CHEMISTRY• Peh Kok Hua• Lee Sze Yien• Chong Boon Weng• Peter YipPre-U TextStrategicSTPM• Purified STPM Syllabus• Scheme of Assessment Starting from Semester 1 STPM Examination 2026SEMESTER2New! Based on> Info > Video> e-QuizDIGITAL RESOURCES2022 – 2024
iiStrategic STPM Semester 2 Chemistry is written based on the purified STPM syllabus introduced bythe Malaysian Examinations Council (MEC), which will be implemented starting from the 2026 STPMSemester 1. This book is carefully designed and well-organised with the following features to enhancestudents’ understanding of the concepts.9 STATES OF MATTERCHAPTERConcept MapVaporisation and vapour pressureLattice structure of crystalline solidAllotrope of carbonBoiling pointsBoyle’s LawCharles’ LawAvogadro’s LawBasic assumptions of kinetic theoryDalton’s law of partial pressuresStates of MatterLiquids Solids Gases1Ideal gas lawChemistry Semester 2 STPM Chapter 9 States of Matter17Chapter9Vapour Pressure and Boiling Point1. The boiling point of a liquid is the temperature at which its vapour pressure equals the external pressure, acting on the liquid surface. 2. At this temperature, the thermal energy of the molecules is great enough for the molecules in the interior of the liquid to break free from their neighbours and enter the gas phase. As a result, bubbles of vapour form within the liquid. 3. The boiling point increases as the external pressure increases. The boiling point of a liquid at 101 kPa pressure is called its normal boiling point. The atmospheric pressure is lower at higher altitudes, so water boils at a temperature lower than 100 °C, and foods generally take longer to cook.Example 9.19Hydrogen peroxide, H2O2, has a normal boiling point of 150 °C. Would you expect hydrogen peroxide to have a higher or lower vapour pressure than water at 25 °C?Solution:Hydrogen peroxide has lower vapour pressure than water at 25 °C.It has higher boiling point than water.It has stronger intermolecular force of attraction than water.Less molecules of H2O2 can break free from their neighbours and enter the gas phase.Quick Check 9.21. Liquids P and Q are poured separately into two beakers with a volume of 10 cm3 each at the same temperature. After an hour, the volumes of liquids P and Q are 2.0 cm3 and 9.0 cm3 respectively. Explain your observation.2. The boiling points of propanone, water and mercury at room conditions are 56.05 °C, 100 °C and 356.7 °C respectively. On the same axis, sketch three graphs to show the changes in vapour pressure with temperature for each of these three liquids.Boiling point is also the temperature at which the liquid is in equilibrium with its vapour at an external pressure of 1 atm.Liquid L vapourExam TipsBoiling point of a liquid depends on the external pressure.Chemistry Semester 2 STPM Chapter 9 States of Matter2Chapter99.1 Gases1. All gases have remarkably similar physical behaviour. (a) Gases are easily compressed, but expand spontaneously to fill whatever container they are in. (b) All gases are composed entirely of non-metallic elements. 2. Furthermore, all have simple molecular formulas.3. The molecules of a gas move chaotically, colliding with each other and with the walls of their container and hence exert pressure.The Gas Laws1. Four variables are needed to define the physical conditions of a gas: (a) Temperature(b) Pressure(c) Volume(d) Amount of gas (number of moles)2. Gas laws are equations that express the relationships among these four variables.Boyle’s Law: The Pressure−Volume Relationship1. Boyle’s law states that: The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure V ∝ 1p2. Mathematical expression: pV = k where k is a constant.3. For a given sample of gas under two different sets of conditions at constant temperature, p1V1 = p2V24. Graphical presentation of Boyle’s law: pVAt constanttemperaturepAt constanttemperatureV1Learning Outcome• State Boyle’s law, Charles’law and Avogadro’s law• Apply Boyle’s law, Charles’law and Avogadro’s law inproblem solving.• Derive the ideal gas law• Apply the ideal gas law inproblem solving includingthe determination of relativemolecular mass and density.• State the basic assumptionsof the kinetic theory.• Explain qualitatively theconditions necessary for agas approaching the idealbehaviour.• State Dalton’s law of partialpressures.• Determine partial pressureof a gas and its compositionusing Dalton’s law of partialpressures.VIDEOConcept of Boyle’s lawPREFACEConcept MapQuick CheckProvides short questions for students to test their understanding of the concepts learnt in the subtopicsLearning OutcomeA list of subtopics that students will learn in each chaptervSTPM Scheme of AssessmentSemesterof StudyCode andPaperNameTheme/ Title Type of test Mark(Weighting) Duration AdministrationSemester1962/1ChemistryPaper 1GeneralChemistry andInorganicChemistryWritten testSection A20 multiple-choicequestionsSection B2 structured questionsSection C2 essay questions60(26.67%)2014261.5 hours CentralassessmentSemester2962/2ChemistryPaper 2PhysicalChemistryWritten testSection A20 multiple-choicequestionsSection B2 structured questionsSection C2 essay questions60(26.67%)2014261.5 hours CentralassessmentSemester3962/3ChemistryPaper 3OrganicChemistryWritten testSection A20 multiple-choicequestionsSection B2 structured questionsSection C2 essay questions60(26.67%)2014261.5 hours Centralassessment962/5ChemistryPaper 5WrittenChemistryPracticalWritten practical test3 structured questions45(20%) 1.5 hours CentralassessmentSemesters1, 2, 3962/4ChemistryPaper 4ChemistryPracticalCoursework15 experiments225To bescaled to 45(20%)Throughout thethreesemestersCourseworkSTPM Scheme of AssessmentLatest STPM Scheme of Assessment starting 2026
vSTPM Scheme of AssessmentSemesterof StudyCode andPaperNameTheme/ Title Type of test Mark(Weighting) Duration AdministrationSemester1962/1ChemistryPaper 1GeneralChemistry andInorganicChemistryWritten testSection A20 multiple-choicequestionsSection B2 structured questionsSection C2 essay questions60(26.67%)2014261.5 hours CentralassessmentSemester2962/2ChemistryPaper 2PhysicalChemistryWritten testSection A20 multiple-choicequestionsSection B2 structured questionsSection C2 essay questions60(26.67%)2014261.5 hours CentralassessmentSemester3962/3ChemistryPaper 3OrganicChemistryWritten testSection A20 multiple-choicequestionsSection B2 structured questionsSection C2 essay questions60(26.67%)2014261.5 hours Centralassessment962/5ChemistryPaper 5WrittenChemistryPracticalWritten practical test3 structured questions45(20%) 1.5 hours CentralassessmentSemesters1, 2, 3962/4ChemistryPaper 4ChemistryPracticalCoursework15 experiments225To bescaled to 45(20%)Throughout thethreesemestersCoursework
viChapter9 States of Matter 19.1 Gases 29.2 Liquids 159.3 Solids 18STPM Practice 9 26e-Quiz 1 QR code 28ChapterEquilibria 29 1010.1 Chemical Equilibria 3010.2 Ionic Equilibria 6410.3 Solubility Equilibria 113STPM Practice 10 130e-Quiz 2 QR code 142ChapterReaction Kinetics 143 1111.1 Rate Law 14411.2 The Effect of Temperature on Reaction Kinetics 15211.3 The Role of Catalysts in Reactions 16111.4 Order of Reactions and Rate Constants 168STPM Practice 11 186e-Quiz 3 QR code 192CONTENTSChapterChemical Energetics 193 1212.1 Enthalpy Changes 19412.2 Hess’s Law 20612.3 Born-Haber Cycle 208STPM Practice 12 211e-Quiz 4 QR code 214ChapterElectrochemistry 215 1313.1 Half-cell and Redox Equations 21613.2 Standard Reduction Potential 23013.3 Non-standard Cell Potentials 24413.4 Electrolysis 26113.5 Application of Electrochemistry 280STPM Practice 13 288e-Quiz 5 QR code 293STPM Model Paper (962/2) 294Answers 299Appendix 334
11 REACTION KINETICSCHAPTERConcept MapRate equation and StoichiometryMaxwell- Boltzmann distribution curveHomogeneous and heterogeneous catalystsExperimental determination of order of reactionReaction mechanismProduction of sulphuric acidReaction KineticsThe role of catalysts in reactionsThe effect of temperature on reaction kineticsOrder of reactions and rate constantsRate law143Rate constantInitial rate methodCollision theoryGraphical methodArrhenius equation
