Chemical Kinetics Class 12 Chemistry NCERT Solutions Download PDF

Structural angle. Order is read straight off the rate law, and the dimensions of k are forced by the requirement that the rate itself carry units of concentration per unit time. So the same two-step procedure works in all four parts: add the exponents to get n; substitute n into the dimension template.

1. The general rate equation r = k [A]^a[B]^b⋯ has order n = a + b + ⋯. Inspect each rate law and add up all the exponents on concentration terms.
2. Solve for k: k = r[A]^a[B]^b⋯ = mol.L^-1.s^-1× (mol.L^-1)^-n = mol^1-n.L^n-1.s^-1.
3. Apply to each part. (i) n=2 gives mol^-1.L.s^-1. (ii) n=2 gives mol^-1.L.s^-1. (iii) n=1.5 gives mol^-1/2.L^1/2.s^-1. (iv) n=1 gives s^-1.

Alternative dimensional route. Instead of memorising the mol^1-n.L^n-1.s^-1 template, you can derive units on the fly. Write rate = k(conc.)ⁿ, transpose to k = rate/(conc.)ⁿ, plug in mol.L^-1.s^-1 for rate and mol.L^-1 for concentration, and let exponents subtract. This ``derive don't memorise'' habit pays off when the problem switches to pressure (units of bar^1-n.t^-1) or to molality.

Concept linkage. Order ≠ molecularity. Molecularity is the number of molecules colliding in an elementary step (always a positive integer), while order is an experimental power that can be zero, fractional, or even negative. The reaction in part (iii) (CH3CHO decomposition) is a complex multi-step process; its n = 3/2 signals a Rice–Herzfeld chain mechanism with a fast pre-equilibrium step.

Cross-check by limiting cases. For n = 0, the template gives mol.L^-1.s^-1 – same as rate, which makes sense because k is the rate when concentration has no effect. For n = 1, the template gives s^-1 – a pure frequency, which matches the radioactive-decay analogue. These two limits anchor the formula in physical meaning.

JEE/NEET relevance. Identifying order from a rate law and writing the corresponding units of k is a one-mark recall question almost every year. The 2022 JEE Main paper, for instance, tested exactly this pattern with n = 5/2. The answer is mechanical once you know the template.

Orders: 2, 2, 1.5, 1. Units of k: mol^-1.L.s^-1, mol^-1.L.s^-1, mol^-1/2.L^1/2.s^-1, s^-1.

Strategic angle. The rate law is given, so the work is pure substitution. The only subtlety is tying Δ[B] to Δ[A] using the 2:1 stoichiometry.

1. Initial substitution. With [A] = 0.1, [B] = 0.2, the product [A][B]² = 0.1 × 0.04 = 0.004, so r₀ = 2.0× 10^-6× 0.004 = 8.0× 10^-9 mol.L^-1.s^-1.
2. Compute the drop in A: Δ[A] = 0.04 mol.L^-1. From 2A : 1B, Δ[B] = 0.02 mol.L^-1, so the remaining [B] = 0.18 mol.L^-1.
3. Final substitution. [A][B]² = 0.06×(0.18)² = 0.06× 0.0324 = 1.944× 10^-3, and r = 2.0× 10^-6× 1.944× 10^-3 = 3.89× 10^-9 mol.L^-1.s^-1.

Alternative ratio method. Skip the absolute substitution. Write r/r₀ = ([A]/[A]₀) ([B]/[B]₀)². Here [A]/[A]₀ = 0.06/0.10 = 0.6 and [B]/[B]₀ = 0.18/0.20 = 0.9. So r/r₀ = 0.6× 0.81 = 0.486, hence r = 0.486× 8.0× 10^-9 = 3.89× 10^-9 mol.L^-1.s^-1. The ratio method is faster when the rate constant is small and you want to avoid multiplying tiny numbers.

Concept linkage. The overall order here is 1 + 2 = 3, so the units of k are mol^-2.L².s^-1 – exactly what the problem provides. Always cross-check k units against the rate law order: a mismatch means the rate law is being misread.

Cross-check by stoichiometric bookkeeping. If Δ[A] = 0.04 in the time the rate is being asked, then by 2A + B, Δ[B] = 0.02 and Δ[A₂ B] = 0.02. Adding A consumed and B consumed in mole units: 0.04 + 0.02 = 0.06 = 2(0.02) moles of nucleotides reacted per litre, matching the 2A + B totals.

JEE/NEET relevance. Rate at a non-initial concentration is a classic two-step JEE pattern: substitute initial, then update from stoichiometry. NEET 2021 used the same idea with 2A + 3B → C, where Δ[B] = (3/2)Δ[A].

r₀ = 8.0× 10^-9 mol.L^-1.s^-1; r_new ≈ 3.89× 10^-9 mol.L^-1.s^-1.

Strategic angle. A zero order rate constant has units of mol.L^-1.s^-1, so k is the rate of reaction. Then multiply by the appropriate stoichiometric ratio for each product.

1. Write the unique rate of reaction so that the stoichiometric factor cancels: r = -12 d[NH3]/dt = d[N2]/dt = 13 d[H2]/dt. For zero order, r = k = 2.5× 10^-4 mol.L^-1.s^-1.
2. Read off product rates. d[N2]/dt = r = 2.5× 10^-4 mol.L^-1.s^-1. d[H2]/dt = 3r = 7.5× 10^-4 mol.L^-1.s^-1.

Alternative ``per-mole'' bookkeeping. Forget the rate-of-reaction definition and just count moles. For every 2 moles of NH3 destroyed, 1 mole of N2 and 3 moles of H2 are made. If NH3 disappears at rate 5.0× 10^-4 mol.L^-1.s^-1 (twice k), then N2 appears at half this rate (2.5× 10^-4) and H2 at 3/2 this rate (7.5× 10^-4). Same answer, no calculus.

Concept linkage. Zero order kinetics on a hot Pt surface is the chemisorption-saturation limit of the Langmuir-Hinshelwood mechanism: _NH3 ≈ 1, so the rate equals k θ ≈ k independent of [NH3]. The same limit appears in catalytic decomposition of HI on Au and of N2O on Pt.

Cross-check by stoichiometry. The numbers must satisfy the unique rate r: r = -12(-5.0× 10^-4) = 2.5× 10^-4, r = +1× d[N2]/dt = 2.5× 10^-4, r = +13× 7.5× 10^-4 = 2.5× 10^-4. All three definitions return the same r.

JEE/NEET relevance. Heterogeneous catalysis questions test either (a) the order being zero, or (b) the role of surface area. Memorise the standard zero-order examples: 2NH3 -> N2 + 3H2 on Pt, 2HI -> H2 + I2 on Au, photochemical reactions at high intensity. NEET 2020 used this exact ammonia decomposition.

d[N2]/dt = 2.5× 10^-4 and d[H2]/dt = 7.5× 10^-4 mol.L^-1.s^-1.

Strategic angle. The numerical work is empty; the question is testing whether the student can carry units cleanly through a fractional order rate law.

1. Recognise that rate, when expressed in pressure units, is dP/dt, so its units are bar/min = bar.min^-1.
2. Form k = Rate/P^3/2. The numerator is bar.min^-1, the denominator is bar^3/2. Subtract exponents of bar: 1 - 3/2 = -1/2. So k has units bar^-1/2.min^-1.

