CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 12

Answer: (c) 7.8 × 10⁻⁴ atm s⁻¹

Explanation:
1. SO₂Cl₂ → SO₂ + Cl₂; Initial pressure: p₀, 0, 0; Pressure at time t: p₀ − p, p, p
2. Let initial pressure p₀ ∝ R₀
3. Pressure at time t, P_t = p₀ − p + p + p = p₀ + p
4. Pressure of reactants at time t, p₀ − p = 2p₀ − P_t ∝ R
5. k = \( \frac{2.303}{t} \) log \( \left(\frac{p_0}{2p_0 - P_{t}}\right) \) = \( \frac{2.303}{100} \) log \( \left(\frac{0.5}{2\times0.5 - 0.6}\right) \) = \( \frac{2.303}{100} \) log 1.25
6. = 2.2318 × 10⁻³ s⁻¹
7. Pressure of SO₂Cl₂ at time t (p_SO2Cl2) = 2p₀ − P_t = 2 × 0.50 − 0.65 atm = 0.35 atm
8. Rate at that time = k × p_SO2Cl2 = (2.2318 × 10⁻³) × (0.35) = 7.8 × 10⁻⁴ atm s⁻¹

Teacher's Note:
1. The rate needs the pressure of the reactant only (0.35 atm), not the total pressure of 0.65 atm.
2. Option (b) is close to k itself, a common trap; rate = k × p_reactant.

Answer: (c) 0.0223 min⁻¹

Explanation:
1. a − x = \( \frac{80}{100} \) × a = 0.8a, t = 10 min
2. k = \( \frac{2.303}{t} \) log \( \frac{a}{a-x} \) = \( \frac{2.303}{10} \) log \( \frac{a}{0.8a} \)
3. k = 0.2303 log \( \frac{10}{8} \) = 0.2303(1 − 0.9030) = 0.2303 × 0.0970 = 0.0223 min⁻¹

Teacher's Note:
1. 20% complete means 80% remains: use a − x = 0.8a, not 0.2a.

Answer: (b) 4.67 × 10⁻³ mol L⁻¹ s⁻¹

Explanation:
1. Average rate during the time interval 30-60 s.
2. Rate = −\( \frac{C_2 - C_1}{t_2 - t_1} \) = −\( \frac{(0.17 - 0.31)}{60 - 30} \) = \( \frac{0.14}{30} \) = 4.67 × 10⁻³ mol L⁻¹ s⁻¹

Teacher's Note:
1. A rate must have units of concentration per time; options in s⁻¹ are values of k, not of rate.

Answer: (a) Zero order and first order

Explanation:
1. Linear plots are obtained in the graph of A_t vs t for zero order reaction and ln(a − x) vs t for first order reaction.

Teacher's Note:
1. Remember the pair: [R] vs t linear → zero order; ln[R] vs t linear → first order. Both have slope −k.

Answer: (a) −\( \frac{1}{3} \)\( \frac{d[H_2]}{dt} \) = +\( \frac{1}{2} \)\( \frac{d[NH_3]}{dt} \)

Explanation:
1. For the reaction N₂ + 3H₂ → 2NH₃,
2. −\( \frac{1}{3} \)\( \frac{d[H_2]}{dt} \) = +\( \frac{1}{2} \)\( \frac{d[NH_3]}{dt} \)

Teacher's Note:
1. The reactant (H₂) gets the minus sign and its coefficient 3; the product (NH₃) is positive with coefficient 2.

Answer: (a) t = \( \frac{[R_0]}{k} \)

Explanation:
1. [R] = [R]₀ − kt
2. For completion of reaction [R] = 0
3. or t = \( \frac{[R]_0}{k} \)

Teacher's Note:
1. A zero order reaction really does finish in a finite time, twice its half-life, unlike a first order reaction.

Answer: (c) 1.5k₁ = 3k₂ = k₃

Explanation:
1. 2NH₃ → N₂ + 3H₂
2. Rate = −\( \frac{1}{2} \)\( \frac{d[NH_3]}{dt} \) = \( \frac{d[N_2]}{dt} \) = \( \frac{1}{3} \)\( \frac{d[H_2]}{dt} \)
3. \( \frac{1}{2} \)k₁[NH₃] = k₂[NH₃] = \( \frac{1}{3} \)k₃[NH₃]
4. 1.5k₁ = 3k₂ = k₃

Teacher's Note:
1. After writing the equal-rate expression, multiply every term by the same number (here 3) to clear the fractions.

