CBSE Class 12 Chemistry Chemical Kinetics MCQs Set 13

Answer: (d) 16

Explanation:
1. When the volume is reduced to 1/4, the concentration of each gas becomes 4 times.
2. New rate = k[4A][4B] = 16 k[A][B], i.e., 16 times the original rate.

Teacher's Note:
1. Reducing the volume increases concentration in the same ratio; then raise that factor to the power of each concentration term in the rate law.

Answer: (c) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction.

Explanation:
1. Order is determined experimentally from the rate law; it need not be equal to the sum of the stoichiometric coefficients in the balanced equation.

Teacher's Note:
1. Order and stoichiometric coefficients match only for elementary reactions – never assume it for an overall equation.

Answer: (a) Order is same as molecularity.

Explanation:
1. For an elementary reaction, the rate law follows directly from the balanced equation, so order is the same as molecularity.
2. Molecularity is the number of reacting species colliding in an elementary step, so it is always a whole number and can never be zero or fractional.

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

Answer: (b) 2 and 1½

Explanation:
1. Molecularity (from the number of reacting molecules, H₂ + Br₂) = 1 + 1 = 2.
2. Order (from the rate law) = 1 + ½ = 1½.

Teacher's Note:
1. Molecularity can never be fractional, so options with 1½ as molecularity can be rejected at once.

Answer: (c) mole litre⁻¹ second⁻¹

Explanation:
1. For a zero order reaction, rate = k, so k has the same unit as rate, i.e., mol L⁻¹ s⁻¹.

Teacher's Note:
1. For zero order, rate and rate constant always have identical units.

Answer: (c) \( \frac{1}{2} \)

Explanation:
1. 2 = 4ⁿ = 2²ⁿ, so 2n = 1 and n = ½.

Teacher's Note:
1. Write the ratio of rates as (ratio of concentrations)ⁿ and express both sides with the same base.

Answer: (b) Decomposition of NH₃

Explanation:
1. Decomposition of NH₃ on a hot platinum surface at high pressure is a zero order reaction, as the metal surface gets saturated with NH₃ molecules.
2. Decomposition of N₂O₅, N₂O and radioactive decay are first order processes.

Teacher's Note:
1. Zero order behaviour usually appears in surface-catalysed reactions where the surface is fully covered, so adding more reactant does not change the rate.

Answer: (a) a first order reaction

Explanation:
1. Rate ∝ [salt]ⁿ; doubling the concentration doubles the rate, so 2 = 2ⁿ and n = 1.

Teacher's Note:
1. Doubling concentration: rate ×1 means zero order, ×2 first order, ×4 second order.

Answer: (c) ln [R]₀

Explanation:
1. ln [R] = −kt + ln [R]₀, which is of the form y = mx + c.
2. Slope = −k and intercept = ln [R]₀.

Teacher's Note:
1. Do not mix up slope and intercept: −k is the slope; for a log₁₀ plot the slope would be −k/2.303.

Answer: (c) 16 min

Explanation:
1. For a first order reaction, t_75% = 2 × t_1/2.
2. t_1/2 = 32/2 = 16 min.

Teacher's Note:
1. 75% completion means two half-lives have passed (100 → 50 → 25% left) – a quick check for first order problems.

Answer: (c) 0.01 s⁻¹

Explanation:
1. k = \( \frac{0.693}{t_{1/2}} \) = \( \frac{0.693}{69.35} \) = 0.01 s⁻¹

Teacher's Note:
1. For first order, half-life is independent of initial concentration, so k follows directly from t_1/2.

Answer: (d) is nearly independent of temperature

Explanation:
1. If E_a = 0, k = Ae⁰ = A; the rate constant no longer depends on the exponential temperature factor, so the rate is nearly independent of temperature.

Teacher's Note:
1. The temperature sensitivity of a rate comes from the e^−Ea/RT factor; with E_a = 0 that factor is 1 at all temperatures.

Answer: (a) ln k vs \( \frac{1}{T} \)

Explanation:
1. ln k = ln A − \( \frac{E_a}{RT} \); a plot of ln k vs 1/T is a straight line with slope = −\( \frac{E_a}{R} \).

Teacher's Note:
1. For a log₁₀ k vs 1/T plot, the slope is −E_a/2.303R – remember the 2.303 when converting.

Answer: (a) Activation energy of forward reaction is E₁ + E₂ and product is less stable than reactant.

Explanation:
1. Activation energy of the forward reaction = energy of activated complex − energy of reactants = E₁ + E₂.
2. Products lie at a higher energy than reactants, so the product is less stable than the reactant (endothermic reaction).

Teacher's Note:
1. Measure activation energy from the starting level up to the peak: forward from reactants, backward from products (here E₁ only).

Answer: (c) activation energy of reaction

Explanation:
1. A catalyst provides an alternate pathway of lower activation energy; it does not change ΔG, ΔH or the equilibrium constant.

