Chemical Kinetics ISC Class 12 Chemistry: Notes and Numericals

What Chemical Kinetics is about

Weightage: Chemical Kinetics is part of the Physical Chemistry group, which carries 25 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Rate of reaction: average and instantaneous

For $aA+bB\to cC+dD$, rate $=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=\frac{1}{c}\frac{d[C]}{dt}=\frac{1}{d}\frac{d[D]}{dt}$. The minus sign makes the rate positive for reactants, which decrease. Dividing by the coefficients gives one rate for the whole reaction. Average rate $=-\frac{\Delta [R]}{\Delta t}$ over an interval; instantaneous rate is the slope of the tangent to the concentration-time curve at a given instant (Δt → 0). Rate is highest at the start and falls as reactants are used up. Units: mol L⁻¹ s⁻¹. Factors affecting rate: nature of reactants (ionic reactions fast, covalent slow), concentration (more collisions), surface area (powders react faster), temperature (more molecules with energy above $E_a$), catalyst (alternative path of lower $E_a$), and radiation (photochemical reactions such as $H_2+C{l}_2$ in sunlight).

Worked example

For $2N_2{O}_5\to 4N{O}_2+O_2$, N₂O₅ disappears at $6.0\times 10^{-4}$ mol L⁻¹ s⁻¹. Find the rate of formation of NO₂ and of O₂, and the rate of reaction.

Try it first, then show the solutionSolution

1. Rate of reaction $=\frac{1}{2}\times 6.0\times 10^{-4}=3.0\times 10^{-4}$ mol L⁻¹ s⁻¹
2. $\frac{d[N{O}_2]}{dt}=4\times 3.0\times 10^{-4}=1.2\times 10^{-3}$; $\frac{d[O_2]}{dt}=3.0\times 10^{-4}$ mol L⁻¹ s⁻¹

Answer: NO₂: $1.2\times 10^{-3}$; O₂: $3.0\times 10^{-4}$; rate of reaction $3.0\times 10^{-4}$ mol L⁻¹ s⁻¹

Law of mass action, rate law and rate constant

Law of mass action: at constant temperature, the rate of a reaction is directly proportional to the product of the active masses of the reactants, each raised to the power of its coefficient. Active mass is the molar concentration (mol L⁻¹) for solutions and gases. In practice the powers must be found by experiment, giving the rate law: rate $=k[A{]}^x[B{]}^y$. $k$ is the rate constant (specific reaction rate): the rate when every concentration is 1 mol L⁻¹. It depends on temperature and catalyst, not on concentration. Finding order from initial-rate data: compare two experiments in which only one concentration changes: if doubling [A] doubles the rate, order in A is 1; quadruples it, order 2; no change, order 0. Then substitute any row into the rate law to find k, with units.

Worked example

For $A+B\to$ products: Exp 1 [A] = 0.10, [B] = 0.10, rate $2.0\times 10^{-3}$; Exp 2 [A] = 0.20, [B] = 0.10, rate $8.0\times 10^{-3}$; Exp 3 [A] = 0.10, [B] = 0.20, rate $2.0\times 10^{-3}$ (mol L⁻¹ s⁻¹). Find the rate law and k.

Try it first, then show the solutionSolution

1. Exp 1→2: [A] doubled, rate × 4, so order in A = 2.
2. Exp 1→3: [B] doubled, rate unchanged, so order in B = 0. Rate = $k[A{]}^2$.
3. $k=\frac{2.0\times 10^{-3}}{(0.10{)}^2}=0.20$ L mol⁻¹ s⁻¹

Answer: Rate = $k[A{]}^2[B{]}^0$; k = 0.20 L mol⁻¹ s⁻¹ (second order overall)

Order and molecularity

Order is the sum of the powers of the concentration terms in the experimentally found rate law. It may be 0, 1, 2, 3 or a fraction, and may differ from the stoichiometric coefficients. Pseudo first order: when one reactant is in large excess its concentration hardly changes, so the reaction appears first order, e.g. acid hydrolysis of an ester in excess water ($C{H}_{3}COO{C}_2{H}_5+H_{2}O\to C{H}_{3}COOH+C_2{H}_{5}OH$, rate = k'[ester]) and inversion of cane sugar. Molecularity is the number of reacting species (atoms, ions, molecules) that collide simultaneously in an elementary step: 1, 2 or rarely 3, never zero or fractional; it is a theoretical concept. Differences: order is experimental, can be zero or fractional, applies to the overall reaction; molecularity is theoretical, a whole number, applies to each elementary step. For an elementary reaction, order = molecularity. Units of k: (mol L⁻¹)${}^{1-n}$ s⁻¹: zero order mol L⁻¹ s⁻¹; first order s⁻¹; second order L mol⁻¹ s⁻¹; third order L² mol⁻² s⁻¹.

