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CBSE Class 12 Chemistry · 11 questions · 27 marks
Thermodynamics tells you whether a reaction can happen; kinetics tells you how fast, and that is often the question that matters. Here you will learn to read a rate law off experimental data, use integrated rate equations for zero- and first-order reactions, and see why a modest rise in temperature can double a rate through the Arrhenius equation.
For a reaction the rate constant has units L mol⁻¹ s⁻¹. The order of the reaction is:
Answer
Second. The general unit of k for an nth-order reaction is (mol L⁻¹)^(1−n) s⁻¹. Setting 1 − n = −1 gives n = 2, which produces mol⁻¹ L s⁻¹, exactly the unit given. A zero-order k has units mol L⁻¹ s⁻¹ and a first-order k has units s⁻¹.
A first-order reaction has a half-life of 20 minutes. What fraction of the reactant remains after 60 minutes?
Answer
1/8. In 60 minutes the reaction passes through 60/20 = 3 half-lives. Each half-life leaves half of what was present, so the fraction remaining is (1/2)³ = 1/8. This works because t½ = 0.693/k is independent of concentration for a first-order reaction.
Which statement about a catalyst is correct?
Answer
It increases the rate by lowering the activation energy through an alternative pathway. A catalyst does not heat the molecules, so the energy distribution is unchanged; instead it offers a route with a lower energy barrier, so a larger fraction of molecules can cross it. Because Ea is lowered equally for the forward and reverse steps, both rates increase by the same factor and the equilibrium position and ΔH stay exactly as they were.
For the reaction 2P + Q → products, doubling [P] at constant [Q] quadruples the rate, while doubling [Q] at constant [P] leaves the rate unchanged. The rate law is:
Answer
Rate = k[P]². Doubling [P] multiplies the rate by 4 = 2², so the order with respect to P is 2. Doubling [Q] gives 2⁰ = 1, no change, so the order with respect to Q is zero and [Q] does not appear in the rate law. The overall order is 2, which need not match the stoichiometric coefficients.
Assertion (A): The acid-catalysed hydrolysis of ethyl ethanoate in dilute aqueous solution follows first-order kinetics. Reason (R): Water is present in such large excess that its concentration remains practically constant, so it is absorbed into the rate constant.
Answer
Both A and R are true and R is the correct explanation of A. The true rate law is Rate = k[ester][H₂O], which is second order overall. Because water is both solvent and reactant, its concentration changes by a negligible fraction, so k[H₂O] can be replaced by a new constant k′ and the observed law becomes Rate = k′[ester]. This is exactly what is meant by a pseudo first-order reaction.
A first-order reaction is 75% complete in 60 minutes. Calculate its rate constant and the time needed for the reaction to be 90% complete.
Answer
Step 1 — 75% complete means [R] = 0.25[R]₀, so [R]₀/[R] = 4. k = (2.303/t) log([R]₀/[R]) = (2.303/60) log 4 = (2.303/60)(0.6021) = 1.3866/60 = 0.0231 min⁻¹. (Shortcut check: 75% complete is exactly two half-lives, so t½ = 30 min and k = 0.693/30 = 0.0231 min⁻¹.) Step 2 — for 90% completion, [R]₀/[R] = 100/10 = 10, so log 10 = 1. t = (2.303/k) log 10 = 2.303/0.0231 = 99.7 min ≈ 100 minutes.
The rate constant of a reaction is 2.0 ×10⁻³ s⁻¹ at 300 K and 8.0 ×10⁻³ s⁻¹ at 320 K. Calculate the activation energy. (R = 8.314 J K⁻¹ mol⁻¹)
Answer
Step 1 — use log(k₂/k₁) = [Ea/(2.303R)] × [(T₂ − T₁)/(T₁T₂)]. Step 2 — k₂/k₁ = 8.0 ×10⁻³ / 2.0 ×10⁻³ = 4, so log 4 = 0.6021. Step 3 — (T₂ − T₁)/(T₁T₂) = 20/(300 × 320) = 20/96000 = 2.083 ×10⁻⁴ K⁻¹. Step 4 — rearrange: Ea = (2.303 × R × 0.6021)/(2.083 ×10⁻⁴) = (2.303 × 8.314 × 0.6021)/(2.083 ×10⁻⁴). Numerator = 19.147 × 0.6021 = 11.53 J mol⁻¹ K⁻¹ ×K. Ea = 11.53/(2.083 ×10⁻⁴) = 55 350 J mol⁻¹ ≈ 55.4 kJ mol⁻¹.
Distinguish between order and molecularity of a reaction, giving one point of difference and one example each.
Answer
Order is the sum of the powers of the concentration terms in the experimentally determined rate law; it is a property of the overall reaction, can be zero or fractional, and cannot be predicted from the balanced equation. Molecularity is the number of reacting species that collide simultaneously in a single elementary step; it is a theoretical quantity, is always a small positive whole number (1, 2 or 3), and has no meaning for a multistep reaction. Example of order: the decomposition of ammonia on a hot platinum surface is zero order, Rate = k. Example of molecularity: the decomposition of N₂O₅ in a single elementary step is unimolecular.