Answer: (b) 6.24 × 10⁻⁴ s⁻¹

Explanation:
1. k = \( \frac{2.303}{(t_2 - t_1)} \) log \( \frac{[A_1]}{[A_2]} \)
2. k = \( \frac{2.303}{(1600 - 800)} \) log \( \frac{1.45}{0.88} \) = \( \frac{2.303}{800} \) × 0.2169 = 6.24 × 10⁻⁴ s⁻¹

Teacher's Note:
1. The first order equation works between any two times: use the time difference (800 s) and the ratio of the two concentrations.

Answer: (d) the rate of slowest intermediate step.

Explanation:
1. The slowest step is rate determining step.

Teacher's Note:
1. Like the narrowest point of a funnel, the slowest step limits how fast products can form overall.

Answer: (b) II only

Explanation:
1. \( \frac{1}{3} \)\( \frac{d[C]}{dt} \) = \( \frac{1}{2} \)\( \frac{d[D]}{dt} \) = \( \frac{-d[A]}{dt} \) = −\( \frac{1}{2} \)\( \frac{d[B]}{dt} \)
2. \( \frac{-d[A]}{dt} \) = −\( \frac{(0.35 - 0.5)}{15} \) = \( \frac{0.15}{15} \) = 0.01 mol/L-s
3. \( \frac{-d[B]}{dt} \) = 0.02 mol/L-s
4. \( \frac{d[C]}{dt} \) = 3 × 0.01 = 0.03 mol/L-s
5. \( \frac{d[D]}{dt} \) = 2 × 0.01 = 0.02 mol/L-s
6. So statement II (0.025 mol/L-s) is incorrect.

Teacher's Note:
1. Find the rate of change of A first, then multiply by each coefficient to get the rates for B, C and D.

Answer: (c) 8 half lives

Explanation:
1. For 99.6% completion, let a = 100, x = 99.6, a − x = 0.4
2. k = \( \frac{2.303}{t_{99.6}} \) log \( \frac{100}{0.4} \) = \( \frac{2.303}{t_{99.6}} \) log 250
3. t_99.6% = \( \frac{2.303}{k} \) log 250 ...(i)
4. k = \( \frac{0.693}{t_{1/2}} \) ...(ii)
5. From equation (i) & (ii), t_99.6% = \( \frac{2.303}{0.693} \) × 2.4 × t_1/2 = 8t_1/2

Teacher's Note:
1. Quick check: after n half-lives (1/2)ⁿ remains; (1/2)⁸ = 1/256 ≈ 0.4%, i.e., 99.6% complete.

Answer: (b) 4.98 × 10⁻⁴ s⁻¹

Explanation:
1. The following data were obtained during the first order thermal decomposition of N₂O_5(g) at constant volume: at t = 0, total pressure = 0.5 atm; at t = 100 s, total pressure = 0.512 atm.
2. Let the pressure of N₂O_5(g) decrease by 2x atm. As two moles of N₂O₅ decompose to give two moles of N₂O_4(g) and one mole of O₂(g), the pressure of N₂O_4(g) increases by 2x atm and that of O_2(g) increases by x atm.
3. 2N₂O_5(g) → 2N₂O_4(g) + O_2(g); at t = 0: 0.5 atm, 0 atm, 0 atm; at time t: (0.5 − 2x) atm, 2x atm, x atm
4. p_t = p_N2O5 + p_N2O4 + p_O2 = (0.5 − 2x) + 2x + x = 0.5 + x
5. x = p_t − 0.5
6. p_N2O5 = 0.5 − 2x = 0.5 − 2(p_t − 0.5) = 1.5 − 2p_t
7. At t = 100 s; p_t = 0.512 atm
8. p_N2O5 = 1.5 − 2 × 0.512 = 0.476 atm
9. k = \( \frac{2.303}{t} \) log \( \frac{p_{t=0}}{p_{t=100}} \) = \( \frac{2.303}{100} \) log \( \frac{0.5\ \text{atm}}{0.476\ \text{atm}} \)
10. = \( \frac{2.303}{100} \) × 0.0214 = 4.92 × 10⁻⁴ s⁻¹ ≈ 4.98 × 10⁻⁴ s⁻¹ (closest option)

Teacher's Note:
1. Total pressure is not the reactant pressure: here p_N2O5 = 1.5 − 2p_t, so 0.512 atm total means 0.476 atm of N₂O₅.

