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State Board · Class 12 · Chemistry

Lesson 6: Chemical Kinetics

7 concepts · 56 practice questions with answers · Free to read

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Rate of reaction

From daily life

Milk boils over fast on a stove but spoils slowly in the fridge — different rates.

Simple definition

Rate of reaction = decrease in reactant (or increase in product) concentration per unit time; unit mol L⁻¹ s⁻¹.

RateRate of reaction: average and instantaneoustime (min)[R] (M)Average rate= –Δ[R]/Δt0 → 20 min= 0.0316 M/minInstantaneous= –d[R]/dtt = 10 min= 0.0303 M/minaA + bB → cC + dD: rate = –(1/a)d[A]/dt = +(1/c)d[C]/dt2N₂O₅ → 4NO₂ + O₂: rate = –½d[N₂O₅]/dt = ¼d[NO₂]/dt = d[O₂]/dt

In the figure : [R] versus t.

Rule 1: Average rate — = –Δ[R]/Δt = +Δ[P]/Δt, the rate over an interval.

  • Easy : [R] 1.00 ⇒ 0.37 M in 20 min ⇒ 0.0316 M/min.
  • Medium : Shrinking the interval gives the instantaneous rate.
  • Exam-type : Minus sign for reactants.

Rule 2: Instantaneous rate — = –d[R]/dt = the slope of the tangent to the curve at t.

  • Easy : t = 10 min ⇒ 0.0303 M/min.
  • Medium : Rate falls with time.
  • Exam-type : Maximum at t = 0.

Rule 3: Link to stoichiometry — 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.

  • Easy : 2N₂O₅ → 4NO₂ + O₂.
  • Medium : d[O₂]/dt = 2.5×10⁻⁴ ⇒ d[NO₂]/dt = 1.0×10⁻³.
  • Exam-type : –d[N₂O₅]/dt = 5.0×10⁻⁴ M/s.

Calculation: rate of NO₂ formation when d[O₂]/dt = 2.5×10⁻⁴ M/s

  • Step 1: rate = d[O₂]/dt = ¼ d[NO₂]/dt.
  • Step 2: d[NO₂]/dt = 4 × 2.5×10⁻⁴.
  • Step 3: = 1.0×10⁻³ M/s; –d[N₂O₅]/dt = 2 × 2.5×10⁻⁴ = 5.0×10⁻⁴.

CET level (MCQ)

  • CET 1) For 2A → B, rate = ? 1) –½d[A]/dt 2) –d[A]/dt 3) –2d[A]/dt 4) d[A]/dt — Answer: 1).
  • CET 2) Unit of rate? 1) mol L⁻¹ s⁻¹ 2) s⁻¹ 3) L mol⁻¹ s⁻¹ 4) mol s — Answer: 1).

Real-life use: Drug shelf life, food preservation, industrial processes.

Remember

Divide by the stoichiometric coefficient; instantaneous = tangent.

Common mistake

Writing rate = –d[A]/dt for 2A → B and forgetting ½.

CET trap

Rate is always positive; the minus sign is for reactants.

Shortcut

Memory clue: "divide by the coefficient".

Spot the difference

RateFormula
Average–Δ[R]/Δt
Instantaneous–d[R]/dt
Unitmol L⁻¹ s⁻¹

Practice questions

Easy
  1. 1Average vs instantaneous rate.
  2. 2Unit of rate.
  3. 3Write the rate of 2N₂O₅ → 4NO₂ + O₂.
Medium
  1. 4Meaning of the tangent.
  2. 5Why does rate fall with time?
  3. 6d[O₂]/dt = 2.5×10⁻⁴ ⇒ NO₂?
Challenging
  1. 7Why the minus sign?
  2. 8Average rate 0 → 20 min.
Show answers (try them yourself first)
  1. 1Interval vs instant.
  2. 2mol L⁻¹ s⁻¹.
  3. 3–½d[N₂O₅]/dt = ¼d[NO₂]/dt = d[O₂]/dt.
  4. 4Slope of the curve.
  5. 5[R] decreases.
  6. 61.0×10⁻³ M/s.
  7. 7A reactant decreases.
  8. 80.0316 M/min.

