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Chemical Kinetics Complete NCERT Resource Guide
Mastering Class 12 Chemistry can feel challenging, but tension lene ki zarurat nahi hai! Chemical Kinetics Class 12 Solutions is here to make your prep incredibly smooth. This chapter is a major scoring area in your CBSE Class 12 Chemistry Chapter 4 syllabus and holds huge weightage in competitive exams like NEET and JEE . In this guide, we provide the Updated NCERT Solutions along with Important Questions and Board Exam Questions 2026 to help you ace your exams with confidence!
Chapter NameChemical Kinetics (Chapter 4)
SubjectChemistry
ClassClass 12
BoardCBSE & Other State Boards
Important TopicsRate of Reaction, Order and Molecularity, Integrated Rate Equations, Half-Life, Arrhenius Equation
Difficulty LevelModerate (Concept + Numerical Based)
Exam WeightageApprox. 7-8 Marks in CBSE
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Learning Objectives
After completing this chapter, students will be able to:
- Define the rate of a chemical reaction and factors influencing it.
- Differentiate clearly between the order and molecularity of a reaction.
- Derive and apply integrated rate equations for zero and first-order reactions.
- Calculate the half-life ($t_{1/2}$) of various chemical reactions.
- Understand the concept of collision theory and calculate activation energy using the Arrhenius Equation.
Key Concepts, Definitions & Formulas
Before diving into the NCERT Solutions, let's quickly refresh the core pillars of Chemical Kinetics:
- Rate of Reaction: The change in concentration of a reactant or product per unit time. $$\text{Average Rate} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta [P]}{\Delta t}$$
- Rate Law: The mathematical expression in which reaction rate is given in terms of molar concentration of reactants raised to some power. For a reaction $A+B \rightarrow C$, $\text{Rate}=k[A]^x[B]^y$, where $(x+y)$ is the Order of Reaction.
- Molecularity: The number of reacting species (atoms, ions, or molecules) taking part in an elementary chemical reaction, which must collide simultaneously in order to bring about a chemical reaction. It is always a whole number.
- Integrated Rate Equation (Zero-Order): $$k = \frac{[R]_0 - [R]}{t}$$
- Integrated Rate Equation (First-Order): $$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$$
- Half-life ($t_{1/2}$): The time required for the concentration of a reactant to decrease to half of its initial value.
- For Zero-Order: $t_{1/2} = \frac{[R]_0}{2k}$
- For First-Order: $t_{1/2} = \frac{0.693}{k}$
- Arrhenius Equation: Shows the effect of temperature on the rate constant. $$k=A e^{-\frac{E_a}{RT}} \quad \implies \quad \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2-T_1}{T_1 T_2} \right]$$
Full NCERT Solutions (Step-by-Step)
Here are the complete step-by-step NCERT Solutions for Class 12 Chemistry Chapter 4 based on the latest textbook exercises.
Question 4.1: From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
1. $3\text{NO}(g) \rightarrow \text{N}_2\text{O}(g)$, $\text{Rate} = k[\text{NO}]^2$
2. $\text{H}_2\text{O}_2(aq) + 3\text{I}^-(aq) + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O}(l) + \text{I}_3^-$, $\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]$
3. $\text{CH}_3\text{CHO}(g) \rightarrow \text{CH}_4(g) + \text{CO}(g)$, $\text{Rate} = k[\text{CH}_3\text{CHO}]^{3/2}$
4. $\text{C}_2\text{H}_5\text{Cl}(g) \rightarrow \text{C}_2\text{H}_4(g) + \text{HCl}(g)$, $\text{Rate} = k[\text{C}_2\text{H}_5\text{Cl}]$
Step 1: Understand the general unit formula. The general unit for a rate constant is given by the formula: $\text{Unit} = (\text{mol L}^{-1})^{1-n} \text{s}^{-1}$, where $n$ is the order of reaction.
Step 2: Analyze Reaction 1. Given: $\text{Rate} = k[\text{NO}]^2$.
Order ($n$): 2 (Second-order reaction).
Dimensions of $k$: $(\text{mol L}^{-1})^{1-2} \text{s}^{-1} = \mathbf{\text{mol}^{-1} \text{L s}^{-1}}$.
Step 3: Analyze Reaction 2. Given: $\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]$.
Order ($n$): $1+1=2$ (Second-order reaction).
Dimensions of $k$: $\mathbf{\text{mol}^{-1} \text{L s}^{-1}}$.
Step 4: Analyze Reaction 3. Given: $\text{Rate} = k[\text{CH}_3\text{CHO}]^{3/2}$.
Order ($n$): $3/2$ or $1.5$.
Dimensions of $k$: $(\text{mol L}^{-1})^{1-1.5} \text{s}^{-1} = (\text{mol L}^{-1})^{-0.5} \text{s}^{-1} = \mathbf{\text{mol}^{-0.5} \text{L}^{0.5} \text{s}^{-1}}$.
Step 5: Analyze Reaction 4. Given: $\text{Rate} = k[\text{C}_2\text{H}_5\text{Cl}]$.
