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Chemical Kinetics

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Chapter Notes

Overview
This chapter deals with the study of reaction rates and their mechanisms, known as chemical kinetics. It explains how fast a reaction occurs, the factors that control the rate (concentration, temperature, catalyst), and the mathematical equations that describe these rates. The one big idea a student must walk away with is that thermodynamics tells if a reaction is possible, but kinetics tells how fast it actually happens, and that the rate depends on the concentration of reactants, temperature, and the presence of a catalyst.
Key Concepts (21)
Chemical kinetics
The branch of chemistry which deals with the study of reaction rates and their mechanisms.
e.g. Studying how fast food spoils or how quickly fuel burns in an engine.
Rate of a reaction
The change in concentration of a reactant or product in unit time.
e.g. If the concentration of a reactant decreases from 0.03 M to 0.02 M in 25 minutes, the average rate is (0.01 M)/(25 min) = 4 × 10⁻⁴ M min⁻¹.
Average rate of a reaction
The rate of a reaction determined over a finite time interval, given by Δ[R]/Δt or Δ[P]/Δt.
e.g. In the hydrolysis of butyl chloride, the average rate over 0-50 s is 1.90 × 10⁻⁴ mol L⁻¹ s⁻¹.
Instantaneous rate of a reaction
The rate of a reaction at a particular moment of time, obtained when the time interval approaches zero, given by -d[R]/dt or d[P]/dt.
e.g. At t = 600 s for the hydrolysis of butyl chloride, the instantaneous rate is 5.12 × 10⁻⁵ mol L⁻¹ s⁻¹.
Rate law or rate expression
The expression in which the reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which must be determined experimentally.
e.g. For the reaction 2NO(g) + O₂(g) → 2NO₂(g), the rate law is Rate = k[NO]²[O₂].
Rate constant (k)
A proportionality constant in the rate law, specific for a given reaction at a given temperature.
e.g. For a first order reaction, k = 2.303/t log([R]₀/[R]) and its unit is s⁻¹.
Order of a reaction
The sum of the powers of the concentration of the reactants in the rate law expression.
e.g. For Rate = k[A]¹/²[B]³/², the overall order is 1/2 + 3/2 = 2 (second order).
Molecularity of a reaction
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously to bring about a chemical reaction.
e.g. Decomposition of NH₄NO₂ → N₂ + 2H₂O is unimolecular; 2HI → H₂ + I₂ is bimolecular.
Elementary reaction
A reaction that takes place in one step.
e.g. H₂O₂ + I⁻ → H₂O + IO⁻ is an elementary bimolecular reaction.
Complex reaction
A reaction that takes place through a sequence of elementary reactions (mechanism).
e.g. Oxidation of ethane to CO₂ and H₂O passes through a series of intermediate steps.
Rate determining step
The slowest step in a reaction mechanism that controls the overall rate of the reaction.
e.g. In the decomposition of H₂O₂ catalysed by I⁻, the first step (H₂O₂ + I⁻ → H₂O + IO⁻) is slow and rate determining.
Half-life of a reaction (t₁/₂)
The time in which the concentration of a reactant is reduced to one half of its initial concentration.
e.g. For a first order reaction, t₁/₂ = 0.693/k, and it is independent of initial concentration.
Pseudo first order reaction
A reaction that is actually of higher order but behaves as first order because one reactant is taken in large excess.
e.g. Hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O (excess) → CH₃COOH + C₂H₅OH, Rate = k[CH₃COOC₂H₅].
Activation energy (Ea)
The minimum energy required to form the activated complex (intermediate) for a reaction to occur.
e.g. For the reaction H₂ + I₂ → 2HI, the energy required to form the activated complex is the activation energy.
Arrhenius equation
An equation showing the temperature dependence of the rate constant: k = A e⁻ᴱᵃ/ᴿᵀ.
e.g. For a reaction, if Ea is 209 kJ/mol at 600 K, the rate constant can be calculated at 700 K using the equation.
Frequency factor (A) or pre-exponential factor
A constant specific to a particular reaction, related to the collision frequency of reactants.
e.g. In the Arrhenius equation, A = 1.61 for a reaction with k = 0.02 s⁻¹ at 500 K.
Catalyst
A substance which increases the rate of a reaction without itself undergoing any permanent chemical change.
e.g. MnO₂ catalyses the decomposition of KClO₃: 2KClO₃ → 2KCl + 3O₂.
Collision theory
A theory based on kinetic theory of gases that states reactant molecules must collide with sufficient kinetic energy (≥ Ea) and proper orientation for a reaction to occur.
e.g. Formation of methanol from bromoethane depends on the proper orientation of reactant molecules; improper orientation leads to no reaction.
Effective collisions
Collisions in which molecules collide with sufficient kinetic energy (threshold energy) and proper orientation to break bonds and form products.
e.g. In the reaction between H₂ and I₂, only collisions with proper orientation and energy above Ea lead to HI formation.
Collision frequency (Z)
The number of collisions per second per unit volume of the reaction mixture.
e.g. For a bimolecular reaction A + B → Products, Rate = ZAB e⁻ᴱᵃ/ᴿᵀ.
Steric factor (P) or probability factor
A factor introduced to account for the fact that not all collisions are effective due to improper orientation.