Alternative via p = cRT. Convert concentration units to pressure units. For order n, the concentration form gives [k]_c = mol^1-n.L^n-1.s^-1. Replacing mol.L^-1 = p/(RT) and substituting s → min multiplies by (RT)^n-1 and converts s^-1 to min^-1. The bare-unit answer in pressure (bar^1-n.min^-1) is what remains – a cleaner route than working out RT numerically.

Concept linkage. For gas-phase reactions in a closed vessel, the partial pressure of each species tracks its concentration linearly (p_i = c_i RT for an ideal gas). So whether you fit kinetics in c or p, the order is invariant – only the units of k change. The same idea is exploited in Q 3.20 and Q 3.21 where pressure measurements stand in for concentrations.

Cross-check by integer cases. For n = 1 (first order), the template gives [k] = bar⁰.min^-1 = min^-1 – the same as s^-1 but in minutes. For n = 2, [k] = bar^-1.min^-1. Our n = 3/2 result (bar^-1/2.min^-1) sits cleanly between these two. The half-power on bar is the hallmark of a 3/2-order reaction.

JEE/NEET relevance. CBSE/NCERT routinely tests pressure-form units of k in one-marker MCQs. The trick is always the same: pull out the order from the rate-law exponent, then apply [k] = [P]^1-n[t]^-1. JEE Main 2023 used this with n = 1/2 for an acid-catalysed gas decomposition.

Rate: bar.min^-1; k: bar^-1/2.min^-1.

Strategic angle. Group the factors by what they change in the collision-theory picture: either the number of effective collisions per second, or the fraction of those collisions that overcome the activation barrier.

1. Number of collisions. Concentration and surface area both raise the collision frequency.
2. Fraction of successful collisions. Temperature raises the fraction of molecules with E ≥ E_a. A catalyst lowers E_a itself, raising the same fraction from the other side.
3. Built-in chemistry. The nature of the reactants (bond energies, ionic vs covalent, gas phase vs solution) sets the baseline k; you cannot change this without changing the reaction.
4. External energy. Light, ultrasound and electric discharge can supply the activation energy directly, bypassing thermal activation.

Alternative classification: ``rate-law'' vs ``rate-constant'' factors. A second, equally useful split is whether a factor moves the concentration term in r = k[A]^a[B]^b or the rate constant k itself. Concentration and surface area change concentration (or effective concentration at the surface). Temperature, catalyst, and nature of reactants change k. Radiation sits separately because it bypasses k entirely. This split is more faithful to the Arrhenius picture.

Concept linkage. Each factor has a quantitative bridge: concentration → rate law (Q 3.2, 3.6); temperature → Arrhenius (Q 3.7, 3.22–3.30); catalyst → effective E_a lowering (covered in Section 3.5); surface area → Langmuir isotherm (advanced). Knowing which bridge to invoke saves time on a JEE/NEET MCQ.

Cross-check using the activated-complex picture. The fraction of molecules with E ≥ E_a at temperature T is e^-E_a/RT. A catalyst lowers E_a from E_a⁰ to E_a^* < E_a⁰. The fraction rises from e^-E_a0/RT to e^-E_a*/RT, multiplying the rate by e^{(E_a0-E_a*)/RT}. For Δ E_a = 20 kJ/mol at 298K, this is a factor of ∼ 3000. That is the engineering power of a catalyst.

JEE/NEET relevance. ``Which factor does not affect rate?'' is a standing NEET MCQ. The answer is always ``the size of the container'' (irrelevant) or ``the catalyst's concentration'' (it is regenerated). The six factors listed here are the official NCERT answer set.

Six main factors: nature of reactants, concentration, temperature, catalyst, surface area, radiation.

Strategic angle. For an n-th order reaction in a single reactant, changing concentration by a factor f changes the rate by fⁿ. Here n = 2.

1. Plug f = 2 into fⁿ = f². The rate goes up by 2² = 4.
2. Plug f = 1/2. The rate goes by (1/2)² = 1/4.

Alternative graphical view. Plot r against [A]: the curve is a parabola through the origin. Move from [A]₀ to 2[A]₀ horizontally; the curve rises four-fold vertically because of the squared dependence. The same parabola shows that going from [A]₀ to [A]₀/2 drops the rate to one quarter. The picture makes the answer obvious without algebra (see the diagram in the main solution).

Concept linkage. The exponent n in r = k[A]ⁿ is also the slope of log r vs log [A]. So if you doubled the concentration and the rate quadrupled, the slope is log 4/log 2 = 2, consistent with n = 2. This is the differential method of determining order, and the second part of Q 3.10 uses exactly this logic.

Cross-check by integration. For second order kinetics in a single reactant, the integrated rate law gives 1/[A] - 1/[A]₀ = kt. Half-life is t_1/2 = 1/(k[A]₀), inversely proportional to [A]₀. So doubling [A]₀ halves t_1/2, consistent with the rate quadrupling: the reaction starts faster and finishes sooner.

JEE/NEET relevance. The pattern ``order n ⇒ factor fⁿ scaling'' is the single most-tested idea in kinetics. JEE Main 2024 asked the same question with f = 3 for a second-order reaction (answer: rate goes up by 9). Memorise the powers of 2 and 3 up to the cube.

Rate → 4r₀ when [A] → 2[A]; rate → r₀/4 when [A] → [A]/2.

Strategic angle. State the empirical fact first (rate roughly doubles per 10), then write down the Arrhenius equation, then show both the linear log form and the two-temperature ratio form that students actually use to solve problems.

1. Temperature coefficient. Define μ = k_T+10/k_T ≈ 2 to 3. This is a rough rule and varies between reactions.
2. Arrhenius equation. k = A e^-E_a/RT. Taking ln, ln k = ln A - E_a/(RT). So a plot of ln k vs 1/T is a straight line of slope -E_a/R.
3. Useful two-temperature form. log(k₂/k₁) = E_a2.303 R·T₂-T₁T₁ T₂. This lets us compute E_a from any pair of (k,T) values, or predict k at a new temperature.

Alternative derivation from collision theory. The fraction of collisions with energy ≥ E_a is e^-E_a/RT (Boltzmann). The total collision frequency Z scales roughly as √T (slow). So k = pZ e^-E_a/RT where p is the steric factor. Folding pZ into A recovers the Arrhenius form. This derivation justifies why temperature dominates: it enters the exponential, while Z enters only as a square root.

Concept linkage. The Arrhenius equation is structurally identical to the radioactive-decay law for fission products (Q 3.14, Q 3.17): in both, the rate is proportional to an exponential of an energy/RT term. This connects chemical kinetics to nuclear physics through a common Boltzmann-distribution origin.

Cross-check by limiting cases. As T → ∞, e^-E_a/RT → 1, so k → A. So A is the rate constant when every collision has enough energy. As T → 0, e^-E_a/RT → 0 and k → 0. Both limits are physically sensible: no collisions are energetic at T = 0; all are at infinite T.

JEE/NEET relevance. Of the 5-6 questions on kinetics that appear annually in JEE Main, 3-4 involve Arrhenius. The two-temperature form is the most-tested formula in the chapter – expect at least one direct numerical, and one conceptual question on its meaning, every year.

Rate constant rises with T; quantitatively, k = A e^-E_a/RT (Arrhenius equation).

Strategic angle. The phrase ``pseudo first order'' is a red herring for this part: we only need the definition of average rate, no rate-law fitting required.