Order ($n$): 1 (First-order reaction).
Dimensions of $k$: $(\text{mol L}^{-1})^{1-1} \text{s}^{-1} = \mathbf{\text{s}^{-1}}$.
Step 2: Analyze Reaction 1. Given: $\text{Rate} = k[\text{NO}]^2$.
Order ($n$): 2 (Second-order reaction).
Dimensions of $k$: $(\text{mol L}^{-1})^{1-2} \text{s}^{-1} = \mathbf{\text{mol}^{-1} \text{L s}^{-1}}$.
Step 3: Analyze Reaction 2. Given: $\text{Rate} = k[\text{H}_2\text{O}_2][\text{I}^-]$.
Order ($n$): $1+1=2$ (Second-order reaction).
Dimensions of $k$: $\mathbf{\text{mol}^{-1} \text{L s}^{-1}}$.
Step 4: Analyze Reaction 3. Given: $\text{Rate} = k[\text{CH}_3\text{CHO}]^{3/2}$.
Order ($n$): $3/2$ or $1.5$.
Dimensions of $k$: $(\text{mol L}^{-1})^{1-1.5} \text{s}^{-1} = (\text{mol L}^{-1})^{-0.5} \text{s}^{-1} = \mathbf{\text{mol}^{-0.5} \text{L}^{0.5} \text{s}^{-1}}$.
Step 5: Analyze Reaction 4. Given: $\text{Rate} = k[\text{C}_2\text{H}_5\text{Cl}]$.
Order ($n$): 1 (First-order reaction).
Dimensions of $k$: $(\text{mol L}^{-1})^{1-1} \text{s}^{-1} = \mathbf{\text{s}^{-1}}$.
Question 4.2: For the reaction: $2A+B \rightarrow A_2B$, the rate law is $\text{Rate} = k[A] [B]^2$ 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 $0.06 \text{ mol L}^{-1}$.
Step 1: Calculate the initial rate.
Given rate law: $\text{Rate} = k[A][B]^2$
$\text{Initial Rate} = (2.0 \times 10^{-6}) \times (0.1) \times (0.2)^2$
$\text{Initial Rate} = 2.0 \times 10^{-6} \times 0.1 \times 0.04 = \mathbf{8.0 \times 10^{-9} \text{ mol L}^{-1} \text{s}^{-1}}$
Step 2: Calculate concentration of B left when A is reduced to 0.06 mol L⁻¹.
Change in concentration of $A = 0.1 - 0.06 = 0.04 \text{ mol L}^{-1}$.
According to the balanced equation, 2 moles of $A$ react with 1 mole of $B$. Therefore, change in concentration of $B = \frac{1}{2} \times \text{Change in } A = \frac{1}{2} \times 0.04 = 0.02 \text{ mol L}^{-1}$.
New concentration of $B$, $[B] = 0.2 - 0.02 = 0.18 \text{ mol L}^{-1}$.
Step 3: Calculate the new rate.
$\text{New Rate} = k[A]_{\text{new}}[B]_{\text{new}}^2$
$\text{New Rate} = (2.0 \times 10^{-6}) \times (0.06) \times (0.18)^2$
$\text{New Rate} = 2.0 \times 10^{-6} \times 0.06 \times 0.0324 = \mathbf{3.89 \times 10^{-9} \text{ mol L}^{-1} \text{s}^{-1}}$
Given rate law: $\text{Rate} = k[A][B]^2$
$\text{Initial Rate} = (2.0 \times 10^{-6}) \times (0.1) \times (0.2)^2$
$\text{Initial Rate} = 2.0 \times 10^{-6} \times 0.1 \times 0.04 = \mathbf{8.0 \times 10^{-9} \text{ mol L}^{-1} \text{s}^{-1}}$
Step 2: Calculate concentration of B left when A is reduced to 0.06 mol L⁻¹.
Change in concentration of $A = 0.1 - 0.06 = 0.04 \text{ mol L}^{-1}$.
According to the balanced equation, 2 moles of $A$ react with 1 mole of $B$. Therefore, change in concentration of $B = \frac{1}{2} \times \text{Change in } A = \frac{1}{2} \times 0.04 = 0.02 \text{ mol L}^{-1}$.
New concentration of $B$, $[B] = 0.2 - 0.02 = 0.18 \text{ mol L}^{-1}$.
Step 3: Calculate the new rate.
$\text{New Rate} = k[A]_{\text{new}}[B]_{\text{new}}^2$
$\text{New Rate} = (2.0 \times 10^{-6}) \times (0.06) \times (0.18)^2$
$\text{New Rate} = 2.0 \times 10^{-6} \times 0.06 \times 0.0324 = \mathbf{3.89 \times 10^{-9} \text{ mol L}^{-1} \text{s}^{-1}}$
Question 4.3: The decomposition of $\text{NH}_3$ on platinum surface is zero order reaction. What are the rates of production of $\text{N}_2$ and $\text{H}_2$ if $k=2.5 \times 10^{-4} \text{ mol}^{-1} \text{ L s}^{-1}$?
