Class 12 Chemistry Ch 3 Chemical Kinetics Exemplar - Download PDF

Correct option: (iii) activation energy of reaction.

Concept used. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. It works by providing an alternative reaction pathway of lower activation energy E_a: the energy barrier between reactants and the transition state (activated complex). A catalyst does not change the energies of the reactants or the products themselves, so it cannot change any state function of the reaction.

1. Recall the thermodynamic identities for a reaction at constant T and p: Δ G = G_products - G_reactants, Δ H = H_products - H_reactants. Both are differences between the products' and the reactants' state functions. A catalyst does not alter either state.
2. The equilibrium constant is fixed by Δ G^∘: Δ G^∘ = -RT ln K_eq. Since Δ G^∘ is unchanged, K_eq is unchanged. So options (i), (ii) and (iv) are eliminated.
3. Only the activation energy can be lowered. With E_a smaller, the Arrhenius rate constant k = A e^-E_a/RT grows because the exponent becomes less negative.

Option (iii): a catalyst changes the activation energy of the reaction.

Correct option: (iii) remains unchanged.

Concept used. The ``heat evolved or absorbed'' at constant pressure is the enthalpy change of the reaction, Δ H_rxn = H_products - H_reactants. Enthalpy is a state function: its value depends only on the initial (reactants) and final (products) states, not on the path between them. A catalyst changes only the path, never the endpoints.

1. Write the energy profile with and without catalyst. Both start at H_reactants and both end at H_products. So the vertical gap Δ H = H_products - H_reactants is the same in both pictures.
2. Hess's law confirms this: any sequence of steps that takes the same reactants to the same products gives the same Δ H. The catalysed pathway is just a different sequence of elementary steps, so its overall Δ H matches the uncatalysed one.
3. Eliminate (i), (ii), (iv): all three claim a change in Δ H, which violates Hess's law.

Option (iii): the heat exchanged is unchanged because Δ H is a state function.

Correct option: (ii) determining the rate constants at two temperatures.

Concept used. The Arrhenius equation k = A e^-E_a/RT links the rate constant k to absolute temperature T through the activation energy E_a and the pre-exponential factor A. Taking natural logarithm: ln k = ln A - E_aR T. For two different temperatures T₁ and T₂ with rate constants k₁ and k₂: ln(k₂k₁) = E_aR(1T₁ - 1T₂).

1. One unknown (E_a) cannot be extracted from a single equation ln k₁ = ln A - E_a/(R T₁) because A is also unknown. Two equations are needed: option (i) gives only one, so it is insufficient.
2. Two rate-constant measurements at T₁ and T₂ give two equations. Subtracting them eliminates A and isolates E_a, hence option (ii) works.
3. Option (iii) (``probability of collision'') corresponds to the steric factor P inside the pre-exponential A, not to E_a. Option (iv) (using a catalyst) changes E_a to a different value E_a', so it cannot measure the original E_a.

Option (ii): measure k at two temperatures and solve for E_a.

Correct option: (i) activation energy of forward reaction is E₁+E₂ and product is less stable than reactant.

Concept used. An energy-profile diagram plots potential energy along the reaction coordinate. The reactant level, product level, and activated-complex (transition state) level appear as horizontal plateaus / a peak. Reading off the figure:

• [leftmargin=*]
• The reactant sits below the dashed line through ``Products''; the product sits above the reactant.
• E₁ is the vertical gap from the dashed product-energy line up to the peak.
• E₂ is the vertical gap from the reactant level up to the same dashed product line.

1. Activation energy of the forward reaction is the height from the reactant level to the peak: E_a,forward = (peak) - (reactant) = E₁ + E₂.
2. Activation energy of the backward reaction is the height from the product level to the peak: E_a,backward = (peak) - (product) = E₁. So the two are not equal: option (iii) is wrong.
3. Stability comparison: the product sits higher than the reactant (by E₂), so the product has more potential energy and is therefore less stable. This rules out options (ii) and (iv).

Option (i): forward E_a = E₁+E₂; product is less stable than reactant.

Correct option: (ii) k=2.303tlogp_i2p_i-p_t.

Concept used. For a first order reaction A -> products the integrated rate law in terms of pressures is k = 2.303t logp_A,0p_A,t, where p_A,0 is the initial partial pressure of A and p_A,t is its partial pressure at time t. We must therefore express p_A,t in terms of the given total pressure p_t and the initial pressure p_i.

1. Set up a pressure table. Initial: p_A=p_i, p_B=0, p_C=0; total =p_i. Let x be the pressure of A that has decomposed by time t. Then at time t: p_A=p_i-x, p_B=x, p_C=x, p_total = p_i-x+x+x = p_i+x.
2. The total pressure at time t is p_t=p_i+x, so x = p_t - p_i. The partial pressure of A is p_A,t = p_i - x = p_i - (p_t-p_i) = 2p_i - p_t.
3. Substitute in the first-order integrated rate law with p_A,0 = p_i: k = 2.303t logp_i2p_i-p_t. This matches option (ii).

Option (ii): k=2.303tlogp_i2p_i-p_t.

Correct option: (i) ln k vs 1/T is a straight line with negative slope and a positive ln k intercept on the vertical axis.

Concept used. Take natural logarithm of the Arrhenius equation k = A e^-E_a/RT: ln k = ln A - E_aR·1T. Comparing with y = mx + c: y = ln k, x = 1/T, slope m = -E_a/R, intercept c = ln A. Since E_a > 0, the slope is negative; since the pre-exponential factor A > 0, the intercept ln A is positive (and the line meets the ln k axis at a non-zero point).

1. Negative slope means ln k decreases as 1/T increases (i.e., as temperature drops). Equivalently, ln k increases when temperature rises. Graphs (ii), (iii), (iv) all show positive slope, so they violate the sign.
2. The intercept on the vertical axis at 1/T=0 is ln A, a finite positive number. Only option (i) shows this: the line crosses the ln k axis at a positive value and then descends with negative slope as 1/T grows.
3. Option (iii) (line through the origin with positive slope) and option (iv) (positive slope, ln k axis intercept positive) both contradict the negative slope.

Option (i): straight line with negative slope -E_a/R and positive intercept ln A.

Correct option: (iv) rate constant increases exponentially with decreasing activation energy and increasing temperature.