Answer: (d) III and IV

Explanation:
1. For zero order reaction, [R] vs t is a straight line plot and half life t_1/2 = \( \frac{[R]_0}{2k} \).
2. For first order reaction ln [R] vs t gives straight line plot and half life = \( \frac{\ln 2}{k} \).
3. So columns III and IV are wrong; columns I, II and V are correct.

Teacher's Note:
1. A common slip is swapping the two straight-line plots; ln[R] vs t belongs to first order, [R] vs t to zero order.

Answer: (c) the rate constant for the given reaction is 2.67 × 10⁵ mol⁻² L² s⁻¹

Explanation:
1. (i) Rate law can be written as follows: Rate = k[A]^p[B]^q
2. From experiments I, II and III: (Rate)₁ = k[2.5 × 10⁻⁴]^p[3.0 × 10⁻⁵]^q = 5.0 × 10⁻⁴ ...(i); (Rate)₂ = k[5.0 × 10⁻⁴]^p[6.0 × 10⁻⁵]^q = 4.0 × 10⁻³ ...(ii); (Rate)₃ = k[1.0 × 10⁻³]^p[6.0 × 10⁻⁵]^q = 1.6 × 10⁻² ...(iii)
3. Dividing equation (iii) by equation (ii), \( \frac{(Rate)_3}{(Rate)_2} \) = \( \frac{(1.0\times10^{-3})^p}{(5.0\times10^{-4})^p} \) = \( \frac{1.6\times10^{-2}}{4.0\times10^{-3}} \) or 2^p = 4 or 2^p = 2², i.e., p = 2
4. Dividing equation (ii) by equation (i), \( \frac{(Rate)_2}{(Rate)_1} \) = \( \frac{(5.0\times10^{-4})^p(6.0\times10^{-5})^q}{(2.5\times10^{-4})^p(3.0\times10^{-5})^q} \) = \( \frac{4.0\times10^{-3}}{5.0\times10^{-4}} \)
5. or 2^p · 2^q = 8 or 2² · 2^q = 8 or 2^q = 8/2² or 2^q = 2¹ or q = 1
6. Thus the rate equation is Rate = k[A]²[B]; order w.r.t. A = 2, order w.r.t. B = 1
7. (ii) Rate constant (k) at 300 K: from experiment 1, 5.0 × 10⁻⁴ mol L⁻¹ s⁻¹ = k(2.5 × 10⁻⁴ mol L⁻¹)²(3.0 × 10⁻⁵ mol L⁻¹)
8. k = \( \frac{5.0\times10^{-4}}{(2.5\times10^{-4})^2(3.0\times10^{-5})} \) = 2.67 × 10⁸ mol⁻² L² s⁻¹
9. So option (c), which gives 2.67 × 10⁵, is incorrect.

Teacher's Note:
1. Pick two experiments in which only one concentration changes (here III and II) to find that order first.
2. The 320 K column is not needed for the orders or k at 300 K; it only shows that the rate rises with temperature.

Answer: (c) order of the reaction

Explanation:
1. For the reaction, xA + yB → Product, r = k[A]^x[B]^y
2. mol L⁻¹ s⁻¹ = k(mol L⁻¹)^x(mol L⁻¹)^y
3. k = \( \frac{mol\ L^{-1}\,s^{-1}}{(mol\ L^{-1})^{x}(mol\ L^{-1})^{y}} \) = (mol L⁻¹)^1−(x+y) s⁻¹
4. where (x + y) = order of the reaction.

Teacher's Note:
1. General unit of k for an nth order reaction: (mol L⁻¹)¹⁻ⁿ s⁻¹. Put n = 0, 1, 2 to check any unit question in seconds.