Rate law, order and rate constant

From daily life

More sugar in tea makes it sweeter — concentration matters. Rates too depend on concentration.

Simple definition

Rate law: rate = k[A]ᵐ[B]ⁿ. Order = m + n (found experimentally). k is the rate constant.

OrderRate vs [A]: identifying the order[A]rZero orderr = kk : mol L⁻¹ s⁻¹[A]rFirst orderr = k[A]k : s⁻¹[A]rSecond orderr = k[A]²k : L mol⁻¹ s⁻¹Initial-rate method: doubling [A] multiplies the rate by 2ⁿe.g. [A] 0.1 → 0.2: rate 2×10⁻³ → 4×10⁻³ ⇒ n = 1[B] 0.1 → 0.2: rate 2×10⁻³ → 8×10⁻³ ⇒ n = 2; k = 2 L² mol⁻² s⁻¹

In the figure : Order and graphs.

Rule 1: Order — 0: rate = k. 1: rate = k[A]. 2: rate = k[A]². Order is not read from stoichiometric coefficients.

  • Easy : 2NO + O₂ → 2NO₂: experimentally 3.
  • Medium : H₂ + I₂ → 2HI: order 2.
  • Exam-type : Order may be fractional.

Rule 2: Unit of k — For order n: (mol L⁻¹)¹⁻ⁿ s⁻¹. Zero: mol L⁻¹ s⁻¹; first: s⁻¹; second: L mol⁻¹ s⁻¹; third: L² mol⁻² s⁻¹.

  • Easy : Unit s⁻¹ ⇒ first order.
  • Medium : L mol⁻¹ s⁻¹ ⇒ second.
  • Exam-type : k depends on temperature.

Rule 3: Method of initial rates — Double one concentration, keep the rest fixed ⇒ rate rises 2ⁿ ⇒ gives n.

  • Easy : [A] 0.1 ⇒ 0.2, rate 2 → 4 ⇒ n = 1.
  • Medium : [B] 0.1 ⇒ 0.2, rate 2 → 8 ⇒ n = 2.
  • Exam-type : k = 2×10⁻³ ÷ (0.1 × 0.01) = 2 L² mol⁻² s⁻¹.

Calculation: rate = k[A][B]², find k

  • Step 1: [A] = 0.1, [B] = 0.1, rate = 2×10⁻³ mol L⁻¹ s⁻¹.
  • Step 2: k = 2×10⁻³ ÷ (0.1 × 0.1²).
  • Step 3: k = 2 L² mol⁻² s⁻¹ (third-order unit).

CET level (MCQ)

  • CET 1) Rate = k[A]²[B] ⇒ overall order? 1) 3 2) 2 3) 1 4) 0 — Answer: 1).
  • CET 2) k has unit s⁻¹ ⇒ order? 1) 1 2) 0 3) 2 4) 3 — Answer: 1).

Real-life use: Dosage, radioactive decay, pollution.

Remember

Order = experiment; unit s⁻¹ = 1; doubling ⇒ 2ⁿ.

Common mistake

Taking stoichiometric coefficients as orders.

CET trap

Order can be fractional, zero or negative; molecularity cannot.

Shortcut

Memory clue: "order comes from experiment".