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics144Chapter1111.1 Rate LawReaction rate and Stoichiometry1. The rate of reaction is defined as the rate of change of the concentration of a substance (be it the reactant or product) with time. The unit is mol dm–3 s–1. Rate = [Inital concentration] − [Final concentration]Time taken = ∆[Concentration]∆t2. In a chemical reaction, the concentration of the reactants decreases with time while the concentration of the products increases with time.3. The rate of reaction can be expressed in two ways:(a) Rate of disappearance of the reactants:Rate = –∆[Reactant]∆t(b) Rate of appearance of the products:Rate = –∆[Product]∆tThe negative sign in front of the rate of disappearance of reactants is to make the rate positive because the concentration of the reactants decreases with time4. Consider a simple reaction: A B (reactant) (product)(a) The rate of disappearance of A = –∆[A]∆t(b) The rate of appearance of B = ∆[B]∆t (c) Since the stoichiometry of the reaction is 1:1, and –∆[A]∆t = ∆[B]∆tthis means that the rate at which A disappears is equal to the rate at which B appears.5. The graph below shows how the concentrations of A and B change over time: Concentration (mol dm–3)Time (s)B: ProductA: ReactantLearning Outcome• State the rate equation.• Determine the rate constantfrom initial rates.• Deduce the initial ratefrom rate equation andexperimental data.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics145Chapter11• The concentration of A decreases with time as it is consumed during the reaction.• The concentration of B increases with time as it is formed.6. For a general reaction: aA + bB → cC + dD The rate of disappearance of reactants = – 1a∆[A]∆t = – 1b∆[B]∆t The rate of appearance of products = 1c∆[C]∆t = 1d∆[D]∆t Hence, – 1a∆[A]∆t = – 1b∆[B]∆t = 1c∆[C]∆t = 1d∆[D]∆t7. Note that a, b, c and d are the stoichiometric coefficients of the substances involved in the balanced equation.Example 11.1Dinitrogen pentoxide decomposes according to the equation:2N2O5(g) → 4NO2(g) + O2(g)At a particular instance, the rate of disappearance of N2O5 is 1.65 × 10–3 mol dm–3 s–1. What is the rate of formation of NO2 and O2 at that instance?Solution:Rate = – 12∆[N2O5]∆t = 14∆[NO2]∆t = ∆[O2]∆tHence, rate of formation of NO2, ∆[NO2]∆t= 4 ×12∆[N2O5]∆t= 2 × (1.65 × 10–3)= 3.30 × 10–3 mol dm–3 s–1The rate of formation of O2,∆[O2]∆t = 14∆[NO2]∆t= 14× (3.30 × 10–3)= 8.25 × 10–4 mol dm–3 s–18. Below is the concentration/time graph for the reaction: A B (reactant) (product)Alternative method:The rate of formation of O2:∆[O2] ——– ∆t = —12∆[N2O5] ———–– ∆t= —12 × (1.65 × 10–3)= 8.25 ×10–4 mol dm–3 s–1
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics146Chapter11 ct1 t2dbaxy0Concentration of A/mol dm–3Time/s(a) The rate of reaction at a particular moment can be obtained from the tangent to the concentration/time graph at that particular point of time.(b) The rates of reaction at time zero, at time t1 and at time t2 are given by:Rate at time zero = yx mol dm–3 s–1Rate at time t1 = ab mol dm–3 s–1Rate at time t2 = cd mol dm–3 s–1The rate at time ‘zero’ is called the initial rate.(c) It can be seen that the rate of reaction at time t2 is less than that at time t1. This is because the concentration of the reactant decreases with time.Rate Equation, Determination of Rate Constant and Initial rate1. One of the main goals in studying chemical kinetics is to find an equation that shows how the rate of a reaction depends on the concentration of the reactants.2. The dependence of the rate of reaction on concentration is expressed in what is called a rate equation or rate law.3. Thus, the rate equation is defined as an equation that shows quantitatively how the rate of reaction is affected by the concentration of reactants.4. The rate equation is different from the stoichiometric equation which shows the reacting mole ratios of the reactants and products.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics147Chapter115. For the general equation: aA + bB → cC + dD If the rate of reaction is proportional to [A]x and rate is also proportional to [B]y, then Rate ∝ [A]x and Rate ∝ [B]y The rate equation is expressed as: Rate = k[A]x[B]y Where k is known as the rate constant x is the rate order with respect to reactant A y is the rate order with respect to reactant B x + y are the overall order of the reaction6. The rate constant, k, is the proportionality constant in the rate equation. It remains constant at a specific temperature. However, when the temperature changes, k will also change. The unit of kdepends on the overall order of the reaction.7. The orders of the reactants, represented by the values of x and y, can only be determined experimentally. They cannot be deduced from the stoichiometric coefficients of the reactants in the balanced chemical equation.8. For the STPM syllabus, we are concerned with three types of order of reactions: First order, second order and zero order.(a) If a reaction is first order with respect to a substance, it means that when the concentration of that substance is changed xtimes, the rate would change by x1 times. (b) If a reaction is second order with respect to a substance, it means that when the concentration of that substance is changed x times, the rate would change by x2 times.(c) If a reaction is zero order with respect to a substance, it means that when the concentration of that substance is changed xtimes, the rate would change by x0 (which is = 1) times. That is, the rate of the reaction is independent of the concentration of that substance. 9. For example, the rate equation for the reaction between carbon monoxide and nitrogen dioxide: CO(g) + NO2(g) → CO2(g) + NO(g) is given by: Rate = k[NO2]2(a) This means that doubling the concentration of NO2 would increase the rate of reaction by 22 or 4 times. Similarly, if the concentration of NO2 is halved, then the rate would decrease to 12 2 or 14 of its original value.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics148Chapter11(b) However, changing the concentration of CO has no effect on the rate of reaction. However, CO must be present for the reaction to occur.Example 11.2The rate equation for the reaction between ethanoic acid and bromine in the presence of hydrogen ions is as follows:Rate = k[CH3COOH][Br2](a) What is the order of reaction with respect to:(i) CH3COOH(ii) Br2(iii) H+(b) What is the overall order of the reaction?Solution:(a) (i) The reaction is first order with respect to CH3COOH(ii) The reaction is first order with respect to Br2(iii) The reaction is zero order with respect to H+(b) Overall order = 1(CH3COOH) + 1(Br2) = 2Example 11.3The following shows rate equation for the following reaction:2A + B + 2C → productsRate = k[A][B]2(a) What is the overall order of the reaction?(b) What is the unit for the rate constant, k?(c) The initial rate of reaction at a certain concentration of A, Band C, is x mol dm–3 s–1. What will the new rate be (at constant temperature) if,(i) the concentration of A is doubled,(ii) the concentration of A is doubled and the concentration of B is halved(iii) the concentration of C is tripled?Solution:(a) Overall order = 1 + 2 = 3(b) k = Rate[A][B]2Unit k = mol dm–3 s–1(mol dm–3)(mol2 dm–6) = dm6 mol–2 s–1(c) Originally: x = k[A][B]2(i) New rate = k[2A][B]2 = 2k[A][B]2 = 2x mol dm–3 s–1
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics149Chapter11(ii) New rate = k[2A]12B2 = 214 k[A][B]2 = 12 k[A][B]2 = 12 x mol dm–3 s–1(iii) x mol dm–3 s–1(Since the reaction is zero order with respect to C, the rate remains the same).10. The rate constant, k, can be found by substituting in the concentrations of the reactants and the initial rate into the rate equation at a specific temperature.11. The value of k depends on the type of reaction, the temperature, and whether a catalyst is used.A larger value of k means the reaction happens faster.12. The unit of k depends on the order of the reaction. Let’s look at each case:(a) Zero-order reaction• The rate equation is:Rate = k[A]0 = k• So, the unit of k is the same as the unit of rate.For example: mol dm–3 s–1(b) First-order reaction• The rate equation is:Rate = k[A]• To find the unit of k:Unit of k = Unit of rateUnit of concentration = mol dm–3 time–1mol dm–3 = time–1• So, for a first-order reaction, the unit of k is just per second (s–1) or per minute (min–1).(c) Second-order reaction• The rate equation might look like:Rate = k[A]2 or Rate = k[A][B]• To find the unit of k:Unit of k = Unit of rate(Unit of concentration)2 = mol dm–3 time–1(mol dm–3)2 = dm3 mol–1 time–1• For example: dm3 mol–1 s–1 or dm3 mol–1 min–113. On the other hand, if we already know the rate equation, the rate constant, and the concentrations of the reactants, we can calculate the initial rate of the reaction. Thisis done by substituting the known values into the rate equation.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics150Chapter1114. Understanding the rate equation allows us to:(a) Adjust the concentration of a specific reactant to obtain a desired reaction rate. This is useful in controlling how fast a reaction proceeds.(b) Gain insight into the reaction mechanism, which refers to the detailed steps involved in the overall reaction. The form of the rate equation can help suggest how many steps occur and which step is the slowest (rate-determining step).Example 11.4The equation for the decomposition of sulphuryl chloride is given below:SO2Cl2(g) → SO2(g) + Cl2(g)The above reaction is of second order.