Answer: (b) Rate = k[A][B]²

Explanation:
1. Suppose order with respect to A and B are x and y respectively. Rate = k[A]^x[B]^y
2. Experiment 1: 0.1 = k(0.3)^x(0.3)^y …(i)
3. Experiment 2: 0.4 = k(0.3)^x(0.6)^y …(ii)
4. Experiment 3: 0.2 = k(0.6)^x(0.3)^y …(iii)
5. Dividing (ii) by (i): \( \frac{0.4}{0.1} \) = \( \frac{(0.6)^{y}}{(0.3)^{y}} \) ∴ y = 2
6. Dividing (iii) by (i): \( \frac{0.2}{0.1} \) = \( \frac{(0.6)^{x}}{(0.3)^{x}} \) ∴ x = 1
7. Rate law: Rate = k[A][B]²

Teacher's Note:
1. Always pick two experiments in which only one concentration changes; doubling it and seeing the rate go up 4 times means order 2.
2. The coefficient 2 of B in the equation is not why the order is 2 – the order comes only from the experimental data.

Answer: (c) remains unchanged

Explanation:
1. There is no effect on heat evolved or absorbed during the reaction in the presence of a catalyst.
2. It is because catalyst influences the rate of reaction and does not participate in the reaction.

Teacher's Note:
1. A catalyst lowers the activation energy of both the forward and the backward reaction equally, so ΔH, ΔG and the equilibrium constant stay the same.

Answer: (c) Average rate up to 40 seconds is \( \frac{V_3}{40} \)

Explanation:
1. Average rate of reaction up to 40 seconds on the basis of the graph is:
2. Average rate of reaction = Change in volume/Change in time interval = ΔV/Δt
3. = \( \frac{V_3-0}{40-0} \) = \( \frac{V_3}{40} \)

Teacher's Note:
1. “Up to 40 s” means from t = 0 (where V = 0) to t = 40 s; the tangent line at 40 s would give the instantaneous rate instead.

Answer: (a) The rate of a reaction decreases with passage of time as the concentration of reactants decreases.

Explanation:
1. The rate of a reaction depends upon the concentration of reactants. As reactants are being consumed to give products, they will be consumed slowly when the reaction is completed.
2. That is why rate of reaction is expressed as decrease in concentration of reactants with the passage of time.

Teacher's Note:
1. Zero order reactions are the one exception: their rate stays constant until the reactant is used up.

Answer: (b) it is an elementary reaction.

Explanation:
1. In case of elementary reaction the rate law can be determined from balanced chemical equation.
2. As, order of the elementary reaction is the sum of the stoichiometric coefficients of the reactants of a reaction.

Teacher's Note:
1. For complex (multi-step) reactions the rate law must be found experimentally; only for a single-step (elementary) reaction does it follow the equation.

Answer: (c) \( \frac{\Delta[Br^-]}{\Delta t}=\frac{5}{6}\frac{\Delta[H^+]}{\Delta t} \)

Explanation:
1. Rate expression for the above equation can be presented as:
2. \( -\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t}=-\frac{\Delta[BrO_3^-]}{\Delta t}=-\frac{1}{6}\frac{\Delta[H^+]}{\Delta t}=+\frac{1}{3}\frac{\Delta[Br_2]}{\Delta t} \)
3. \( \Rightarrow \frac{\Delta[Br^-]}{\Delta t}=5\frac{\Delta[BrO_3^-]}{\Delta t}=\frac{5}{6}\frac{\Delta[H^+]}{\Delta t} \)
4. \( \frac{\Delta[Br^-]}{\Delta t}=\frac{5}{6}\frac{\Delta[H^+]}{\Delta t} \)

Teacher's Note:
1. Divide each rate of change by its stoichiometric coefficient; then equate the Br⁻ and H⁺ terms and rearrange – the signs cancel because both are reactants.

Answer: (d) 1/2

Explanation:
1. Rate (r) ∝ [X]ⁿ [where n = order of reaction]
2. If the concentration of X is increased by 4 times, X′ = 4X, then r′ ∝ [X′]ⁿ
3. \( \frac{r'}{r} \) = \( \frac{[4X]^{n}}{[X]^{n}} \) = 2 ⇒ 4ⁿ = 2 ∴ n = \( \frac{1}{2} \)
4. Order of reaction = 1/2

Teacher's Note:
1. Set up (concentration factor)ⁿ = (rate factor): here 4ⁿ = 2, so n = ½. Fractional orders are allowed for complex reactions.

Answer: (d) \( \frac{[R]_0}{2k} \)

Explanation:
1. Half life period of a zero order reaction = \( \frac{[R]_0}{2k} \)
2. where [R]₀ = initial concentration of reactant and k = rate constant.