Concept used. In the Arrhenius equation k = A e^-E_a/RT, the exponent is the dimensionless ratio -E_a/(RT). Two ways to make the exponent less negative (and therefore k larger):

• [leftmargin=*]
• Lower the numerator E_a (smaller barrier).
• Raise the denominator RT (more thermal energy per mole).

1. Hold T fixed and lower E_a: As E_a decreases, -E_a/(RT) becomes less negative, k = A e^-E_a/RT rises. So a decrease in E_a raises k. This rules out options (i) and (ii), which claim E_a increasing raises k.
2. Hold E_a fixed and raise T: As T increases, -E_a/(RT) approaches 0, k = A e^-E_a/RT rises. So an increase in T raises k. This rules out options (i) and (iii), which claim T decreasing raises k.
3. Combining both directions, option (iv) is the only one consistent with the formula.

Option (iv): k rises as E_a falls and as T rises.

Correct option: (iii) V₃40.

Concept used. The average rate of a reaction over a time interval [t_a, t_b] is r̄ = Δ(measured quantity)Δ t = Q(t_b) - Q(t_a)t_b - t_a. ``Average rate upto 40 s'' means the time interval starts at t=0 (initial) and ends at t=40 s, so t_a=0, t_b=40.

1. Read from Fig. 4.2: at t=0 no hydrogen has been released yet, so V(0)=0. At t=40 s the curve reads V₃.
2. Apply the average-rate formula: r̄ = V(40) - V(0)40 - 0 = V₃ - 040 = V₃40. This is option (iii).
3. Why the other options fail. Options (i), (ii), (iv) use time intervals 0→ 30, 30 → 40 and 20 → 40 respectively for the numerator while keeping the denominator as 40 or 40-30 or 40-20. None of these is the ``up to 40 s'' interval [0,40] that starts at the origin.

Option (iii): r̄=V₃40.

Correct option: (iii) the order is not always equal to the sum of stoichiometric coefficients (so this statement is incorrect).

Concept used. The order of a reaction is the experimentally measured exponent in the rate law. For a rate law of the form r = k [A]^x [B]^y, the order is x+y. Two crucial facts:

• [leftmargin=*]
• Order need not equal molecularity. Molecularity comes from the balanced elementary step; order comes from the empirical rate law of the overall reaction.
• For complex (multi-step) reactions, the rate law is governed by the slowest (rate-determining) elementary step, so the stoichiometric coefficients of the overall balanced equation are usually irrelevant.

1. Check (i): order can be fractional (e.g. half-order in H2 for the H2 + Br2 reaction). True statement.
2. Check (ii): order is determined experimentally from kinetic data (initial-rate method, half-life method, integrated-rate-equation fits). True statement.
3. Check (iii): for KClO3 + 6FeSO4 + 3H2SO4 -> KCl + 3H2O + 3Fe2(SO4)3, the stoichiometric sum is 1+6+3=10, but the reaction is experimentally second order. So the claim that order always equals stoichiometric sum is false. Statement (iii) is incorrect ⇒ this is the answer.
4. Check (iv): for any rate law r = k [A]^x [B]^y … the order is x + y + ⋯. True statement.

Option (iii) is the incorrect statement; order need not equal the sum of stoichiometric coefficients.

Correct option: (ii) V₄-V₂50-30.

Concept used. The instantaneous rate at time t is the slope of the tangent to the volume-vs-time curve at that point. Numerically, it is best estimated by a small symmetric interval centred at the time in question, [t-Δ t, t+Δ t], so the approximation is r_inst(t) ≈ V(t+Δ t) - V(t-Δ t)2Δ t. The closer and the more symmetric the interval, the better the estimate.

1. Read Fig. 4.2's time-axis ticks: 20, 30, 40, 50. The volume readings on the curve are V₁ at 20 s, V₂ at 30 s, V₃ at 40 s, V₄ at 50 s on the curve, and V₅ at 50 s on the tangent line.
2. Build symmetric chords centred at t=40 s:
   • [leftmargin=*]
   • [30, 50] on the tangent → (V₅-V₂)/(50-30), option (i). The tangent line itself, so this gives the true slope at t=40.
   • [30, 40] on the curve → (V₃-V₂)/(40-30), option (iii). Asymmetric one-sided chord, but a reasonable secant-slope estimate.
   • [20, 40] on the curve → (V₃-V₁)/(40-20), option (iv). Symmetric chord of width 20 across t=30? Actually it is centred at t=30, not at t=40.
   Re-reading the Exemplar's intent: options (i), (iii), (iv) each represent a legitimate way to read off the rate at t=40 from Fig. 4.2 (tangent, forward chord, backward chord). Option (ii) uses V₄ which lies on the curve at 50 s (not on the tangent line), so (V₄-V₂)/(50-30) is the average rate from 30 s to 50 s, not the instantaneous rate at 40 s.
3. Therefore, option (ii) does not represent the instantaneous rate at t = 40 s; it is an average rate.

Option (ii): (V₄-V₂)/(50-30) is the average rate from 30 to 50 s, not the instantaneous rate at 40 s.

Correct option: (i) the rate of a reaction decreases with passage of time as the concentration of reactants decreases.

Concept used. The rate law for any non-zero order reaction makes rate depend on the instantaneous concentration of reactants: r = k [A]^x [B]^y ⋯ As the reaction proceeds, reactants are consumed and their concentrations [A], [B], … decrease. With x, y > 0, the rate r also decreases. Two more facts:

• [leftmargin=*]
• Rate depends on temperature through k = A e^-E_a/RT (Arrhenius), so it is not independent of T.
• Higher reactant concentration ⇒ more collisions per unit volume per unit time ⇒ higher rate, not lower.

1. Check (i): rate ∝ [A]^x with x > 0 in nearly every reaction (zero-order is the special exception). As [A] drops, r drops. True.
2. Check (ii): claim is that rate is constant in time. False except for zero-order reactions, and the question is asking about a general reaction. So (ii) is incorrect.
3. Check (iii): Arrhenius equation makes k depend on T. So rate does depend on temperature. (iii) is incorrect.
4. Check (iv): higher [A] raises r, not lowers it. Statement (iv) reverses the direction. Incorrect.

Option (i) is the correct statement.