e.g. For complex molecules, the rate constant is given by Rate = P ZAB e⁻ᴱᵃ/ᴿᵀ.
Formulas & Equations (14)
Rate = -Δ[R]/Δt = +Δ[P]/Δt
Reads as: Rate equals negative change in concentration of reactant over change in time, or positive change in concentration of product over change in time.
Variables: [R] = concentration of reactant (mol L⁻¹), [P] = concentration of product (mol L⁻¹), t = time (s or min)
Use when: Calculating average rate of a reaction over a time interval.
⚠ Forgetting the negative sign for reactant concentration, leading to a negative rate value.
Instantaneous rate = -d[R]/dt = d[P]/dt
Reads as: Instantaneous rate equals negative derivative of reactant concentration with respect to time, or derivative of product concentration with respect to time.
Variables: d[R] = infinitesimal change in reactant concentration (mol L⁻¹), dt = infinitesimal change in time (s)
Use when: Finding the rate at a specific instant from a concentration vs time graph by drawing a tangent.
⚠ Using average rate instead of instantaneous rate when the question asks for rate at a particular time.
For reaction aA + bB → cC + dD: Rate = -(1/a) d[A]/dt = -(1/b) d[B]/dt = (1/c) d[C]/dt = (1/d) d[D]/dt
Reads as: Rate equals negative one over stoichiometric coefficient of A times derivative of A, and so on.
Variables: a, b, c, d = stoichiometric coefficients (dimensionless)
Use when: Expressing rate for reactions where stoichiometric coefficients are not equal to one.
⚠ Not dividing by the stoichiometric coefficient, leading to different rate values for different species.
Rate = k [A]ˣ [B]ʸ
Reads as: Rate equals rate constant times concentration of A raised to power x times concentration of B raised to power y.
Variables: k = rate constant (units vary), x = order with respect to A, y = order with respect to B, x+y = overall order
Use when: Writing the rate law for a reaction based on experimental data.
⚠ Assuming x and y are equal to the stoichiometric coefficients from the balanced equation.
For zero order reaction: k = ([R]₀ - [R]) / t
Reads as: Rate constant equals initial concentration minus concentration at time t divided by time.
Variables: [R]₀ = initial concentration (mol L⁻¹), [R] = concentration at time t (mol L⁻¹), t = time (s)
Use when: Calculating rate constant for a zero order reaction.
⚠ Using this formula for first order reactions.
For first order reaction: k = (2.303/t) log([R]₀/[R])
Reads as: Rate constant equals 2.303 divided by time times log of initial concentration divided by concentration at time t.
Variables: k = rate constant (s⁻¹), t = time (s), [R]₀ = initial concentration (mol L⁻¹), [R] = concentration at time t (mol L⁻¹)
Use when: Calculating rate constant or concentration for a first order reaction.
⚠ Using natural log (ln) instead of log base 10 without the factor 2.303.
For first order reaction: [R] = [R]₀ e⁻ᵏᵗ
Reads as: Concentration at time t equals initial concentration times e to the power of negative kt.
Variables: [R] = concentration at time t (mol L⁻¹), [R]₀ = initial concentration (mol L⁻¹), k = rate constant (s⁻¹), t = time (s)
Use when: Finding concentration after a given time for a first order reaction.
⚠ Forgetting the negative sign in the exponent.
For first order gas phase reaction: k = (2.303/t) log(pᵢ / (2pᵢ - pₜ))
Reads as: Rate constant equals 2.303 divided by time times log of initial pressure divided by twice initial pressure minus total pressure.
Variables: pᵢ = initial pressure of reactant (atm), pₜ = total pressure at time t (atm), t = time (s)
Use when: Calculating rate constant for a first order gas phase reaction where total pressure is measured.
⚠ Using pₜ directly instead of calculating the pressure of the reactant as (2pᵢ - pₜ).
Half-life for zero order: t₁/₂ = [R]₀ / 2k
Reads as: Half-life equals initial concentration divided by twice the rate constant.
Variables: t₁/₂ = half-life (s), [R]₀ = initial concentration (mol L⁻¹), k = rate constant (mol L⁻¹ s⁻¹)
Use when: Calculating half-life for a zero order reaction.
⚠ Forgetting that t₁/₂ depends on initial concentration for zero order.
Half-life for first order: t₁/₂ = 0.693 / k
Reads as: Half-life equals 0.693 divided by the rate constant.
Variables: t₁/₂ = half-life (s), k = rate constant (s⁻¹)
Use when: Calculating half-life for a first order reaction.
⚠ Using 0.693 as 2.303 or confusing it with the formula for zero order.
Arrhenius equation: k = A e⁻ᴱᵃ/ᴿᵀ
Reads as: Rate constant equals frequency factor times e to the power of negative activation energy divided by R times T.
Variables: k = rate constant (units vary), A = frequency factor (same units as k), Ea = activation energy (J mol⁻¹), R = gas constant (8.314 J mol⁻¹ K⁻¹), T = temperature (K)
Use when: Expressing temperature dependence of rate constant.
⚠ Using Ea in kJ without converting to J, or using Celsius instead of Kelvin.
ln k = -Ea/(RT) + ln A
Reads as: Natural log of k equals negative activation energy divided by R T plus natural log of A.
Variables: k = rate constant, Ea = activation energy (J mol⁻¹), R = 8.314 J mol⁻¹ K⁻¹, T = temperature (K), A = frequency factor
Use when: Plotting ln k vs 1/T to find Ea (slope = -Ea/R) and A (intercept = ln A).
⚠ Plotting log k vs 1/T and using the same slope formula without the 2.303 factor.