Worked example

What is the order of a reaction whose rate constant has units mol⁻¹ L s⁻¹? If rate = $k[A{]}^{1/2}[B]$, what is the overall order?

Try it first, then show the solutionSolution

1. (mol L⁻¹)${}^{1-n}$ s⁻¹ = mol⁻¹ L s⁻¹ means 1 − n = −1, so n = 2.
2. Overall order = 1/2 + 1 = 3/2.

Answer: Second order; 1.5 (a fractional order is possible because order is experimental)

Mechanism and the rate-determining step

An elementary reaction takes place in a single step. A complex reaction happens through a series of elementary steps called its mechanism; adding the steps gives the overall reaction. The slowest step controls the overall rate and is the rate-determining step: the rate law contains only species that take part in (or before) this step. Example: $2N{O}_2+F_2\to 2N{O}_{2}F$ has rate = $k[N{O}_2][F_2]$. Proposed mechanism: (1) $N{O}_2+F_2\to N{O}_{2}F+F$ (slow); (2) $N{O}_2+F\to N{O}_{2}F$ (fast). The slow step is bimolecular, involving one NO₂ and one F₂, matching the rate law. Species formed in one step and used in a later one (F) are intermediates. A mechanism is acceptable only if its steps add to the overall equation and its slow step gives the observed rate law.

Worked example

For $2H_2{O}_2\to 2H_{2}O+O_2$ in the presence of I⁻, the rate law is rate = $k[H_2{O}_2][I^{-}]$. Propose a two-step mechanism and identify the slow step, the intermediate and the catalyst.

Try it first, then show the solutionSolution

1. Step 1 (slow): $H_2{O}_2+I^{-}\to H_{2}O+I{O}^{-}$; Step 2 (fast): $H_2{O}_2+I{O}^{-}\to H_{2}O+I^{-}+O_2$
2. Adding gives $2H_2{O}_2\to 2H_{2}O+O_2$. The slow step involves one H₂O₂ and one I⁻, matching the rate law.
3. IO⁻ is formed then used (intermediate); I⁻ is used then regenerated (catalyst).

Answer: Slow step $H_2{O}_2+I^{-}\to H_{2}O+I{O}^{-}$; intermediate IO⁻; catalyst I⁻

Zero order reactions

Rate = $k[A{]}^0=k$: the rate does not depend on concentration. Integrating $-\frac{d[A]}{dt}=k$ from $[A{]}_0$ at t = 0 gives $[A]=[A{]}_0-kt$, so $k=\frac{[A{]}_0-[A]}{t}$. A plot of [A] against t is a straight line of slope −k. Half-life: $t_{1/2}=\frac{[A{]}_0}{2k}$, proportional to initial concentration. Units of k: mol L⁻¹ s⁻¹. The reaction finishes completely at $t=[A{]}_0/k$. Examples: decomposition of NH₃ on a hot platinum surface at high pressure (the surface is saturated, so rate is fixed); decomposition of HI on gold; some enzyme reactions at high substrate concentration.

Worked example

A zero order reaction has k = 0.020 mol L⁻¹ min⁻¹ and $[A{]}_0$ = 0.80 mol L⁻¹. Find the half-life and the time for complete reaction.