1. Identify the interval and the corresponding concentrations: [A]₃₀ = 0.31, [A]₆₀ = 0.17, both in mol.L^-1.
2. Compute the drop: 0.31 - 0.17 = 0.14 mol.L^-1. Divide by 30s: r_av = 0.14/30 = 4.67× 10^-3 mol.L^-1.s^-1.

Alternative test of first order behaviour. If you compute ln[A] at each time point and check whether it falls linearly with t, you can verify the order while you are at it. Doing this: ln 0.55 = -0.598, ln 0.31 = -1.171, ln 0.17 = -1.772, ln 0.085 = -2.465. The differences over each 30s interval are -0.573, -0.601, -0.693. Roughly constant, so first order is plausible – and the slope gives k ≈ 0.60/30 = 0.020 s^-1.

Concept linkage. The average rate over 30 to 60 s is a secant slope on the [A] vs t curve. The instantaneous rate -d[A]/dt at any t in this interval is the tangent slope and is in general different. As Δ t → 0, average rate → instantaneous rate.

Cross-check by mid-point estimate. The instantaneous rate at the midpoint t = 45 s should approximately equal the average rate. Using k = 0.020 s^-1 and a midpoint [A] ≈ (0.31 + 0.17)/2 = 0.24 mol.L^-1, r_inst = k[A] = 0.020× 0.24 = 4.8× 10^-3 mol.L^-1.s^-1. Matches our 4.67× 10^-3 within rounding.

JEE/NEET relevance. Average-rate calculations are routine one-mark NEET items. The trap is sign: Δ[A] = [A]₂ - [A]₁ is negative for a reactant, and the formula uses -Δ[A]/Δ t to make rate positive. Always write the minus sign explicitly.

r_av ≈ 4.67 × 10^-3 mol.L^-1.s^-1.

Strategic angle. For any rate law r = k[A]^a[B]^b, the effect of multiplying [A] by f_A and [B] by f_B is to multiply the rate by f_A^a f_B^b. With a = 1 and b = 2:

1. Part (i): r = k[A][B]² by definition.
2. Part (ii): f_A = 1, f_B = 3: 1¹· 3² = 9.
3. Part (iii): f_A = 2, f_B = 2: 2¹· 2² = 2· 4 = 8.

Alternative log-based verification. Take logs: log r = log k + log[A] + 2log[B]. Each unit increase in log[B] (i.e. a factor of 10 in [B]) raises log r by 2. Each unit increase in log[A] raises log r by 1. So tripling [B] gives Δ log r = 2log 3 = 0.954, hence r' = 10^0.954 r = 9 r.

Concept linkage. The overall order a + b = 3 is the molecularity of an elementary termolecular step. Real termolecular elementary steps are rare (three molecules colliding at once is improbable). Most order-3 rate laws arise from multi-step mechanisms with a fast pre-equilibrium. NO+O₂ kinetics is the classical textbook example.

Cross-check by extreme limits. If f_A = 0 (i.e. A removed), r' = 0 – the reaction stops. If f_B = 0, r' = 0 – the reaction also stops. Both reactants are needed; neither is in a zero-order regime. Contrast with Q 3.10 part (i) where B has order zero and removing B does not stop the reaction.

JEE/NEET relevance. Effect-of-concentration MCQs are the single most common kinetics format. The pattern: read off orders, compute f_A^a f_B^b, pick the matching option. JEE Main 2022 asked the same question with orders (2, 1) and factors (3, 2) giving 9 × 2 = 18.

(i) r = k[A][B]²; (ii) × 9; (iii) × 8.

Strategic angle. The two rates labelled 5.07× 10^-5 are a strong hint: when [B] changes from 0.30 to 0.10 at fixed [A], nothing happens to the rate. So B does not appear in the rate law.

1. Order in B: rates 1 and 2 are equal ⇒ B has order zero.
2. Order in A: from experiments 1 and 3 with [A] doubled and [B] now irrelevant, 2.82 = 2^x, giving x = log 2.82/log 2 ≈ 1.5.
3. Overall order = 1.5; rate law is r = k[A]^3/2.

Alternative simultaneous-equations route. Without the ``constant [A]'' shortcut, set up r_i = k [A]_i^x [B]_i^y for each row and take ratios: r₃r₁ = (0.400.20)^x(0.050.30)^y, r₂r₁ = (0.200.20)^x(0.100.30)^y. The second equation gives 1 = (1/3)^y, so y = 0. Substituting into the first gives 2.82 = 2^x, so x = 1.5. Same answer, slightly more algebra. Useful when no two experiments hold the same [A] or [B].

Concept linkage. Half-integer orders like 1.5 are the fingerprint of a multi-step mechanism with a fast equilibrium of the form A2 <=> 2A where the actual reactant is the atom A derived from a dimer A₂. The square root of the dimer concentration enters the rate law and gives the half-power. Decomposition of CH3CHO (Q 3.1 part iii) is a textbook 3/2-order example.

Cross-check by computing k. Using experiment 1 ([A] = 0.20, r = 5.07× 10^-5): k = r/[A]^1.5 = 5.07× 10^-5/(0.20)^1.5 = 5.07× 10^-5/0.0894 = 5.67× 10^-4 mol^-1/2.L^1/2.s^-1. Check with experiment 3: r = k(0.40)^1.5 = 5.67× 10^-4 × 0.2530 = 1.43× 10^-4.

JEE/NEET relevance. ``Find the order from initial-rate data'' is a 4-mark NCERT/Board favourite. The recipe: hold one species constant, take a ratio of rates, equate to the concentration ratio raised to the order. NEET 2024 used three experiments to test the same skill.

Order with respect to A = 1.5, with respect to B = 0.

Strategic angle. The data are arranged neatly so that pairs of experiments differ in only one concentration. That makes the ratio approach a quick win.

1. Pair II→III: only [B] changes by × 2; rate changes by × 4 = 2². So order in B is 2.
2. Pair I→IV: only [A] changes by × 4; rate changes by × 4 = 4¹. So order in A is 1.
3. Rate law: r = k[A][B]². Use Experiment I to find k = 6.0× 10^-3/(0.1· 0.01) = 6.0 mol^-2.L².min^-1.

Alternative consistency check across all rows. Recompute k using each experiment:

• Exp III: k = 0.288/(0.3× 0.16) = 0.288/0.048 = 6.0.
• Exp IV: k = 0.024/(0.4× 0.01) = 0.024/0.004 = 6.0.

Concept linkage. Order ≠ molecularity here: the stoichiometry 2A + B predicts ``order 2 in A, order 1 in B'' for an elementary step. Experiment gives ``order 1 in A, order 2 in B'' – the opposite assignment. The reaction must therefore go through a multi-step mechanism with B doubling up in a rate-determining step.

Cross-check by overall-order units. Overall order is 1 + 2 = 3. Units of k should be mol^1-3.L^3-1.min^-1 = mol^-2.L².min^-1. The number 6.0 in these units predicts rates in mol.L^-1.min^-1.

JEE/NEET relevance. The four-experiment, four-row table is a CBSE 3-mark format. The recipe is fixed: (i) find y from a row pair with fixed [A], (ii) find x from a pair with fixed [B], (iii) compute k from any single row. JEE Main 2021 used a nearly-identical table.

r = k[A][B]²; k = 6.0 mol^-2.L².min^-1.