log(k₂/k₁) = (Ea/(2.303 R)) × (1/T₁ - 1/T₂)
Reads as: Log of ratio of rate constants equals activation energy divided by 2.303 R times the difference of reciprocals of temperatures.
Variables: k₁, k₂ = rate constants at T₁ and T₂, Ea = activation energy (J mol⁻¹), R = 8.314 J mol⁻¹ K⁻¹, T₁, T₂ = temperatures (K)
Use when: Calculating activation energy or rate constant at a different temperature.
⚠ Writing (T₂ - T₁) instead of (1/T₁ - 1/T₂) in the formula.
For bimolecular elementary reaction: Rate = ZAB e⁻ᴱᵃ/ᴿᵀ
Reads as: Rate equals collision frequency of A and B times e to the power of negative activation energy divided by R T.
Variables: ZAB = collision frequency of reactants A and B, Ea = activation energy (J mol⁻¹), R = 8.314 J mol⁻¹ K⁻¹, T = temperature (K)
Use when: Understanding collision theory for simple bimolecular reactions.
⚠ Forgetting that this applies only to elementary reactions, not complex ones.
How It Works
Determining instantaneous rate from a graph
  1. Plot concentration of reactant or product vs time.
  2. Draw a tangent line that touches the curve at the specific time of interest.
  3. Calculate the slope of the tangent line (Δy/Δx).
  4. For reactant, the instantaneous rate is the negative of the slope; for product, it is the slope.
💡 The tangent must touch the curve at exactly one point; a common mistake is drawing a secant (line through two points) instead of a tangent.
Deriving integrated rate equation for zero order reaction
  1. Start with differential rate law: -d[R]/dt = k [R]⁰ = k.
  2. Rearrange: d[R] = -k dt.
  3. Integrate both sides: ∫ d[R] = -k ∫ dt, giving [R] = -kt + I.
  4. Apply initial condition: at t=0, [R] = [R]₀, so I = [R]₀.
  5. Final equation: [R] = -kt + [R]₀, or k = ([R]₀ - [R])/t.
💡 The plot of [R] vs t is a straight line with slope = -k and intercept = [R]₀.
Deriving integrated rate equation for first order reaction
  1. Start with differential rate law: -d[R]/dt = k [R].
  2. Rearrange: d[R]/[R] = -k dt.
  3. Integrate both sides: ∫ d[R]/[R] = -k ∫ dt, giving ln[R] = -kt + I.
  4. Apply initial condition: at t=0, [R] = [R]₀, so I = ln[R]₀.
  5. Final equation: ln[R] = -kt + ln[R]₀, or k = (2.303/t) log([R]₀/[R]).
💡 The plot of ln[R] vs t is a straight line with slope = -k and intercept = ln[R]₀.
Deriving rate constant for first order gas phase reaction using total pressure
  1. For reaction A(g) → B(g) + C(g), let initial pressure of A be pᵢ.
  2. At time t, let x be the decrease in pressure of A, so pA = pᵢ - x, pB = x, pC = x.
  3. Total pressure pₜ = (pᵢ - x) + x + x = pᵢ + x, so x = pₜ - pᵢ.
  4. Pressure of A at time t: pA = pᵢ - (pₜ - pᵢ) = 2pᵢ - pₜ.
  5. Substitute into first order equation: k = (2.303/t) log(pᵢ / (2pᵢ - pₜ)).
💡 The pressure of the reactant is not directly given; it must be calculated from total pressure using stoichiometry.
Effect of temperature on reaction rate (Arrhenius equation)
  1. Write Arrhenius equation: k = A e⁻ᴱᵃ/ᴿᵀ.
  2. Take natural log: ln k = -Ea/(RT) + ln A.
  3. Plot ln k vs 1/T; the graph is a straight line.
  4. Slope of the line = -Ea/R, from which Ea can be calculated.
  5. Intercept = ln A, from which A can be calculated.
  6. For two temperatures, use: log(k₂/k₁) = (Ea/(2.303 R)) × (1/T₁ - 1/T₂).
💡 A 10°C rise in temperature approximately doubles the rate constant for many reactions.
Effect of catalyst on activation energy
  1. Catalyst provides an alternate pathway or reaction mechanism.
  2. It forms temporary bonds with reactants to form an intermediate complex.
  3. The intermediate complex decomposes to yield products and the catalyst.
  4. This alternate pathway has a lower activation energy than the uncatalysed reaction.
  5. Lower activation energy leads to a higher rate constant (from Arrhenius equation).
💡 A catalyst does not change ΔG or equilibrium constant; it only helps attain equilibrium faster.
Collision theory of chemical reactions
  1. Reactant molecules are assumed to be hard spheres.
  2. Reaction occurs when molecules collide with each other.
  3. The number of collisions per second per unit volume is called collision frequency (Z).
  4. Only collisions with kinetic energy equal to or greater than activation energy (Ea) are effective.
  5. Additionally, molecules must have proper orientation for bond breaking and formation.
  6. The rate constant is given by: Rate = P ZAB e⁻ᴱᵃ/ᴿᵀ, where P is the steric factor.
💡 Not all collisions lead to reaction; only effective collisions (sufficient energy + proper orientation) do.
Exam Traps — Don't Make These Mistakes
⚠ Confusing order with molecularity
✗ Order of a reaction is the number of molecules colliding in the rate-determining step.
→ Order is an experimental quantity determined from rate law; molecularity is the number of reacting species in an elementary step. Order can be zero or fractional, but molecularity cannot.
✓ Order of a reaction is the sum of powers of concentration terms in the experimentally determined rate law. Molecularity is the number of reacting species that must collide simultaneously in an elementary reaction.
⚠ Using stoichiometric coefficients in rate law
✗ For the reaction 2NO + O₂ → 2NO₂, the rate law is Rate = k[NO]²[O₂] because the coefficients are 2 and 1.