Step 1: Write the chemical equation and rate relationships.
The chemical equation for the decomposition of ammonia is: $2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2$.
The relationship between the rates of consumption of reactants and production of products is: $$\text{Rate} = \frac{1}{2} \frac{d[\text{NH}_3]}{d|t|} = \frac{d[\text{N}_2]}{d|t|} = \frac{1}{3} \frac{d[\text{H}_2]}{d|t|}$$
Step 2: Apply zero-order kinetics conditions.
Since the reaction is of zero order: $\text{Rate} = k = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$.
Step 3: Calculate individual production rates.
Rate of production of $\text{N}_2$:
$\frac{d[\text{N}_2]}{d|t|} = \text{Rate} = \mathbf{2.5 \times 10^{-4} \text{ mol L}^{-1} \text{s}^{-1}}$
Rate of production of $\text{H}_2$:
$\frac{d[\text{H}_2]}{d|t|} = 3 \times \text{Rate} = 3 \times 2.5 \times 10^{-4} = \mathbf{7.5 \times 10^{-4} \text{ mol L}^{-1} \text{s}^{-1}}$
The chemical equation for the decomposition of ammonia is: $2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2$.
The relationship between the rates of consumption of reactants and production of products is: $$\text{Rate} = \frac{1}{2} \frac{d[\text{NH}_3]}{d|t|} = \frac{d[\text{N}_2]}{d|t|} = \frac{1}{3} \frac{d[\text{H}_2]}{d|t|}$$
Step 2: Apply zero-order kinetics conditions.
Since the reaction is of zero order: $\text{Rate} = k = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$.
Step 3: Calculate individual production rates.
Rate of production of $\text{N}_2$:
$\frac{d[\text{N}_2]}{d|t|} = \text{Rate} = \mathbf{2.5 \times 10^{-4} \text{ mol L}^{-1} \text{s}^{-1}}$
Rate of production of $\text{H}_2$:
$\frac{d[\text{H}_2]}{d|t|} = 3 \times \text{Rate} = 3 \times 2.5 \times 10^{-4} = \mathbf{7.5 \times 10^{-4} \text{ mol L}^{-1} \text{s}^{-1}}$
Question 4.4: The decomposition of dimethyl ether leads to the formation of $\text{CH}_4$, $\text{H}_2$ and $\text{CO}$ and the reaction rate is given by $\text{Rate}=k[\text{CH}_3\text{OCH}_3]^{3/2}$. If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constant?
Step 1: Determine the Unit of Rate.
Since pressure replaces concentration, the unit of rate is $\mathbf{\text{bar min}^{-1}}$.
Step 2: Determine the Unit of Rate Constant ($k$).
$\text{Rate} = k(P_{\text{CH}_3\text{OCH}_3})^{3/2} \implies k = \frac{\text{Rate}}{(P_{\text{CH}_3\text{OCH}_3})^{3/2}}$
$\text{Unit of } k = \frac{\text{bar min}^{-1}}{(\text{bar})^{3/2}} = \mathbf{\text{bar}^{-1/2} \text{min}^{-1}}$
Since pressure replaces concentration, the unit of rate is $\mathbf{\text{bar min}^{-1}}$.
Step 2: Determine the Unit of Rate Constant ($k$).
$\text{Rate} = k(P_{\text{CH}_3\text{OCH}_3})^{3/2} \implies k = \frac{\text{Rate}}{(P_{\text{CH}_3\text{OCH}_3})^{3/2}}$
$\text{Unit of } k = \frac{\text{bar min}^{-1}}{(\text{bar})^{3/2}} = \mathbf{\text{bar}^{-1/2} \text{min}^{-1}}$
Question 4.5: Mention the factors that affect the rate of a chemical reaction.
Step 1: Identify key factors. The rate of a chemical reaction depends upon the following key factors:
1. Concentration of Reactants: Higher concentration increases collision frequency, increasing the rate.
2. Temperature: Generally, increasing temperature increases kinetic energy and the reaction rate.
3. Presence of a Catalyst: A catalyst lowers the activation energy barrier, speeding up the process.
4. Surface Area of Reactants: Finely divided solids react faster due to increased surface contact.
5. Nature of Reactants: Reactions involving ionic species are typically faster than those involving covalent bonds.
1. Concentration of Reactants: Higher concentration increases collision frequency, increasing the rate.
2. Temperature: Generally, increasing temperature increases kinetic energy and the reaction rate.
3. Presence of a Catalyst: A catalyst lowers the activation energy barrier, speeding up the process.
4. Surface Area of Reactants: Finely divided solids react faster due to increased surface contact.
5. Nature of Reactants: Reactions involving ionic species are typically faster than those involving covalent bonds.
Question 4.6: A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled, (ii) reduced to half?
Step 1: Define initial conditions.
Let the initial rate expression be: $\text{Rate}_1 = k[A]^2$.
Step 2: Evaluate doubling concentration.