(a) Write the rate equation for the above reaction.(b) At 320 °C, when the initial concentration of SO2Cl2 is 0.040 mol dm–3, the rate of decomposition of SO2Cl2 is 6.0 ×10–5 mol dm–3 s–1. Calculate the rate constant at 320 °C.(c) Predict the rate of decomposition of SO2Cl2 at 320 °C when the initial concentration of SO2Cl2 is 0.060 mol dm–3.Solution:(a) For a second order reaction:Rate = k[SO2Cl2]2(b) Rate = k[SO2Cl2]26.0 × 10–5 = k(0.040)2k = 6.0 × 10–5(0.040)2 = 3.75 × 10−2 dm3 mol–1 s–1(c) Rate = k[SO2Cl2]2= 3.75 × 10−2 × (0.060)2= 1.35 × 10−4 mol dm–3 s–1Example 11.5In the gas-phase decomposition of ozone, the reaction proceeds asfollows:2O3(g) → 3O2(g)The reaction is first order with respect to ozone.(a) Write the rate equation for the reaction. What is the overall order of the reaction?(b) Calculate the rate of reaction at 25 °C if the rate constant is k = 2.0 × 10–3 s–1, and the concentration of ozone is 0.080 mol dm–3.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics151Chapter11(c) What is the rate of:(i) loss of ozone (O3), and(ii) formation of oxygen gas (O2)?Solution:(a) Rate equation:Rate = k[O3]Overall order = 1(b) Rate = k[O3]= 2.0 × 10−3 × 0.080= 1.6 × 10−4 mol dm–3 s–1(c) (i) From the rate expression: − 12d[O3]dt = k[O3] − 12d[O3]dt = (−2) × 1.6 × 10−4 = −3.2 × 10−4 mol dm–3 s–1 Rate of loss of ozone = 3.2 × 10−4 mol dm–3 s–1(ii) 13d[O2]dt = k[O3] d[O2]dt = 3k[O3] d[O3]dt = (3) × 1.6 × 10−4 = 4.8 × 10−4 mol dm–3 s–1 Rate of formation of oxygen =4.8 × 10−4 mol dm–3 s–1Quick Check 11.11. The reaction taking place in the Haber process is: N2(g) + 3H2(g) → 2NH3(g)At a particular, the rate is given as:–∆[N2]∆T = 1.5 × 10–3 mol dm–3 s–1What is the rate of reaction express in terms of:(a) H2, and (b) NH3? 2. Consider the following reaction: 2Cu2+(aq) + 6CN–(aq) → 2[Cu(CN)2 ]–(aq) + (CN)2(aq) (a) Express the rate of reaction in terms of the four species in the equation. (b) At a particular moment, the rate of formation of [Cu(CN)2]– is 5.60 × 10–2 mol dm–3 s–1. What is the rate express in terms of (i) CN–, and (ii) (CN)2?
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics152Chapter113. The reaction between fluorine and chlorine dioxide is as follows: F2(g) + 2ClO2(g) → 2FClO2(g) If the rate of disappearance of F2 is 0.095 g s–1, what is the rate of formation of FClO2 in mol dm–3 s–1?4. Consider the following reaction: A + B → product The reaction is first order with respect to A and second order with respect to B. (a) Write the rate equation for the reaction. (b) What is the unit of the rate constant? (c) The rate of reaction at temperature T is x mol dm–3 s–1. Find, in terms of x, the rates of reaction at temperature T when (i) the concentration of A is halved. (ii) the concentration of B is halved. (iii) the concentration of A is doubled and the concentration of B tripled.11.2 The Effect of Temperature on Reaction KineticsCollision Theory1. The qualitative effect of temperature on the rate of reaction can be explained using collision theory.2. Collision theory explains how chemical reactions occur by focusing on the collisions between reactant particles. 3. A reaction takes place when reactant particles collide with one another, but not all collisions lead to product formation. 4. Only those collisions that satisfy certain conditions are effective in producing chemical change.5. Effective collisions are collisions that result in the formation of products, while ineffective collisions do not lead to any reaction. 6. The rate of reaction is determined by the frequency of effective collisions. 7. Hence, any factor that increases the number of effective collisions will also increase the rate of reaction.8. For a collision to be effective, the reacting particles must possess enough energy to overcome the repulsion between their electron clouds. This minimum energy is known as the activation energy (Ea).Learning Outcome• Explain qualitatively theeffect of temperature onthe rate of reaction by usingcollision theory.• Explain the distribution ofmolecular energy at differenttemperatures by usingthe Maxwell-Boltzmanndistribution curve.• Explain the relationshipbetween rate constantwith activation energy andtemperature using Arrheniusequation.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics153Chapter11 Insufficient energyIneffectivecollisionRepulsionWith sufficient energyEffectivecollisionAttraction9. If the colliding particles do not have sufficient energy to overcome this barrier, they will simply repel each other and no reaction will occur. 10. The activation energy is defined as the minimum energy that must be overcome by reactant molecules before a reaction can take place.11. The energy profile of a typical chemical reaction is shown below: [E1 is the activation energy for the forward reaction. E2 is the activation energy for the reverse reaction. ∆H is the enthalpy change of reaction.]EnergyReaction pathwayReactantsΔE1E2ProductsH[Activated complex]12. The activation energies for some reactions are given in the table below:Reaction Activation energy/ kJ2N2O(g) → 2N2(g) + O2(g) 482.02HI(g) → H2(g) + I2(g) 390.0CH3Cl(g) + H2O(l) → CH3OH(aq) + HCl(aq) 115.02N2O5(g) → 2N2O4(g) + O2(g) 217.013. Let’s consider an example:(a) Hydrogen iodide is formed when hydrogen and iodine molecules collide. H2(g) + I2(g) → 2HI(g) (b) However, at 450 °C, theoretical calculations show approximately 1 × 1028 collisions occur every second, yet only around 1 × 1015 molecules are formed per second. This huge difference indicates that only a very small fraction of collisions is effective.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics154Chapter11(c) This inefficiency arises because, as H2 and I2 molecules approach each other, their electron clouds repel one another. Only molecules with sufficient kinetic energy can overcome this repulsion, allowing the electron clouds to interact and new bonds to form, resulting in the formation of HI.RepulsionElectron cloud H Electron cloud 2 I214. Besides energy, the orientation of the collision is also important. Even if particles collide with sufficient energy, they must do so in the correct direction to allow bond formation. 15. For example, consider the reaction between a methyl free radical and chloroform:H3C• + CHCl3 → CH4 + Cl3C•(a) If the methyl radical approaches the hydrogen atom of chloroform with the right orientation and sufficient energy, the reaction proceeds and methane is formed:H3C•+ H—CCl3 → CH4 + Cl3C•H • H3C CClClCl Products(b) However, if the methyl radical collides with a chlorine atom instead, no reaction occurs, even if the energy is sufficient. H3C•+ Cl—CHCl2 → No products formed HC • H3C ClClCl No products formed16. Therefore, for a chemical reaction to occur, two main conditions must be met:(a) The colliding particles must have energy equal to or greater than the activation energy.(b) The particles must collide with the correct orientation.17. Temperature has a significant effect on the rate of reaction. (a) An increase in temperature raises the average kinetic energy of the particles, which leads to more frequent collisions. (b) More importantly, it increases the fraction of particles that have enough energy to overcome the activation energy barrier. (c) This leads to a higher number of effective collisions, and thus a higher reaction rate.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics155Chapter11Energy Distribution and Temperature (MaxwellBoltzmann Curve)1. The Maxwell-Boltzmann distribution curve is used to explain howenergy is distributed among particles in a sample. 2. It also shows the relationship between particle energy and temperature, and how temperature changes affect the number of particles with energy equal to or greater than the activation energy.3. From the curve below, it is observed that at any given temperature, only a small fraction of molecules possess energy equal to or greater than the activation energy. This explains why only a small portion of the total collisions result in a successful reaction. Energy Ea Number of particlesThese particles do nothave enough energyto reactOnly these particleshave enough energyto react4. For most reactions, the rate of reaction increases exponentially with increasing (regardless of whether the reaction is endothermic or exothermic) temperature. Rate of reactionTemperature * Exponential relationship between rate of reaction and temperature.5. As temperature increases, the average kinetic energy of the molecule increases, resulting in more collisions per unit time. 6. At the same time, the number of molecules having energy equal to or greater than the activation energy also increases. This is illustrated by the Maxwell-Boltzmann distribution curve below: Fraction of moleculesMolecular energyT1T1T2T2 >Ea * Number of particles having energy Ea increases with temperature.