→ Rate law cannot be predicted from the balanced chemical equation; it must be determined experimentally. In this case, the experimental data matches the coefficients, but that is not always true (e.g., CHCl₃ + Cl₂ → CCl₄ + HCl has Rate = k[CHCl₃][Cl₂]¹/²).
✓ Rate law for a reaction is determined experimentally. For the given reaction, experimental data shows Rate = k[NO]²[O₂], but this is not a general rule.
⚠ Forgetting the negative sign in rate expression
✗ Rate = Δ[R]/Δt
→ The concentration of reactant decreases with time, so Δ[R] is negative. Without the negative sign, the rate would be negative, which is not meaningful. The rate must be a positive quantity.
✓ Rate = -Δ[R]/Δt (for reactant) or Rate = +Δ[P]/Δt (for product).
⚠ Mixing up units of rate constant for different orders
✗ For a first order reaction, the unit of k is mol L⁻¹ s⁻¹.
→ The unit of k depends on the overall order. For first order, k has unit s⁻¹ (or time⁻¹). For zero order, it is mol L⁻¹ s⁻¹. For second order, it is L mol⁻¹ s⁻¹.
✓ For a first order reaction, the unit of k is s⁻¹ (time⁻¹). For zero order, it is mol L⁻¹ s⁻¹. For second order, it is L mol⁻¹ s⁻¹.
⚠ Using log instead of ln in integrated rate equation
✗ k = (1/t) ln([R]₀/[R])
→ The integrated rate equation for first order is ln[R] = -kt + ln[R]₀, which gives k = (1/t) ln([R]₀/[R]). However, if using log base 10, the formula is k = (2.303/t) log([R]₀/[R]). Students often mix the two forms.
✓ If using natural log: k = (1/t) ln([R]₀/[R]). If using log base 10: k = (2.303/t) log([R]₀/[R]).
⚠ Assuming half-life is always constant
✗ Half-life of a reaction is always constant and independent of initial concentration.
→ Half-life is constant only for first order reactions. For zero order reactions, t₁/₂ = [R]₀/(2k), which depends on initial concentration.
✓ For a first order reaction, half-life is constant (t₁/₂ = 0.693/k) and independent of initial concentration. For zero order, half-life is directly proportional to initial concentration.
Compare & Contrast
Definition and applicability
Order of a reaction
Molecularity of a reaction
Order is the sum of powers of concentration terms in the rate law (experimental), can be zero, fractional, or integer, and applies to both elementary and complex reactions. Molecularity is the number of reacting species in an elementary step, is always a positive integer (≥1), and applies only to elementary reactions.
Dependence of rate on concentration
Zero order reaction
First order reaction
For zero order, rate is independent of concentration (Rate = k). For first order, rate is directly proportional to the first power of concentration (Rate = k[R]). Half-life for zero order depends on initial concentration (t₁/₂ = [R]₀/2k), while for first order it is constant (t₁/₂ = 0.693/k).
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1 short answer (2-3 marks) on rate expression or half-life, 1 long answer (5 marks) on integrated rate equation, Arrhenius equation, or numerical problem. Total: 7-10 marks.
Most asked: Order and molecularity of reactions · Integrated rate equations for zero and first order reactions · Half-life calculations · Arrhenius equation and activation energy · Effect of temperature and catalyst on rate · Pseudo first order reactions · Rate law from experimental data
Don't skip: Numerical problems on first order rate constant using integrated rate equation or half-life; Derivation of integrated rate equations for zero and first order reactions; Numerical on Arrhenius equation to find Ea or k at different temperatures; Distinguishing between order and molecularity with examples
One-Line Revision (read aloud before the exam)
  1. Chemical kinetics is the study of reaction rates and their mechanisms.
  2. Average rate of a reaction is Δ[R]/Δt or Δ[P]/Δt over a finite time interval.
  3. Instantaneous rate is -d[R]/dt or d[P]/dt at a specific moment.
  4. Rate law must be determined experimentally; it cannot be predicted from the balanced equation.
  5. Order of a reaction is the sum of exponents of concentration terms in the rate law.
  6. Molecularity is the number of reacting species in an elementary step; it is always a positive integer.
  7. For a zero order reaction, rate is constant and independent of concentration.
  8. For a first order reaction, rate is proportional to the first power of concentration.
  9. Integrated rate equation for first order: k = (2.303/t) log([R]₀/[R]).
  10. Half-life for first order: t₁/₂ = 0.693/k, independent of initial concentration.
  11. Half-life for zero order: t₁/₂ = [R]₀/(2k), depends on initial concentration.
  12. Pseudo first order reactions occur when one reactant is in large excess.
  13. Arrhenius equation: k = A e⁻ᴱᵃ/ᴿᵀ, shows temperature dependence of rate constant.
  14. Activation energy (Ea) is the minimum energy required for a reaction to occur.
  15. A catalyst lowers activation energy and provides an alternate pathway.
  16. Collision theory states that molecules must collide with sufficient energy and proper orientation.
  17. Effective collisions have energy ≥ Ea and correct orientation.
  18. A catalyst does not change ΔG or equilibrium constant of a reaction.
  19. For a first order gas phase reaction, k = (2.303/t) log(pᵢ/(2pᵢ - pₜ)).
  20. The slowest step in a mechanism is the rate determining step.