Spot the difference

OrderRateUnit of k
0kmol L⁻¹ s⁻¹
1k[A]s⁻¹
2k[A]²L mol⁻¹ s⁻¹

Practice questions

Easy
  1. 1What is a rate law?
  2. 2Why is order experimental?
  3. 3Find the unit of k.
Medium
  1. 4Method of initial rates.
  2. 5Rate = k[A][B]² ⇒ overall order?
  3. 6[A] doubled, rate ×4 ⇒ ?
Challenging
  1. 7What does k depend on?
  2. 8Rate constant vs rate.
Show answers (try them yourself first)
  1. 1Rate in terms of concentrations.
  2. 2Not from coefficients.
  3. 3(mol L⁻¹)¹⁻ⁿ s⁻¹.
  4. 4Double one at a time.
  5. 53.
  6. 6n = 2.
  7. 7Temperature and catalyst.
  8. 8k does not change with concentration.

Molecularity and reaction mechanism

From daily life

In traffic, the slowest signal sets the whole trip time — as the slow step does in a reaction.

Simple definition

Molecularity is the number of species colliding in an elementary step; the slow step sets the rate.

MechanismMolecularity and reaction mechanismMolecularitymolecules colliding in an elementary step: 1, 2, 3; whole numbersOrderfound by experiment; may be 0, fractional or negativeSlow stepthe rate-determining stepe.g. 2NO₂ + F₂ → 2NO₂FStep 1 (slow): NO₂ + F₂ → NO₂F + FStep 2 (fast): NO₂ + F → NO₂Frate = k[NO₂][F₂] (from the slow step); order = 2The reaction is only as fast as its slowest step

In the figure : Molecularity and mechanism.

Rule 1: Molecularity vs order — Molecularity: theoretical, whole number 1, 2, 3 (3 rare). Order: experimental, any number.

  • Easy : For an elementary step order = molecularity.
  • Medium : Not so for a complex reaction.
  • Exam-type : Molecularity is never zero.

Rule 2: Mechanism — Several elementary steps; the slowest is rate-determining and gives the rate law.

  • Easy : 2NO₂ + F₂ ⇒ 2NO₂F.
  • Medium : Slow: NO₂ + F₂.
  • Exam-type : Rate = k[NO₂][F₂].

Rule 3: Intermediates — Species formed and used up in the steps (F); they do not appear in the rate law.

  • Easy : F is used in the fast step.
  • Medium : Absent from the overall equation.
  • Exam-type : Unlike a catalyst, which is used first and regenerated.

Reasoning: rate law for H₂O₂ + 2I⁻ + 2H⁺ → 2H₂O + I₂

  • Step 1: Slow: H₂O₂ + I⁻ → H₂O + IO⁻.
  • Step 2: Two species in the slow step ⇒ molecularity 2.
  • Step 3: rate = k[H₂O₂][I⁻]; order = 2.

CET level (MCQ)

  • CET 1) Molecularity cannot be? 1) 0 2) 1 3) 2 4) 3 — Answer: 1).
  • CET 2) The rate is set by? 1) slow step 2) fast step 3) last step 4) first step — Answer: 1).

Real-life use: Drug metabolism, atmospheric chemistry.

Remember

Molecularity = theory, whole number; slow step = rate.

Common mistake

Treating molecularity as fractional or zero.

CET trap

Order can be fractional for complex reactions; molecularity cannot.

Shortcut

Memory clue: "order = experiment; molecularity = head-count".

Spot the difference

FeatureOrderMolecularity
Sourceexperimenttheory
Valueany1, 2, 3
Applies tooverallelementary

Practice questions

Easy
  1. 1What is molecularity?
  2. 2Order vs molecularity.
  3. 3What is the slow step?
Medium
  1. 4What is an intermediate?
  2. 5Intermediate vs catalyst.
  3. 6Rate law for NO₂ + F₂.
Challenging
  1. 7Order of an elementary step?
  2. 8Why not zero molecularity?
Show answers (try them yourself first)
  1. 1Species in an elementary step.
  2. 2Experiment vs theory.
  3. 3The rate-determining step.
  4. 4Formed and consumed in steps.
  5. 5A catalyst is regenerated.
  6. 6k[NO₂][F₂].
  7. 7Equal to molecularity.
  8. 8No collision is possible.