Teacher's Note:
1. A catalyst speeds up the forward and backward reactions equally, so equilibrium is reached faster but its position does not change.

Answer: (a) the same.

Explanation:
1. Rate constant of a reaction does not depend upon the concentration of the reactants.

Teacher's Note:
1. Doubling [B] doubles the rate, not the rate constant – students often mix up the two. k changes only with temperature (or a catalyst).

Answer: (b) Catalyst raises the activation energy.

Explanation:
1. (i) When catalyst is added in reaction medium, the rate of reaction increases by decreasing activation energy of the molecule. Hence, it follows an alternative pathway.
2. (ii) Catalyst does not change the enthalpy change of reaction. Energy of reactant and product remain same in both catalysed and uncatalysed reaction.
3. Therefore, (b) is the incorrect statement.

Teacher's Note:
1. A catalyst always lowers E_a by providing a new pathway; it never raises it (a substance that slows a reaction is called an inhibitor and works differently).

Answer: (b) determining the rate constants at two temperatures.

Explanation:
1. Activation energy of a chemical reaction is linked to the rate constants of a reaction at two different temperatures, that is k₁ and k₂:
2. \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \)
3. where E_a = activation energy, T₂ = higher temperature, T₁ = lower temperature, k₁ = rate constant at temperature T₁, k₂ = rate constant at temperature T₂.
4. This equation is known as the Arrhenius equation.

Teacher's Note:
1. With one rate constant you still have two unknowns (A and E_a); a second temperature lets A cancel out.

Answer: (b) It alters Gibbs energy (ΔG) of the reaction.

Explanation:
1. The characteristics of catalysts are given below:
2. (i) It catalyses the forward and backward reactions to the same extent as it decreases energy of activation, hence increases the rate of both the reactions.
3. (ii) Catalyst does not alter Gibbs free energy, as the energies of the reactants and products remain the same.
4. (iii) It does not alter equilibrium of reaction, as equilibrium constant is related to ΔG°, which is unchanged.
5. (iv) It provides an alternate mechanism by reducing activation energy between reactants and product.

Teacher's Note:
1. A catalyst helps a system reach equilibrium faster but does not shift the position of equilibrium.

Answer: (b) 6.93 × 10⁻³ s⁻¹

Explanation:
1. For a first order reaction, k = \( \frac{0.693}{t_{1/2}} \) = \( \frac{0.693}{100} \) = 6.93 × 10⁻³ s⁻¹

Teacher's Note:
1. Only for first order reactions is t_1/2 = 0.693/k, independent of concentration; do not use it for zero order reactions.

Answer: (b) \( \frac{-E_a}{2.303R} \)

Explanation:
1. log K = log A − \( \frac{E_a}{2.303R} \) × \( \frac{1}{T} \), which is of the form y = c + mx.
2. Hence, slope = −\( \frac{E_a}{2.303R} \).

Teacher's Note:
1. With ln K instead of log K, the slope would be −E_a/R; check which logarithm the graph uses.

Answer: (b) first

Explanation:
1. When C is in large excess, [C] remains practically constant and gets merged with K.
2. Rate = K′[D], so the reaction behaves as a first order (pseudo first order) reaction.

Teacher's Note:
1. A reactant present in large excess drops out of the observed order – this is the idea behind pseudo first order reactions.

Answer: (b) 3.2 kJ

Explanation:
1. ΔH = E_a(forward) − E_a(reverse)
2. E_a(reverse) = 57.2 − 54 = 3.2 kJ

Teacher's Note:
1. For an endothermic reaction (ΔH positive), the reverse reaction always has the smaller activation energy.

Answer: (c) the absolute temperature is infinity

Explanation:
1. k = Ae^−E_a/RT. When T → ∞, E_a/RT → 0 and e⁰ = 1, so k = A.
2. If E_a = RT, k = A/e, which is not equal to A.

Teacher's Note:
1. k equals A only when the exponential factor becomes 1, i.e. when E_a/RT becomes zero.

Answer: (a) zero order

Explanation:
1. A straight line with negative slope for [A] vs time follows [A] = [A]₀ − kt, which is the integrated rate equation of a zero order reaction.

Teacher's Note:
1. For a first order reaction, [A] vs time is a curve; it is ln[A] vs time that is a straight line.

Answer: (a) Rate = K[A][B]

Explanation:
1. For an elementary reaction, the powers in the rate law equal the stoichiometric coefficients: Rate = K[A]¹[B]¹, order = 2.

Teacher's Note:
1. Only for elementary (single-step) reactions can the rate law be written directly from the equation.

Answer: (b) 0, 2

Explanation:
1. The photochemical combination of H₂ and Cl₂ is a zero order reaction (its rate depends on light intensity, not on concentration).
2. Two molecules take part in the reaction, so molecularity = 2.

Teacher's Note:
1. Molecularity can never be zero, so options (a) and (d) can be ruled out at once.