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics156Chapter117. An increase of 10 °C usually doubles the rate of reaction. However, calculations from the kinetic theory show that the rate of collision only increases about 0.016 times when the temperature increases by 10 °C. (Refer to Example 11.6) 8. Look at the Maxwell-Boltzmann distribution curves below attemperature T K and (T + 10) K Fraction of moleculesEnergyTT + 10DCB AEEaΔADE ≈ 2 × ΔABC(a) The total number of particles having energy equal or greater than the activation energy, at temperature T, is given the area of triangle ABC(b) While the total number of particles having energy equal or greater than the activation energy, at temperature (T + 10), is given the area of triangle BDE, which is about two times the area of triangle ABC. 9. However, for biological reactions involving enzymes, the rate of reaction will increase with temperature until a certain point (the optimum temperature), beyond which the rate decreases as the enzymes (which are protein molecules) get denatured. Rate of reaction involving enzymesOptimumtemperatureTemperatureExample 11.6According to the kinetic theory, the kinetic energy (K.E.) of a particle is directly proportional to its absolute temperature (T).K.E. ∝ TCalculate the relative speed of the particle at 310 K and 300 K.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics157Chapter11Solution:Kinetic energy = 12 mv 2Since 12 mv 2 ∝ TTherefore, v 2 ∝ THence, v22v12 ∝ T2T1 [Where T2 = 310 K and T1 = 300 K]By substitution:v2v1 = T2T1= 310300 = 1.016Hence, the molecular speed at 310 K is 1.016 times the molecular speed at 300 K. In other words, the molecular speed increases only 0.016 times when the temperature increases by 10 K from 300 K to 310 K.Effect of Temperature on Reaction Kinetics (Arrhenius Equation)1. The quantitative effect of temperature on the rate constant, and hence the rate of reaction, can be explained using the Arrhenius Equation.2. The relationship between the rate constant and temperature is an exponential one. It is given by the Arrhenius equation: A = Ae –EaRT3. Taking natural log throughout: ln k = ln k – –EaRT or ln k = ln A – –Ea2.303RT where: k = Rate constant A = Constant Ea = Activation energy R = Universal gas constant (R = 8.31 J mol–1 K–1) T = Absolute temperature (K)4. The constant A is known as the frequency factor. It composes two components: (a) The probability factor, which is the fraction of collisions with the correct orientation. (b) The collision frequency, which is the number of collisions per unit time.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics158Chapter115. The e–EaRT term refers to the fraction of particles having energy equal to or greater than the activation energy. 6. From the above equation, it can be seen that the rate constant would increase (a) When the activation energy is decreased (by addition of suitable catalyst).(b) When the temperature increases.7. A plot of ln k against 1T gives a straight line with the slope equals to –EaR . Gradient =In k–EaR1T8. A plot of log k against 1T gives a straight line with the slope equals to –Ea2.303R . Gradient =log k1T2.303–EaR9. If the rate constant of a particular reaction is k1 at temperature T1, and is k2 at temperature T2, then, log k2k1 = Ea2.303R1T1 – 1T2 Similarly, ln k2k1 = EaR 1T1 – 1T2
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics159Chapter11Example 11.7Reaction A has Ea = 50 kJ mol−1, while reaction B has Ea = 80 kJ mol−1. Which reaction is faster at 298 K?Solution:At the same temperature, lower Ea leads to a higher rate constant (k).e–EaRT is larger for smaller Ea. So, reaction A is faster than reaction Bat 298 K.Example 11.8The variation in the rate constant, k, at two different temperatures for the reaction2NO(g) + O2(g) → 2NO2(g)is given in the following table: Temperature (°C) k (dm3 mol–1 s–1)300 1.20 × 10−4350 3.60 × 10−3(a) Calculate the activation energy (Ea) of the reaction in kJ mol–1.(b) Calculate the rate constant, k at 320 °C, and comment on your answers obtained.Solution:(a) Step 1: Convert temperatures to KelvinT1 = 300 + 273 = 573 KT2 = 350 + 273 = 623 KStep 2: Use the Arrhenius equation in logarithmic form: ln k2k1 = EaR 1T1 – 1T2ln 3.60 × 10−31.2 × 10−4 = Ea8.31 1573 – 1623 Ea = 202 kJ mol–1(b) Substituting T = 320 + 273 = 593 , Ea = 202 000 J mol–1ln kk1 = EaR 1T1 – 1T ln k3.6 × 10−3 = Ea8.31 1623 – 1593 k = 5.03 × 10–4 dm3 mol–1 s–1Comment: The rate constant increases significantly as temperature increases, consistent with the Arrhenius theory. When the temperature increases 20 °C
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics160Chapter11Example 11.9The rate constant for a reaction is measured at various temperatures:Temperature (K) Rate constant, k (s–1)300 2.3 × 10−5310 6.7 × 10−5320 1.8 × 10−4Determine the activation energy of the reaction in kJ mol–1Solution:Step 1: Prepare the table with ln k and 1T values.T (K) k (s–1) ln k 1T (K–1)300 2.3 × 10−5 –10.68 0.00333310 6.7 × 10−5 –9.61 0.00323320 1.8 × 10−4 –8.62 0.00313Step 2: Plot ln k (y-axis) against 1T (x-axis).Determine the slope of the line:In k1(0.00333, –10.68)(0.00313, –8.62)–9.0–9.5–10.0–10.50.00315 0.00320 0.00325 0.00330 T (K–1)Gradient = (−10.68) − (−8.62)(0.00333 − 0.00313) =−10300Gradient = –EaR =−10300 Ea =10300(8.31) = 85593 = 8.6 kJ mol−1Quick Check 11.21. The rate constant for a reaction is, k = 2.3 × 10−4 s−1 at 298 K. The activation energy, Ea is 75.0 kJ mol−1. Calculate the frequency factor A.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics161Chapter112. In an investigation, the A is 2.00 × 1012 s−1 and Ea = 95.0 kJ mol−1. What temperature is needed for k = 1.00 × 104 s−1?3. The reaction between hydrogen and iodine as below: H2(g) + I2(g) L 2HI(g) The following data are obtained:Temperature (K) k (mol–1 dm³ s–1)500 2.5 × 10–3550 1.24 × 10–2(a) Find the activation energy, Ea, in kJ mol–1.(b) Estimate the rate constant at 520 K.4. The rate constant for a reaction is measured at various temperatures.Temperature (K) Rate Constant, k (s–1)300 2.3 × 10−5310 6.7 × 10−5320 1.8 × 10−4 Determine graphically the activation energy of the reaction in kJ mol–1.5. The variation of the rate constants with temperatures for a reaction is given in the table below.Temperature (oC) 283 302 427 518Rate constant, k 4.05 × 10–7 1.40 × 10–6 1.33 × 10–3 4.49 × 10–2Plot a suitable graph to determine the activation energy in kJ mol–1.11.3 The Role of Catalysts in ReactionsEffect of Catalysts on the Rate of Reaction1. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed by the reaction. 2. A catalyst reacts with the reactants to form an intermediate but is regenerated at the end of the reaction. 3. During the course of the reaction, the chemical nature of the catalyst might change, but remains unchanged chemically at the end of the reaction. 4. A catalyst alters, (a) The rate and rate constants of a reaction(b) The mechanism of reaction(c) The order of reaction(d) The activation energy of the reactionLearning Outcome• Explain the effect ofcatalysts on the rate ofreactions.• Describe the productionof sulphuric acid using thefollowing catalysts(i) oxides of nitrogen(ii) vanadium(V) oxide.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics162Chapter115. A catalyst does not affect (a) The yield(b) The overall stoichiometry(c) The enthalpy change of the reaction. 6. The rate of reaction is proportional to the amount of catalyst used. However, since they are regenerated they are usually used in small quantities because catalysts are expensive. 7. A catalyst alters the rate of reaction by allowing the reaction to proceed by an alternative pathway with lower activation energy. By doing so, more particles will have enough energy to overcome the lower activation energy and increases the number of effective collisions per second as illustrated by the Maxwell-Boltzmanndistribution curve below: Fraction of moleculeEnergyEa = Activation energy without catalystEaʹ = Activation energy in the presence of a catalystEaʼ Ea8. The energy profile for a typical chemical reaction in the presence and in the absence of a catalyst is shown below. Activation energywithout catalystActivation energywith catalystReactantsProductsEnergyProgress of reaction9. The action of catalysts is very specific. A catalyst that catalyses one reaction might not catalyse another reaction.10. There are two types of catalysts:(a) A catalyst which is in the same physical state as the reactants is known as a homogeneous catalyst.