Correct option: (iii) Δ[Br^-]Δ t=56Δ[H⁺]Δ t.

Concept used. For a balanced reaction aA + bB + cC → dD + eE, the unique rate of reaction r is obtained by dividing each species' rate of change by its stoichiometric coefficient and giving the reactants a minus sign: r = -1aΔ[A]Δ t = -1bΔ[B]Δ t = 1dΔ[D]Δ t ⋯ Two of these forms can be equated, giving relations between Δ[A]/Δ t and Δ[B]/Δ t.

1. Read coefficients of Br^- and H⁺ from the balanced equation: 5 and 6 respectively. Both are reactants.
2. Write the common rate using each: r = -15Δ[Br^-]Δ t = -16Δ[H⁺]Δ t.
3. Eliminate the common r to relate Δ[Br^-] and Δ[H⁺]: 15Δ[Br^-]Δ t = 16Δ[H⁺]Δ t ⇒ Δ[Br^-]Δ t = 56Δ[H⁺]Δ t. (Both reactants disappear, so both Δ/Δ t are negative; the ratio 5/6 is positive, matching option (iii).)

Option (iii): Δ[Br^-]Δ t=56Δ[H⁺]Δ t.

Correct option: (i) (a) only.

Concept used. A reaction is exothermic if the product level lies below the reactant level on the energy-coordinate plot (so Δ H = H_products - H_reactants < 0, heat is released). It is endothermic if the product level is above the reactant level (Δ H > 0). Reading the three graphs:

1. Graph (a): the curve starts at the ``Reactants'' plateau, rises to the ``Activated complex'' peak, then falls to ``Products'' below the starting level. Products are lower than reactants ⇒ Δ H < 0 ⇒ exothermic.
2. Graph (b): the curve starts at ``Reactants'' lower, rises to the peak, then descends to ``Products'' above the starting level. Products higher ⇒ Δ H > 0 ⇒ endothermic.
3. Graph (c): the curve starts and ends at the same level (Reactants and Products at the same height). Products equal reactants ⇒ Δ H = 0 (thermoneutral, neither exothermic nor endothermic).

Only (a) shows the product level below the reactant level. So the answer is (i): (a) only.

Option (i): graph (a) alone is exothermic.

Correct option: (i) the same.

Concept used. The rate constant k in a rate law is a property of the reaction itself and the temperature: it does not depend on reactant concentrations. The rate r varies with [A] and [B], but the proportionality constant k stays fixed at a given T.

1. Write the rate law: r = k [A] [B]. The two factors that change with experiment are [A] and [B]. The factor k is fixed by the reaction conditions (T, catalyst, solvent), not by how much A or B is in the flask.
2. Doubling [B] changes r: r_new = k [A] (2[B]) = 2 k [A] [B] = 2r. So rate doubles. But k itself is unchanged because we did not change T or add a catalyst.
3. Options (ii)–(iv) confuse rate (which doubles) with rate constant (which is unchanged).

Option (i): k is unchanged; only the rate doubles when [B] doubles.

Correct option: (iii) (this is the incorrect statement).

Concept used. The collision theory of bimolecular reactions makes the following claims:

• [leftmargin=*]
• Molecules behave like hard spheres (no internal structure).
• A reaction occurs only when colliding molecules have energy ≥ the threshold (activation) energy E_a AND collide in a proper orientation.
• Rate ∝ Z_AB e^-E_a/RT P, where Z_AB is the collision frequency and P is the steric (orientation) factor.

1. Check (i): collision theory models molecules as hard spheres and ignores structural details. True.
2. Check (ii): rate = number of effective collisions per unit volume per unit time. True.
3. Check (iii): claims that energy alone is sufficient. This omits the orientation requirement. False.
4. Check (iv): both energy and orientation are required. True (this is the statement (iii) gets wrong).

Option (iii) is the incorrect statement; energy alone is not enough.

Correct option: (iv) infinite.

Concept used. For a first order reaction, the integrated rate law is [R]_t = [R]₀ e^-kt ⇔ t = 1k ln[R]₀[R]_t. At 100% completion, all reactant is consumed: [R]_t = 0. The ratio [R]₀/[R]_t diverges to infinity, and so does ln of it. Hence t → ∞. A first-order reaction approaches completion asymptotically and never literally completes in finite time.

1. Start from the integrated form, [R]_t = [R]₀ e^-kt. At ``100% completion'', [R]_t = 0, so 0 = [R]₀ e^-kt ⇒ e^-kt=0 ⇒ kt = ∞ ⇒ t = ∞.
2. Sanity check using t_1/2. For 50% completion at t_1/2 = 1.26× 10¹⁴ s, k = 0.693t_1/2 = 0.6931.26× 10¹⁴ = 5.50 × 10^-15 s^-1. Substitute into the formula for completion of fraction f: t = 2.303klog11-f. At f → 1, log[1/(1-f)] → ∞, so t → ∞.
3. Options (i)–(iii) give finite numbers; only (iv) matches.

Option (iv): a first-order reaction needs infinite time for 100% completion.

Correct option: (ii) Rate = k [A] [B]².

Concept used. The initial-rates method: compare two experiments in which only one reactant concentration is changed, keeping the other(s) constant. The ratio of rates determines the order in the changing reactant via r₂r₁ = ([X]₂[X]₁)^x, where x is the order in X and [X] is the changing concentration.

1. Determine order in B. Compare experiments 1 and 2: [A] is held at 0.30, [B] doubles from 0.30 to 0.60, and rate goes from 0.10 to 0.40 (a 4-fold rise). r₂r₁ = 0.400.10 = 4 = 2^y ⇒ y = 2. Order in B is 2.
2. Determine order in A. Compare experiments 1 and 3: [B] is held at 0.30, [A] doubles from 0.30 to 0.60, and rate goes from 0.10 to 0.20 (a 2-fold rise). r₃r₁ = 0.200.10 = 2 = 2^x ⇒ x = 1. Order in A is 1.
3. Combine: r = k [A]¹ [B]² = k [A] [B]², which matches option (ii). Overall order = 1 + 2 = 3.

Option (ii): Rate = k [A] [B]², overall order 3.

Correct option: (ii) (this is the incorrect statement; a catalyst does not alter Δ G).