Teacher's Note:
1. Zero order t_½ grows with [R]₀; first order t_½ = 0.693/k does not depend on [R]₀ at all.

Answer: (b) −k

Explanation:
1. For zero order reaction, r = k[A]⁰, so [R] = −kt + [R]₀.
2. Graphically, the plot of [R] vs time is a straight line with slope = −k (intercept = [R]₀).

Teacher's Note:
1. The 2.303 factor appears only when the y-axis is log₁₀[R] (first order); a plain [R] vs t plot has slope −k.

Answer: (d) Infinite

Explanation:
1. In a chemical reaction, rate of a reaction is calculated in specified time such as half life of the reaction for 50% completion of reaction, but it is impossible to calculate time for 100% completion of reaction.
2. Thus, it is assumed that reaction will be 100% complete only after infinite time.

Teacher's Note:
1. Do not double the half-life: after two half-lives a first order reaction is only 75% complete, not 100%.

Answer: (b) k = \( \frac{2.303}{t} \) log \( \frac{p_i}{2p_i-p_t} \)

Explanation:
1. The initial pressure of the system before decomposition of A is p_i. After time t, the pressure of A decreases by x atm.
2. A(g) → B(g) + C(g); Initial: p_i, 0, 0; After time t: (p_i − x), x, x
3. p_t = (p_i − x) + x + x = p_i + x ⇒ x = p_t − p_i
4. p_A = p_i − (p_t − p_i) = 2p_i − p_t
5. Rate constant, k = \( \frac{2.303}{t} \) log \( \frac{[A]_0}{[A]} \) = \( \frac{2.303}{t} \) log \( \frac{p_i}{2p_i-p_t} \)

Teacher's Note:
1. The rise in total pressure (x) equals the fall in pressure of A, because each A gives two gas molecules; so p_A = p_i − x = 2p_i − p_t.
2. Inside the log the initial value always goes on top; a ratio less than 1 would give a negative k.

Answer: (c) 1/2

Explanation:
1. Rate of reaction r ∝ [A]ⁿ …(i)
2. If concentration of A is increased by nine times, rate becomes three times: r′ = 3r, A′ = 9A, 3r ∝ [9A]ⁿ …(ii)
3. From (i) and (ii): \( \frac{r}{3r} \) = \( \frac{[A]^{n}}{[9A]^{n}} \) ⇒ \( \frac{1}{3} \) = \( \left(\frac{1}{3}\right)^{2n} \)
4. 1 = 2n ⇒ n = 1/2 ∴ Order of reaction = ½

Teacher's Note:
1. Write 9 as 3²: then 3²ⁿ = 3 immediately gives n = ½.

Answer: (b) depends on the concentration of reactants present in excess.

Explanation:
1. Rate constant of a pseudo-first-order reaction depends on the concentration of reactants present in excess. In a chemical reaction, A + B → C, rate of reaction R = k[A][B].
2. But if concentration of B is in excess, then concentration of B will remain constant. The chemical reaction will be termed as pseudo-first order reaction such that R = k′[A], where k′ = k[B] is the new rate constant value.

Teacher's Note:
1. In acid hydrolysis of ethyl acetate, k′ = k[H₂O]; using a different (still large) amount of water changes k′.

Answer: (c) −k

Explanation:
1. For a first order reaction, r = k[R]
2. ln[R] = −kt + C (C = ln[R]₀)
3. So the plot of ln[R] vs time is a straight line with slope = −k.

Teacher's Note:
1. Natural log gives slope −k; log₁₀[R] vs t gives slope −k/2.303. Read the axis label carefully before choosing.

Answer: (d) \( \frac{d[Br_2]}{dt} \) = \( -\frac{3}{5} \) \( \frac{d[Br^-]}{dt} \)

Explanation:
1. Rate = \( -\frac{1}{5}\frac{d[Br^-]}{dt} = +\frac{1}{3}\frac{d[Br_2]}{dt} \)
2. Hence, \( \frac{d[Br_2]}{dt} = -\frac{3}{5}\frac{d[Br^-]}{dt} \)

Teacher's Note:
1. Divide each rate by its own stoichiometric coefficient; the minus sign appears because Br⁻ is consumed while Br₂ is formed.