(b) A catalyst which is not in the same physical state as the reactants is known as a heterogeneous catalyst.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics163Chapter11Homogeneous Catalyst1. Homogeneous catalysis is usually explained using the intermediateproduct theory.2. During the reaction, the catalyst forms an intermediate compound, which then reacts to regenerate the catalyst and produce the final products.3. Consider the reaction between aqueous potassium iodide and potassium persulphate: 2KI(aq) + K2S2O8(aq) → 2K2SO4(aq) + I2(aq) The ionic equation for the reaction is: 2I–(aq) + S2O82–(aq) → I2(aq) + 2SO42–(aq)4. The activation energy of this reaction is expected to be high because it involves collisions between negatively charged particles, which tend to repel one another. Hence, the rate is low. I–RepulsionS2O82–5. However, the rate increases in the presence of a little aqueous iron(III) chloride as a catalyst.6. In the presence of Fe3+(aq) ions, the reaction does not involve direct collisions between the negatively charged I– and S2O82– ions. 7. Instead, the Fe3+ ions collide with the I– ions to produce iodine, and Fe3+ is converted into Fe2+ (the intermediate product). 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(aq) I–AttractionFe3+8. The Fe2+ ions then collide with S2O82– ions to produce SO42– ion and Fe3+ is regenerated. 2Fe2+(aq) + S2O82–(aq) → 2Fe3+(aq) + 2SO42–(aq) S2O82–AttractionFe2+9. The two steps can be combined as follows: Step 1 : 2Fe3+(aq) + 2I–(aq) → 2Fe2+(aq) + I2(aq) Step 2 : 2Fe2+(aq) + S2O82–(aq) → 2Fe3+(aq) + 2SO42–(aq) Overall reaction: S2O82–(aq) + 2I–(aq) → 2SO42–(aq) + I2(aq) Where:Fe3+ = catalystFe2+ = intermediate compound
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics164Chapter1110. The catalysed steps are expected to have lower activation energies because they involve collisions between opposite charged particles. EnergyProgress of reactionEaUncatalysedreactionCatalysedreactionE ʼa11. Thus, in the catalysed mechanism of homogeneous catalysis:(a) The catalyst undergoes a temporary chemical change, forming an intermediate, which then reacts to regenerate the original catalyst.(b) There is typically a change in oxidation state of the catalyst during the formation and breakdown of the intermediate. Transition elements or their compounds are good homogeneous catalysts.(c) The sum of all the elementary steps in the mechanism is equal to the overall stoichiometric equation of the reaction.12. Other examples of homogeneous catalysis are:Reaction Catalyst2SO2(g) + O2(g) → 2SO3(g) NO2 (g)CH3COOH(I) + C2H5OH(I) → CH3COOC2H5(I) + H2O(l) H2SO4 (aq)2KCIO3(s) → 2KCI(s) + 3O2(g) MnO2 (s)2MnO4–(aq) + 5C2O42–(aq) + 16H+(aq) →2Mn2+(aq) + 10CO2(g) + 8H2O(l) Mn2+(aq)CH3COCH3 (aq) + l2 (aq) → CH3COCH2l(aq) + HI(aq) H+(aq)Example 11.10Using balanced equations, illustrate the action of the following catalyst in the reactions concerned.(a) 2H2O2(aq) → 2H2O(l) + O2(g) [Catalyst: I–(aq)](b) 2KClO3(s) → 2KCl(s) + 3O2(g) [Catalyst: MnO2(s)]Solution:(a) H2O2(aq) + I−(aq) → H2O(l) + IO−(aq)H2O2(aq) + IO−(aq) → H2O(l) + O2(g) + I−(aq)(b) 2KClO3(s) + 4MnO2(s) → 2KCl(s) + 2Mn2O7(s)2Mn2O7(s) → 4MnO2(s) + 3O2(g)
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics165Chapter11Heterogeneous Catalyst1. The adsorption theory is usually used to explain the action of heterogeneous catalyst. 2. One must differentiate between adsorption and absorption. 3. Adsorption occurs on the surface of a substance while absorption occurs in the body of a substance. Adsorption Absorption4. An example of heterogeneous catalysis is the reaction between hydrogen gas and iodine vapour at 450 °C in the presence of solid nickel as catalyst. Ni(s) H2(g) + I2(g) ⎯⎯⎯→2HI(g)5. The H2 and I2 molecules get adsorbed on the surface of nickel. This is done by the formation of temporary bonds between the vacant or half-filled 3d orbitals of nickel.Surface of NiHNiHNiINiINiHNiHNiINiINi6. The adsorption serves three important purposes: (a) It brings the H2 and I2 molecules closer to one another. (b) It weakens the covalent bonds in the H2 and I2 molecules. (c) It holds the reactant molecules in the right orientation for new bonds to be formed. Bond in the processof formingHNiHNiINiINiHNiHNiINiINiBond in the process ofbreaking7. Eventually, the old bonds are broken completely and new bonds are formed. The HI molecules are then released from the nickel surface (this is called desorption) providing a new surface for the adsorption of other reactant molecules. HNiHNiINiINiHNiHNiINiINi8. However, there are some heterogeneous catalysis that can be explained through the intermediate product theory.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics166Chapter119. Examples of heterogeneous catalysis are:Industrial process Reaction CatalystHydrogenation ofalkenes CH2=CH2(g) + H2 (g) → CH3CH3(g) Ni(s) or Pt(s)Dehydration ofethanol C2H5OH(g) → C2H4(g) + H2O(l) Al2O3(s) or SiO2(s)Haber process N2(g) + 3H2(g) → 2NH3(g) Fe(s)Ostwald process 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) Pt(s) or Rh(s)Industrial Production of Sulphuric Acid1. The use of catalysts in sulphuric acid production, covering both the Lead Chamber Process (historical method) and the Contact Process (modern method)2. It involves the catalytic oxidation of sulphur dioxide (SO2) to sulphur trioxide (SO3).3. Key Reaction: catalyst 2SO2(g) + O2(g) ⎯⎯⎯→ 2SO3(g)Use of Oxides of Nitrogen in Sulphuric Acid Production1. In the historical Lead Chamber Process, oxides of nitrogen such as nitrogen monoxide (NO) and nitrogen dioxide (NO2) were used as homogeneous catalysts to assist in the oxidation of sulphur dioxide (SO2) to sulphur trioxide (SO3).2. In this process, sulphur dioxide reacts with nitrogen dioxide in the following way: Step 1: SO2(g) + NO2(g) → SO3(g) + NO(g)3. The nitrogen monoxide (NO) formed is not wasted. It is reoxidised by atmospheric oxygen to regenerate nitrogen dioxide: Step 2: 2NO(g) + O2(g) → 2NO2(g)4. This NO/NO2 cycle acts as a catalyst system, continuously converting SO2 into SO3 without being consumed in the overall reaction. Overall reaction: 2SO2(g) + O2(g) → 2SO3(g)5. Since both NO and NO2 are gases, this is an example of homogeneous catalysis, where the catalyst is in the same phase as the reactants.6. However, the Lead Chamber Process is less efficient, produces lower yields, and is no longer used in modern industry. 7. It has been largely replaced by the Contact Process, which is more effective and economical.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics167Chapter11Use of Vanadium(V) Oxide in Sulphuric Acid Production1. In the modern industrial production of sulphuric acid, the Contact Process is used. 2. A key step in this process is the oxidation of sulphur dioxide (SO2) to sulphur trioxide (SO3), which is catalysed by vanadium(V) oxide (V2O5).3. This is an example of heterogeneous catalysis, where the solid catalyst facilitates a reaction between gaseous reactants.4. Overall reaction: V2O5(s) 2SO2(g) + O2(g) ⎯⎯⎯→2SO3(g)5. The action of vanadium(V) oxide as a catalyst can be explained using the intermediate product theory, which suggests that the catalyst forms a temporary intermediate that later decomposes to regenerate the original catalyst.(a) Step 1: Formation of intermediate• Sulphur dioxide is adsorbed onto the surface of vanadium(V)oxide and is oxidised.• SO2(g) + V2O5(s) → SO3(g) + V2O4(s)• In this step, SO2 is oxidised to SO3• Vanadium(V) is reduced to vanadium(IV) in the form ofV2O4.(b) Step 2: Regeneration of catalyst• The vanadium(IV) oxide (V2O4) is then reoxidised by oxygen in the air, regenerating vanadium(V) oxide.• 2V2O4(s) + O2(g) → 2V2O5(s) • This completes the catalytic cycle. The catalyst V2O5 is regenerated and not consumed in the overall reaction.(c) The intermediate product theory explains how V2O5 is temporarily converted to V2O4, then regenerated.(d) The reaction occurs on the surface of the catalyst through adsorption, facilitating bond breaking and formation.6. Operating Conditions:• Temperature: ~450 °C• Pressure: ~1–2 atm (moderate pressure)• Excess oxygen: Shifts equilibrium to increase SO3 yield (exothermic reaction)7. Advantages of using V2O5:• High efficiency and high yield of SO3.• Catalyst is not consumed and remains active over a long period.• The process is cost-effective and suitable for large-scale industrialproduction of sulphuric acid.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics168Chapter11Quick Check 11.31. Hydrogen peroxide decomposes slowly at room temperature according to the following reaction: 2H2O2(aq) → 2H2O(l) + O2(g) However, the reaction is significantly faster in the presence of hydrogen bromide (HBr), which acts as a catalyst.