Concept used. A catalyst lowers E_a for both forward and backward elementary steps by the same amount, so both k_f and k_b increase by the same factor. The equilibrium constant K_eq = k_f/k_b and the Gibbs energy Δ G = -RTln K_eq are therefore unchanged. So catalysts:

• [leftmargin=*]
• Accelerate forward and reverse equally (true).
• Leave Δ G, Δ H, K_eq unchanged (true).
• Provide an alternate, lower-E_a pathway (true).

1. Check (i): same lowering of E_a on both sides of the barrier, so the rate-constant ratio k_f/k_b is unchanged and both sides speed up by the same factor. True.
2. Check (ii): claims Δ G changes. Since Δ G depends only on the reactant and product free energies, which the catalyst leaves alone, Δ G is unchanged. Statement (ii) is incorrect.
3. Check (iii): with Δ G unchanged, K_eq = e^{-Δ G∘/(RT)} is unchanged. True.
4. Check (iv): this is the standard textbook definition. True.

Option (ii) is the incorrect statement; a catalyst does not alter Δ G.

Correct option: (ii) depends on the concentration of reactants present in excess.

Concept used. A pseudo first order reaction is one that is intrinsically of higher order (often second order overall), but is run with one reactant in such large excess that its concentration is essentially constant during the reaction. Example: CH3COOC2H5 + H2O H⁺ CH3COOH + C2H5OH, true rate = k [ester] [H2O]. Since water is the solvent ([H2O] ≈ 55.5 M, hardly changing during the reaction), the empirical rate r = k [H2O] [ester] = k' [ester], with k' = k [H2O] behaving as a (pseudo) first-order rate constant.

1. True rate law (second order): r = k [A] [B].
2. With B in excess and roughly constant: k_obs = k' = k [B]_excess. The observed rate constant therefore includes the concentration of B in excess.
3. So k_obs depends on the concentration of the excess reactant. If you change [B]_excess (say run the experiment in a different solvent ratio), the pseudo first-order rate constant k' changes.

Option (ii): the pseudo first-order rate constant absorbs the concentration of the excess reactant.

Correct option: (ii).

Concept used. For the reversible first-order reaction A B starting from pure A, both [A] and [B] relax exponentially towards their equilibrium values: [A](t) = [A]_eq + ([A]₀ - [A]_eq) e^{-(k_f+k_b) t}, [B](t) = [B]_eq (1 - e^{-(k_f+k_b) t}). Hence [A] starts at [A]₀ and decays exponentially to [A]_eq (a horizontal plateau), while [B] starts at 0 and rises exponentially to [B]_eq. The two curves cross once and then flatten.

1. Identify the qualitative features. (a) [A] starts high, decays monotonically to a constant; (b) [B] starts at 0, rises monotonically to a constant; (c) at long times both curves flatten as the system reaches equilibrium.
2. Compare with the four sub-figures: option (ii) is the only one where [A] falls exponentially to a positive plateau (not to zero, because the reaction is reversible) and [B] rises exponentially to the same plateau. The two curves meet at the intermediate level of equilibrium.
3. Rule out the others. Option (i) shows [B] rising sharply at long times, not flattening. Option (iii) shows [B] rising linearly (not exponentially). Option (iv) shows [A] rising and [B] falling, which reverses the labels.

Option (ii): both [A] and [B] approach the equilibrium level exponentially.

Correct options: (i), (iii), (iv).

Concept used. The rate law can be predicted from the balanced equation only if the reaction is a single elementary step. In every other case, the rate law must be measured experimentally. The three situations where direct prediction fails:

• [leftmargin=*]
• Reverse reaction matters: the net rate is r = k_f [A]^a [B]^b - k_b [C]^c [D]^d, not just the forward power product.
• Multi-step (complex) mechanism: the slowest step controls the rate law, and the slowest step may involve fewer or more species than the overall equation.
• Excess reactant: the apparent order collapses (pseudo-order behaviour). E.g., a true second-order reaction looks first-order if one reactant is in large excess.

1. Check (i): if reverse reaction is fast and competitive, the net rate involves k_b and product concentrations. The balanced equation alone gives only the forward stoichiometry. So (i) is correct (rate law cannot be read off).
2. Check (ii): for an elementary reaction, the rate law's exponents do equal the stoichiometric coefficients, so it can be predicted directly. (ii) is wrong (i.e., not a case where prediction fails).
3. Check (iii): for multi-step (complex) reactions, the slow step determines the rate law. The overall equation is the sum of all steps, so it does not reveal the slow step. (iii) is correct.
4. Check (iv): when one reactant is in excess, its concentration is constant and absorbed into k_obs, hiding the true order in that species. The balanced equation gives no clue about which one is in excess. (iv) is correct.

Options (i), (iii), (iv).

Correct options: (i), (iv).

Concept used. For an elementary reaction (one single step), order and molecularity coincide: both equal the number of molecules colliding in that step. Molecularity is defined only for elementary steps and must be a positive integer (1, 2, or rarely 3); it cannot be zero because at least one molecule must be involved in the collision.

1. Statement (i): order = molecularity for elementary reactions. True.
2. Statements (ii) and (iii): order may differ from molecularity only for complex reactions, not elementary. So these do not apply.
3. Statement (iv): molecularity is the number of molecules colliding, hence ≥ 1. Cannot be zero. True.

Options (i) and (iv).

Correct options: (i), (ii).

Concept used. A unimolecular reaction has molecularity 1 in the rate-determining step: a single molecule undergoes the change. For an elementary unimolecular step, order = molecularity = 1, so the rate is r = k [A].

1. Statement (i): unimolecular ⇒ one species in the rate-determining step. True.
2. Statement (ii): for the rate-determining step, order = molecularity = 1 (both equal). True.
3. Statement (iii): claims molecularity = 1 and order = 0. Contradicts (ii); for an elementary unimolecular step both are 1. False.
4. Statement (iv): says molecularity = order = 1 for the whole reaction. For complex reactions, even if the slow step is unimolecular, the overall reaction may include pre-equilibrium steps that change the overall order. So (iv) is not always true. The most defensible answer is just (i) and (ii), referring to the slow step.

Options (i), (ii).

Correct options: (i), (iv).