MCQs — 10 Questions with Answers

Q1. For the reaction: N₂O₅ → N₂O₄ + ½O₂, if the rate of disappearance of N₂O₅ is 1.4 × 10⁻⁵ mol L⁻¹ s⁻¹, what is the rate of appearance of O₂?

  • A. 0.7 × 10⁻⁵ mol L⁻¹ s⁻¹ ✓
  • B. 1.4 × 10⁻⁵ mol L⁻¹ s⁻¹
  • C. 2.8 × 10⁻⁵ mol L⁻¹ s⁻¹
  • D. 0.35 × 10⁻⁵ mol L⁻¹ s⁻¹

Answer: A — From stoichiometry, 1 mole N₂O₅ produces 0.5 mole O₂, so rate of O₂ = 0.5 × 1.4 × 10⁻⁵ = 0.7 × 10⁻⁵ mol L⁻¹ s⁻¹.

Q2. Which statement correctly distinguishes thermodynamics from chemical kinetics?

  • A. Thermodynamics predicts speed; kinetics predicts feasibility
  • B. Thermodynamics predicts feasibility (ΔG); kinetics determines rate at which reaction proceeds ✓
  • C. Both predict the same information about reactions
  • D. Kinetics uses entropy; thermodynamics uses concentration

Answer: B — Thermodynamics answers 'will it happen?' (ΔG < 0); kinetics answers 'how fast will it happen?' — they are complementary.

Q3. From the data below for reaction A → B, calculate average rate between t = 50s and t = 100s: t(s): 0, 50, 100, 150 [A](mol L⁻¹): 1.00, 0.80, 0.64, 0.51

  • A. 1.6 × 10⁻³ mol L⁻¹ s⁻¹
  • B. 3.2 × 10⁻³ mol L⁻¹ s⁻¹ ✓
  • C. 6.4 × 10⁻³ mol L⁻¹ s⁻¹
  • D. 3.2 × 10⁻² mol L⁻¹ s⁻¹

Answer: B — rav = Δ[A]/Δt = (0.80 – 0.64)/(100 – 50) = 0.16/50 = 0.0032 = 3.2 × 10⁻³ mol L⁻¹ s⁻¹.

Q4. Why is a negative sign used in the expression –Δ[R]/Δt for the rate of disappearance of reactant R?

