NCERT Solutions for Class 12 Chemistry Chapter 3 Chemical Kinetics
NCERT Solutions for Class 12 Chemistry Chapter 3 Chemical Kinetics provides comprehensive coverage of reaction rates, rate laws, order and molecularity of reactions, integrated rate equations, half-life periods, Arrhenius equation, and activation energy.
Overview of Topics & Subtopics (Click to Jump)
1. Rate of Reaction, Factors & Order/Molecularity
Short Note:
• Rate of Reaction: Change in concentration of reactants or products per unit time ($r = -\frac{\Delta [R]}{\Delta t} = +\frac{\Delta [P]}{\Delta t}$). Unit: $\text{mol L}^{-1}\text{s}^{-1}$.
• Rate Law & Rate Constant ($k$): Expression relating reaction rate to concentration of reactants: $r = k[A]^x [B]^y$. $k$ is characteristic for a specific reaction at a given temperature.
• Order of Reaction ($n = x + y$): Sum of powers of reactant concentrations in experimentally determined rate law. Can be 0, fractional, or integer.
• Molecularity: Number of reacting species taking part in an elementary reaction step simultaneously. Always a positive whole integer ($1, 2, 3$). Cannot be zero or fractional.
2. Integrated Rate Equations, Half-Life & Pseudo First Order
Short Note:
• Zero Order Reaction: Rate is independent of concentration.
- Integrated equation: $k = \frac{[R]_0 - [R]}{t}$
- Half-life: $t_{1/2} = \frac{[R]_0}{2k}$ (Directly proportional to initial concentration $[R]_0$).
• First Order Reaction: Rate depends linearly on reactant concentration.
- Integrated equation: $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$
- Half-life: $t_{1/2} = \frac{0.693}{k}$ (Independent of initial concentration $[R]_0$).
• Pseudo First Order Reaction: A higher-order reaction that behaves as a first-order reaction because one reactant is in excess (e.g., acid hydrolysis of ethyl acetate, inversion of cane sugar).
3. Temperature Dependence, Arrhenius Equation & Collision Theory
Short Note:
• Temperature Coefficient: Most chemical reactions double their rate for every $10^\circ\text{C}$ rise in temperature.
• Arrhenius Equation: $k = A e^{-E_a / RT}$
- Logarithmic Form: $\log k = \log A - \frac{E_a}{2.303 R T}$
- Two-Temperature Form: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
• Activation Energy ($E_a$): Minimum energy required by reactant molecules to form an activated complex and turn into products.
• Collision Theory: Effective collisions depend on two factors: Energetic factor (Energy $\ge E_a$) and Orientational factor (Proper spatial alignment of molecules during collision).
Intext & Conceptual Questions
| Feature | Order of Reaction | Molecularity of Reaction |
|---|---|---|
| Definition | Sum of exponents of concentration terms in experimental rate law expression. | Number of species colliding simultaneously in an elementary step. |
| Nature | Experimental value. | Theoretical value. |
| Values Possible | Can be zero, integer, or fractional. | Always a positive whole integer ($1, 2, 3$). Never zero or fractional. |
| Applicability | Applicable to elementary as well as complex reactions. | Applicable only to elementary steps of a reaction. |
• Rate equation for the given reaction: $r_1 = k [A]^1 [B]^2$
• When concentration of B is tripled ($[B'] = 3[B]$):
$r_2 = k [A]^1 (3[B])^2 = k [A]^1 \times 9[B]^2 = 9 \times r_1$
• Therefore, the rate of reaction increases by 9 times.
Textbook Chapter End Exercise Questions
Given:
• Rate constant $k = 1.15 \times 10^{-3}\text{ s}^{-1}$
• Initial concentration $[R]_0 = 5\text{ g}$
• Remaining concentration $[R] = 3\text{ g}$
Step 1: Apply First-Order Integrated Rate Formula
$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$
Step 2: Substitute Values
$t = \frac{2.303}{1.15 \times 10^{-3}} \log \left(\frac{5}{3}\right)$
$t = 2002.6 \times \log(1.667)$
$t = 2002.6 \times 0.2219 \approx 444.38\text{ s}$
• The time required for reduction is 444.4 seconds.
Given:
• $T_1 = 500\text{ K}$, $k_1 = 0.02\text{ s}^{-1}$
• $T_2 = 700\text{ K}$, $k_2 = 0.07\text{ s}^{-1}$
• $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$
Step 1: Calculate Activation Energy ($E_a$)
$\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$
$\log \left(\frac{0.07}{0.02}\right) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{700 - 500}{500 \times 700} \right]$
$\log(3.5) = \frac{E_a}{19.147} \times \left[ \frac{200}{350000} \right]$
$0.5441 = \frac{E_a}{19.147} \times \left(5.714 \times 10^{-4}\right)$
$E_a = \frac{0.5441 \times 19.147}{5.714 \times 10^{-4}} \approx 18231\text{ J mol}^{-1} = \mathbf{18.23\text{ kJ mol}^{-1}}$
Step 2: Calculate Frequency Factor ($A$) using $T_1 = 500\text{ K}$
$\log k_1 = \log A - \frac{E_a}{2.303 R T_1}$
$\log(0.02) = \log A - \frac{18231}{2.303 \times 8.314 \times 500}$
$-1.6990 = \log A - 1.9043$
$\log A = -1.6990 + 1.9043 = 0.2053$
$A = 10^{0.2053} \approx \mathbf{1.603\text{ s}^{-1}}$
• Activation Energy ($E_a$) = 18.23 kJ mol⁻¹
• Pre-exponential factor ($A$) = 1.60 s⁻¹
Read More & Chapter Resources
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