Concept used. For a complex (multi-step) reaction, the rate is controlled by the slowest elementary step. The molecularity of that slow step (an integer ≥ 1) equals the order of the overall reaction in most simple cases; complex mechanisms with pre-equilibria can shift this, but typically the question is asking about the standard case. Molecularity, being a count of colliding molecules, cannot be zero or fractional.

1. Statement (i): in the absence of pre-equilibrium gymnastics, the overall order = molecularity of the rate-determining step. True for standard mechanisms.
2. Statements (ii), (iii): not generally true; they would require complications beyond the standard kinetic picture. Skip.
3. Statement (iv): molecularity is a count of reactant molecules in an elementary step. It is always 1, 2, or 3; never 0 and never fractional. True.

Options (i) and (iv).

Correct options: (i), (iii), (iv).

Concept used. A zero-order reaction has rate independent of reactant concentration: r = k [A]⁰ = k. This happens here because the platinum catalyst surface is saturated with ammonia at high pressure: every active site is occupied, and the rate is set by the (constant) surface decomposition step.

1. Statement (i): for zero order, rate = k. True.
2. Statement (ii): rate is independent of [NH3] because the order is zero. False.
3. Statement (iii): as the reaction proceeds, [NH3] decreases but the rate stays at k (the saturated surface keeps re-filling until ammonia runs out). True.
4. Statement (iv): the zero-order behaviour holds only within the surface-saturated regime. A further change in pressure that takes the system out of that regime (e.g. at very low coverage, or when the surface chemistry changes) will alter the rate. So pressure does change the rate in general. True.

Options (i), (iii) and (iv).

Correct options: (i), (iv).

Concept used. The activated complex sits at the peak of the energy-profile diagram, higher than both reactants and products. When it decomposes, it can go either way: ``downhill'' back to reactants, or ``downhill'' on to products. Either way, the descent from the peak releases energy.

1. From the peak (H_*), descending to reactants (H_R < H_*) releases energy H_* - H_R = E_a,forward.
2. From the peak, descending to products (H_P < H_*) releases H_* - H_P = E_a,backward.
3. In both branches, ``energy is released'' (statement i). Going back to reactants is option (iv). So (i) and (iv) are both correct.

Options (i) and (iv).

Correct options: (i), (iii).

Concept used. The Maxwell-Boltzmann (M-B) distribution of molecular kinetic energies has a single peak at the most probable energy E_mp = 12kT per molecule (in 1D; (3/2)kT for the mean kinetic energy in 3D). As temperature rises:

• [leftmargin=*]
• The peak shifts to higher energy (most probable energy increases).
• The peak flattens and broadens (so the fraction of molecules at the most probable energy decreases; the total area is conserved).

1. Statement (i): fraction at E_mp decreases at higher T. True (curve flattens).
2. Statement (ii): claims fraction increases. False.
3. Statement (iii): E_mp shifts to the right. True.
4. Statement (iv): false.

Options (i) and (iii).

Correct options: (i), (iv).

Concept used. The M-B distribution is a normalised probability density: ₀^∞ f(E) dE = 1 regardless of temperature. The total area under the curve is therefore conserved (equal to 1) at every T. As T rises, the peak shifts to higher E (right) and broadens (more spread).

1. Total probability = area = 1, fixed. So (i) is correct, (ii) and (iii) are wrong.
2. Peak shift right and broadening: (iv) is correct.

Options (i) and (iv).

Correct options: (i), (ii).

Concept used. Arrhenius: k = A e^-E_a/RT. Increase T ⇒ exponent less negative ⇒ k rises ⇒ rate rises. Decrease E_a ⇒ exponent less negative ⇒ k rises ⇒ rate rises.

1. (i): T↑ raises rate. True.
2. (ii): E_a↓ raises rate. True.
3. (iii): claims k decreases with T. Opposite of Arrhenius. False.
4. (iv): claims rate falls when E_a falls. Opposite. False.

Options (i) and (ii).

Correct (incorrect) options: (ii), (iv).

Concept used. A catalyst (a) provides a new lower-E_a pathway and (b) leaves all thermodynamic state functions (including Δ H) unchanged. So a statement is incorrect if it claims the catalyst raises E_a or alters Δ H.

1. (i) Correct: catalyst provides an alternative pathway. So statement (i) is true (not incorrect).
2. (ii) Wrong: catalyst lowers, not raises, E_a. Mark as incorrect.
3. (iii) Correct: catalyst lowers E_a. Not incorrect.
4. (iv) Wrong: Δ H is a state function, unchanged by catalyst. Mark as incorrect.

Options (ii) and (iv) are incorrect.

Correct options: (i), (iv).

Concept used. For a zero-order reaction:

• [leftmargin=*]
• Rate is independent of [R]: r = k (rate vs time is a horizontal line, then drops sharply when [R] runs out).
• Concentration falls linearly with time: [R] = [R]₀ - kt, a straight line with slope -k.

1. Graph (i): reaction rate vs time is a horizontal plateau, then drops to zero when [R] is exhausted. This is the defining ``rate = k until reactant runs out'' picture. Correct for zero order.
2. Graph (ii): concentration vs time shows an exponential decay. That is first order, not zero order. Incorrect.
3. Graph (iii): rate vs time is rising, which makes no physical sense for a reaction whose reactant is being consumed. Incorrect.
4. Graph (iv): concentration vs time is a straight line with negative slope -k. This is exactly [R] = [R]₀ - kt. Correct for zero order.

Options (i) and (iv).

Correct options: (i), (iv).

Concept used. Two signatures of a first order reaction:

• [leftmargin=*]
• Half-life is independent of initial concentration: t_1/2 = 0.693/k does not depend on [R]₀. So t_1/2 vs [R]₀ is a horizontal line.
• log([R]₀/[R]) vs time is a straight line with slope k/2.303, passing through the origin (since at t=0 the log is 0).

1. (i) t_1/2 vs [R]₀ horizontal: signature of first order. Correct.
2. (ii) t_1/2 linearly decreasing with [R]₀: this is the signature of second order (t_1/2 = 1/(k[R]₀) is a hyperbola; linear decrease is also wrong, but more importantly, not first order). Incorrect.
3. (iii) Concentration vs time as exponential decay: that does describe first order, but the option's label is ``molar concentration [P]'' (product), and the curve is a decaying exponential, which is wrong (product should increase). Incorrect.
4. (iv) log([R]₀/[R]) vs time as straight line through origin with slope k/2.303: signature of first order. Correct.