First order reactions and half-life

From daily life

Radioactive decay and the fall of a drug level in the body are first order.

Simple definition

First order: rate = k[R]; k = (2.303/t) log([R]₀/[R]); half-life t½ = 0.693/k is independent of [R]₀.

First orderFirst order: [R] and half-lifetime[R]1t½½2t½¼3t½⅛[R] = [R]₀ e⁻ᵏᵗk = (2.303/t) log([R]₀/[R])t½ = 0.693 / kindependent of [R]₀log[R] vs tstraight line, slope = –k/2.303k = 0.0693 min⁻¹ ⇒ t½ = 10 min; after 30 min (½)³ = 12.5 % remains80 % complete in 40 min ⇒ k = 2.303/40 × log 5 = 0.0402 min⁻¹

In the figure : [R] versus t.

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First order reactions and half-life

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Rule 1: Integrated rate law — ln[R] = ln[R]₀ – kt ⇒ log[R] vs t is a straight line with slope –k/2.303.

  • Easy : [R] = [R]₀ e⁻ᵏᵗ.
  • Medium : 80 % complete in 40 min ⇒ k = 0.0402 min⁻¹.
  • Exam-type : Unit s⁻¹ or min⁻¹.

Rule 2: Half-life — t½ = 0.693/k. Every half-life halves it: 50 %, 25 %, 12.5 %.

  • Easy : k = 0.0693 min⁻¹ ⇒ t½ = 10 min.
  • Medium : 30 min ⇒ 12.5 % remains.
  • Exam-type : 99.9 % complete ≈ 10 t½.

Rule 3: Examples of first order — N₂O₅ decomposition, radioactive decay, SO₂Cl₂ decomposition, H₂O₂ decomposition.

  • Easy : 2N₂O₅ → 4NO₂ + O₂: first order.
  • Medium : Carbon-14 decay.
  • Exam-type : Sucrose inversion is pseudo-first.

Calculation: 80 % in 40 min ⇒ k and t½

  • Step 1: [R]₀ = 100, [R] = 20.
  • Step 2: k = (2.303/40) × log 5 = 0.0402 min⁻¹.
  • Step 3: t½ = 0.693 ÷ 0.0402 = 17.2 min.

CET level (MCQ)

  • CET 1) t½ for first order? 1) 0.693/k 2) 1/k[R]₀ 3) [R]₀/2k 4) k/0.693 — Answer: 1).
  • CET 2) t½ = 20 min ⇒ left after 1 h? 1) 12.5 % 2) 25 % 3) 50 % 4) 6.25 % — Answer: 1).

Real-life use: Carbon dating, drug dosage interval.

Remember

t½ = 0.693/k; n half-lives ⇒ (½)ⁿ.

Common mistake

Thinking first-order t½ depends on [R]₀.

CET trap

Remaining = 100 – percent complete.

Shortcut

Memory clue: "half-life = 0.693 over k".

Spot the difference

Half-livesLeft
150 %
225 %
312.5 %

Practice questions

Easy
  1. 1First-order equation.
  2. 2Derive t½.
  3. 3k = 0.0693 min⁻¹ ⇒ t½?
Medium
  1. 4How much remains after 30 min?
  2. 580 % complete ⇒ k?
  3. 6What is the graph?
Challenging
  1. 7Unit of k?
  2. 8Give an example.
Show answers (try them yourself first)
  1. 1k = (2.303/t) log([R]₀/[R]).
  2. 2At [R] = [R]₀/2, t = 0.693/k.
  3. 310 min.
  4. 412.5 %.
  5. 50.0402 min⁻¹.
  6. 6log[R] vs t is a straight line.
  7. 7s⁻¹ / min⁻¹.
  8. 8N₂O₅ decomposition.

Zero order and pseudo-first order reactions

From daily life

NH₃ on hot platinum: the surface is saturated so the rate is constant.

