JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
આ MCQ મોડ્યુલ આના પર આધારિત છે: Half Life Pseudo
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Half Life Pseudo
🎓 Class 12
Chemistry
CBSE
Theory
Ch 3 – Chemical Kinetics
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Half Life Pseudo
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Half Life Pseudo
3.6 First Order Gas-Phase Reactions and Partial Pressures
For a gas-phase reaction \( A(g) \to B(g) + C(g)\) the concentration of A is proportional to its partial pressure \(p_A\). Suppose initially only A is present at total pressure \(p_i\). At time \(t\), \(p_A\) drops by an amount \(x\); pressure of B and C each rise by \(x\). The total pressure is
\[ p_t \;=\; (p_i - x) + x + x \;=\; p_i + x \quad\Rightarrow\quad x = p_t - p_i \]
So \( p_A = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t \). Substituting in the first-order equation
\[ k = \frac{2.303}{t}\,\log\frac{p_i}{p_A} \;=\; \frac{2.303}{t}\,\log\frac{p_i}{2p_i - p_t} \]
Worked Example 3.9 L3 Apply
The thermal decomposition of \(N_2O_5(g)\) is first order: \(2 N_2O_5(g) \to 2 N_2O_4(g) + O_2(g)\). At constant volume and temperature, the following data were measured for an analogous reaction \(N_2O_5 \to 2NO_2 + ½ O_2\) starting at \(p_i\):
| Time / s | Total pressure / atm |
|---|---|
| 0 | 0.5 |
| 100 | 0.512 |
Calculate the rate constant.
Using \(p_A = 2p_i - p_t = 2(0.5) - 0.512 = 0.488\) atm? — careful: stoichiometry here gives 2.5 mol gas per 1 mol reactant.
Using NCERT formula derived for \(2N_2O_5 \to 2N_2O_4 + O_2\): \( p_{A} = p_i - 2(p_t-p_i)\). With \(p_i = 0.5, p_t = 0.512\):
\(p_A = 0.5 - 2(0.012) = 0.476\) atm.
\(k = \dfrac{2.303}{100}\log\dfrac{0.5}{0.476} = \dfrac{2.303}{100}(0.0216) = 4.98 \times 10^{-4}\) s⁻¹.
3.7 Half-Life of a Reaction
The half-life of a reaction, denoted \(t_{1/2}\), is the time required for the concentration of a reactant to fall to half of its initial value.
Half-Life of a Zero Order Reaction
For zero order: \([R] = [R]_0 - kt\). Setting \([R] = [R]_0/2\) at \(t = t_{1/2}\):
\[ \frac{[R]_0}{2} = [R]_0 - k\,t_{1/2} \quad\Rightarrow\quad \boxed{\; t_{1/2} \;=\; \frac{[R]_0}{2k} \;} \]
For zero order, \(t_{1/2}\) is directly proportional to the initial concentration.
Half-Life of a First Order Reaction
For first order: \(k = \dfrac{2.303}{t}\log\dfrac{[R]_0}{[R]}\). Setting \([R] = [R]_0/2\) at \(t = t_{1/2}\):
\[ k = \frac{2.303}{t_{1/2}}\log 2 \quad\Rightarrow\quad \boxed{\; t_{1/2} \;=\; \frac{0.693}{k} \;} \]
For first order, \(t_{1/2}\) is constant — independent of starting concentration. This is why radioactive nuclei have a single quoted half-life.
Summary of Integrated Rate Equations
| Order | Rate Law | Integrated form | Linear plot | Slope | t₁/₂ | Units of k |
|---|---|---|---|---|---|---|
| 0 | Rate = k | [R] = [R]₀ − kt | [R] vs t | −k | [R]₀/2k | mol L⁻¹ s⁻¹ |
| 1 | Rate = k[R] | ln[R] = ln[R]₀ − kt | ln[R] vs t | −k | 0.693/k | s⁻¹ |
Worked Example 3.10 L3 Apply
A first-order reaction is found to have a rate constant \(k = 5.5 \times 10^{-14}\) s⁻¹. Find the half-life of the reaction.
\(t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{5.5 \times 10^{-14}\text{ s}^{-1}} = 1.26 \times 10^{13}\) s.
Worked Example 3.11 L3 Apply
Show that for a first-order reaction the time required for 99% completion is twice the time required for 90% completion.