(i) When the concentration is doubled ($[A]'=2[A]$):
$\text{Rate}_2 = k(2[A])^2 = 4k[A]^2 = 4 \times \text{Rate}_1$.
The rate of reaction becomes 4 times the original rate.
Step 3: Evaluate halving concentration.
(ii) When the concentration is reduced to half ($[A]''=\frac{1}{2}[A]$):
$\text{Rate}_3 = k\left(\frac{1}{2}[A]\right)^2 = \frac{1}{4}k[A]^2 = \frac{1}{4} \times \text{Rate}_1$.
The rate of reaction is reduced to one-fourth of the original rate.
Let the initial rate expression be: $\text{Rate}_1 = k[A]^2$.
Step 2: Evaluate doubling concentration.
(i) When the concentration is doubled ($[A]'=2[A]$):
$\text{Rate}_2 = k(2[A])^2 = 4k[A]^2 = 4 \times \text{Rate}_1$.
The rate of reaction becomes 4 times the original rate.
Step 3: Evaluate halving concentration.
(ii) When the concentration is reduced to half ($[A]''=\frac{1}{2}[A]$):
$\text{Rate}_3 = k\left(\frac{1}{2}[A]\right)^2 = \frac{1}{4}k[A]^2 = \frac{1}{4} \times \text{Rate}_1$.
The rate of reaction is reduced to one-fourth of the original rate.
Question 4.7: What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Step 1: State the general temperature effect.
Generally, the rate constant of a reaction increases rapidly with a rise in temperature. For most chemical reactions, the rate constant becomes nearly doubled for every $10^\circ\text{C}$ rise in temperature.
Step 2: Provide the quantitative representation.
Quantitatively, this effect is accurately expressed by the Arrhenius Equation: $$k=A e^{-\frac{E_a}{RT}}$$
Where:
$k =$ Rate constant
$A =$ Arrhenius pre-exponential factor (Frequency factor)
$E_a =$ Activation energy
$R =$ Gas constant
$T =$ Absolute temperature (in Kelvin)
Generally, the rate constant of a reaction increases rapidly with a rise in temperature. For most chemical reactions, the rate constant becomes nearly doubled for every $10^\circ\text{C}$ rise in temperature.
Step 2: Provide the quantitative representation.
Quantitatively, this effect is accurately expressed by the Arrhenius Equation: $$k=A e^{-\frac{E_a}{RT}}$$
Where:
$k =$ Rate constant
$A =$ Arrhenius pre-exponential factor (Frequency factor)
$E_a =$ Activation energy
$R =$ Gas constant
$T =$ Absolute temperature (in Kelvin)
Question 4.8: In a pseudo first-order hydrolysis of ester in water, the following results were obtained: Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Step 1: Extract data for the required interval.
For the interval $t_1 = 30\text{ s}$ to $t_2 = 60\text{ s}$:
$C_1 = 0.31 \text{ mol L}^{-1}$
$C_2 = 0.17 \text{ mol L}^{-1}$
Step 2: Apply the average rate formula.
The formula for average rate is: $\text{Average Rate} = -\frac{C_2-C_1}{t_2-t_1}$
$\text{Average Rate} = -\frac{0.17-0.31}{60-30} = \frac{0.14}{30} = \mathbf{4.67 \times 10^{-3} \text{ mol L}^{-1} \text{s}^{-1}}$
For the interval $t_1 = 30\text{ s}$ to $t_2 = 60\text{ s}$:
$C_1 = 0.31 \text{ mol L}^{-1}$
$C_2 = 0.17 \text{ mol L}^{-1}$
Step 2: Apply the average rate formula.
The formula for average rate is: $\text{Average Rate} = -\frac{C_2-C_1}{t_2-t_1}$
$\text{Average Rate} = -\frac{0.17-0.31}{60-30} = \frac{0.14}{30} = \mathbf{4.67 \times 10^{-3} \text{ mol L}^{-1} \text{s}^{-1}}$
Question 4.9: A reaction is first order in $A$ and second order in $B$. (1) Write the differential rate equation. (2) How is the rate affected on increasing the concentration of $B$ three times? (3) How is the rate affected when the concentration of both $A$ and $B$ are doubled?
Step 1: Differential Rate Equation.
$\frac{d[R]}{dt} = k[A]^1[B]^2$
Step 2: Calculate rate change when B is tripled.
When concentration of $B$ is tripled ($[B]'=3[B]$):
$\text{Rate}' = k[A][3[B]]^2 = 9k[A][B]^2 = \mathbf{9 \times \text{Rate}}$. The rate increases 9 times.
Step 3: Calculate rate change when both are doubled.
When both $[A]$ and $[B]$ are doubled ($[A]'=2[A]$, $[B]'=2[B]$):
$\text{Rate}'' = k[2[A]][2[B]]^2 = k(2[A])(4[B]^2) = 8k[A][B]^2 = \mathbf{8 \times \text{Rate}}$. The rate increases 8 times.
$\frac{d[R]}{dt} = k[A]^1[B]^2$
Step 2: Calculate rate change when B is tripled.