(a) Suggest a two-step mechanism to explain how HBr catalyses the decomposition of hydrogenperoxide.(b) Sketch an energy profile diagram for the reaction with and without the catalyst. Clearly label.2. The industrial synthesis of ammonia in the Haber Process involves the reaction between nitrogen and hydrogen gases. N2(g) + 3H2(g) → 2NH3(g) Experimental data for the reaction with and without a catalyst are given below:Condition Enthalpy change, ΔH(kJ mol–1)Activation energy, Ea(kJ mol–1)Without catalyst –92.0 230.0In presence of iron (Fe) –92.0 120.0(a) Explain the role of iron as a catalyst in the Haber Process.(b) Based on the data above, sketch a labelled energy profile diagram for the reaction in the presence and absence of the catalyst.11.4Order of Reactions and Rate ConstantsThe Initial Rate Method1. Initial rate is the rate of reaction at time zero where the concentration of the reactants is maximum. 2. It can be obtained by drawing tangent to the concentration/time graph at time = 0. xy0Initial rate = yx Concentration of A/mol dm–3TimeLearning Outcome• Deduce the order of areaction and the rateconstant by the initial ratemethod and graphicalmethods;• Propose the reactionmechanism with theobserved kinetics;• Calculate half-life (t 12) andrate constant for first-orderand second-order reactions
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics169Chapter113. For a general reaction: aA + bB → products In order to determine the order with respect to substance A, two or more experiments are conducted using varying amounts of Abut keeping the concentration of B constant throughout. Hence, whatever changes in the rates are due to A and not B. 4. To determine the order with respect to B, we keep the concentration of A constant while varying the concentration of B.Example 11.11The data below is the result of the kinetic study of the following reaction:2A + B → ProductsExperiment [A] (mol dm–3)[B] (mol dm–3)Initial rate (mol dm–3 min–1)I 0.12 0.26 0.514II 0.12 0.52 2.06III 0.24 0.26 1.03(a) Determine,(i) the order of reaction with respect to A and B(ii) what is the overall order?(b) Write the rate equation for the reaction.(c) Using experiment (I), calculate the rate constant and state its unit.(d) Calculate the rate when the concentrations of A and B are 0.34 mol dm–3 and 0.52 mol dm–3 respectively.Solution:(a) Let the rate equation be:Rate = k[A]x[B]y(i) Experiment IIExperiment I : 2.060.514 = 0.520.26 y 4 = 2y y = 2 Experiment IIIExperiment I : 1.030.514 = 0.240.12 x 2 = 2x x = 1(ii) Overall order = (2 + 1) = 3(b) Rate = k[A][B]2(c) 0.514 = k(0.12)(0.26)2 k = 63.4 dm6 mol–2 min–1(d) Rate = 63.4 × (0.34)(0.52)2 = 5.83 mol dm–3 min–1
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics170Chapter11Quick Check 11.41. The data below refers to the reaction at 300 K. 2NO(g) + O2(g) → 2NO2(g)Experiment [NO]/mol dm–3 [O2]/mol dm–3 Relative initial rateI 0.58 0.29 1.0II 1.16 0.29 4.0III 1.16 0.58 8.0 Determine the rate equation for the reaction at 300 K2. Consider the data below regarding the reaction: A + 2B → productsExperiment [A]/mol dm–3 [B]/mol dm–3 Initial rate/ mol dm–3 s–1I 1.0 1.0 1.44 × 10–2II 1.0 8.0 9.12 × 10–1III 4.0 1.0 5.88 × 10–2 Determine the rate equation for the reaction.3. Chlorine dioxide disproportionate in alkaline solution according to the equation: 2ClO2(aq) + 2OH–(aq) → ClO3–(aq) + ClO2–(aq) + H2O(l)The kinetic results of the above reaction are given in the table below:Experiment [CIO2]/mol dm–3 [OH–]/mol dm–3 Initial rate/ M s–1I 0.10 0.184 0.032II 0.14 0.184 0.063III 0.14 0.368 0.126(a) Determine the oxidation number of chlorine in (i) ClO2(ii) ClO3–(iii) ClO2–(b) Determine the order of reaction with respect to (i) ClO2(ii) OH–(c) Calculate the rate constant and state its unit. (d) Calculate the rate of reaction when the concentrations of ClO2– and OH– are 0.23 mol dm–3 and 0.37 mol dm–3 respectively.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics171Chapter11Graphical Method (Integrated Rate Equation Method)First Order Reaction1. The differential equation for a first order reaction is: A → products Rate = k[A]1 – dCdt = kC (where C is the concentration of A at time t) – dCC = k dt Integrating the equation:C t ∫ dCC = – ∫ k dtC0 0 ln C – ln C0 = –kt (where C0 is the concentration of A at time ‘zero’)∴ In C = –kt + In C0 Or: log C = – kt2.303 + log C0 Or: In C0C = kt2. A plot of ln C against t gives a straight line with the slope equals to –k. tGradient = –kIn C3. Similarly, a plot of ln C0C against t would give a straight line with the slope equal to k. Gradient = ktInC0C
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics172Chapter114. From the integrated rate equation we can obtain another important quantity of the reaction, which is the half-life, t 12. The half-life is defined as the time taken for the initial amount(concentration, mass, moles etc) of a reactant to decrease to half its value.5. The half-life can be determined graphically from the concentrationtime graph as shown below: tC0C02[A]t 216. From the first order equation :ln C = –kt + ln C0When t = t 12 , C = C02Substituting into the above equation:ln C02 = –kt 12 + ln C0Rearranging:t 12 = ln 2k7. The expression above shows that the half-life of a first orderreaction is independent of the initial amount of the reactant.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics173Chapter118. The concentration/time graph for a typical first order reaction is shown below:1.00.5(t1)(t2)(t3)02ndhalf life3rdhalf life1sthalf lifeTime/minute5 10 15 20 25[A]The half-life for the reaction is 7.0 minutes. The time taken for theamount of A to decrease from 1.0 to 0.5, and from 0.5 to 0.25, and from 0.25 to 0.125 is the same.∴ t1= t2 =t3Example 11.12The half-life of a first order reaction is 15 minutes.(a) Calculate the time taken for the original amount of A to 116 decrease to its original amount.(b) Calculate the rate constant.Solution:(a) 1 → 12 → 14 → 18 → 116 The time required = 4 half-lives = 4 × 15 = 60 minutes(b) Using the formula: t 12= ln 2k k = ln 215 = 4.62 × 10–2 min–1
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics174Chapter11Example 11.13The following data is obtained for a reactant A in a particular reaction.Time/min 0.0 10.0 20.0 30.0 40.0[A]/mol dm–3 1.00 0.94 0.88 0.83 0.78Plot a graph of log[A] against time. (a) Use your graph to determine the rate constant.(b) Calculate the half-life.Solution:Time/min 0 10 20 30 40log [A] 0 –0.027 –0.056 –0.081 –0.11–10–1210 20 30 40 50Time/min–8–6–4–2Iog [A] (× 10–2)Gradient ==11.0 × 10–240 –– 2.75 × 10–3(a) Gradient of graph = – k2.303 = – 2.75 × 10–3 ∴ k = 6.33 × 10–3 min–1(b) Half-life = 0.693k = 0.6936.33 × 10–3 = 109.5 min
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics175Chapter11Second Order Reaction1. The differential equation for a second order reaction is: A → products Rate = k[A]2 – dCdt = kC2 – dCC2 = k dt Integrating the equation:C t ∫ dCC2 = – ∫ k dt C0 0 – 1C + 1C0 = – kt∴1C = kt + 1C02. For a second order reaction, a plot of 1C against t gives a straight line with the slope equals to k.3. The half-life for the second order reaction is obtained as follows:1C = kt + 1C0When t = t 12, C = C02Thus, 1C02 = kt 12+ 1C0t 12 = 1kC0 ∴ The higher the concentration, the shorter the half-life.4. The half-life of a second order reaction is inversely proportionalto the initial concentration of the reactant. If the initial amount is high, the half-life will be shorter and vice versa.Gradient = kt1CThe higher the concentration, the shorter the half-life
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics176Chapter111.00.50100 200 300 4001sthalf life2ndhalf life3rdhalf lifeTime/s[A](t1)(t2)(t3) ∴ t2 = 2t1 t3 = 2t2Example 11.14The data below refers to the following reaction:A → products.Time (s) 0 40 80 120 160[A]/ mol dm–3 16.1 9.10 6.41 4.95 4.02Plot a graph of 1[A] against time.(a) Use your graph to determine the rate constant.(b) Calculate the half-life of the reaction when the concentration ofA is(i) 1.00 mol dm–3(ii) 0.5000 mol dm–3Solution:Time / min 0 40 80 120 1601[A] (× 10–2)(dm3 mol–1)6.21 11.0 20.2 24.9 4.02From the graph:1st half-life = 5 min2nd half-life = 10 min3rd half-life = 20 min
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics177Chapter1120 40 60 80 100 120 140 160 08101224614161820222426Gradient ==(22.6 – 10.0) ×10–1140 – 321.17 × 10–2(× 10–2)/mol–1 dm3[A]1t/s(a) Gradient = k = 1.17 × 10–2 dm3 mol–1 s–1(b) Using the formula: t 12 = 1kC0(i) When C0 = 1.00 mol dm–3 t 12= 1(1.17 × 10–2) (1.00) = 85.5 s(ii) When C0 = 0.50 mol dm–3 t 12= 1(1.17 × 10–2) (0.500) = 171 sZero Order Reaction1. The differential equation for a zero order reaction is: A → products Rate = k[A]0 Or: Rate = k – dCdt = k – dC = k dt Integrating the equation: C t ∫ dC = – ∫ k dt C0 0 C – C0 = – kt∴ C = – kt + C0