  • A. Because concentration of R always increases
  • B. Because Δ[R] is negative (concentration decreases), the negative sign converts it to a positive rate value ✓
  • C. Because it distinguishes reactants from products
  • D. Because kinetics always uses negative numbers

Answer: B — Reactant concentration decreases, so Δ[R] is negative; multiplying by –1 ensures the reported rate is positive, matching convention that reaction rates are positive quantities.

Q5. Which of the following is NOT a correct statement about instantaneous rate?

  • A. It is obtained when Δt approaches zero
  • B. It is always equal to the average rate over any time interval ✓
  • C. It is represented mathematically as d[C]/dt
  • D. It can be determined graphically as the slope of a tangent to the curve

Answer: B — Instantaneous rate equals average rate only by coincidence; average rate is constant over its interval, while instantaneous rate varies at each moment.

Q6. For a reaction at 25°C, the concentration of reactant changes from 0.500 mol L⁻¹ to 0.250 mol L⁻¹ in 200 seconds. What is the average rate of reaction?

  • A. 1.25 × 10⁻³ mol L⁻¹ s⁻¹ ✓
  • B. 0.250 mol L⁻¹ s⁻¹
  • C. 2.5 × 10⁻³ mol L⁻¹ s⁻¹
  • D. 100 mol L⁻¹ s⁻¹

Answer: A — rav = Δ[C]/Δt = (0.500 – 0.250)/200 = 0.250/200 = 1.25 × 10⁻³ mol L⁻¹ s⁻¹.

Q7. The concentration-time graph for a reaction shows that the average rate decreases with successive time intervals. What does this indicate about the reaction?

  • A. The reaction has stopped proceeding
  • B. Reactant concentration is depleting, so change in concentration per unit time becomes smaller ✓
  • C. The reaction is becoming faster over time
  • D. Temperature is increasing

Answer: B — As reactants are consumed, their concentration decreases, making Δ[R] smaller in each successive interval; thus average rate = Δ[R]/Δt also decreases.

Q8. Consider two reactions: (I) Diamond → Graphite (thermodynamically favourable but occurs imperceptibly slowly) and (II) AgNO₃ + NaCl → AgCl (occurs instantaneously). Both statements below are given: Statement 1: Reaction (I) has negative ΔG but very high activation energy; Reaction (II) has negative ΔG and low activation energy. Statement 2: Kinetics determines the observable speed regardless of thermodynamic feasibility. (A) Both statements are correct (B) Statement 1 is correct, Statement 2 is incorrect (C) Statement 1 is incorrect, Statement 2 is correct (D) Both statements are incorrect

  • A. Both statements are correct ✓
  • B. Statement 1 is correct, Statement 2 is incorrect
  • C. Statement 1 is incorrect, Statement 2 is correct
  • D. Both statements are incorrect

Answer: A — Statement 1: Correct — high Ea makes reaction slow despite favourable ΔG; low Ea makes reaction fast. Statement 2: Correct — kinetics explains observable rates; thermodynamics alone cannot.

Q9. If the instantaneous rate of a reaction at t = 300s is calculated from the tangent to a [C] vs t curve, and the tangent passes through points (250s, 0.80 mol L⁻¹) and (350s, 0.60 mol L⁻¹), what is the instantaneous rate at 300s?

  • A. 0.002 mol L⁻¹ s⁻¹ ✓
  • B. 0.001 mol L⁻¹ s⁻¹
  • C. 0.02 mol L⁻¹ s⁻¹
  • D. 0.2 mol L⁻¹ s⁻¹

Answer: A — Slope of tangent = Δ[C]/Δt = (0.60 – 0.80)/(350 – 250) = –0.20/100 = –0.002; instantaneous rate = |–0.002| = 0.002 mol L⁻¹ s⁻¹.

Q10. Which of the following correctly ranks the three time intervals in Example 3.1 (C₄H₉Cl hydrolysis) in order of decreasing average rate?

  • A. 0–50s > 50–100s > 300–400s ✓
  • B. 300–400s > 50–100s > 0–50s
  • C. 0–50s > 300–400s > 50–100s
  • D. 50–100s > 0–50s > 700–800s

Answer: A — From Table 3.1: 0–50s has rate 1.90 × 10⁻⁴, 50–100s has 1.70 × 10⁻⁴, and 300–400s has 1.10 × 10⁻⁴ mol L⁻¹s⁻¹; early intervals show higher rates as reactant concentration is higher.

Flashcards

What is the fundamental difference between thermodynamics and chemical kinetics?

Thermodynamics predicts whether a reaction is feasible (ΔG < 0), while kinetics determines how fast that reaction actually proceeds.

Define average rate of a reaction with proper sign convention.

Average rate = –Δ[R]/Δt for reactants or +Δ[P]/Δt for products, where negative sign ensures rate is expressed as a positive quantity.

What is instantaneous rate and how is it obtained graphically?

Instantaneous rate is the reaction rate at a specific moment, obtained by drawing a tangent to the concentration-time curve at that instant and calculating its slope.