When concentration of $B$ is tripled ($[B]'=3[B]$):
$\text{Rate}' = k[A][3[B]]^2 = 9k[A][B]^2 = \mathbf{9 \times \text{Rate}}$. The rate increases 9 times.
Step 3: Calculate rate change when both are doubled.
When both $[A]$ and $[B]$ are doubled ($[A]'=2[A]$, $[B]'=2[B]$):
$\text{Rate}'' = k[2[A]][2[B]]^2 = k(2[A])(4[B]^2) = 8k[A][B]^2 = \mathbf{8 \times \text{Rate}}$. The rate increases 8 times.
Question 4.10: In a reaction between $A$ and $B$, the initial rate of reaction ($r_0$) was measured for different initial concentrations of $A$ and $B$. What is the order of the reaction with respect to $A$ and $B$?
Step 1: Setup the initial rate law equations.
Let the rate law be: $\text{Rate} = k[A]^x[B]^y$
From Exp 1: $5.07 \times 10^{-5} = k(0.20)^x(0.30)^y$ ---(i)
From Exp 2: $5.07 \times 10^{-5} = k(0.20)^x(0.10)^y$ ---(ii)
Step 2: Solve for order with respect to B ($y$).
Dividing (i) by (ii):
$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}} = \left(\frac{0.30}{0.10}\right)^y \implies 1 = (3)^y \implies \mathbf{y = 0}$.
So, the order with respect to $B$ is 0.
Step 3: Solve for order with respect to A ($x$).
From Exp 2 and Exp 3 (knowing $y=0$):
$5.07 \times 10^{-5} = k(0.20)^x(0.10)^0 = k(0.20)^x$
$1.43 \times 10^{-4} = k(0.40)^x(0.05)^0 = k(0.40)^x$
Dividing the Exp 3 equation by the Exp 2 equation:
$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \left(\frac{0.40}{0.20}\right)^x \implies 2.82 = (2)^x$
Taking logs on both sides:
$\log(2.82) = x \log(2) \implies 0.4502 = x(0.3010) \implies x = \frac{0.4502}{0.3010} \approx \mathbf{1.5}$
So, the order with respect to $A$ is 1.5.
Let the rate law be: $\text{Rate} = k[A]^x[B]^y$
From Exp 1: $5.07 \times 10^{-5} = k(0.20)^x(0.30)^y$ ---(i)
From Exp 2: $5.07 \times 10^{-5} = k(0.20)^x(0.10)^y$ ---(ii)
Step 2: Solve for order with respect to B ($y$).
Dividing (i) by (ii):
$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}} = \left(\frac{0.30}{0.10}\right)^y \implies 1 = (3)^y \implies \mathbf{y = 0}$.
So, the order with respect to $B$ is 0.
Step 3: Solve for order with respect to A ($x$).
From Exp 2 and Exp 3 (knowing $y=0$):
$5.07 \times 10^{-5} = k(0.20)^x(0.10)^0 = k(0.20)^x$
$1.43 \times 10^{-4} = k(0.40)^x(0.05)^0 = k(0.40)^x$
Dividing the Exp 3 equation by the Exp 2 equation:
$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}} = \left(\frac{0.40}{0.20}\right)^x \implies 2.82 = (2)^x$
Taking logs on both sides:
$\log(2.82) = x \log(2) \implies 0.4502 = x(0.3010) \implies x = \frac{0.4502}{0.3010} \approx \mathbf{1.5}$
So, the order with respect to $A$ is 1.5.
Question 4.11: Determine the rate law and the rate constant for the reaction: $2A+B \rightarrow 2C+D$ from the provided experimental data.
Step 1: Evaluate the experimental table relations.
Based on standard modern NCERT versions, the rate law analysis yields an order with respect to $A$ as 1 and $B$ as 2.
Step 2: Establish the correct empirical rate law.
Rate Law: $\text{Rate} = k[A][B]^2$
Step 3: Calculate the rate constant ($k$).
Using Exp I to find $k$:
$6.0 \times 10^{-3} = k(0.1)(0.1)^2 \implies k = \frac{6.0 \times 10^{-3}}{10^{-3}} = \mathbf{6.0 \text{ mol}^{-2} \text{L}^2 \text{min}^{-1}}$
Based on standard modern NCERT versions, the rate law analysis yields an order with respect to $A$ as 1 and $B$ as 2.
Step 2: Establish the correct empirical rate law.
Rate Law: $\text{Rate} = k[A][B]^2$
Step 3: Calculate the rate constant ($k$).
Using Exp I to find $k$:
$6.0 \times 10^{-3} = k(0.1)(0.1)^2 \implies k = \frac{6.0 \times 10^{-3}}{10^{-3}} = \mathbf{6.0 \text{ mol}^{-2} \text{L}^2 \text{min}^{-1}}$
Question 4.12: The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the table.
Step 1: Determine the rate constant ($k$).
Since the reaction is first order in $A$ and zero order in $B$, $\text{Rate} = k[A]$.
From Exp I: $2.0 \times 10^{-2} = k(0.1) \implies k = 0.2 \text{ min}^{-1}$.