Options (i) and (iv).

Concept used. A bimolecular reaction has rate law r = k [A] [B] (two molecules in the rate-determining step). If one reactant is in such large excess that its concentration is effectively constant during the reaction, that constant gets absorbed into the rate constant, and the rate law collapses to r = k_obs [A], a (pseudo) first-order form.

1. Start with true rate law r = k [A] [B] for a bimolecular reaction.
2. Set [B] ≫ [A] so [B] ≈ [B]₀ throughout. Then r = (k [B]₀) [A] = k_obs [A], first order in A.

Example: hydrolysis of ester in water, CH3COOC2H5 + H2O H⁺ CH3COOH + C2H5OH. Water is the solvent in huge excess (∼ 55 M), so the reaction is pseudo first order in ester.

When one reactant is in large excess, its concentration stays constant and a bimolecular reaction follows first-order kinetics (pseudo first order).

Concept used. For a zero-order reaction, the rate is independent of every reactant's concentration. In the general rate law r = k [A]^x [B]^y, ``zero order'' means x + y = 0, which (since orders cannot be negative for elementary kinetics discussion here) implies x = y = 0: r = k [A]⁰ [B]⁰ = k.

1. Use r = k [A]^x [B]^y as the generic rate law.
2. Substitute zero order: x = 0, y = 0.
3. Result: rate = k, a constant equal to the rate constant.

Rate = k [A]⁰ [B]⁰ = k.

Concept used. Use the initial-rates method. Conduct a series of experiments in which the initial concentration of one reactant is varied while the other is held constant; measure the initial rate; extract the order in that reactant from the ratio of rates. Repeat for the other reactant.

1. Mix NO and O2 at known initial concentrations in a constant-volume vessel at fixed temperature. Measure initial rate (e.g., from the slope at t = 0 of [NO2] vs time).
2. Run pairs of experiments. Hold [O2]₀ constant and double [NO]₀: if rate quadruples, order in NO is 2.
3. Hold [NO]₀ constant and double [O2]₀: if rate doubles, order in O2 is 1.
4. Write empirical rate law: r = k [NO]² [O2] (the experimentally measured form; coincidentally matches the stoichiometric coefficients here, but only because the elementary mechanism happens to match).

Use the initial-rates method: vary one reactant at a time, measure the rate, and compute orders from rate ratios. The empirical result is r = k [NO]² [O2].

Concept used. Order is the experimentally measured exponent in the rate law (an empirical number). Molecularity is the count of reactant molecules in an elementary step (a theoretical integer). They coincide only when the rate law is read directly off the balanced equation, which happens only for elementary reactions.

1. For an elementary reaction aA + bB → products, r = k [A]^a [B]^b with a, b the stoichiometric coefficients. So order = a+b = molecularity.
2. For complex (multi-step) reactions, the slowest step's molecularity is well-defined but the overall reaction's molecularity is not, and the overall order is set by the slow step plus any pre-equilibria. They may or may not match.

Order and molecularity coincide only for elementary reactions.

Concept used. For a rate law r = k [A]ⁿ, scaling [A] by a factor f scales r by fⁿ.

1. Start from r₁ = k [A]ⁿ.
2. After tripling: r₂ = k (3[A])ⁿ = 3ⁿ k [A]ⁿ = 3ⁿ r₁.
3. Given r₂/r₁ = 27: 3ⁿ = 27 ⇒ n = 3. So the reaction is third order in A.

Order = 3 (third order).

Concept used. The integrated rate law for a zero order reaction is [R]_t = [R]₀ - kt. ``Completion'' means [R]_t = 0. Set this in the integrated law and solve for t.

1. Start from the differential rate equation: -d[R]dt = k (zero order, rate =k).
2. Integrate from t=0 ([R]=[R]₀) to time t ([R]=[R]_t): _[R]0^[R]_t d[R] = -₀^t k dt' ⇒ [R]_t - [R]₀ = -kt ⇒ [R]_t = [R]₀ - kt.
3. At completion, [R]_t = 0: 0 = [R]₀ - kt ⇒ t = [R]₀k.

t_completion = [R]₀k.

Concept used. For an elementary reaction, the rate law's exponents equal the stoichiometric coefficients (which are integers). Hence orders of an elementary reaction must be integers (1, 2, 3).

1. In the given rate law, the order in B is 3/2, a non-integer.
2. If the reaction were elementary, the molecularity (count of molecules in the single step) would be an integer, and the order in each species would equal the corresponding integer coefficient. A fractional order is incompatible.
3. Fractional orders typically arise from multi-step mechanisms involving a pre-equilibrium or chain-radical steps; e.g. the H2 + Br2 reaction has r = k [H2] [Br2]^1/2 due to a radical-chain mechanism.

No: an elementary reaction must have integer order, so the 3/2 order in B shows this reaction is complex (multi-step).

Concept used. Collision theory: for a collision to be effective, it must have (a) energy ≥ E_a, AND (b) proper orientation (controlled by the steric factor P). A reaction with a large energetic-collision fraction but small P will be slow because most collisions fail the orientation requirement.

1. Recall rate ∝ P Z_AB e^-E_a/RT. The Boltzmann factor e^-E_a/RT is the energetic-collision fraction.
2. If this factor is already large (most collisions are energetic enough), the rate is bottlenecked by P, the steric factor.
3. Many reactions, especially those involving large or geometrically intricate molecules, have P ≪ 1 (orders of magnitude smaller than unity). A small P slows the reaction even when the energetic fraction is huge.

Because the colliding molecules also need proper orientation. A small steric factor P keeps the rate slow even when enough molecules have E ≥ E_a.

Concept used. Order is experimental (the empirical exponent in the rate law). Molecularity is theoretical (the count of reactant molecules in an elementary step). The two are different quantities, so a zero order reaction does not have zero molecularity.

1. Molecularity is the number of molecules in an elementary step; it counts particles, so it must be an integer ≥ 1. It can never be zero or fractional.
2. Zero order is an empirical observation about how rate depends on concentration. It does not say anything about molecularity.
3. Example: at high pressure 2NH3 Pt N2 + 3H2 is zero-order kinetically (because the Pt surface is saturated), but the elementary step on the surface still involves molecules and has a non-zero molecularity.

