + x$
Therefore, $x = P_t - P_i$.

Pressure of A at time $t$ ($P_A$) $= P_i - x = P_i - (P_t - P_i) = 2P_i - P_t$.

Using first order equation:
$k = \frac{2.303}{t} \log \frac{\text{Initial Pressure}}{\text{Pressure of A at t}}$
$\boxed{k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}}$

Q4: Conceptual & Graphs (2 Marks)

Question:
(a) Plot a graph showing variation of potential energy with reaction coordinate for a reaction with a catalyst and without a catalyst.
(b) What is the effect of a catalyst on $\Delta G$ (Gibbs Energy)?

Answer:
(a) The graph should show two humps. The path with catalyst has a smaller hump (lower Activation Energy $E_a'$). The path without catalyst has a higher hump (higher $E_a$). Reactant and Product energy levels remain unchanged.
(b) A catalyst has NO effect on $\Delta G$. It only changes the speed of reaching equilibrium, not the thermodynamics.

🚀 Final Revision: The "Do Not Forget" List

Before you enter the exam hall, memorize these:

Log values: $\log 2 = 0.3010$, $\log 3 = 0.4771$, $\log 5 = 0.6990$, $\log 7 = 0.8450$.
Unit Conversion: $R = 8.314 \, \text{J/K mol}$. If $E_a$ is in kJ, convert $R$ to kJ ($8.314 \times 10^{-3}$) or $E_a$ to Joules.
Temperature: Always convert Celsius to Kelvin ($+273$).
Slope check: For first order $\log[R]$ vs $t$, slope is negative. For Arrhenius $\log k$ vs $1/T$, slope is negative.
Time fraction: $t_{99.9\%} \approx 10 \times t_{1/2}$ for first order reactions. (Great for checking MCQ answers).

Prepared specifically for CBSE Class 12 Board Exams | Chemical Kinetics Mastery

Good Luck! You got this! 🎓