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics178Chapter112. For a zero order reaction, a plot of C against t gives a straight line with a slope equals to –k. Gradient = ktC3. The half-life for a zero order reaction is:t 12 = C02k4. The half-life of a zero order reaction is directly proportional to theinitial concentration of the reactant. That means the half-life islonger if the initial amount of reactant is higher.00.51.02 4 6 8 10 12 14 Time/min1st half life2ndhalf life3rdhalf life[A]The higher the concentration,the longer the half-life.From the graph:1st half-life = 6.5 min2nd half-life = 3.25 min3rd half-life = 1.63 min
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics179Chapter115. The table below summarises the integrated rate equation and the half-life for zero-, first- and second-order reactions.Order Rate equation Integrated rate equation Half-life, t 120 Rate = k C = – kt + C0C02k1 Rate = k[A]ln C = – kt + ln C0lnC0C = ktIn 2k2 Rate = k[A]2 1C = kt + 1C01kC06. The concentration/time graph for the zero, first and second order reactions are shown in the plot below for comparison. Zero orderFirst orderSecond orderTimeConcentrationRate/Concentration Graph1. For a first order reaction, a plot of rate against concentration, [A] gives a straight line passing through the origin with the gradient equal to the rate constant. Gradient = k[A]Rate2. For a second order reaction, a plot of rate against concentration, [A]2gives a straight line passing through the origin with the gradient equal to the rate constant. Gradient = k[A]2Rate
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics180Chapter113. The plot of rate against concentration, [A] gives a curve, specifically a parabola starting from the origin. This is because rate increases with the square of concentration, the curve becomes steeper as [A] increases. [A]Rate4. For a zero order reaction, a plot of rate against concentration, [A] gives a line parallel to the concentration axis. The intercept on the rate axis gives the rate constant. k[A]RateQuick Check 11.51. The table below refers to the reaction: X → Y + ZTime/min 0 10 20 30 40 50[X]/mol dm–3 1.00 0.75 0.56 0.45 0.38 0.25(a) Plot a suitable graph to determine the order of reaction with respect to X.(b) Calculate the rate constant for the reaction.2. The data below refers to the hydrolysis of sucrose.Time/min 0 20 50 65 150[Sucrose]/mol dm–3 0.38 0.27 0.18 0.15 0.087(a) Plot a suitable graph to determine the order of reaction with respect to X.(b) Calculate the rate constant for the reaction.3. Consider the data below:Time/min 0.0 5.0 10.0 15.0 20.0 25.0[A]/mol dm–3 1.00 0.88 0.75 0.62 0.50 0.37 Plot a graph to determine the order with respect to A.For a second order reaction
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics181Chapter114. The decomposition of dinitrogen pentoxide follows a second order kinetic. 2N2O5(g) → 2N2O4(g) + O2(g)[N2O5]/mol dm–3 1.00 1.49 1.90 2.13Initial rate/mol dm–3 s–1 1.89 2.45 2.90 3.15(a) Plot a suitable graph to show that the reaction is first order with respect to N2O5.(b) Determine the rate constant.5. The decomposition of hydrogen peroxide is first order with respect to [H2O2]. 2H2O2(aq) → 2H2O(l) + O2(g) Sketch the graph you would obtain for the following:(a) Rate against [H2O2] (b) Rate against [H2O2]2 (c) [H2O2] against timeRate Equation and Reaction Mechanism1. One-step reactions are called elementary reactions. Very fewchemical reactions occur in only one step. 2. Most chemical reactions take place in several consecutive steps. These sequence of steps is called the reaction mechanism.3. Reaction mechanism is usually proposed on the basis of experimental results which are provided by the rate equation.4. The sum of the individual steps in the reaction mechanism gives the overall stoichiometric equation for the reaction.5. In a multiple steps reaction, one of the steps will be the slowest (i.e. the one that has the highest activation energy). This step forms the ‘bottle neck’ of the reaction. It is called the rate-determining step.6. The overall rate of reaction is then dependent on the ratedetermining step. The rate equation would contain only the species that are involved in the rate-determining step.7. For example, A + B + C → products Assuming the reaction mechanism is: A + B Slow X X + C Fast products8. The overall rate of reaction depends only on the rate of formation of the intermediate, X. Hence, the rate equation is given by: Rate = k[A][B]9. Increasing the concentration of C does not affect the rate of reaction because C cannot take part in the reaction until ‘X’ is formed.Sum of all the individual stepsmust give the overallbalanced stoichiometricequation.Exam TipsThe rate-determining step isthe step with the highestactivation energy.Exam Tips'X' is the intermediate.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics182Chapter1110. The energy profile for such a reaction is as shown below: EnergyReaction pathwayA + B + CProductshigh Ealow Ea‘X’ + CAlkaline Hydrolysis of 2-methyl-2-indopropane1. The equation of hydrolysis is: CH3 CH3 | | CH3—C—CH3 + OH– → CH3—C—CH3 + I– | | I OH Or: (CH3)3C—I + OH– → (CH3)3C—OH + I–2. Kinetic studies of the reaction show that the reaction is first order with respect to (CH3)3CI and zero order with respect to OH–. The rate equation is given by:Rate = k[(CH3)3CI]3. This shows that the reaction must involve more than one step. The rate determining step would involve only (CH3)3CI.4. The proposed mechanism is as follows: (CH3)3C—I Slow (CH3)3C+ + I– (CH3)3C+ + OH– Fast (CH3)3C—OH5. The proposed mechanism is consistent with the rate equation. Furthermore, the first step would require high activation energy because it involves the breaking of covalent bonds.6. The second step is expected to be fast as it involves the combination of opposite charged particles.7. The energy profile is as shown in the diagram. EnergyReaction pathway(CH3)3COH + I–(CH3)3C+ + OH– + I–(CH3)3Cl + OH–This is a SN2 mechanism.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics183Chapter11Alkaline Hydrolysis of 1-iodobutane1. The equation for the hydrolysis is: CH3CH2CH2CH2I + OH– → CH3CH2CH2CH2OH + I–2. Kinetic studies show that the reaction is first order with respect to both 1-iodobutane and OH– ions as shown in the rate equation below. Rate = k[CH3CH2CH2CH2I][OH–]3. In this case, the species appearing in the rate equation are the same as those in the stoichiometric equation. This is an example of a onestep (or elementary) reaction.4. The proposed mechanism is: Direct collision between CH3CH2CH2CH2I and OH– to form an intermediate (or an activated complex). H H CH3CH2CH2CH2I + OH– Slow HO.......C.......I | CH2 | CH2|CH3Activated complex5. In the activated complex, the carbon-oxygen bond is in the processof forming, while the carbon-iodine bond is in the process ofbreaking.6. The activated complex then decomposes rapidly to yield the product. CH3CH2CH2CH2(OH)I– Fast CH3CH2CH2CH2OH + I–7. The energy profile takes the following form. EnergyReaction pathwayCH3CH2CH2CH2I + OH–CH3CH2CH2CH2OH + I–H HHO C ICH2CH2CH3–– –The energy profile for aSN1 mechanism has two'humps', while that for a SN2 mechanism has one 'hump'.Exam Tips