Try it first, then show the solutionSolution

1. $t_{1/2}=\frac{0.80}{2\times 0.020}=20$ min
2. Complete reaction: $t=\frac{[A{]}_0}{k}=\frac{0.80}{0.020}=40$ min (just two half-lives)

Answer: 20 min; 40 min

First order reactions and half-life

Rate = $k[A]$. Derivation: $-\frac{d[A]}{dt}=k[A]$ gives $\frac{d[A]}{[A]}=-kdt$. Integrating from $[A{]}_0$ at t = 0 to [A] at t: $\ln \frac{[A{]}_0}{[A]}=kt$, so $k=\frac{2.303}{t}\log \frac{[A{]}_0}{[A]}$ (or with a = initial amount and x reacted, $k=\frac{2.303}{t}\log \frac{a}{a-x}$). A plot of log[A] against t is a straight line of slope $-\frac{k}{2.303}$. Half-life: put $[A]=[A{]}_0/2$: $t_{1/2}=\frac{2.303\log 2}{k}=\frac{0.693}{k}$. Characteristics: units of k are time⁻¹ (s⁻¹), so k does not depend on the concentration unit; $t_{1/2}$ is independent of initial concentration; the reaction never strictly finishes; after n half-lives the fraction left is $(1/2{)}^n$. Examples: decomposition of N₂O₅, H₂O₂; radioactive decay; hydrolysis of esters in excess water (pseudo first order).

Worked example

A first order reaction is 80% complete in 40 minutes. Calculate k, the half-life and the time for 99.9% completion.

Try it first, then show the solutionSolution

1. $k=\frac{2.303}{40}\log \frac{100}{20}=\frac{2.303\times 0.699}{40}=0.0402$ min⁻¹
2. $t_{1/2}=\frac{0.693}{0.0402}=17.2$ min
3. $t_{99.9\%}=\frac{2.303}{0.0402}\log 1000=171.7$ min (about 10 half-lives)

Answer: k = 0.0402 min⁻¹; 17.2 min; 172 min

Activation energy, energy profiles and catalysts

Reactants must pass over an energy barrier before forming products. The minimum energy colliding molecules must have to react is the threshold energy; the extra energy above the reactants' average energy needed to reach it is the activation energy $E_a$ (= threshold energy − average energy of reactants). At the top of the barrier an unstable, high-energy activated complex forms, in which old bonds are partly broken and new ones partly formed. Exothermic reactions: products lower than reactants (ΔH negative). Endothermic: products higher (ΔH positive). For an endothermic reaction $E_a$(forward) = $E_a$(backward) + ΔH. Low $E_a$ means a fast reaction. A catalyst provides an alternative path with lower activation energy, so more collisions succeed; it does not change ΔH, the equilibrium constant or the position of equilibrium; it speeds forward and backward reactions equally.

Worked example

A catalyst lowers the activation energy of a reaction from 100 kJ mol⁻¹ to 75 kJ mol⁻¹ at 300 K. By what factor does the rate increase (A unchanged)?

Try it first, then show the solutionSolution

1. $\frac{k_{cat}}{k}=e^{(E_a-E_a^{\prime })/RT}=e^{25000/(8.314\times 300)}$
2. $=e^{10.02}\approx 2.25\times 10^4$

Answer: About 22 500 times faster

Collision theory and the Arrhenius equation

Collision theory (qualitative): molecules must collide to react; only effective collisions lead to products. A collision is effective if the molecules have (1) energy at least equal to the threshold energy and (2) the proper orientation, so the right atoms come into contact. Raising the temperature increases the collision frequency slightly but greatly increases the fraction of molecules with energy above $E_a$ (the high-energy tail of the distribution widens), so the rate roughly doubles for every 10 K rise. Arrhenius equation: $k=A{e}^{-E_a/RT}$. A is the frequency (pre-exponential) factor, related to collision frequency and orientation; $e^{-E_a/RT}$ is the fraction of molecules with energy ≥ $E_a$. Log form: $\log k=\log A-\frac{E_a}{2.303RT}$. A graph of log k against 1/T is a straight line with slope $-\frac{E_a}{2.303R}$ and intercept log A. Two temperatures: $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$, with R = 8.314 J K⁻¹ mol⁻¹.

Worked example

The rate constant of a reaction is 0.030 s⁻¹ at 400 K and 0.120 s⁻¹ at 500 K. Calculate $E_a$.