For 99% completion: \([R]/[R]_0 = 0.01\). Then \(t_{99} = \dfrac{2.303}{k}\log\dfrac{1}{0.01} = \dfrac{2.303}{k}(2) = \dfrac{4.606}{k}\).
For 90% completion: \([R]/[R]_0 = 0.10\). Then \(t_{90} = \dfrac{2.303}{k}\log\dfrac{1}{0.10} = \dfrac{2.303}{k}(1) = \dfrac{2.303}{k}\).
Therefore \(t_{99}/t_{90} = 4.606/2.303 = 2\). Hence \( t_{99} = 2 \, t_{90}\). Q.E.D.
Worked Example 3.12 L3 Apply
The half-life for radioactive decay of \(^{14}C\) is 5730 years. An archaeological artefact contains wood that has only 80% of the \(^{14}C\) found in a living tree. Estimate the age of the sample.
For first order decay, \(k = 0.693/t_{1/2} = 0.693/5730 = 1.21 \times 10^{-4}\) yr⁻¹.
Time taken for [R]/[R]₀ to fall to 0.80:
\(t = \dfrac{2.303}{k}\log\dfrac{1}{0.80} = \dfrac{2.303}{1.21 \times 10^{-4}}(0.0969) = 1845\) years.
The sample is approximately 1845 years old.
Interactive: Half-Life Visualizer (First Order)
Watch a population of reactant atoms decay through successive half-lives.
Number of half-lives elapsed = 2.0
Fraction of [R] remaining = 0.250 (= 25.0%)
Rate constant k = 0.00693 s⁻¹
3.8 Pseudo First Order Reactions
Reactions that look like they should be of higher order but turn out to follow first-order kinetics under certain conditions are called pseudo first order reactions.
Hydrolysis of Ester
Consider \( CH_3COOC_2H_5 + H_2O \to CH_3COOH + C_2H_5OH\). The true rate law would be Rate = \(k\,[CH_3COOC_2H_5][H_2O]\) (overall order 2). However, when carried out in dilute aqueous solution, water is in vast excess:
- [Ester] ≈ 0.01 M
- [H₂O] ≈ 55.5 M (almost pure water)
The change in \([H_2O]\) during the entire reaction is negligible — perhaps 0.01 M out of 55.5 M, i.e. 0.018%. So [H₂O] is treated as a constant:
\[ \text{Rate} = k\,[\text{ester}][H_2O] = k'[\text{ester}], \quad\text{where } k' = k[H_2O] \]
This is now a first order rate law in ester, with a pseudo first order rate constant \(k'\).
Inversion of Cane Sugar
Another classic example:
\[ C_{12}H_{22}O_{11} + H_2O \xrightarrow{H^+} C_6H_{12}O_6 + C_6H_{12}O_6 \]
(sucrose hydrolysing to glucose + fructose). Again [H₂O] is enormously in excess. The reaction follows first-order kinetics with a half-life independent of sucrose concentration.
| Time / min | [Sucrose]/mol L⁻¹ |
|---|---|
| 0 | 0.500 |
| 30 | 0.451 |
| 60 | 0.407 |
| 90 | 0.367 |
Plotting \(\log[\text{sucrose}]\) against time gives a straight line, confirming pseudo first order kinetics.
Activity 3.3 — Verify Pseudo First Order Kinetics for Sugar Inversion
Setup: Take 25 mL of 1.0 M sucrose solution and 25 mL of 1.0 M HCl. Mix and immediately use a polarimeter to measure the angle of optical rotation at intervals of 10 min for one hour. Sucrose rotates plane-polarised light to the right (+); the products (glucose + fructose, 'invert sugar') rotate to the left (−). Compute \(\log(r_t - r_\infty)\) where \(r\) is the rotation.
Predict: What kind of plot of \(\log(r_t - r_\infty)\) versus time will you observe if the inversion is pseudo first order?
You will obtain a straight line with negative slope. Slope = \(-k/2.303\) gives the pseudo first order rate constant directly. Since [H₂O] is virtually constant (~55 M against ~0.5 M sugar), the apparent first-order behaviour is observed even though the molecular act involves both sugar and water.
Historically this experiment by Wilhelmy in 1850 was the very first quantitative kinetic study and gave birth to chemical kinetics as a discipline.
Interactive: Why Water in Excess Means First Order
Drag the slider for the [water]:[ester] ratio and see how the pseudo rate constant \(k' = k[H_2O]\) and the % change in [water] during the reaction respond.