Strategic angle. Once the orders are stated, the rate law becomes r = k[A], a single-variable formula. Find k first, then use it as a converter between r and [A] in every other row.

1. From row I, k = r/[A] = 0.02/0.1 = 0.2 min^-1.
2. Row II: [A] = r/k = 0.04/0.2 = 0.2 mol.L^-1.
3. Row III: r = k [A] = 0.2 × 0.4 = 0.08 = 8.0× 10^-2 mol.L^-1.min^-1.
4. Row IV: [A] = 0.02/0.2 = 0.1 mol.L^-1.

Alternative ratio-only method. Skip computing k. Use r_i/r_j = [A]_i/[A]_j. Then [A]_II/[A]_I = r_II/r_I = 0.04/0.02 = 2, so [A]_II = 2× 0.1 = 0.2 mol.L^-1. For row III: r_III/r_I = [A]_III/[A]_I = 4, so r_III = 4× 0.02 = 0.08 mol.L^-1.min^-1. For row IV: same as row I, so [A]_IV = 0.1 mol.L^-1. Faster when you don't need k for downstream calculations.

Concept linkage. A reaction first order in A and zero order in B is most commonly an acid-catalysed reaction where B is the catalyst (e.g. H+): the concentration of B controls k via the effective rate constant k_eff = k_true[B]⁰→ constant. Iodination of acetone by I2 in acid is the classic example: zero order in I2 because halogenation is fast after the rate-determining enolisation step.

Cross-check by units of k. Overall order is 1 + 0 = 1, so [k] = min^-1. Our k = 0.2 min^-1 matches. The boxed-zero column of [B] in the table is a red herring – intentional, because rows II and IV with different [B] both give the same answer for [A] as if [B] never appeared.

JEE/NEET relevance. Fill-in-the-blanks tables on rate-law data are a NEET staple. Once orders are known, the problem reduces to two divisions and one multiplication. NEET 2022 used a similar table with order 2 in A, order 1 in B.

II: 0.2 mol.L^-1; III: 8.0× 10^-2 mol.L^-1.min^-1; IV: 0.1 mol.L^-1.

Strategic angle. One formula, three substitutions. The only thing to watch is that the time units in t_1/2 are the inverse of the units in k.

1. Part (i): t_1/2 = 0.693/200 = 3.465× 10^-3 s.
2. Part (ii): t_1/2 = 0.693/2 = 0.3465 min ≈ 20.79 s.
3. Part (iii): t_1/2 = 0.693/4 = 0.1733 year.

Alternative derivation from the integrated rate law. For first order, ln([R]₀/[R]) = kt. At half-life, [R] = [R]₀/2, so ln 2 = k t_1/2, giving t_1/2 = ln 2/k = 0.693/k. The constant ln 2 = 0.6931 appears because the natural log of 2 governs all exponential halving processes – chemical, nuclear, even the doubling time of bacterial cultures.

Concept linkage. For first order kinetics, t_1/2 is independent of [A]₀. For zero order, t_1/2 = [A]₀/(2k): proportional to [A]₀. For second order, t_1/2 = 1/(k[A]₀): inversely proportional to [A]₀. Recognising the order-dependence pattern of t_1/2 is itself a way to identify the order from experimental data.

Cross-check by mean lifetime. The mean lifetime is τ = 1/k. For (i), τ = 1/200 = 5.0× 10^-3 s. Note τ = t_1/2/ln 2 ≈ t_1/2/0.693 = t_1/2× 1.443. Check: 3.465× 10^-3× 1.443 = 5.00× 10^-3.

JEE/NEET relevance. Half-life numericals are NCERT favourites because the formula is trivial. The trick is unit-matching: k in s^-1 gives t_1/2 in seconds; k in year^-1 gives years. NEET 2023 tested exactly this with three sub-parts.

(i) 3.465× 10^-3 s; (ii) 0.3465 min; (iii) 0.1733 year.

Strategic angle. Two-step calculation. Convert the half-life to a decay constant, then read the time from the integrated first order rate law.

1. k = 0.693/5730 = 1.2094× 10^-4 year^-1.
2. t = (2.303/k) log([R]₀/[R]) = (2.303/1.2094× 10^-4)× log(1/0.8) = 1.9043× 10⁴× 0.0969 ≈ 1845 year.

Alternative direct formula. Skip the conversion to k and use the half-life formula directly: t = t_{1/2 2}([R]₀/[R]) = t_1/2 log(1/0.8)/log 2 = 5730 × 0.0969/0.3010 = 5730 × 0.3219 = 1844 year. Saves one step, no decay constant needed.

Concept linkage. The integrated rate law of any first order process is [R] = [R]₀ 2^-t/t_1/2. Solving for t gives t/t_1/2 = ₂([R]₀/[R]). So the ``number of half-lives'' elapsed is simply the log base 2 of the depletion factor. This view treats t_1/2 as the natural unit of time for the process.

Cross-check by half-life count. The wood has 80% of its starting ¹⁴C. After exactly one half-life, 50% would remain; after exactly 0.5 half-lives, √0.5 = 70.7% would remain. Our 80% corresponds to fewer than 0.5 half-lives, specifically ₂(1/0.8) = 0.322 half-lives. In years: 0.322 × 5730 = 1845.

JEE/NEET relevance. Carbon-14 dating is the showcase application of first-order kinetics in NCERT. Both NEET and JEE test variations on this every 2-3 years. The trap is the depletion ratio: ``80% remaining'' means [R]/[R]₀ = 0.80, not 0.20.

Age ≈ 1845 years.

Strategic angle. Take the data, log-transform it, fit a straight line, read off slope and intercept. This is the standard recipe for first order kinetics and works on any data set.

1. Compute log[N2O5] at each t (table in main solution). The points fall on a straight line.
2. Slope between t = 0 and t = 3200 s: (-2.456 + 1.788)/3200 = -2.09× 10^-4 s^-1. Therefore k = -2.303× slope = 4.81× 10^-4 s^-1.
3. t_1/2 = 0.693/k = 0.693/(4.81× 10^-4) ≈ 1440 s. Graphical estimate ≈ 1500 s matches within experimental error.

Alternative point-by-point k estimation. Compute k at each non-zero data point using k = (2.303/t)log([R]₀/[R]):

• t = 400: k = (2.303/400)log(1.63/1.36) = (5.76× 10^-3)(0.0786) = 4.53× 10^-4.
• t = 800: k = (2.303/800)log(1.63/1.14) = (2.88× 10^-3)(0.1553) = 4.47× 10^-4.
• t = 1600: k = (2.303/1600)log(1.63/0.78) = (1.44× 10^-3)(0.3197) = 4.60× 10^-4.
• t = 2400: k = (2.303/2400)log(1.63/0.53) = (9.60× 10^-4)(0.4882) = 4.69× 10^-4.
• t = 3200: k = (2.303/3200)log(1.63/0.35) = (7.20× 10^-4)(0.6686) = 4.81× 10^-4.

Concept linkage. For 2N2O5 -> 4NO2 + O2, the stoichiometric coefficient 2 in front of N2O5 means the rate of disappearance of N2O5 is twice the rate of reaction: -d[N2O5]/dt = 2 r where r is the unique rate. The k we found is the rate constant for the disappearance of N2O5, which is what the data measures directly.