Simple definition

Zero order: rate = k; [R] = [R]₀ – kt; t½ = [R]₀/2k. Pseudo-first order: one reactant in large excess behaves as first order.

Zero/pseudoZero order and pseudo-first ordertime[R][R] = [R]₀ – ktslope = –kt½ = [R]₀ / 2kZero-order examplesNH₃ on hot PtH₂ + Cl₂ (light)Pseudo-first orderCH₃COOC₂H₅ + H₂OH₂O in excess ⇒ r = k′[ester]sucrose inversion[R]₀ = 0.5 M, k = 0.02 M/s ⇒ t½ = 12.5 s; complete at 25 sZero order: t½ ∝ [R]₀; first order: t½ constant

In the figure : Zero and pseudo-first.

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Zero order and pseudo-first order reactions

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Rule 1: Zero order — [R] vs t is a straight line with slope –k; t½ ∝ [R]₀; completion time = [R]₀/k.

  • Easy : [R]₀ = 0.5 M, k = 0.02 M/s ⇒ t½ = 12.5 s.
  • Medium : Complete = 25 s.
  • Exam-type : NH₃ on Pt ⇒ N₂ + H₂.

Rule 2: Pseudo-first order — CH₃COOC₂H₅ + H₂O ⇒ acid + alcohol: water in excess ⇒ rate = k′[ester], k′ = k[H₂O].

  • Easy : Sucrose + H₂O ⇒ inversion.
  • Medium : Molecularity 2, order 1.
  • Exam-type : Acid as catalyst.

Rule 3: Comparison — Zero: t½ depends on [R]₀; first: t½ constant. [R] is linear in zero order and exponential in first.

  • Easy : Zero-order k: mol L⁻¹ s⁻¹.
  • Medium : First-order k: s⁻¹.
  • Exam-type : Pseudo = apparent order 1.

Calculation: zero order, [R]₀ = 0.5 M, k = 0.02 M/s

  • Step 1: t½ = [R]₀ / 2k.
  • Step 2: = 0.5 ÷ (2 × 0.02) = 12.5 s.
  • Step 3: Completion takes [R]₀ ÷ k = 25 s.

CET level (MCQ)

  • CET 1) [R] vs t for zero order? 1) straight line 2) curve 3) point 4) hyperbola — Answer: 1).
  • CET 2) Pseudo-first-order example? 1) ester hydrolysis 2) N₂O₅ decomposition 3) NH₃/Pt 4) H₂ + Cl₂ — Answer: 1).

Real-life use: Catalyst surfaces, drug formation.

Remember

Zero: t½ = [R]₀/2k; pseudo: water in excess.

Common mistake

Taking t½ constant for zero order.

CET trap

In zero order t½ ∝ [R]₀.

Shortcut

Memory clue: "zero = straight line; first = curve".

Spot the difference

FeatureZeroFirst
t½[R]₀/2k0.693/k
[R] vs tlineexp.
kmol L⁻¹ s⁻¹s⁻¹

Practice questions

Easy
  1. 1Zero-order equation.
  2. 2Zero-order t½.
  3. 3What is pseudo-first order?
Medium
  1. 4Why is ester hydrolysis pseudo?
  2. 5Completion time for 0.5 M, 0.02 M/s?
  3. 6Example of zero order.
Challenging
  1. 7Zero vs first comparison.
  2. 8Order of sucrose inversion?
Show answers (try them yourself first)
  1. 1[R] = [R]₀ – kt.
  2. 2[R]₀ / 2k.
  3. 3One reactant in excess.
  4. 4Water is in excess.
  5. 525 s.
  6. 6NH₃ on Pt.
  7. 7t½ dependent vs constant.
  8. 8Pseudo-first.

Collision theory, activation energy and catalysis

From daily life

A match does not light until you strike it — activation energy must be supplied.