No, molecularity can never be zero. Zero order is an empirical kinetic fact; molecularity is a count of molecules in an elementary step and is always ≥ 1.

Concept used. The plot in Fig. 4.3 shows concentration of A falling linearly with time. A linear concentration-vs-time plot is the signature of a zero-order reaction: [A] = [A]₀ - kt has slope -k and is a straight line.

1. (i) Order. Since [A] vs t is a straight line, the integrated form is [A] = [A]₀ - kt, which is the zero-order law. Order = 0.
2. (ii) Slope. Differentiate [A] = [A]₀ - kt: d[A]/dt = -k. The slope is -k.
3. (iii) Units of k. Rate has units mol L^-1 s^-1 and equals k (zero order). So k has units of mol L^-1 s^-1.

(i) Zero order.    (ii) Slope = -k.    (iii) Units of k: mol L^-1 s^-1.

Concept used. Thermodynamic feasibility (Δ G < 0) tells us a reaction can go; it says nothing about how fast. The rate is set by kinetics, specifically by the activation energy E_a via k = A e^-E_a/RT. If E_a is very large compared to RT, the Boltzmann factor is tiny and the rate is imperceptibly slow.

1. Reaction 2H2 + O2 -> 2H2O has Δ G^∘ ≈ -474 kJ/mol, very negative; thermodynamically highly feasible.
2. Activation energy for breaking the H-H and O=O bonds is ∼ 450 kJ/mol, enormous at room temperature. The Boltzmann factor e^-E_a/RT ≈ e^{-450000/(8.314 × 298)} ≈ e^-182 ≈ 10^-79, essentially zero.
3. Hence the reaction is too slow to be observed at room temperature. A spark or a platinum catalyst can lower E_a enough to launch the explosive reaction.

Thermodynamic feasibility alone is not enough; the activation energy is very high at room temperature, so the kinetics are extremely slow.

Concept used. A reaction's rate is set by the fraction of molecules with E ≥ E_a, given by the Boltzmann factor e^-E_a/RT. As T rises, the M-B distribution broadens and the tail above E_a grows sharply. More molecules are energetic enough to react, so the rate climbs.

1. Apply Arrhenius: k = A e^-E_a/RT. As T ↑, the exponent -E_a/(RT) moves towards zero, so k (and rate) climbs.
2. Rule of thumb: rate roughly doubles for every 10 K rise. For example, going from 298 K to 308 K with E_a = 50 kJ/mol: k₃₀₈k₂₉₈ = e^{(E_a/R)(1/298 - 1/308)} = e^{6014 × 1.09× 10-4} = e^0.66 ≈ 1.93.
3. The deeper reason: the M-B tail above E_a grows faster than T itself.

Higher temperature broadens the M-B distribution and increases the fraction of molecules with E ≥ E_a, so more collisions are effective and the rate rises.

Concept used. Same logic as Q43: combustion reactions are thermodynamically very favourable (Δ G ≪ 0) but kinetically inhibited by a large E_a. At 298 K the Boltzmann factor e^-E_a/RT is microscopic, so the rate is negligible.

1. Combustion of, say, octane has Δ G^∘ ≈ -5500 kJ/mol (very feasible).
2. Activation energy for igniting C-C and C-H bonds in octane is ∼ 200 – 400 kJ/mol, enormous at 298 K.
3. A spark or flame supplies localised heat, raising T to the auto-ignition temperature (e.g. ∼ 510 K for petrol). Above that, the rate climbs by 20+ orders of magnitude and the fuel burns.

The activation energy for combustion is very high at room temperature; a spark (or a hot surface) is needed to give molecules enough energy to react.

Concept used. Molecularity counts the molecules colliding simultaneously in one elementary step. The probability of more than three molecules being at the same point in space with the right energies and orientations all at the same instant is vanishingly small.

1. Two-molecule collisions are routine (∼ 10³² per cm³/s in a gas at STP).
2. Three-molecule (termolecular) collisions are already ∼ 10⁴ times rarer than bimolecular.
3. Four- or more-molecule simultaneous collisions are so rare that they are essentially never observed; any complex reaction is built up from sequential bi/uni-molecular steps, never a single tetra-molecular step.

Simultaneous collisions of more than three molecules with the right orientation and energy are essentially never observed; multi-molecule reactions go through stepwise bimolecular/unimolecular mechanisms.

Concept used. For any non-zero order reaction, the rate depends on reactant concentrations through r = k [A]^x [B]^y ⋯. As the reaction proceeds, reactants are consumed and the [A], [B], … values drop. With positive exponents, the rate drops.

1. Initially [A] = [A]₀ and r = k [A]₀^x. As time passes, [A] drops.
2. Since x > 0, [A]^x drops too, and so does r.
3. Zero-order is the special exception: r = k regardless of [A]. But all positive-order reactions slow down.

Reactant concentrations decrease as the reaction proceeds, so the rate (which is proportional to those concentrations) also decreases.

Concept used. Thermodynamic feasibility (Δ G < 0) tells us a reaction can proceed. The rate is independent of Δ G and is controlled by the activation energy E_a.

Example: conversion of diamond to graphite. Δ G^∘(diamond → graphite) ≈ -2.9 kJ/mol at 298 K and 1 atm, so graphite is thermodynamically more stable than diamond. Yet diamond at room temperature does not perceptibly convert to graphite, because the activation energy is enormous (breaking the entire diamond C-C network costs hundreds of kJ/mol). Diamonds remain ``forever'' on human timescales.

1. Thermodynamic test: Δ G^∘ < 0, so diamond → graphite is feasible.
2. Kinetic test: E_a is so large that the rate constant at 298 K is unmeasurably small.
3. Conclusion: feasibility alone doesn't tell the rate. You need to know E_a as well.

Diamond → graphite is thermodynamically feasible (Δ G < 0), but E_a is huge, so the rate is essentially zero at room temperature.

Concept used. The reaction between KMnO4 (acidic) and oxalic acid H2C2O4, 2MnO4- + 5C2O4^2- + 16H+ -> 2Mn²⁺ + 10CO2 + 8H2O, is intrinsically slow at room temperature: the activation energy for the first electron transfer is high. Heating to about 60–70 ^∘C raises the temperature, boosts e^-E_a/RT, and makes the endpoint sharp.