Cross-check half-lives within the data. The concentration falls from 1.63× 10^-2 at t = 0 to roughly 0.82× 10^-2 between t = 1200 and t = 1600 s, then to 0.41× 10^-2 between t = 2400 and t = 2800 s. The two half-lives are both ≈ 1400-1500 s. Constancy confirms first order.

JEE/NEET relevance. The N2O5 data set is a textbook problem used in nearly every kinetics chapter. JEE Main 2020 asked exactly this question, providing k and requiring the student to compute t_1/2 – a one-step inversion of the half-life formula.

k ≈ 4.81 × 10^-4 s^-1; t_1/2 ≈ 1440 s (within rounding).

Strategic angle. Two routes: direct formula, or counting half-lives. Both should give the same answer; the second is faster.

1. Direct. t = (2.303/k)log 16 = (2.303/60)(4log 2) = (2.303/60)(1.204) = 4.62× 10^-2 s.
2. Half-life count. t_1/2 = 0.693/60 = 1.155× 10^-2 s. Falling to 1/16 means 4 half-lives, so t = 4× 1.155× 10^-2 = 4.62× 10^-2 s.

Alternative ``natural log'' shortcut. Using natural logs instead of base 10: kt = ln([R]₀/[R]) = ln 16 = 4ln 2 = 2.773. So t = 2.773/60 = 4.62× 10^-2 s. Same answer, slightly fewer arithmetic steps because ln 2 = 0.693 is already the numerator of t_1/2. Most modern textbooks prefer this ln form; older ones use ₁₀ and absorb 2.303.

Concept linkage. The factor 1/16 = (1/2)⁴ links to the binary half-life count: 4 doublings of the depletion factor correspond to 4 half-lives. The same logic gives ``time to 1/32'' as 5 t<sub>1/2</sub>, ``time to 1/64'' as 6 t_1/2, and so on. Powers of 2 map directly to half-life counts.

Cross-check using e-folds. The mean lifetime τ = 1/k governs 1/e ≈ 36.8% depletion. So t/τ = ln 16 = 2.773. With τ = 1/60 = 1.667× 10^-2 s, t = 2.773 × 1.667× 10^-2 = 4.62× 10^-2 s. Three different time scales – direct formula, half-life count, e-fold count – all agree.

JEE/NEET relevance. ``Falling to 1/16'', ``1/32'', ``1/128'' are NCERT favourites because the answer is an integer multiple of t_1/2 – no calculator needed. NEET 2019 used 1/16 exactly; JEE Main 2024 used 1/64.

t = 4 t_1/2 ≈ 4.62 × 10^-2 s.

Strategic angle. Compute k once, then plug each requested time into log([R]₀/[R]) = kt/2.303.

1. k = 0.693/28.1 = 0.02466 year^-1.
2. t = 10 year: log([R]₀/[R]) = 0.02466× 10/2.303 = 0.1071; [R]₀/[R] = 1.279; [R] = 1/1.279 = 0.782 mg.
3. t = 60 year: log([R]₀/[R]) = 0.02466× 60/2.303 = 0.6425; [R]₀/[R] = 4.391; [R] = 1/4.391 = 0.228 mg.

Alternative direct exponential. Skip the log step and use [R] = [R]₀ (1/2)^t/t_1/2 directly. For t = 10: (1/2)^10/28.1 = (1/2)^0.3559 = 2^-0.3559 = 0.782, so [R] = 0.782 mg. For t = 60: (1/2)^60/28.1 = (1/2)^2.135 = 2^-2.135 = 0.2278, so [R] = 0.228 mg. The exponential form uses no k at all, just the half-life and elapsed time – often faster for radioactive-decay problems.

Concept linkage. ⁹⁰Sr is a β^- emitter (⁹⁰Sr -> ⁹⁰Y + e^-). Its decay is first order because each nucleus decays independently with probability k dt per infinitesimal time. This is the same first-order machinery that governs N2O5 decomposition (Q 3.15) and sucrose inversion (Q 3.25). All first order processes share the same mathematics.

Cross-check by half-life accounting. 10 year/28.1 year = 0.356 half-lives; remaining fraction 2^-0.356. Use 2^-0.356 = e^{-0.356ln 2} = e^-0.247 = 0.781. 60 year/28.1 = 2.135 half-lives. After 2 half-lives, 25% remains (0.25 mg); we expect a bit less than 25% at 2.135. Indeed 0.228 < 0.25 – consistent.

JEE/NEET relevance. Multi-time radioactive-decay questions test whether students can re-use the same k across multiple time points. The integrated rate law is unchanged; only t varies. JEE Main 2023 used ¹⁴C at 5, 11,460, and 17,190 years (half-life multiples of 0, 2, 3).

∼ 0.782 mg after 10 years; ∼ 0.228 mg after 60 years.

Strategic angle. Compute both times in terms of k, then take the ratio. The factor k cancels, leaving an integer.

1. t_90% = (2.303/k)log(1/0.1) = (2.303/k)log 10 = 2.303/k.
2. t_99% = (2.303/k)log(1/0.01) = (2.303/k)log 100 = (2.303/k)· 2.
3. Ratio t_99%/t_90% = 2.

Alternative ``decades'' interpretation. Each factor-of-10 depletion takes the same time t<sub>10</sub> = 2.303/k = (ln 10)/k, call it ``one decade''. 90% completion (1/10 remaining) is one decade. 99% completion (1/100 remaining) is two decades. 99.9% completion (1/1000 remaining) is three decades. The integer ratio follows from the fact that 1/10, 1/100, 1/1000 are successive powers of 10, and the depletion in log space is linear.

Concept linkage. The same idea – ``equal depletion factors take equal time'' – is what makes t_1/2 concentration-independent for first order kinetics. Halving and tenthing are both depletions by fixed factors, and both take a time that depends only on k, not on [R]₀.

Cross-check by t_99%/t_50%. t_50% = (2.303/k)log 2 = 0.693/k = t_1/2. t_99% = 2.303log 100/k = 4.606/k. Ratio t_99%/t_50% = 4.606/0.693 = 6.64. So 99% completion takes about 6.64 half-lives. Equivalently 6.64log 2 = 2.00 decades, matching the two-decade count from the original derivation.

JEE/NEET relevance. ``Prove t<sub>99%</sub> = 2 t<sub>90%</sub>'' is a frequent CBSE board question; the general statement ``t_{(1 - 10^-n)%} scales linearly with n'' is the JEE-level extension. Both ask for the same insight: factor-of-10 depletions form a linear ladder in log space.

t_99% = 2 t_90%.

Strategic angle. Use the 30%-decomposed data to find k in one step; then apply t_1/2 = 0.693/k.

1. log([R]₀/[R]) = log(100/70) = log(10/7) = 1 - 0.8451 = 0.1549.
2. k = (2.303/t)log([R]₀/[R]) = (2.303/40)(0.1549) = 0.05758× 0.1549 = 8.92× 10^-3 min^-1.
3. t_1/2 = 0.693/(8.92× 10^-3) = 77.7 min.

Alternative ratio-form derivation. Skip k entirely by relating the two times directly: t_1/2/t = log 2/log(100/70). So t_1/2 = 40 min× log 2/log(10/7) = 40× 0.3010/0.1549 = 40 × 1.943 = 77.7 min. This shortcut is faster when the rate constant itself is not required for any downstream question.