Step 2: Solve for Blank (i) - Experiment II.
$\text{Rate} = k[A] \implies 4.0 \times 10^{-2} = 0.2 \times [A] \implies [A] = \mathbf{0.2 \text{ mol L}^{-1}}$.
Step 3: Solve for Blank (ii) - Experiment III.
$\text{Rate} = 0.2 \times 0.4 = \mathbf{8.0 \times 10^{-2} \text{ mol L}^{-1} \text{min}^{-1}}$.
Step 4: Solve for Blank (iii) - Experiment IV.
$2.0 \times 10^{-2} = 0.2 \times [A] \implies [A] = \mathbf{0.1 \text{ mol L}^{-1}}$.
Since the reaction is first order in $A$ and zero order in $B$, $\text{Rate} = k[A]$.
From Exp I: $2.0 \times 10^{-2} = k(0.1) \implies k = 0.2 \text{ min}^{-1}$.
Step 2: Solve for Blank (i) - Experiment II.
$\text{Rate} = k[A] \implies 4.0 \times 10^{-2} = 0.2 \times [A] \implies [A] = \mathbf{0.2 \text{ mol L}^{-1}}$.
Step 3: Solve for Blank (ii) - Experiment III.
$\text{Rate} = 0.2 \times 0.4 = \mathbf{8.0 \times 10^{-2} \text{ mol L}^{-1} \text{min}^{-1}}$.
Step 4: Solve for Blank (iii) - Experiment IV.
$2.0 \times 10^{-2} = 0.2 \times [A] \implies [A] = \mathbf{0.1 \text{ mol L}^{-1}}$.
Question 4.13: Calculate the half-life of a first order reaction from their rate constants given below: (i) $200 \text{ s}^{-1}$, (ii) $2 \text{ min}^{-1}$, (iii) $4 \text{ years}^{-1}$.
Step 1: Setup formula.
For a first-order reaction, $t_{1/2} = \frac{0.693}{k}$.
Step 2: Calculate for (i).
$k = 200 \text{ s}^{-1}$:
$t_{1/2} = \frac{0.693}{200} = \mathbf{3.46 \times 10^{-3} \text{ s}}$
Step 3: Calculate for (ii).
$k = 2 \text{ min}^{-1}$:
$t_{1/2} = \frac{0.693}{2} = \mathbf{0.3465 \text{ min}}$
Step 4: Calculate for (iii).
$k = 4 \text{ years}^{-1}$:
$t_{1/2} = \frac{0.693}{4} = \mathbf{0.173 \text{ years}}$
For a first-order reaction, $t_{1/2} = \frac{0.693}{k}$.
Step 2: Calculate for (i).
$k = 200 \text{ s}^{-1}$:
$t_{1/2} = \frac{0.693}{200} = \mathbf{3.46 \times 10^{-3} \text{ s}}$
Step 3: Calculate for (ii).
$k = 2 \text{ min}^{-1}$:
$t_{1/2} = \frac{0.693}{2} = \mathbf{0.3465 \text{ min}}$
Step 4: Calculate for (iii).
$k = 4 \text{ years}^{-1}$:
$t_{1/2} = \frac{0.693}{4} = \mathbf{0.173 \text{ years}}$
Question 4.14: The half-life for radioactive decay of $^{14}\text{C}$ is 5730 years. An archaeological artifact containing wood had only $80\%$ of the $^{14}\text{C}$ found in a living tree. Estimate the age of the sample.
Step 1: Find the rate constant ($k$).
Radioactive decay follows first-order kinetics.
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} = 1.21 \times 10^{-4} \text{ year}^{-1}$
Step 2: Find the age ($t$).
Let initial concentration $[A]_0 = 100$, then remaining concentration $[A] = 80$.
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} = \frac{2.303}{1.21 \times 10^{-4}} \log \left(\frac{100}{80}\right)$
$t = \frac{2.303}{1.21 \times 10^{-4}} \log(1.25) = \frac{2.303 \times 0.0969}{1.21 \times 10^{-4}} \approx \mathbf{1845 \text{ years}}$
The estimated age of the wood sample is 1845 years.
Radioactive decay follows first-order kinetics.
$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} = 1.21 \times 10^{-4} \text{ year}^{-1}$
Step 2: Find the age ($t$).
Let initial concentration $[A]_0 = 100$, then remaining concentration $[A] = 80$.
$t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} = \frac{2.303}{1.21 \times 10^{-4}} \log \left(\frac{100}{80}\right)$
$t = \frac{2.303}{1.21 \times 10^{-4}} \log(1.25) = \frac{2.303 \times 0.0969}{1.21 \times 10^{-4}} \approx \mathbf{1845 \text{ years}}$
The estimated age of the wood sample is 1845 years.
Question 4.15: The experimental data for decomposition of $\text{N}_2\text{O}_5$ in gas phase at $318\text{ K}$ are given. Show that it follows first order reaction.
Step 1: Establish the condition for first-order kinetics.
For a first-order reaction, the rate constant $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$ must be constant at all time intervals.