Try it first, then show the solutionSolution

1. $\log \frac{0.120}{0.030}=\log 4=0.602$
2. $E_a=\frac{2.303\times 8.314\times 0.602\times 400\times 500}{500-400}$
3. $E_a=23050$ J mol⁻¹ ≈ 23.1 kJ mol⁻¹

Answer: About 23.1 kJ mol⁻¹

Chemistry in real life: Chemical Kinetics

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Rate Law, Order and Molecularity
• Integrated Rate Equations and Half-Life
• Arrhenius Equation and Activation Energy

Chemical Kinetics formula sheet

Rate in terms of each species

$rate=-\frac{1}{a}\frac{d[A]}{dt}=\frac{1}{c}\frac{d[C]}{dt}$

Divide by the coefficient.

Rate law

$rate=k[A{]}^x[B{]}^y,order=x+y$

x and y from experiment only.

Units of k

$(mol{L}^{-1}{)}^{1-n}{s}^{-1}$

0: mol L⁻¹ s⁻¹; 1: s⁻¹; 2: L mol⁻¹ s⁻¹; 3: L² mol⁻² s⁻¹.

Zero order

$[A]=[A{]}_0-kt,t_{1/2}=\frac{[A{]}_0}{2k}$

Half-life ∝ initial concentration.

First order

$k=\frac{2.303}{t}\log \frac{[A{]}_0}{[A]}$

Also $k=\frac{2.303}{t}\log \frac{a}{a-x}$.

First order half-life

$t_{1/2}=\frac{0.693}{k}$

Independent of $[A{]}_0$; after n half-lives, $(1/2{)}^n$ remains.

Arrhenius equation

$k=A{e}^{-E_a/RT},\log k=\log A-\frac{E_a}{2.303RT}$

Slope of log k vs 1/T = $-E_a/2.303R$.

Arrhenius at two temperatures

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$

R = 8.314 J K⁻¹ mol⁻¹, never 0.0821.

Pseudo first order hydrolysis

$C{H}_{3}COO{C}_2{H}_5+H_{2}O\to C{H}_{3}COOH+C_2{H}_{5}OH$

Water in excess: rate = k'[ester].

Decomposition of hydrogen peroxide

$2H_2{O}_2\to 2H_{2}O+O_2$

First order; catalysed by I⁻.

Chemical Kinetics: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Chemical Kinetics

The question patterns to expect, and the method that works for each.

Tricks to remember Chemical Kinetics

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Units tell the order

  Count the 'mol' power in k: mol¹ → zero, mol⁰ → first, mol⁻¹ → second, mol⁻² → third.

  Use it when: Order from units of k.

• 0.693 is half

  $t_{1/2}=0.693/k$ for first order: no concentration in it, so no dependence on it.

  Use it when: Half-life one-markers.

• OMEN

  Order: experimental, may be fractional or zero. Molecularity: whole Number, Elementary step only.

  Use it when: Order vs molecularity.

• k's in the log, T's in the bracket

  $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\cdot \frac{T_2-T_1}{T_1{T}_2}$.

  Use it when: Arrhenius two-temperature problems.

• Slow step rules

  Whatever is in the slow step is in the rate law.

  Use it when: Mechanism questions.

Chemical Kinetics: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Chemical Kinetics

The slips examiners see most often, and what to write instead.

• ✗Mistake: Misreading order from the units of k

  CISCE examiners' comments, 2024 and 2025: for k = 0.05 s⁻¹, candidates wrote 'zero' order and 'dependent' instead of first order and 'independent'; for units mol⁻² L² s⁻¹, they answered second order instead of third.

  Wrong working

  mol⁻² L² s⁻¹ → second order.

  Right working

  (mol L⁻¹)${}^{1-n}$ = mol⁻² L²: 1 − n = −2, n = 3.

  ✓Do this: Solve 1 − n = power of mol every time.

  Usually costs: 1 mark.

• ✗Mistake: Leaving k out of the rate law or getting the overall order wrong

  CISCE examiners' comments, 2024 and 2025: from rate data candidates gave the overall order as second instead of third, wrote R = [A]²[B] without k, could not find the order with respect to NO, and gave a wrong value or unit of k.

  Wrong working

  Rate = [A]²[B]; k = 28.4 (no unit).

  Right working

  Rate = k[A]²[B]; overall order 3; k with units L² mol⁻² min⁻¹.

  ✓Do this: Write the rate law with k, add the powers, then get units from the order.