Concept linkage. The result t_1/2/t_30% ≈ 1.94 matches the general pattern t_p% = (2.303/k)log(100/(100-p)). So the ratio of any two completion times for the same first-order reaction depends only on the fractions, never on k itself. This is one of the cleanest signatures of first-order kinetics.

Cross-check by half-life count. If t_1/2 = 77.7 min, then in 40 min the reaction is (40/77.7) = 0.515 half-lives along. Fraction remaining: (1/2)^0.515 = 2^-0.515 = 0.699. So fraction decomposed = 1 - 0.699 = 0.301 = 30.1%. Matches the given 30% within rounding.

JEE/NEET relevance. ``Find t_1/2 from a single time-fraction pair'' is the most common 3-mark CBSE format for first-order numericals. The recipe is fixed: convert the fraction to a depletion ratio, compute k, then 0.693/k. NEET 2018 used 25% decomposition in 20 minutes.

t_1/2 ≈ 77.7 min.

Strategic angle. Convert each total-pressure reading to a reactant partial pressure using the stoichiometric link P_A = 2P₀ - P_t, then read k from the standard first order integrated form.

1. At t = 360 s, P_A = 2(35) - 54 = 16 mm Hg; k = (2.303/360)log(35/16) = (2.303/360)(0.3400) = 2.175× 10^-3 s^-1.
2. At t = 720 s, P_A = 2(35) - 63 = 7 mm Hg; k = (2.303/720)log(35/7) = (2.303/720)(0.6990) = 2.236× 10^-3 s^-1.
3. Mean k ≈ 2.21× 10^-3 s^-1.

Alternative two-point regression. Take the ratio of the two data points directly: log(P_A,1/P_A,2)/(t₂ - t₁) = k/2.303. log(16/7) = log 2.286 = 0.3590. t₂ - t₁ = 720 - 360 = 360 s. k = 2.303× 0.3590/360 = 2.296× 10^-3 s^-1. This skips the initial pressure P₀ entirely and gives k from just the two later points – useful when P₀ is uncertain.

Concept linkage. The trick P_A = 2P₀ - P_t comes from the mole balance for A → B + C (one mole in, two moles out, net +1 mole gas per mole reacted). For a generic reaction aA -> products with mole increase Δ n per mole of A, P_A = P₀ - (P_t - P₀)/Δ n. Here Δ n = 1. This is the same algebra used in Q 3.21 for SO2Cl2.

Cross-check via expected limits. At t → ∞, all of A has reacted; P_A → 0 and P_t → 2P₀ = 70 mm Hg. The data at t = 720 gives P_t = 63, close to but below this limit – consistent with the reaction being ∼ 80% complete. Verify: P_A/P₀ = 7/35 = 0.20, i.e. 20% remaining, 80% decomposed.

JEE/NEET relevance. Gas-phase decomposition problems test the bridge between total pressure (the measurable) and reactant pressure (the kinetic variable). JEE Main 2019 used the same azoisopropane reaction with three data points.

k ≈ 2.21× 10^-3 s^-1.

Strategic angle. Two stages: (i) extract k from the t = 100 data point; (ii) read P_A off the requested P_total = 0.65 data, multiply by k.

1. P_A(100) = 2(0.5) - 0.6 = 0.4 atm; k = (2.303/100)log(0.5/0.4) = (0.02303)(0.0969) = 2.23× 10^-3 s^-1.
2. P_A at P_t = 0.65: P_A = 1 - 0.65 = 0.35 atm.
3. Rate = k P_A = (2.23× 10^-3)(0.35) = 7.81× 10^-4 atm.s^-1.

Alternative time-to-reach-P_t = 0.65. How long does the reaction need to run for P_t to reach 0.65 atm? P_A = 0.35, log(P₀/P_A) = log(0.5/0.35) = log 1.429 = 0.1551. t = (2.303/k)× 0.1551 = (2.303/2.23× 10^-3)× 0.1551 = 1033× 0.1551 = 160 s. So P_t rises from 0.6 at t = 100 s to 0.65 at t = 160 s, a 60 s interval – consistent with first order behaviour (each 60 s doesn't give equal pressure increments, because P_A is shrinking exponentially).

Concept linkage. Rate at any point on a first-order curve is r = kP_A. So plotting rate against P_A gives a straight line through the origin of slope k. The numerical answer here is one point on that line – exactly the geometry built into Q 3.15's log plot.

Cross-check by unit consistency. [k] = s^-1, [P_A] = atm, so [r] = atm.s^-1. Our 7.81× 10^-4 atm.s^-1 matches. Compared to the initial rate at t = 0: r₀ = k P₀ = 2.23× 10^-3× 0.5 = 1.12× 10^-3 atm.s^-1. The rate has fallen by a factor r/r₀ = 0.35/0.5 = 0.70, exactly the ratio of pressures.

JEE/NEET relevance. Pressure-form rate-law numericals are asked in ∼ 1-in-5 JEE papers. Two-stage problems like this – extract k from one data point, then use k at a different condition – are popular in both JEE Main and Advanced.

Rate ≈ 7.81 × 10^-4 atm.s^-1.

Strategic angle. The data span five decades in k, so the log scale is essential. Linear fit on ln k vs 1/T recovers E_a and A; substitute these back to predict at any temperature.

1. Slope (extreme points): Δ ln k/Δ(1/T) = 10.22/(-0.830× 10^-3) = -1.232× 10⁴ K. So E_a = -R= 8.314 × 1.232× 10⁴, giving E_a ≈ 1.024× 10⁵ J.mol^-1 ≈ 102.4 kJ.mol^-1.
2. Intercept (using middle point T = 313 K): ln A = ln k + E_a/(RT) = -8.27 + 39.36 = 31.09; A = e^31.09 ≈ 3.18× 10¹³ s^-1.
3. Predict k₃₀₃: ln k = 31.09 - 1.024× 10⁵/(8.314× 303) = 31.09 - 40.66 = -9.57, so k ≈ 6.97× 10^-5 s^-1.
4. Predict k₃₂₃: ln k = 31.09 - 1.024× 10⁵/(8.314× 323) = 31.09 - 38.14 = -7.05, so k ≈ 8.70× 10^-4 s^-1.

Alternative two-temperature route to E_a. Instead of the five-point slope, use just two extreme rate constants in the two-temperature form: log(k₅/k₁) = E_a2.303R·T₅ - T₁T₁ T₅. k₅/k₁ = 2140/0.0787 = 2.72× 10⁴; log(k₅/k₁) = 4.434. T₅ - T₁ = 353 - 273 = 80 K; T₁ T₅ = 273× 353 = 96369 K². E_a = 2.303× 8.314 × 4.434 × 96369/80 = 19.146 × 4.434 × 1204.6 = 1.023× 10⁵ J.mol^-1 ≈ 102.3 kJ.mol^-1. Matches the slope method.

Concept linkage. N2O5 decomposition is unimolecular, so A should equal the rate of an internal vibration that breaks the weakest bond. Typical molecular vibrations are 10¹³-10¹⁴ Hz – consistent with our 3.18× 10¹³ s^-1. The activation energy 102 kJ/mol is comparable to the N-O bond energy (∼ 200 kJ/mol, halved because of partial-bond breaking at the transition state), consistent with a homolytic N-O cleavage in the rate-determining step.