Answer: (d) CH₃COOC₂H₅ + H₂O \( \xrightarrow{H^+} \) CH₃COOH + C₂H₅OH

Explanation:
1. In acid hydrolysis of ethyl acetate, water is in large excess and its concentration stays constant.
2. Rate = K[CH₃COOC₂H₅], so the reaction is pseudo first order.

Teacher's Note:
1. Alkaline hydrolysis of an ester (with NaOH) is a true second order reaction – do not confuse it with acid hydrolysis.

Answer: (b) Increase in the number of activated molecules

Explanation:
1. A rise in temperature increases the fraction of molecules having energy equal to or more than the activation energy (activated molecules), so the rate increases sharply.

Teacher's Note:
1. Temperature does not lower E_a – only a catalyst does that. Collision frequency also rises, but only slightly.

Answer: (b) 1

Explanation:
1. The half-life remains practically constant (about 30 s) whatever the initial concentration.
2. Half-life independent of initial concentration is a property of first order reactions.

Teacher's Note:
1. Remember: t_1/2 ∝ [R]₀ for zero order, constant for first order, and ∝ 1/[R]₀ for second order.

Answer: (d) E_a = E′_a

Explanation:
1. Activation energy is the energy barrier between reactants and the activated complex; it does not depend on how the chemical equation is written (its stoichiometric coefficients).

Teacher's Note:
1. Unlike ΔH, activation energy is not multiplied when the equation is multiplied by a number.

Answer: (c) 48.8%

Explanation:
1. For a first order reaction, the same fraction reacts in each equal time interval. Fraction left after 1 h = 0.8.
2. Fraction left after 3 h = (0.8)³ = 0.512, so extent of reaction = 1 − 0.512 = 0.488 = 48.8%.

Teacher's Note:
1. A first order reaction does not complete 20% per hour in a linear way, so 3 × 20% = 60% is a trap.

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

Explanation:
1. For a first order reaction, t_1/2 = ln 2/k, where half-life period is independent of initial concentration of the reacting species and is calculated from the rate constant.

Teacher's Note:
1. The reason is just the definition of half-life; the real explanation is the formula t_½ = 0.693/k, which has no [R]₀ in it.

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

Explanation:
1. Instantaneous rate is the rate at a particular moment of time (slope of the tangent), so A is true.
2. Average rate is a single value for the chosen interval, so R is also true, but it does not explain the meaning of instantaneous rate.

Teacher's Note:
1. Average rate = Δ[R]/Δt over an interval; instantaneous rate is its limit as Δt → 0, i.e., d[R]/dt.

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

Explanation:
1. The rate law of this reaction is determined experimentally and is rate = k[CHCl₃][Cl₂]^1/2, so A is true.
2. Rate law cannot be predicted from the balanced equation (here the power of Cl₂ is ½, not 1), so R is false.

Teacher's Note:
1. The fractional power on Cl₂ is itself the proof that rate laws come from experiment, not from coefficients.

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

Explanation:
1. Order and molecularity are different: order is experimental and can be fractional, molecularity is theoretical and always a whole number, so A is false.
2. R correctly describes how each is obtained, so R is true.

Teacher's Note:
1. They become numerically equal only for elementary reactions; in general they are different quantities.

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

Explanation:
1. Inversion of cane sugar is a pseudo first order reaction, so A is true.
2. Water is present in large excess, so its concentration remains almost constant during the reaction; R is false.

Teacher's Note:
1. Pseudo first order arises when one reactant is in large excess, so its concentration is effectively constant and merges into k.

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

Explanation:
1. For most reactions, the rate constant nearly doubles for a 10° rise in temperature, so A is true.
2. The energy distribution of gas molecules (Maxwell–Boltzmann distribution) depends on temperature, so R is true; but R by itself does not explain the doubling of the rate constant.

Teacher's Note:
1. The ratio k_T+10/k_T (about 2 to 3) is called the temperature coefficient.

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

Explanation:
1. If E_a = 0, then k = Ae⁰ = A, so k becomes independent of temperature; A is true.
2. A lower activation energy means a larger fraction of molecules can cross the energy barrier, so the reaction is faster; R is also true. But R does not explain why temperature has no effect when E_a = 0.

Teacher's Note:
1. Temperature affects k only through the factor e^−Ea/RT; with E_a = 0 this factor is 1 at every temperature.

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

Explanation:
1. The Arrhenius equation, with experimentally found A and E_a, gives fairly accurate rate constants for reactions of simple as well as complex molecules, so A is true.
2. Only those collisions in which molecules have proper orientation (and sufficient energy) lead to chemical change, so R is false.

Teacher's Note:
1. Proper orientation is the reason collision theory adds the steric (probability) factor P: rate = PZ_ABe^−Ea/RT.

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

Explanation:
1. A catalyst provides a new path through a different activated complex with lower activation energy.
2. The energies of reactants and products are unchanged, so ΔH remains the same. R correctly explains A.

Teacher's Note:
1. ΔH = energy of products − energy of reactants; a catalyst changes only the height of the hill between them.