If 100% of ester reacts, fractional change in [H₂O] = 0.018 %
Verdict: [H₂O] is essentially constant → reaction follows first order kinetics in ester.
Worked Example 3.13 L3 Apply
Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The volume of NaOH at infinite time is 17.4 mL. At time t the volume used was 10.6 mL after 25 min. Calculate the pseudo first order rate constant.
NaOH volume at \(t = \infty\) corresponds to total ester originally present, i.e. \([R]_0 \propto V_\infty = 17.4\) mL.
NaOH volume at time t corresponds to ester reacted, so unreacted ester \([R] \propto (V_\infty - V_t) = 17.4 - 10.6 = 6.8\) mL.
\(k' = \dfrac{2.303}{t}\log\dfrac{V_\infty}{V_\infty - V_t} = \dfrac{2.303}{25}\log\dfrac{17.4}{6.8}\)
\(= \dfrac{2.303}{25}(0.4079) = 0.0376\) min⁻¹.
Competency-Based Questions
Q1. The half-life of a first order reaction is 30 min. The fraction of reactant remaining after 90 min is: L3
(c) 1/8. 90 min = 3 half-lives. Fraction remaining = (1/2)³ = 1/8.
Q2. For a zero order reaction with k = 0.020 mol L⁻¹ min⁻¹ and [R]₀ = 0.20 mol L⁻¹, the half-life is: L3
(a) 5 min. \(t_{1/2} = [R]_0/(2k) = 0.20/(2 \times 0.020) = 5\) min.
Q3. (Short answer) Why is the hydrolysis of cane sugar in dilute aqueous acid called pseudo first order? L4
Because the true rate law involves [sucrose] and [H₂O] (overall order 2), but [H₂O] is in vast excess (≈ 55 M vs 0.5 M sucrose) and changes negligibly. The observable rate law reduces to Rate = k′[sucrose] — apparent first order.
Q4. (True/False) The half-life of a first order reaction increases as initial concentration increases. L2
False. For first order, \(t_{1/2} = 0.693/k\), independent of [R]₀. (For zero order, however, \(t_{1/2} \propto [R]_0\).)
Q5. (Long answer) A radioactive isotope has a half-life of 10 days. What percentage will remain undecayed after 30 days, and how long will it take for only 10% to remain? L5
After 30 days = 3 half-lives → fraction remaining = (1/2)³ = 1/8 = 12.5%.
For 10% remaining: \(k = 0.693/10 = 0.0693\) day⁻¹. \(t = (2.303/k)\log(100/10) = (2.303/0.0693) \times 1 = 33.2\) days.
For 10% remaining: \(k = 0.693/10 = 0.0693\) day⁻¹. \(t = (2.303/k)\log(100/10) = (2.303/0.0693) \times 1 = 33.2\) days.
Assertion–Reason Questions
(A) Both true & R explains A. (B) Both true but R does not explain A. (C) A true, R false. (D) A false, R true.
Assertion: The half-life of a first order reaction is independent of initial concentration.
Reason: Half-life depends only on the rate constant which is itself temperature-dependent only.
(A) Both true; R explains A. \(t_{1/2} = 0.693/k\) — and k depends only on T.
Assertion: Hydrolysis of an ester in dilute aqueous solution is pseudo first order.
Reason: Esters react with water in a 1:1 stoichiometry.
(B) Both true, but R does not explain A. The pseudo first order behaviour is due to large excess of water making [H₂O] constant — not because of stoichiometry.
Assertion: For a zero order reaction half-life increases linearly with initial concentration.
Reason: \(t_{1/2} = [R]_0/2k\).
(A) Both true; R is the formula for zero order half-life.
Frequently Asked Questions - Half Life Pseudo
What is the main concept covered in Half Life Pseudo?
In NCERT Class 12 Chemistry Chapter 3 (Chemical Kinetics), "Half Life Pseudo" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Half Life Pseudo useful in real-life or applied chemistry?
Real-life applications of "Half Life Pseudo" from NCERT Class 12 Chemistry Chapter 3 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Half Life Pseudo?
Key reactions in "Half Life Pseudo" (NCERT Class 12 Chemistry Chapter 3 Chemical Kinetics) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 3?
NCERT Class 12 Chemistry Chapter 3 (Chemical Kinetics) is structured so each part builds chemical understanding sequentially. "Half Life Pseudo" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Half Life Pseudo?
CBSE board questions from "Half Life Pseudo" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Half Life Pseudo" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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