Why does the average rate of a reaction decrease over time?

As reactant concentration decreases with time, the change in concentration over successive time intervals becomes smaller, making the average rate lower.

What are the units of reaction rate when concentration is in mol L⁻¹ and time in seconds?

Units are mol L⁻¹ s⁻¹, derived from [concentration]/[time] = (mol L⁻¹)/(s).

For the reaction R → P with equal stoichiometry, how do the rates of disappearance and appearance compare?

The magnitude of the rate of disappearance of R equals the rate of appearance of P, though expressed with opposite signs in equations.

Why is it impossible to predict reaction rate at a specific moment using only average rate?

Average rate is constant over its defined time interval and does not reflect changes in concentration at individual moments; instantaneous rate is needed for that.

What does the slope of a tangent line on a [concentration] vs time graph represent?

The slope represents the instantaneous rate of change of concentration at the exact point where the tangent touches the curve.

State the relationship between kinetic studies and practical applications like food preservation or fuel combustion control.

Kinetics helps determine how to alter reaction conditions (concentration, temperature, catalyst) to speed up or slow down reactions for practical purposes.

Why must we use a negative sign when expressing the rate of disappearance of a reactant?

Reactant concentration decreases (negative change), so the negative sign converts the negative Δ[R] into a positive reaction rate value.

Important Board Questions

Define average rate and instantaneous rate of a reaction. Write the mathematical expression for instantaneous rate. [2 marks]

State that average rate = Δ[C]/Δt over a finite interval Δt, and instantaneous rate = d[C]/dt as Δt → 0 (limit of derivative). Mention that instantaneous rate is obtained from tangent slope.

For the reaction 2A → B, the concentration of A decreases from 0.80 mol L⁻¹ to 0.60 mol L⁻¹ in 10 minutes. Calculate (i) the average rate of disappearance of A, and (ii) the average rate of appearance of B. Show all working steps. [5 marks]

Use rav = –Δ[A]/Δt for disappearance (ensure negative sign); apply stoichiometry (2 moles A produce 1 mole B) so rate of B formation = 0.5 × rate of A disappearance. Convert time to seconds if asked for mol L⁻¹ s⁻¹.

Explain why kinetic studies are as important as thermodynamic studies for understanding chemical reactions. Support your answer with a real example, showing how a reaction can be thermodynamically favourable but kinetically too slow to observe. Discuss how chemists use kinetic knowledge in practical applications (food preservation, drug design, industrial catalysts). [6 marks]

Contrast ΔG < 0 (feasible, thermodynamics) with rate of reaction (kinetics). Use diamond-to-graphite example: ΔG < 0 but extremely slow. Explain activation energy barrier; discuss catalyst role; mention concentration, temperature effects on observable speed; link to real-world applications like decomposition prevention or reaction acceleration.

Assertion-Reason Questions for Chemical Kinetics

Q1. Assertion (A):

The rate of disappearance of a reactant is expressed with a negative sign because concentration decreases over time.

Reason (R):

To make the reaction rate a positive quantity, the negative change in reactant concentration is multiplied by –1, giving rav = -Δ[R]/Δt.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true and the reason correctly explains why the negative sign is necessary. The negative sign convention ensures rate values are positive for easier interpretation and comparison.

Q2. Assertion (A):

Thermodynamics predicts whether a reaction is feasible, while Chemical Kinetics determines how fast the reaction proceeds.

Reason (R):

Diamond converts to graphite because thermodynamic data show this conversion is favorable.

Answer: C — A is true but R is false — The assertion is correct: thermodynamics (ΔG < 0) predicts feasibility; kinetics predicts rate. However, the reason is misleading. While diamond-to-graphite conversion is thermodynamically favorable, it is kinetically hindered (extremely slow), so it is not observed under normal conditions. The reason does not correctly explain the assertion.

Q3. Assertion (A):

Average rate is constant for a given time interval, while instantaneous rate is the rate at a specific moment in time.

Reason (R):

Instantaneous rate is calculated by taking the average rate over an infinitesimally small time interval (Δt → 0), which is obtained graphically as the slope of a tangent to the concentration-time curve.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true. The assertion distinguishes the two concepts correctly, and the reason properly explains how instantaneous rate is derived from average rate (as a limit) and how it is determined graphically.

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Case-Based Questions (CBQ) for Chemical Kinetics

Case Study:

A pharmaceutical company is developing a medicine for wound healing and needs to understand how quickly the drug reacts in the human body. Researchers conducted an experiment where they measured the concentration of the active ingredient over 800 seconds. The concentration data showed: at t=0 s, [drug] = 0.100 mol L⁻¹; at t=50 s, [drug] = 0.0905 mol L⁻¹; at t=100 s, [drug] = 0.0820 mol L⁻¹; and at t=400 s, [drug] = 0.0439 mol L⁻¹. The scientists plotted this data and observed that the curve became progressively flatter, indicating the drug concentration was decreasing at a slowing rate.