Simple definition

Collision theory: a reaction needs a collision with energy ≥ Ea and proper orientation; the fraction of effective collisions is f = e^(–Ea/RT).

Activation energyActivation energy and catalysisEnergyreaction pathReactantsProductsEa(uncatalysed)Ea′catalysedΔHCatalyst: lowers Ea; ΔH and the equilibrium constant do not changeCollision theory: enough energy (≥ Ea) + proper orientation; f = e^(–Ea/RT)

In the figure : Energy profile.

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Collision theory, activation energy and catalysis

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Rule 1: Activation energy — Ea = energy of the activated complex – energy of the reactants; higher Ea ⇒ slower reaction.

  • Easy : Ea hardly changes with temperature.
  • Medium : ΔH ≠ Ea.
  • Exam-type : A catalyst lowers Ea.

Rule 2: Proper orientation — Steric factor p: only correctly oriented collisions succeed; rate = p · Z · e^(–Ea/RT).

  • Easy : CH₃Br + OH⁻: backside attack.
  • Medium : Collision count ≠ rate.
  • Exam-type : Z = collision frequency.

Rule 3: Catalyst — Provides an alternative path with lower Ea; ΔH, equilibrium constant and position are unchanged; it cannot start a non-spontaneous reaction.

  • Easy : H₂ + O₂ on Pt.
  • Medium : Enzymes are biological catalysts.
  • Exam-type : Both forward and reverse reactions speed up.

Reasoning: why does a catalyst not shift equilibrium?

  • Step 1: It lowers Ea for both directions by the same amount.
  • Step 2: Both kf and kr rise by the same factor.
  • Step 3: K = kf/kr is unchanged.

CET level (MCQ)

  • CET 1) A catalyst changes? 1) Ea 2) ΔH 3) K 4) position of equilibrium — Answer: 1).
  • CET 2) Fraction of effective collisions? 1) e^(–Ea/RT) 2) e^(Ea/RT) 3) Ea/RT 4) RT/Ea — Answer: 1).

Real-life use: Catalytic converters, Haber process, enzymes.

Remember

Collision + energy ≥ Ea + orientation; catalyst = lower Ea, same K.

Common mistake

Saying a catalyst changes ΔH.

CET trap

A catalyst changes only the rate, not the equilibrium yield.

Shortcut

Memory clue: "catalyst = Ea down, everything else same".

Spot the difference

QuantityWith catalyst
Ealower
ΔHsame
Ksame

Practice questions

Easy
  1. 1State collision theory.
  2. 2What is activation energy?
  3. 3Why orientation?
Medium
  1. 4How does a catalyst work?
  2. 5Energy profile with a catalyst.
  3. 6Meaning of f = e^(–Ea/RT).
Challenging
  1. 7Why K is unchanged?
  2. 8ΔH vs Ea.
Show answers (try them yourself first)
  1. 1Collision, energy, orientation.
  2. 2Minimum extra energy.
  3. 3Steric factor p.
  4. 4Via a lower-Ea path.
  5. 5Lower Ea, same ΔH.
  6. 6Fraction of successful collisions.
  7. 7Both kf and kr rise.
  8. 8Ea is the barrier; ΔH the difference.

Arrhenius equation and effect of temperature

From daily life

Milk spoils faster in summer — higher temperature speeds reactions.

Simple definition

Arrhenius: k = A e^(–Ea/RT); ln k = ln A – Ea/RT. A 10 K rise raises the rate about 2–3 times.

ArrheniusArrhenius equation1/Tln kslope = –Ea/Rln Ak = A e^(–Ea/RT)ln k = ln A – Ea/RTlog(k₂/k₁) =Ea/(2.303R) ·(1/T₁ – 1/T₂)R = 8.314 J K⁻¹ mol⁻¹Ea = 50 kJ/mol, 300 → 310 K ⇒ k₂/k₁ = 1.91 (≈ 2×)k doubles (300 → 310 K) ⇒ Ea ≈ 53.6 kJ/mol10 K rise ⇒ rate ≈ 2–3× (temperature coefficient)

In the figure : ln k versus 1/T.