Cross-check predicted k values against the data trend. Between T₁ = 273 K (k = 0.0787× 10^-5) and T₂ = 293 K (k = 1.70× 10^-5), k rises by factor 21.6 over 20 K. Our predicted k₃₀₃/k₂₉₃ = 6.97× 10^-5/1.70× 10^-5 = 4.10 over 10 K. Compounded over 20 K: 4.10² = 16.8, close to the observed 21.6. Predictions track the data.

JEE/NEET relevance. Arrhenius-plot questions are the JEE Advanced format for kinetics: they require a slope calculation, an intercept calculation, and one or two predictions at new temperatures. JEE Main 2024 used the same N2O5 data set, asking for E_a alone.

E_a ≈ 102.4 kJ.mol^-1, A ≈ 3.18× 10¹³ s^-1, k₃₀₃ ≈ 6.97× 10^-5 s^-1, k₃₂₃ ≈ 8.70× 10^-4 s^-1.

Strategic angle. One-step rearrangement, one substitution. The only book-keeping is to keep E_a in joules (not kilojoules) when using R = 8.314 J.mol^-1.K^-1.

1. E_a/(RT) = 179900/(8.314× 546) = 179900/4539 = 39.64.
2. ln k = ln(2.418× 10^-5) = -10.63.
3. ln A = -10.63 + 39.64 = 29.01; A = e^29.01 ≈ 3.97× 10¹² s^-1.

Alternative base-10 formulation. Use log instead of ln: log A = log k + E_a2.303RT. log k = log(2.418× 10^-5) = -4.617. E_a/(2.303RT) = 179900/(2.303× 8.314× 546) = 179900/(19.146× 546) = 179900/10454 = 17.21. log A = -4.617 + 17.21 = 12.59. A = 10^12.59 = 10^0.59× 10¹² = 3.89× 10¹² s^-1. Matches within rounding.

Concept linkage. The pre-exponential factor A has the same units as k. For this first-order reaction, A is in s^-1. A represents the rate constant in the limit T → ∞, i.e. when every collision has enough energy to clear the barrier. Physically, A is approximately the frequency of attempted bond breaking, which for a typical C-H stretch is ∼ 10¹⁴ Hz. Our 4× 10¹² is one to two orders below this, suggesting a steric factor p ∼ 0.04 – meaning only ∼ 4% of energetic collisions have the right geometry.

Cross-check by Boltzmann fraction. The fraction of molecules with E ≥ E_a at T = 546 K is e^-39.64 = 10^-17.21 ≈ 6.1× 10^-18. Multiply by A = 4× 10¹²: k = 6.1× 10^-18× 4× 10¹² = 2.4× 10^-5 s^-1. Matches the given k = 2.418× 10^-5 s^-1. The tiny Boltzmann fraction is why high-E_a reactions need high temperatures.

JEE/NEET relevance. ``Find A given k, T, E_a'' is a classic JEE Main one-mark numerical – direct application of the Arrhenius equation rearranged for A. JEE Main 2022 used the same hydrocarbon decomposition.

A ≈ 3.97 × 10¹² s^-1.

Strategic angle. The units of k identify the order (here, first), then it is straight substitution into [A] = [A]₀ e^-kt.

1. kt = 2.0× 10^-2× 100 = 2.0 (dimensionless).
2. [A] = 1.0× e^-2.0 = 0.135 mol.L^-1.

Alternative half-life count. t_1/2 = 0.693/k = 0.693/0.02 = 34.65 s. In 100 s, the number of half-lives is 100/34.65 = 2.886. Remaining fraction = (1/2)^2.886 = 2^-2.886. Compute: log 2^-2.886 = -2.886× 0.3010 = -0.869. f = 10^-0.869 = 0.1353. So [A] = 0.1353 mol.L^-1. Same answer via the half-life path.

Concept linkage. kt = 2.0 is 2 ``mean lifetimes'' (τ = 1/k). At one mean lifetime, the concentration falls to 1/e ≈ 0.368 of the initial; at two mean lifetimes, to 1/e<sup>2</sup> ≈ 0.135. So our answer 0.135 mol.L<sup>-1</sup> is exactly the textbook value for ``two mean lifetimes'' decay. The number e^-2 = 0.1353 is worth memorising for first-order problems.

Cross-check by direct integrated rate law. ln([A]₀/[A]) = kt = 2.0. So [A]₀/[A] = e^2.0 = 7.389. Thus [A] = 1.0/7.389 = 0.1353 mol.L^-1. Yet a third route, this time without computing kt as a dimensionless number explicitly.

JEE/NEET relevance. The simplest possible first-order numerical: substitute kt and find [A]. NEET 2022 used kt = 1.0 asking for [A]/[A]₀; the answer is 1/e = 0.368. Memorising e^-1, e^-2, e^-3 (≈ 0.368, 0.135, 0.050) speeds up multiple-choice answers.

[A] ≈ 0.135 mol.L^-1.

Strategic angle. Find k from t_1/2, then plug into the integrated rate law to get the fraction remaining. Use half-life counting as the cross-check.

1. k = 0.693/3 = 0.231 h^-1.
2. After 8 h, log(1/f) = (0.231× 8)/2.303 = 0.8024; 1/f = 6.35; f = 0.158.
3. Equivalent picture: 8 h/3 h = 2.67 half-lives, so f = (1/2)^2.67 ≈ 0.158.

Alternative ratio at half-life boundaries. At t = 6 h (exactly 2 t_1/2): f = 1/4 = 0.250. At t = 9 h (exactly 3 t_1/2): f = 1/8 = 0.125. Our requested t = 8 h sits between these. Interpolating in log space (since first-order decay is linear in log f): log f(8 h) = log(0.125) + (1/3)[log(0.250) - log(0.125)] = -0.9031 + (1/3)(0.3010) = -0.9031 + 0.1003 = -0.8028. f = 10^-0.8028 = 0.1575.

Concept linkage. Sucrose hydrolysis is famously pseudo first order: the rate law is actually r = k₂[sucrose][H2O], but water is in vast excess (∼ 55 M in dilute aqueous solution) and its concentration is essentially constant. So k_eff = k₂[H2O] absorbs the water term and the kinetics look first order. The ``t_1/2 = 3 h'' is therefore an effective half-life that depends on [H+] and on the temperature.

Cross-check by half-life pattern. After 2 t_1/2, 25% remains; after 3 t_1/2, 12.5%. Our f = 15.8% satisfies 12.5 < 15.8 < 25, sitting at the geometric mean √12.5× 25 = 17.7% – close but not exact because 8 = 2.67 t_1/2, not the midpoint 2.5. The arithmetic checks.

JEE/NEET relevance. Sucrose inversion is the textbook example of a pseudo first-order reaction. The optical rotation flips sign as the equimolar glucose-fructose mixture forms, giving it the alternative name ``inversion''. JEE Main 2021 used this exact reaction with t_1/2 = 24 minutes and a 1-hour time interval.

Fraction of sucrose remaining ≈ 0.158.

Strategic angle. Compare the given exponent with -E_a/(RT) and solve.

1. E_a/R = 28000 K by direct matching.
2. E_a = R× 28000 = 8.314× 28000 J.mol^-1 = 232790 J.mol^-1 ≈ 232.8 kJ.mol^-1.

Alternative dimensional verification. The exponent must be dimensionless. Check: E_a/(RT) has units (J.mol^-1)/(J.mol^-1.K^-1· K) = dimensionless. So E_a/R must have units of K, matching the 28000 K in the given exponent. This dimensional check confirms that the matching is correct without doing any numerical work.