Q1. What is the average rate of drug decomposition between t=0 and t=100 s?

  • A. 8.0 × 10⁻⁴ mol L⁻¹ s⁻¹
  • B. 9.0 × 10⁻⁵ mol L⁻¹ s⁻¹
  • C. 1.80 × 10⁻⁴ mol L⁻¹ s⁻¹ ✓
  • D. 0.0180 mol L⁻¹ s⁻¹

Answer: C

Q2. Why does the average rate of drug decomposition decrease from the 0-50 s interval (1.90 × 10⁻⁴ mol L⁻¹ s⁻¹) to the 300-400 s interval (1.10 × 10⁻⁴ mol L⁻¹ s⁻¹)?

  • A. The drug becomes thermodynamically unstable over time
  • B. The concentration of the active ingredient decreases, so less drug is available to react per unit time ✓
  • C. The temperature of the body decreases as time progresses
  • D. The volume of distribution in the body increases, diluting the drug

Answer: B

Q3. If a researcher wants to determine the instantaneous rate of drug decomposition at exactly t=200 s, which method should they use?

  • A. Calculate the average rate between t=0 and t=400 s
  • B. Read the concentration value directly from the y-axis at t=200 s
  • C. Plot the concentration vs. time data, draw a smooth curve, and then draw a tangent to the curve at t=200 s and calculate its slope ✓
  • D. Use the formula rav = Δ[drug]/Δt with the nearest two data points

Answer: C

Q4. What does the observation that the curve becomes flatter as time increases reveal about the relationship between drug concentration and reaction rate?

  • A. The reaction rate is independent of drug concentration
  • B. As drug concentration decreases, the instantaneous rate of reaction decreases ✓
  • C. The flattening curve indicates the drug is becoming more stable
  • D. The reaction will eventually stop regardless of how much drug remains

Answer: B

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Previous Year Questions (PYQ) for Chemical Kinetics

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 56(B) · 1 mark

The rate of a reaction is doubled when the concentration of the reactant is doubled. The order of the reaction is :

  • 0
  • 1
  • 2
  • 0·5

CBSE 2026 · Set 56(B) · 1 mark

Assertion (A) : Rate of reaction increases with an increase in temperature. Reason (R) : Number of effective collisions decreases with an increase in temperature.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56(B) · 2 marks

Write any two differences between the order of a reaction and molecularity of a reaction.

CBSE 2026 · Set 56(B) · 3 marks

A first order reaction is 50% complete in 20 minutes. Calculate the time for 75% completion of the reaction. [log 4 = 0·6]

CBSE 2026 · Set 56/1/1 · 1 mark

Which of the following curve represents the first order reaction?

  • A
  • B
  • C
  • D

CBSE 2026 · Set 56/1/1 · 2 marks

Following reaction takes place in one step: 2A + B → 2C. How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume? Will there be any change in the order of reaction with the reduced volume?

CBSE 2026 · Set 56/1/1 · 3 marks

For the first order thermal decomposition reaction, following data was obtained: C2H5Cl(g) → C2H4(g) + HCl(g). S. No. Time(s) Total Pressure (atm) 1 0 0.30 2 30 0.50. Calculate rate constant. [Given: log 3 = 0.48]

CBSE 2026 · Set 56/1/2 · 1 mark

For a reaction: N2 + 3H2 → 2NH3, the rate of reaction with respect to NH3 is

  • +1/3 d[NH3]/dt
  • -1/2 d[NH3]/dt
  • +1/4 d[NH3]/dt
  • +1/2 d[NH3]/dt

CBSE 2026 · Set 56/1/3 · 1 mark

Which of the following curve represents the first order reaction ?

  • [Diagram/Map-based question — skipped]
  • [Diagram/Map-based question — skipped]
  • [Diagram/Map-based question — skipped]
  • [Diagram/Map-based question — skipped]

CBSE 2026 · Set 56/1/3 · 1 mark

The activation energy of a reaction can be determined from the slope of which of the following curves ?

  • lnk vs. T
  • lnk vs. 1/T
  • ln k/T vs. T
  • T/lnk vs. 1/T

CBSE 2026 · Set 56/1/3 · 2 marks

Following reaction takes place in one step : 2A + B → 2C. How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume ? Will there be any change in the order of reaction with the reduced volume ?

CBSE 2026 · Set 56/1/3 · 3 marks

For the first order thermal decomposition reaction, following data was obtained : C 2H5Cl(g) → C2H4(g) + HCl(g). S. No. Time(s) Total Pressure (atm) 1 0 0.30 2 30 0.50. Calculate rate constant. [Given : log 3 = 0.48]

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Class 12 Chemistry Formula Sheet

Every formula and named reaction from all 10 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Chemistry Previous Year Questions, Chapter-wise

All 1241 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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