  Usually costs: 1-2 marks.

• ✗Mistake: Swapping k and T in the Arrhenius equation or using the wrong R

  CISCE examiners' comments, 2024: candidates substituted temperatures in place of rate constants, used R = 0.0821 L atm instead of 8.314 J K⁻¹ mol⁻¹, and could find k but not the percentage of substance left after a given time.

  Wrong working

  $\log \frac{700}{500}=\dots$; R = 0.0821.

  Right working

  $\log \frac{k_2}{k_1}$ on the left; R = 8.314 J K⁻¹ mol⁻¹; then % left = $(1/2{)}^n\times 100$ or from $\log \frac{a}{a-x}$.

  ✓Do this: Write the formula with symbols first, then substitute.

  Usually costs: 1-2 marks.

• ✗Mistake: Order versus molecularity confusion

  CISCE examiners' comments, 2025: candidates could not state a correct difference between order and molecularity, could not give molecularity 2 for the step 2A → B + C though they found the order, and failed to give the pseudo order when [NO] was in large excess.

  Wrong working

  Rate = $k[A{]}^{3/2}$, so molecularity = 3/2.

  Right working

  Order = 3/2 (from the rate law); molecularity of the elementary step $2A\to$ is 2.

  ✓Do this: Molecularity comes from the step's equation; order from the rate law.

  Usually costs: 1 mark.

Practice questions on Chemical Kinetics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The half-life of a zero order reaction is:
   (a) independent of $[A{]}_0$
   (b) directly proportional to $[A{]}_0$
   (c) inversely proportional to $[A{]}_0$
   (d) proportional to $[A{]}_0^2$

   Hint

   $t_{1/2}=[A{]}_0/2k$.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   Which cannot be fractional?
   (a) order
   (b) molecularity
   (c) rate constant
   (d) rate

   Hint

   Count of colliding particles.

   Answer

   (b) molecularity

3. Q3 · 1 mark · Assertion-reasonBoard level

   Assertion: A catalyst increases the rate of a reaction without changing ΔH.
   Reason: A catalyst lowers the activation energy by providing an alternative path.
   (a) Both true, Reason explains Assertion
   (b) Both true, Reason does not explain Assertion
   (c) Assertion true, Reason false
   (d) Both false

   Hint

   ΔH is a state function.

   Answer

   (a). The lower barrier explains the faster rate; ΔH depends only on reactants and products.

4. Q4 · 1 mark · Very short answerBoard level

   Define activation energy.

   Hint

   Threshold minus average.

   Answer

   The extra energy that reactant molecules must absorb, over their average energy, to reach the threshold energy and form the activated complex.

5. Q5 · 2 marks · Short answerHarder board level

   Show that for a first order reaction the time for 99.9% completion is about ten times the half-life.

   Hint

   log 1000 = 3.

   Answer

   $t_{99.9}=\frac{2.303}{k}\log 1000=\frac{6.909}{k}$; $t_{1/2}=\frac{0.693}{k}$; ratio = 9.97 ≈ 10.

6. Q6 · 3 marks · Short answerHarder board level

   A first order reaction is 20% complete in 10 min. How long will it take to be 75% complete?

   Hint

   Find k first.

   Answer

   k = (2.303/10) log(100/80) = 0.02232 min⁻¹. t = (2.303/0.02232) log 4 = 62.1 min.

7. Q7 · 3 marks · Short answerHarder board level

   The rate constant of a reaction is $1.0\times 10^{-3}$ s⁻¹ at 300 K. If $E_a$ = 60 kJ mol⁻¹, calculate k at 320 K.

   Hint

   Two-temperature Arrhenius.

   Answer

   log(k₂/k₁) = 60 000 × 20/(2.303 × 8.314 × 300 × 320) = 0.6533; k₂/k₁ = 4.50; k₂ = $4.5\times 10^{-3}$ s⁻¹.

8. Q8 · 2 marks · Short answerHarder board level

   For $A\to B$, a graph of log[A] against t is a straight line of slope $-3.0\times 10^{-3}$ s⁻¹. State the order and calculate k and $t_{1/2}$.

   Hint

   Slope = −k/2.303.