Here, $[A]_0 = 1.63 \times 10^{-2} \text{ mol L}^{-1}$.
Step 2: Calculate $k$ at $t=400\text{ s}$.
$k_1 = \frac{2.303}{400} \log \left(\frac{1.63}{1.24}\right) = \frac{2.303}{400} (0.1186) = 6.82 \times 10^{-4} \text{ s}^{-1}$
Step 3: Calculate $k$ at $t=800\text{ s}$.
$k_2 = \frac{2.303}{800} \log \left(\frac{1.63}{0.92}\right) = \frac{2.303}{800} (0.2484) = 7.15 \times 10^{-4} \text{ s}^{-1}$
Step 4: Draw Conclusion.
Since the values of $k$ are very close to each other (around $7 \times 10^{-4} \text{ s}^{-1}$), the reaction strictly follows first-order kinetics.
For a first-order reaction, the rate constant $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$ must be constant at all time intervals.
Here, $[A]_0 = 1.63 \times 10^{-2} \text{ mol L}^{-1}$.
Step 2: Calculate $k$ at $t=400\text{ s}$.
$k_1 = \frac{2.303}{400} \log \left(\frac{1.63}{1.24}\right) = \frac{2.303}{400} (0.1186) = 6.82 \times 10^{-4} \text{ s}^{-1}$
Step 3: Calculate $k$ at $t=800\text{ s}$.
$k_2 = \frac{2.303}{800} \log \left(\frac{1.63}{0.92}\right) = \frac{2.303}{800} (0.2484) = 7.15 \times 10^{-4} \text{ s}^{-1}$
Step 4: Draw Conclusion.
Since the values of $k$ are very close to each other (around $7 \times 10^{-4} \text{ s}^{-1}$), the reaction strictly follows first-order kinetics.
Extra Important Questions (Board Style)
Section A: Multiple Choice Questions (MCQs)
Q1. Which of the following expressions represents the safe unit of rate constant for a third-order reaction?
A) $\text{mol L}^{-1}\text{s}^{-1}$
B) $\text{mol}^{-2}\text{L}^2\text{s}^{-1}$
C) $\text{s}^{-1}$
D) $\text{mol}^{-1}\text{Ls}^{-1}$
Correct Answer Choice: B
Step 1: Formula Execution. Using $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$ for $n=3$, we get $(\text{mol L}^{-1})^{-2}\text{s}^{-1} = \text{mol}^{-2}\text{L}^2\text{s}^{-1}$.
Step 1: Formula Execution. Using $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$ for $n=3$, we get $(\text{mol L}^{-1})^{-2}\text{s}^{-1} = \text{mol}^{-2}\text{L}^2\text{s}^{-1}$.
Q2. The half-life period of a first-order reaction is 69.3 seconds. What is its rate constant?
A) $10^{-2} \text{ s}^{-1}$
B) $10^{-4} \text{ s}^{-1}$
C) $10^{-1} \text{ s}^{-1}$
D) $10^{-3} \text{ s}^{-1}$
Correct Answer Choice: A
Step 1: Calculation. $k = 0.693 / t_{1/2} = 0.693 / 69.3 = 0.01 = 10^{-2} \text{ s}^{-1}$.
Step 1: Calculation. $k = 0.693 / t_{1/2} = 0.693 / 69.3 = 0.01 = 10^{-2} \text{ s}^{-1}$.
Section B: Assertion-Reason Section
Q3. Assertion: Order of a reaction can be zero or fractional.
Reason: Molecularity of a reaction can never be zero or fractional.
Correct Answer Choice: B (Both are true, but Reason is not the correct explanation).
Step 1: Verification. Both statements are true facts. Order is experimentally determined (can be 0 or fraction), while molecularity is theoretical (count of reactant molecules) and must be positive integers. However, the reason does not explain why the order can be fractional.
Step 1: Verification. Both statements are true facts. Order is experimentally determined (can be 0 or fraction), while molecularity is theoretical (count of reactant molecules) and must be positive integers. However, the reason does not explain why the order can be fractional.
Section C: Short Answer & Long Answer Questions
Q4. Distinguish between order and molecularity of a reaction.
Step 1: Contrast the definitions. Order is the sum of powers of concentration terms in the rate law expression. Molecularity is the number of reacting species colliding simultaneously in an elementary step.
Step 2: Contrast numerical possibilities. Order can be zero, fractional, or integer. Molecularity is always a whole number (never zero or fractional).
Step 3: Contrast derivation methods. Order is determined experimentally. Molecularity is a theoretical concept calculated from the mechanism.
Step 2: Contrast numerical possibilities. Order can be zero, fractional, or integer. Molecularity is always a whole number (never zero or fractional).
Step 3: Contrast derivation methods. Order is determined experimentally. Molecularity is a theoretical concept calculated from the mechanism.
Q5. A first-order reaction takes 40 minutes for $30\%$ decomposition. Calculate its $t_{1/2}$.
Step 1: Identify given parameters. $t=40 \text{ min}$, $[A]_0=100$, $[A]=100-30=70$.