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics184Chapter11Example 11.15Nitrogen monoxide reacts with chlorine in the gaseous phase according to the equation:2NO(g) + Cl2(g) → 2NOCl(g)The experimental rate equation is:Rate = k[NO][Cl2]Propose a mechanism for the reaction.Solution:NO + Cl2 Slow NOCl2NOCl2 + NO Fast 2NOClSummary1. The rate of reaction is defined as the rate of change of the concentration of a substance with time.2. The instantaneous rate is obtained by drawing tangents to the concentration/ time graph at a particular moment.3. Activation energy is the minimum energy that has to be overcome before a reaction can occur.4. Effective collisions are collisions that have enough energy to overcome the activation energy, and also in the correct orientation.5. The factors that affect the rates of reactions are:(a) Concentration(b) Pressure(c) Temperature(d) Catalyst6. Increasing temperature increases the fraction of particles having energy equal to or greater than the activation energy.7. A catalyst increases the rate of reaction by providing an alternative pathway, with lower activation energy, for the reaction to occur.8. Homogeneous catalyst is a catalyst that is in the same physical state as the reactants.9. Heterogeneous catalyst is a catalyst that is in a different physical state as the reactants.10. The order of reaction with respect to a substance is the power to which the molar concentration of that reactant is raised in an experimentally determined rate equation.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics185Chapter1111. Half-life is the time taken for the amount of a reactant to decrease to half of its original quantity.12. Summaries the integrated rate equation and the half-life for zero-order, first-order and second-orderreactions.Order Rate equationIntegrated rate equationHalf-life, t 120 Rate = k C = –kt + C0C02k1 Rate = k[A]ln C = –kt + ln C0lnC0C = ktIn 2k2 Rate = k[A]2 1C = kt +1kC01kC013. The Arrhenius equation is given by: k = Ae–EaRT14. The rate constant of a reaction increases with temperature irrespective of whether the reaction is endothermic or exothermic.15. The rate constant is dependent on temperature and the presence of a catalyst.16. Reaction mechanism describes the sequence of steps leading to the formation of products.17. The rate-determining step is the step with the highest activation energy (or the slowest step) in the reaction mechanism.18. The rate equation only involves species in the rate-determining step.19. Change Effect on rate of reaction Effect on rate constantIncreasing concentration Increases No effectDecreasing concentration Decreases No effectIncreasing pressure Increases No effectDecreasing pressure Decreases No effectIncreasing temperature Increases IncreasesDecreasing temperature Decreases Decreasesaddition of catalyst Increases Increases
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics186Chapter11STPM PRACTICE 11Objective Questions1. Which of the following statements about the rate constant for a first order reaction is not correct?A It has a unit of s–1.B It is a constant at constant temperature.C It does not change even if a catalyst is added.D It is independent of the concentrating of the reactants.2. Which of the following rate equations is correct for the following reaction?3O2(g) → 2O3(g)A ∆[O2]∆t = ∆[O3]∆tB – ∆[O2]∆t = ∆[O3]∆tC ∆[O2]∆t = ∆[O3]2∆tD – ∆[O2]∆t = 32∆[O3]2∆t3. The Maxwell distribution curves for a gaseous reaction under different conditions are shown below.Energy Number of moleculesXYWhich of the following is true?A The curve X is obtained at a higher temperature.B The curve X is obtained at a higher pressure.C The total area of curve X is the same as curve Y.D Curve Y is one with the presence of a catalyst.4. In acidic solution, bromate(V) ions oxidise bromide ions to bromine.BrO3– + 6H+ + 5Br– → 3Br2 + 3H2OGiven the following data:Experiment [BrO3–]/ mol dm–3[Br–]/ mol dm–3[H+]/ mol dm–3Relative rate1 0.40 0.28 0.031 12 0.60 0.28 0.031 1.53 0.60 0.56 0.031 34 0.40 0.28 0.062 4Determine the overall order of the reaction.A 2B 3C 4D Insufficient data5. Which of the following statements regarding catalyst is not true?A A catalyst alters the mechanism of a reaction.B A catalyst does not take part chemically in the reaction.C The amount of catalyst used affects the rate of a reaction.D The same amount of products would be obtained from a particular reaction irrespective of whether a catalyst is present or not.6. In general, as the temperature increases,A the rate will increase if the reaction is endothermic.B the rate will decrease if the reaction is exothermic.C the rate will increase regardless of whether the reaction is endothermic or exothermic.D the rate will remain the same if the reaction is zero order.
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics187Chapter117. The acid catalysed reaction between propanone and iodine is: CH3COCH3 + I2H+ CH2ICOCH3 + HI The result of an experiment is shown below: Time [I2]/mol dm–3 From this, it can be concluded thatA the acid is not involved in the reaction.B it is a third order reaction overall.C the rate of reaction does not depend on the concentration of iodine.D the rate-determining step involves only iodine.8. The mechanism for the reaction: P + 2Q → PQ2 is shown below:P + Q → PQ slow PQ + Q → PQ2 fastWhich of the following rate equations is consistent with the above mechanism?A rate = k[PQ] C rate = k[P][Q]B rate = k[PQ][Q] D rate = k[PQ2]9. Combustion of ethanol in pure oxygen is faster than in air becauseA oxygen is a catalyst for the combustion process.B oxygen lowers the activation energy.C the concentration of oxygen is higher.D nitrogen from the air hinders the combustion process.10. The reaction between iodine and propanone in the presence of dilute sulphuric acid is given by the equation below: I2(aq) + CH3COCH3(aq) →H+(aq) + I–(aq) + CH2ICOCH3 The reaction is zero order with respect to iodine. It can be deduced thatA the reaction is first order with respect to propanone.B sulphuric acid acts as a catalyst.C iodine is involved in the rate-determiningstep.D the rate equation is: rate = k[CH3COCH2][I2][H+]11. Consider the reaction: 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) Which of the following species in the above reaction appears or disappears the fastest?A NH3B O2C NO2D H2O12. The initial rate of decomposition of a compound X was measured at two initial concentration of Xwith the following results.[X]/mol dm–3 0.20 0.30Initial rate/mol dm–3 min–1 0.68 1.52 What is the order of reaction with respect to substance X?A Zero C 1.5B 1.0 D 213. The rate of a chemical reaction is roughly doubled by increasing the temperature by 10 °C. This is becauseA the bond energy of the particles is halved by raising the temperature by 10 °C.B increasing the temperature by 10 °C doubles the rate of collision.C increasing the temperature by 10 °C doubles the number of particles having energy greater than the activation energy.D increasing the temperature by 10 °C doubles the average kinetic energy of the particles.14. Consider the reaction: 2A → B + 2C Time In[A ] The gradient of the line is equal toA kB k–1C –kD ln k
Chemistry Semester 2 STPM Chapter 11 Reaction Kinetics188Chapter1115. Which of the following is necessarily true of an overall second order reaction?A the half-life is a constant.B the rate of reaction would increase four folds if the concentration of the reactants is doubled.C the reaction occurs in two steps.D it must involve two substances.16. A reaction is represented by equation below.aA + bB → cC ΔH = 0.00 The rate of reaction with respect to A is x mol dm–3 s–1. Which of the following statement is true of the reaction?A The rate of reaction is independent of temperature.B The rate of reaction with respect to B is bxmol dm–3 s–1.C The activation energy of the forward reaction is the same as the activation energy of the reverse reaction.D Catalyst does not affect the activation energy of the reaction.17. A catalystA increases the kinetic energy of the reactant particles.B increases the fraction of particles having energy greater than the activation energy.C increases the rate of collisions between the particles.D increases the intensity of the collisions between the particles.18. Which of the following statements correctly explains why a small increase in temperature usually leads to a large increase in the rate of a gaseous reaction?A The number of collisions increases at higher temperature.B The activation energy decreases at higher temperature.C The number of collisions with energy greater than the activation energy increases.D The gas occupies a larger volume at higher temperature.19. Consider the reaction below. 2P + 2Q → R The rate of reaction increases eight times when the concentrations of P and Q are doubled. Which of the following rate law is not correct?A Rate = k[R][P][Q]B Rate = k[P][Q]2C Rate = k[P]3D Rate = k[Q]320. Which of the following statements is not true?A The rate of a chemical reaction does not depend on the amount of catalyst used.B The rate of exothermic reaction increases with increasing temperature.C All endothermic reactions require heating.D All exothermic reactions can occur spontaneously.21. In a first order reaction, it takes 8.2 minutes for the concentration of the reactant to decrease from 0.085 mol dm–3 to 0.055 mol dm–3. The half-life ofthe reaction isA 6.2 minutesB 8.2 minutesC 13.0 minutesD Insufficient data22. The half-life of a first order reaction 2.3 × 103 s. Which statement is true about the reaction? A The activation energy for the reaction is high.B The rate constant is 3.0 × 10–4 s–1.C The rate is 3.0 × 10–8 mol dm–3 s–1 when the concentration is 0.010 mol dm–3.D The rate constant is independent of temperature.23. The graphs of initial rate versus concentration for the reaction W + X → products are shown below: Rate[W] Rate[X] What is the order of reaction with respect to Wand X?W XA 1 1B 1 2C 0 2D 2 1