Answer: (b) \( \frac{V_4 - V_2}{50 - 30} \)

Explanation:
1. Instantaneous rate at 40 s is the slope of the tangent drawn to the curve at t = 40 s.
2. The points (20, V₁), (30, V₂), (40, V₃) and (50, V₅) lie on this tangent, but V₄ lies on the curve, not on the tangent. So \( \frac{V_4 - V_2}{50 - 30} \) does not give the instantaneous rate.

Teacher's Note:
1. Any two points on the tangent give the same slope – check which value is read from the tangent and which from the curve.

Answer: (d) 3

Explanation:
1. Unit of k = (mol L⁻¹)¹⁻ⁿ s⁻¹ = mol⁻² L² s⁻¹
2. 1 − n = −2, so n = 3.

Teacher's Note:
1. Use the general unit (mol L⁻¹)¹⁻ⁿ s⁻¹ and compare the power of mol – it is quicker than memorising each order separately.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. In a heterogeneous system, the reactant is adsorbed on the surface of a solid catalyst. The fraction of the surface of the catalyst covered by the reactant is proportional to its concentration at low values and the rate of reaction is first order.
2. However at higher concentration, the surface of catalyst is fully covered and the reaction rate becomes independent of concentration and it becomes zero order reaction.
3. The reason is a correct definition of first order but does not explain the change of order.

Teacher's Note:
1. The change to zero order happens because the metal surface is saturated, so adding more NH₃ cannot raise the rate.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Instantaneous rate of a reaction is equal to small change in concentration (dx) during a small interval of time (dt) at that particular instant of time divided by the time interval.

Teacher's Note:
1. Graphically, the instantaneous rate is the slope of the tangent to the concentration–time curve at that instant.

Answer: (c) A is true, but R is false.

Explanation:
1. For a first order reaction, Rate₁ = k[A₁]; if [A₂] = 2[A₁], Rate₂ = k[2A₁] = 2 Rate₁.
2. For a given reaction, rate constant is constant and independent of the concentration of the reactant.

Teacher's Note:
1. Rate changes with concentration; the rate constant k changes only with temperature (or a catalyst) – never with concentration.

Answer: (c) A is true, but R is false.

Explanation:
1. As the temperature of a reaction is increased, the rate of the reaction increases because the reactant molecules collide more frequently and with greater energy per collision.
2. Such effective collisions will increase the conversion of reactants into products in a short period of time.

Teacher's Note:
1. Higher temperature means faster molecules, so collisions become more frequent and more of them have energy above the activation energy.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Dust particles of large surface area released from wheat, millet, barley are suspended in air.
2. Then, in a grain elevator, friction between particles with the elevator in the presence of oxygen creates a spark in the fuel (dust). Thus, explosion takes place due to increase in temperature of particles and elevator created by friction.

Teacher's Note:
1. Finely divided solids react much faster than lumps because more surface is exposed to the other reactant – surface area is one of the factors affecting rate.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state.
2. Molecularity is the number of molecules that participate in the reaction and it is calculated as every single step in a reaction whether it is elementary or complex, while order is an experimental quantity. Thus, in an elementary reaction both molecularity and order are same.

Teacher's Note:
1. Molecularity is always a whole number (1, 2 or 3) and is defined only for an elementary step; order can be zero or fractional.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. CH₃COOC₂H₅ (ester) + H₂O (excess) → CH₃COOH + C₂H₅OH
2. Hydrolysis of an ester follows first order kinetics as [H₂O] remains nearly constant during the course of the reaction. It is pseudo first order reaction.

Teacher's Note:
1. This is the standard example of a pseudo first order reaction: bimolecular, but first order because water is in large excess.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. For a complex reaction, order of overall reaction is an experimental quantity that is equal to the molecularity of the slowest step of the elementary reaction.
2. As rate of overall reaction depends upon the total number of molecules involved in the slowest step of the reaction, for a complex reaction molecularity and order are not same. Order of reaction may be any integer or rational number whereas molecularity can be a whole number 1, 2, 3.

Teacher's Note:
1. For a complex reaction, molecularity has meaning only for each individual step; the overall order is decided by the slowest (rate determining) step.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Order of a reaction may be zero or fractional. It can be determined through the rate law expression by the sum of powers of reactants as order of reaction.
2. It is an experimental quantity which cannot be analysed from the balanced chemical equation.

Teacher's Note:
1. Order comes from the experimentally found rate law, never from the coefficients of the overall equation.