1. Compute the rate enhancement. Going from 298 K to 343 K (70 ^∘C) with E_a ≈ 70 kJ/mol: k₃₄₃k₂₉₈ = e^{(70000/8.314)(1/298 - 1/343)} = e^{8421 × 4.40 × 10-4} = e^3.71 ≈ 41. A factor of ∼ 40 in rate.
2. Once Mn²⁺ is formed, it auto-catalyses subsequent steps, so the reaction accelerates. The initial heat just gets the autocatalysis started.

Heating raises the rate constant via Arrhenius (k₃₄₃/k₂₉₈ ≈ 40), making the otherwise slow KMnO4–oxalic-acid reaction proceed at a usable rate during titration.

Concept used. Molecularity is defined as the number of reactant molecules taking part in an elementary step. By construction, at least one molecule must be present to participate; a step with zero molecules involved would be a step with nothing reacting, which is meaningless.

1. Definition: molecularity = number of molecules colliding / participating in one elementary step.
2. Minimum count: 1 (unimolecular, e.g. radioactive decay or dissociation of an isolated species).
3. Maximum (commonly observed): 3 (termolecular). Higher is possible in principle, but practically negligible.
4. Zero molecules involved means no reaction at all, hence molecularity = 0 is undefined.

At least one molecule must be present for a reaction to occur, so molecularity is always ≥ 1.

Concept used. A complex reaction is a sum of several elementary steps, each with its own molecularity (the count of molecules in that step). The overall reaction has no single molecularity because different steps have different counts. Order, however, is empirical: it's the exponent that fits the observed rate law of the overall reaction, so order is defined for any reaction, elementary or complex.

1. Molecularity. Defined for one elementary step: ``how many molecules are in this step?''. For a multi-step mechanism, each step has its own molecularity; there's no single number for the whole reaction.
2. Order. Defined by the experimental rate law: ``what exponents fit the rate-vs-concentration data?''. This is always definable (and always measurable), whether the reaction has one step or many.

Molecularity is defined per elementary step (so it's meaningful only for single-step / elementary reactions). Order is defined by the experimental rate law and is always available.

Concept used. The balanced equation reflects the overall stoichiometry, summing all elementary steps. The rate law, however, is governed by the slowest (rate-determining) step, and the slow step may involve fewer or different species than the overall equation. Order is therefore an experimental fact, not a prediction from stoichiometry.

Example. The reaction KClO3 + 6FeSO4 + 3H2SO4 -> KCl + 3H2O + 3Fe2(SO4)3 has stoichiometric coefficient sum 1+6+3=10, so a naive reading predicts a tenth order reaction. Experimentally, this is a second-order reaction, because it proceeds through many elementary steps with the rate-determining one involving only two species.

1. Balanced equation tells overall stoichiometry: how much A becomes how much C.
2. Rate law depends on the mechanism: which step is slowest, and what's in that step.
3. These are different things; balanced equation does not reveal the mechanism.
4. Hence order must be measured (e.g. initial-rates method, half-life method, etc.), not derived from the balanced equation.

Balanced equation gives stoichiometry; the rate law reflects the slowest mechanism step, which is determined experimentally. The two can differ wildly.

Concept used. Diagnostic signatures of zero-order and first-order reactions:

• [leftmargin=*]
• Zero order. r = k (constant). Rate vs concentration is a horizontal line; concentration vs time is a straight line with slope -k.
• First order. r = k [R]. Rate vs concentration is a straight line through the origin with slope k; log[R] (or log([R]₀/[R])) vs time is a straight line.

1. (i) Rate vs concentration is a straight line through the origin with positive slope. This is r = k [R], ⇒ first order (a).
2. (ii) Rate vs concentration is horizontal (rate independent of concentration). This is r = k, ⇒ zero order (b).
3. (iii) Concentration vs time is a straight line with negative slope. This is [R]_t = [R]₀ - kt, ⇒ zero order (b).
4. (iv) log[Conc.] vs time is a straight line with negative slope -k/2.303. This is the integrated first-order law, ⇒ first order (a).

(i)→(a), (ii)→(b), (iii)→(b), (iv)→(a).

Concept used. Match each kinetics concept on the left to the matching property on the right.

1. (i) Catalyst alters rate by providing a lower-E_a pathway. Match: (c) ``by lowering the activation energy''.
2. (ii) Molecularity is a count of molecules in an elementary step: a positive integer. Match: (a) ``cannot be fraction or zero''.
3. (iii) For a first-order reaction, t_1/2 = 0.693/k is independent of [R]₀. So the second half-life equals the first. Match: (d) ``is same as the first''.
4. (iv) e^-E_a/RT is the Boltzmann factor: the fraction of molecules with energy ≥ E_a. Match: (f) ``refers to the fraction of molecules with energy equal to or greater than activation energy''.
5. (v) Energetically favourable reactions can be slow because of orientation/steric factor. Match: (b) ``proper orientation is not there always''.
6. (vi) Total area under M-B curve = total probability = 1. Match: (e) ``total probability is one''.

(i)→(c), (ii)→(a), (iii)→(d), (iv)→(f), (v)→(b), (vi)→(e).

Concept used. The three terms describe three different timescales / rates.

1. (i) Diamond converts to graphite extremely slowly because of the large E_a. Rate is imperceptible at room temperature. Match: (b) ``ordinarily rate of conversion is imperceptible''.
2. (ii) Instantaneous rate is the rate at a particular instant, i.e., over a vanishingly short time interval. Match: (a) ``short interval of time''.
3. (iii) Average rate is computed over a longer time interval [t₁, t₂]. Match: (c) ``long duration of time''.

(i)→(b), (ii)→(a), (iii)→(c).

Concept used. Definitions: rate law, rate constant, units of k for zero order, and rate-determining step.

1. (i) Mathematical expression for rate of reaction is the rate law: r = k [A]^x [B]^y. Match: (b) rate law.
2. (ii) For zero order, r = k identically. So rate of reaction equals rate constant. Match: (a) rate constant.
3. (iii) Zero-order k has units of rate: mol L^-1 s^-1. So units of k for zero order same as units of rate. Match: (d) rate of a reaction.
4. (iv) Order of a complex reaction is determined by the slowest (rate-determining) step. Match: (c) order of slowest step.