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Arrhenius equation and effect of temperature

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Rule 1: Graph — ln k vs 1/T is a straight line with slope –Ea/R and intercept ln A.

  • Easy : Ea = –slope × R.
  • Medium : R = 8.314 J K⁻¹ mol⁻¹.
  • Exam-type : A = frequency (pre-exponential) factor.

Rule 2: At two temperatures — log(k₂/k₁) = Ea/(2.303 R) · (1/T₁ – 1/T₂) = Ea (T₂ – T₁) / (2.303 R T₁T₂).

  • Easy : Ea = 50 kJ/mol, 300 → 310 K ⇒ k₂/k₁ = 1.91.
  • Medium : k doubles ⇒ Ea ≈ 53.6 kJ/mol.
  • Exam-type : Larger Ea ⇒ more temperature-sensitive.

Rule 3: Catalyst and temperature — Higher T raises the fraction of molecules with energy ≥ Ea; a catalyst lowers Ea, increasing k.

  • Easy : The e^(–Ea/RT) fraction.
  • Medium : f rises with T.
  • Exam-type : Lower Ea ⇒ larger f.

Calculation: k doubles from 300 to 310 K; find Ea

  • Step 1: log 2 = 0.3010.
  • Step 2: Ea = 2.303 × 8.314 × 0.3010 × (300 × 310 ÷ 10).
  • Step 3: Ea ≈ 53.6 kJ/mol.

CET level (MCQ)

  • CET 1) Slope of ln k vs 1/T? 1) –Ea/R 2) Ea/R 3) –R/Ea 4) A — Answer: 1).
  • CET 2) A 10 K rise raises the rate about? 1) 2–3× 2) 10× 3) halves 4) unchanged — Answer: 1).

Real-life use: Food storage, drug expiry, industrial temperature.

Remember

ln k vs 1/T; slope –Ea/R; 10 K ⇒ ≈ 2×.

Common mistake

Mixing kJ and J (R is in J).

CET trap

R = 8.314 J, so use Ea in J/mol.

Shortcut

Memory clue: "downward slope = –Ea/R".

Spot the difference

EquationUse
k = A e^(–Ea/RT)value of k
ln k = ln A – Ea/RTgraph
log(k₂/k₁)two temperatures

Practice questions

Easy
  1. 1Write the Arrhenius equation.
  2. 2Ea from the graph?
  3. 3Two-temperature formula.
Medium
  1. 4Why does rate rise with T?
  2. 5300 → 310 K, Ea = 50 kJ ⇒ k₂/k₁?
  3. 6k doubles ⇒ Ea?
Challenging
  1. 7What is A?
  2. 8Temperature coefficient.
Show answers (try them yourself first)
  1. 1k = A e^(–Ea/RT).
  2. 2–slope × R.
  3. 3log(k₂/k₁) = Ea(T₂–T₁)/(2.303RT₁T₂).
  4. 4More molecules exceed Ea.
  5. 51.91.
  6. 6≈ 53.6 kJ/mol.
  7. 7Frequency factor.
  8. 8k₂/k₁ for a 10 K rise ≈ 2–3.

Quick revision of the whole lesson

ConceptKey formula / fact
Rate–(1/a)d[A]/dt
Orderexperimental; unit of k
Molecularitywhole number 1, 2, 3
First ordert½ = 0.693/k
Zero ordert½ = [R]₀/2k
Catalystlower Ea; same K
Arrheniusln k = ln A – Ea/RT
Temperature10 K ⇒ ≈ 2–3×
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🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Rate.
  • Rate law, order.
  • Molecularity, mechanism.
  • First order, t½.
  • Zero, pseudo-first.
  • Collision, Ea, catalyst.
  • Arrhenius.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.