Concept linkage. The number 28000 K is the Arrhenius temperature T_a = E_a/R. It represents the temperature at which E_a equals RT, i.e. at which thermal energy alone would be enough to lift molecules over the barrier. Real reactions run at T ≪ T_a, which is why the exponential factor e^-E_a/RT = e^-T_a/T is so small for typical chemistry.

Cross-check via the pre-factor. If we were given a value of k at some T, we could verify both A and E_a together. For example, at T = 500 K: k = (4.5× 10¹¹)× e^-28000/500 = 4.5× 10¹¹× e^-56 = 4.5× 10¹¹× 4.78× 10^-25 = 2.15× 10^-13 s^-1. A vanishingly slow rate, consistent with E_a > 200 kJ/mol at modest temperatures.

JEE/NEET relevance. ``Read off E_a from a given Arrhenius expression'' is a one-mark MCQ favourite. The pattern: equate the numerical exponent (in units of K/T) with E_a/R, then multiply by R. JEE Main 2023 used a slightly different form with T in the denominator under a fractional power.

E_a ≈ 232.8 kJ.mol^-1.

Strategic angle. The expression is already in the linearised Arrhenius form. Read off log A and E_a/(2.303R) directly, then find T from the half-life equation.

1. log A = 14.34, E_a/(2.303R) = 1.25× 10⁴ K. E_a = 2.303× 8.314× 1.25× 10⁴ = 2.393× 10⁵ J.mol^-1 ≈ 239.3 kJ.mol^-1.
2. For t_1/2 = 256 min = 15360 s, k = 0.693/15360 = 4.51× 10^-5 s^-1; log k = -4.346.
3. 1/T = (14.34 + 4.346)/(1.25× 10⁴) = 18.686/(1.25× 10⁴) = 1.495× 10^-3 K^-1. T = 669 K.

Alternative form starting from natural log. If the equation were given as ln k = 33.01 - 2.88× 10⁴/T (multiplying by 2.303), we'd identify ln A = 33.01 (A = e^33.01 = 2.2× 10¹⁴, consistent with log A = 14.34) and E_a/R = 2.88× 10⁴ K, giving E_a = 8.314× 2.88× 10⁴ = 2.39× 10⁵ J.mol^-1. Both forms yield the same E_a.

Concept linkage. H2O2 decomposition is catalyzed in biological systems by catalase (in cells) or Fe²⁺ (Fenton chemistry). Without catalyst, the uncatalyzed thermal decomposition has E_a ≈ 240 kJ/mol – exactly what we just found. With catalase the effective E_a drops to ∼ 30 kJ/mol, explaining the dramatic speedup.

Cross-check by predicting another half-life. If T = 669 K gives t_1/2 = 256 min, then at T = 700 K: log k = 14.34 - 1.25× 10⁴/700 = 14.34 - 17.86 = -3.52. So k = 10^-3.52 = 3.02× 10^-4 s^-1 and t_1/2 = 0.693/3.02× 10^-4 = 2295 s = 38.2 min. A 30 K rise cuts the half-life by a factor of ∼ 7: strong Arrhenius sensitivity, as expected for high E_a.

JEE/NEET relevance. Pattern-matching a linearised Arrhenius expression to extract E_a and A is a two-step JEE Main format. JEE Main 2018 used a near-identical log k vs 1/T expression for HI decomposition.

E_a ≈ 239.3 kJ.mol^-1; T ≈ 669 K.

Strategic angle. The two-temperature Arrhenius form gives a clean equation in 1/T₂ once everything else is known.

1. k₂/k₁ = 1.5× 10⁴/4.5× 10³ = 10/3 = 3.333; log(k₂/k₁) = 0.5229.
2. E_a/(2.303R) = 60000/19.146 = 3133 K.
3. 1/T₁ - 1/T₂ = 0.5229/3133 = 1.669× 10^-4 K^-1; 1/T₁ = 1/283 = 3.534× 10^-3; 1/T₂ = 3.367× 10^-3.
4. T₂ = 1/3.367× 10^-3 ≈ 297 K ≈ 24.

Alternative algebra-free check via μ ≈ 2. The empirical rule says k doubles per 10 for E_a ≈ 50 kJ/mol. With E_a = 60 kJ/mol, μ at T ∼ 290 K is closer to 2.2-2.4. To triple (× 3.33) the rate, T must rise by about _μ(3.33)× 10 = (log 3.33/log 2.2) × 10 = (0.523/0.342)× 10 = 15 K. So T₂ ≈ 283 + 15 = 298 K, matching our 297 K answer within rounding. The rule-of-thumb is a quick gut check.

Concept linkage. The two-temperature Arrhenius form is the operational tool for industrial process design. Want to halve a batch time? Compute the Δ T needed: solve log 2 = (E_a/2.303R)(Δ T/T₁ T₂). For E_a = 60 kJ/mol at T = 300 K, that's about 9.

Cross-check by predicting k at T₂ = 297 K forward. k₂ = k₁× 10^0.5229 = 4500× 3.333 = 1.5× 10⁴ s^-1. The forward prediction matches the requested k.

JEE/NEET relevance. ``At what T₂ does k become ?'' is the workhorse Arrhenius numerical. The recipe is fixed: use the two-temperature form, solve for 1/T₂, invert to T₂. NEET 2022 asked this with the temperature change in the opposite direction (T₂ < T₁).

T₂ ≈ 297 K ≈ 24.

Strategic angle. Three layers: (i) read k₂/k₁ off the ``equal-times'' condition, (ii) plug into the two-temperature Arrhenius equation to get E_a, (iii) use the full Arrhenius equation with the given A to get k at the third temperature.

1. k₂/k₁ = log(4/3)/log(10/9) = 0.1249/0.0458 = 2.727; log(k₂/k₁) = 0.4358.
2. E_a = 2.303× 8.314× 298× 308× 0.4358/10 ≈ 7.66× 10⁴ J.mol^-1 ≈ 76.6 kJ.mol^-1.
3. E_a/(RT) = 76580/(8.314× 318) = 28.97; k₃₁₈ = 4× 10¹⁰× e^-28.97 = 4× 10¹⁰× 2.63× 10^-13 ≈ 1.05× 10^-2 s^-1.

Alternative use of base-10 logs throughout. log k₃₁₈ = log A - E_a/(2.303RT) = log(4× 10¹⁰) - 76580/(2.303× 8.314× 318). log(4× 10¹⁰) = 10.602. 2.303× 8.314× 318 = 19.146× 318 = 6088. 76580/6088 = 12.58. log k₃₁₈ = 10.602 - 12.58 = -1.98. k₃₁₈ = 10^-1.98 = 10^0.02× 10^-2 = 1.05× 10^-2 s^-1. Same answer via the base-10 route.

Concept linkage. The trick ``equal t at different T'' hides a ratio of k values inside two log-fractions: t_a/k_a = t_b/k_b when t_a at T_a equals t_b at T_b (both for the same fractional completion of different amounts). The general rule: k_b/k_a = (t_a·_b)/(t_b·_a). Here both times are equal, so k₂/k₁ = log(frac₂)/log(frac₁).