(i) Derive the integrated rate equation for a first-order reaction and show that its half-life is independent of the initial concentration. (ii) For the gas-phase decomposition of a compound X the following initial-rate data were collected at 400 K. Determine the order and the value of k. Experiment 1: [X] = 0.10 mol L⁻¹, rate = 4.0 ×10⁻⁴ mol L⁻¹ s⁻¹. Experiment 2: [X] = 0.20 mol L⁻¹, rate = 8.0 ×10⁻⁴ mol L⁻¹ s⁻¹. Experiment 3: [X] = 0.40 mol L⁻¹, rate = 1.6 ×10⁻³ mol L⁻¹ s⁻¹.
Answer
(i) For R → P with Rate = k[R], write −d[R]/dt = k[R]. Separating variables: d[R]/[R] = −k dt. Integrating from [R]₀ at t = 0 to [R] at time t: ln([R]/[R]₀) = −kt, i.e. ln[R] = ln[R]₀ − kt. Converting to base 10: k = (2.303/t) log([R]₀/[R]). A plot of log[R] against t is a straight line of slope −k/2.303, confirming first order. Half-life: put [R] = [R]₀/2 and t = t½, so k = (2.303/t½) log 2 = 0.693/t½, giving t½ = 0.693/k. Since [R]₀ has cancelled out, the half-life of a first-order reaction depends only on k (and hence temperature), not on how much reactant you started with. (ii) Comparing experiments 1 and 2, [X] doubles from 0.10 to 0.20 and the rate doubles from 4.0 ×10⁻⁴ to 8.0 ×10⁻⁴, a factor of 2 = 2¹. Comparing 2 and 3, [X] doubles again and the rate again doubles, factor 2 = 2¹. Therefore rate ∝ [X]¹ and the reaction is first order, Rate = k[X]. k from experiment 1: k = rate/[X] = 4.0 ×10⁻⁴/0.10 = 4.0 ×10⁻³ s⁻¹. Check with experiment 3: k = 1.6 ×10⁻³/0.40 = 4.0 ×10⁻³ s⁻¹, consistent. The unit s⁻¹ independently confirms first order.
(i) State the Arrhenius equation and explain the physical meaning of A and Ea. (ii) Draw and describe the potential-energy profile of an exothermic reaction, marking Ea(forward), Ea(reverse) and ΔH, and state the relation between them. (iii) Explain, using the Maxwell–Boltzmann distribution, why a rise of only 10 °C near room temperature can roughly double a reaction rate.
Answer
(i) The Arrhenius equation is k = A e^(−Ea/RT), or in logarithmic form log k = log A − Ea/(2.303RT). A is the frequency (pre-exponential) factor: it measures how often molecules collide with the correct orientation, and it has the same units as k. Ea is the activation energy, the minimum extra energy above the average that colliding molecules must possess for the collision to be effective; the term e^(−Ea/RT) is the fraction of molecules with at least that energy. (ii) The profile plots potential energy against reaction coordinate. Reactants sit at a certain energy; the curve rises to a peak (the activated complex or transition state) and then falls to the products, which for an exothermic reaction lie lower than the reactants. Ea(forward) is the height from reactants to peak; Ea(reverse) is the height from products to the same peak; ΔH is the vertical gap between reactants and products and is negative here. The relation is ΔH = Ea(forward) − Ea(reverse), so for an exothermic reaction Ea(forward) is smaller than Ea(reverse). (iii) The Maxwell–Boltzmann curve plots the fraction of molecules against kinetic energy. At the higher temperature the curve flattens and its maximum shifts to higher energy, so the peak is lower but the tail beyond the activation energy is much fatter. The number of molecules with energy greater than Ea is proportional to the area under that tail, and because the dependence is exponential, e^(−Ea/RT), a small rise in T produces a large proportional increase in that area. Near 300 K a 10 °C rise typically roughly doubles the number of effective collisions, so the rate roughly doubles even though the average molecular speed increases by only a few per cent.
A pharmacy stores a liquid medicine whose active ingredient decomposes by a first-order process. At 25 °C the rate constant for decomposition is 3.85 ×10⁻³ day⁻¹. The medicine must be discarded once 50% of the active ingredient has decomposed. On a hot day the storage room reaches 35 °C, where the rate constant is measured as 7.70 ×10⁻³ day⁻¹. (a) Calculate the shelf life (half-life) of the medicine at 25 °C. (b) Calculate the shelf life at 35 °C. (c) By what factor does the rate constant change over this 10 °C rise, and what does that tell you about Ea? (d) A chemist suggests adding a catalyst to make the medicine last longer. Comment on this suggestion.
Answer
(a) For a first-order process t½ = 0.693/k = 0.693/(3.85 ×10⁻³) = 180 days. The medicine keeps for about 180 days at 25 °C. (b) t½ = 0.693/(7.70 ×10⁻³) = 90 days, so the shelf life halves to about 90 days at 35 °C. (c) k₂/k₁ = 7.70 ×10⁻³ / 3.85 ×10⁻³ = 2, so the rate constant doubles for a 10 °C rise. This is the typical temperature coefficient near room temperature and corresponds to a moderate activation energy: using log 2 = [Ea/(2.303 × 8.314)][10/(298 × 308)] gives Ea ≈ 5.3 ×10⁴ J mol⁻¹, about 53 kJ mol⁻¹. (d) The suggestion is wrong. A catalyst lowers the activation energy of the decomposition, so it would make the medicine decompose faster, not slower. The correct approach is to slow the reaction by lowering the temperature — refrigerated storage — since k falls exponentially as T falls, or to use an inhibitor or protective packaging that removes a reactant such as oxygen or moisture.
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