Step 2: Calculate rate constant ($k$).
$k = \frac{2.303}{40} \log \left(\frac{100}{70}\right) = \frac{2.303}{40} \times 0.1549 = 0.00892 \text{ min}^{-1}$
Step 3: Calculate half-life.
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.00892} \approx \mathbf{77.7 \text{ minutes}}$
Step 2: Calculate rate constant ($k$).
$k = \frac{2.303}{40} \log \left(\frac{100}{70}\right) = \frac{2.303}{40} \times 0.1549 = 0.00892 \text{ min}^{-1}$
Step 3: Calculate half-life.
$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.00892} \approx \mathbf{77.7 \text{ minutes}}$
Q6. Explain Arrhenius equation. A reaction rate quadruples when temperature changes from $293\text{ K}$ to $313\text{K}$. Calculate the activation energy ($E_a$).
Step 1: Set up the formula. Given: $T_1 = 293\text{ K}$, $T_2 = 313\text{ K}$, $\frac{k_2}{k_1} = 4$.
Formula: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2-T_1}{T_1 T_2} \right]$
Step 2: Substitute values and evaluate.
$\log(4) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{313-293}{293 \times 313} \right]$
$0.6021 = \frac{E_a}{19.147} \left[ \frac{20}{91709} \right]$
Step 3: Solve for $E_a$.
$E_a = \frac{0.6021 \times 19.147 \times 91709}{20} = \mathbf{52863 \text{ J mol}^{-1} = 52.86 \text{ kJ mol}^{-1}}$
Teacher's Tip: Include a graph showing a linear plot with a negative slope equal to $-E_a/ 2.303R$ to help students visualize temperature dependence.
Formula: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left[ \frac{T_2-T_1}{T_1 T_2} \right]$
Step 2: Substitute values and evaluate.
$\log(4) = \frac{E_a}{2.303 \times 8.314} \left[ \frac{313-293}{293 \times 313} \right]$
$0.6021 = \frac{E_a}{19.147} \left[ \frac{20}{91709} \right]$
Step 3: Solve for $E_a$.
$E_a = \frac{0.6021 \times 19.147 \times 91709}{20} = \mathbf{52863 \text{ J mol}^{-1} = 52.86 \text{ kJ mol}^{-1}}$
Teacher's Tip: Include a graph showing a linear plot with a negative slope equal to $-E_a/ 2.303R$ to help students visualize temperature dependence.
Common Mistakes Students Make
- Confusing Units: Writing the wrong unit for rate constants. Always re-verify using $(\text{mol L}^{-1})^{1-n}\text{s}^{-1}$.
- Log Calculation Errors: Mixing up natural log ($\ln$) and base-10 log ($\log$). Remember: $\ln x = 2.303 \log x$.
- Molecularity of Complex Reactions: Trying to find the molecularity of a complex multi-step reaction. Remember, molecularity is only defined for elementary steps!
Exam Preparation Tips
- Focus on Graphs: Make sure you can draw and recognize plots for zero-order ($\text{[A]}$ vs $t$) and first-order ($\log \text{[A]}$ vs $t$) reactions.
- Formula Sheet: Keep a sticky note of half-life formulations handy.
- Solve Numericals First: This chapter is highly analytical. Practice writing the given data step-by-step to score partial marks even if your final calculation goes off. Exam mein steps ke direct marks milte hain!
Frequently Asked Questions (FAQs)
1. Is Chemical Kinetics important for CBSE Class 12 Boards?
Yes! It is one of the high-weightage chapters in Physical Chemistry, typically contributing 7 to 8 marks.
2. What is a pseudo first-order reaction example?
Inversion of cane sugar or acid-catalyzed hydrolysis of an ester are classic examples where one reactant is present in large excess (usually water).
3. Where can I find updated NCERT solutions?
Right here on examspark.in! We continuously update our resources according to the newest CBSE 2026-27 curriculums.
4. How does a catalyst change reaction rate?
A catalyst provides an alternative pathway with a lower activation energy, which increases the fraction of effective collisions.
5. Can molecularity be zero?
No, at least one molecule must be present to collide and react, so molecularity is always a positive integer.
Conclusion: Lucky, mastering Chemical Kinetics is highly logical and scoring if you understand the core mechanics of Rate Laws and Integrated Equations. Make sure you practice old question papers alongside these NCERT Solutions to solidify your concepts. For high-quality notes, test series, and comprehensive guides, stay tuned to examspark.in!
More Class 12 Chemistry Chapters
Ch 1: The Solid State
Ch 2: Solutions
Ch 3: Electrochemistry
Ch 4: Chemical Kinetics
Ch 5: Surface Chemistry
Ch 6: Isolation of Elements
Ch 7: The p-Block Elements
Ch 8: The d- & f-Block Elements
Ch 9: Coordination Compounds
Ch 10: Haloalkanes
Ch 11: Alcohols & Phenols
Ch 12: Aldehydes & Ketones
Ch 13: Amines
Ch 14: Biomolecules
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