   Answer

   First order. k = 2.303 × 3.0 × 10⁻³ = $6.91\times 10^{-3}$ s⁻¹; $t_{1/2}$ = 0.693/6.91 × 10⁻³ = 100 s.

9. Q9 · 2 marks · Short answerHOTS / competency

   The reaction $2N{O}_2+F_2\to 2N{O}_{2}F$ has rate = $k[N{O}_2][F_2]$. Suggest a two-step mechanism consistent with this and name the intermediate.

   Hint

   One NO₂ and one F₂ in the slow step.

   Answer

   Slow: $N{O}_2+F_2\to N{O}_{2}F+F$; fast: $N{O}_2+F\to N{O}_{2}F$. Intermediate: F atom.

10. Q10 · 2 marks · Short answerHarder board level

    Explain, with collision theory, why (a) not every collision causes reaction; (b) powdered marble reacts faster with acid than lumps.

    Hint

    Energy, orientation, surface.

    Answer

    (a) Only collisions with energy ≥ threshold and proper orientation are effective. (b) Powder has greater surface area, so more acid particles collide with it per second.

11. Q11 · 5 marks · Long answerHarder board level

    (a) Derive $[A]=[A{]}_0-kt$ for a zero order reaction and sketch [A] against t. [2]
    (b) Why is the decomposition of NH₃ on hot platinum zero order at high pressure? [1]
    (c) A first order reaction has $t_{1/2}$ = 69.3 min. Find the time for the concentration to fall to one-tenth. [2]

    Hint

    k = 0.693/69.3.

    Answer

    (a) $-\frac{d[A]}{dt}=k$; integrate: $[A]=-kt+C$; at t = 0, C = $[A{]}_0$. Graph: straight line falling with slope −k, intercept $[A{]}_0$. (b) The surface is saturated with NH₃, so rate does not depend on gas concentration. (c) k = 0.0100 min⁻¹; t = (2.303/0.0100) log 10 = 230 min.

12. Q12 · 3 marks · Short answerHOTS / competency

    Two reactions have the same A but $E_a$ values of 50 and 60 kJ mol⁻¹. Which is faster at 300 K, and by roughly what factor? Which is more sensitive to temperature?

    Hint

    Ratio $e^{\Delta E_a/RT}$.

    Answer

    The 50 kJ reaction is faster by $e^{10000/(8.314\times 300)}=e^{4.01}\approx 55$ times. The 60 kJ reaction is more sensitive to temperature (steeper Arrhenius plot).

13. Q13 · 2 marks · Short answerHOTS / competency

    In the reaction $2NO+O_2\to 2N{O}_2$, rate = $k[NO{]}^2[O_2]$. How does the rate change if the volume of the vessel is halved?

    Hint

    Every concentration doubles.

    Answer

    Each concentration doubles: rate × 2² × 2 = 8 times.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State the law of mass action.Show answer

   Rate is proportional to the product of active masses (molar concentrations) of reactants, each raised to its coefficient.

2. 2.Units of k for a first order reaction?Show answer

   s⁻¹ (time⁻¹).

3. 3.First order half-life formula?Show answer

   $t_{1/2}=0.693/k$

4. 4.What is the rate-determining step?Show answer

   The slowest elementary step of a mechanism, which controls the overall rate.

5. 5.Two conditions for an effective collision?Show answer

   Energy ≥ threshold energy and proper orientation.

6. 6.Write the Arrhenius equation.Show answer

   $k=A{e}^{-E_a/RT}$

7. 7.Slope of log k against 1/T?Show answer

   $-E_a/2.303R$

8. 8.Give an example of a pseudo first order reaction.Show answer

   Acid hydrolysis of ethyl acetate in excess water (or inversion of cane sugar).

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write the rate of reaction in terms of each species?
• ☐Can you get orders, the rate law and k with units from a data table?
• ☐Can you give two differences between order and molecularity?
• ☐Can you derive the first order rate equation and its half-life?
• ☐Can you find % remaining after a time using half-lives?
• ☐Can you draw and label an energy profile with a catalyst?
• ☐Can you find $E_a$ from two rate constants with R = 8.314?
• ☐Can you explain why a 10 K rise roughly doubles the rate?

Chemical Kinetics: frequently asked questions