CHEMSHEETS.co.uk 23-Feb-2016 Chemsheets A2 1001 Page
www.CHEMSHEETS.co.uk23-Feb-2016Chemsheets A2 1001Page 1
SECTION 1 – Recap of AS KineticsWhat is reaction rate? The rate of a chemical reaction is a measure of how fast a reaction takes place. It is defined as the change in concentration of a reactant or product per unit time. For example, in a reaction between magnesium and hydrochloric acid, the rate could be measured in terms ofthe change in concentration of the hydrochloric acid per second.Collision theory For particles to react they must collide with sufficient energy to react and at the correct orientation. The minimum energy particles need to react is called the activation energy.Maxwell-Boltzmann distribution The particles in substance do not all have thesame amount of energy. The energy of the particles in a gas is shown bya Maxwell-Boltzmann distribution.Numberofmolecules There are no particles with no energy (i.e. they allhave some energy). The area under the curve represents the totalnumber of particles. The peak of the curve gives the energy that moreparticles have than any other energy (sometimeslabelled Emp, i.e. most probable energy).EmpEaEnergy The activation energy (Ea) can be shown on the axis. Often only a small proportion of molecules have energygreater than or equal to the activation energy. When particles collide with each other they gain or lose energy, and so the energy of each individual particle isconstantly changing. This means that a particle that does not have enough energy to react may gain enough ina collision so that it can react in a further collision. www.CHEMSHEETS.co.uk23-Feb-2016Chemsheets A2 1001Page 2
As the temperature changes, the averageenergy of the particles increases and so thedistribution curve changes. The total areaunder the curve stays the same as there arethe same number of particles, and so the peakwill move to higher energy but less particles willhave that energy.NumberofmoleculesLower temperatureHigher temperatureEmp EmpLower HigherTTEaEnergyFactors affecting ratesFactorEffectReasonConcentration of asolutionThe higher the concentration, thefaster the reaction (unless thatreagent is zero order)Particles are closer together and so thereare more frequent successful collisionsSurface area of a solidThe more pieces a solid is brokenup into, the greater the surfacearea (a powder has a massivesurface area).There are more particles exposed at thesurface that can be collided with, and sothere are more frequent successful collisionsThe greater the surface area, thefaster the reaction.Pressure of gases(gases that arereactants)The higher the pressure, the fasterthe reaction (unless that reagent iszero order)Particles are closer together and so thereare more frequent successful collisionsTemperatureThe higher the temperature, thefaster the reaction.Particles have more energy and so a greaterproportion of the collisions are successful.The particles also move faster and socollisions are more frequent. Therefore thereare more frequent successful collisions.CatalystA catalyst is a substance thatincreases the rate of a reaction butis not used up.It provides an alternative route / mechanismwith a lower activation energy and thereforea greater proportion of the collisions aresuccessful. www.CHEMSHEETS.co.uk23-Feb-2016Chemsheets A2 1001Page 3
SECTION 2 – Finding rates using gradients Graphs can be plotted to show how the concentration of a reactant or product changes during the course of areaction. The gradient of this graph at any time equals the rate of reaction at that time. The steeper the line, the faster the reaction. We can calculate the gradient by drawing a tangent to the line at any point (tip – make the gradient line quitelong – if it is too short then there will be a larger degree of error in the gradient calculation stemming fromreading the values) In the graph below, the rate at 0 seconds (the rate at the start – the initial rate) is 0.019 mol dm s . The rate-3 -1after 40 seconds is 0.0048 mol dm s .-3-1 Most reactions start faster and slow down due to there being less reactant particles as time goes on and soless frequent successful collisions between reactant particles.0.60.5 Gradient (at time 0 s)Concentration (mol/dm3) change in vertical 0.5 0.019 mol dm-3 s-1change in horizontal 260.40.3Gradient (at time 40 s) change in vertical 0.28 0.0048 mol dm-3 s-1change in horizontal 580.20.1 00102030405060708090100110120time (s) www.CHEMSHEETS.co.uk23-Feb-2016Chemsheets A2 1001Page 4
TASK 1 – Finding rates using gradients1 Magnesium reacts with hydrochloric acid to form hydrogen gas. The data below shows how the hydrochloricacid concentration varied with time.Time (s)-3[HCl] (mol dm 0.06a) Plot a graph of acid concentration against time.b) By drawing gradients, calculate the rate at 0, 10, 20 and 30 seconds.c) Explain why the rate changes during the reaction as it does.2 Sodium reacts with water to form sodium hydroxide solution and hydrogen gas. The data below shows howthe concentration of sodium hydroxide solution varies with time.Time (s)-3[NaOH] (mol dm 0.56a) Plot a graph of sodium hydroxide concentration against time.b) By drawing gradients, calculate the rate at 0, 20 and 40 seconds. www.CHEMSHEETS.co.uk23-Feb-2016Chemsheets A2 1001Page 5
SECTION 3 – Introduction to rate equations and ordersConsider any reaction where one of the reactants is A. By experiment the rate of the reaction can be related tothe concentration of A:rate [A]nwhere [] means concentration in mol dm-3n is called the order of reaction with respect to that reactant, and is usually 2, 1 or 0OrderZero orderFirst orderSecond ordern012The effectthe reaction rate is proportional0to [A] ( 1) and so the rate isnot affected by changes in [ A]the reaction rate is proportionalto [A]the reaction rate is proportional2to [A][A] x 2rate unchangedrate x 2rate x 2 (i.e. x 4)[A] x 3rate unchangedrate x 3rate x 3 (i.e. x 9)[A] x 10rate unchangedrate x 10rate x 10 (i.e. x 100)[A] is 2rate unchangedrate 2rate 2 (i.e. 4)2222x A y B productsConsider a reaction between A and B:For any reaction the effects of the concentration of each individual reactant can be combined into a rateequation, which for the above reaction would be of the form:reaction rate [A] [B]mmnreaction rate k [A] [B]n(k rate constant)e.g. for the above rate equation: order of reaction with respect to:A is m, B is n,overall order is (m n)The order of reaction with respect to a given reactant is the power to which the concentration of thereactant is raised in the rate equation.The overall order of reaction is the sum of the powers of the concentration terms in the rateequation. The values of m and n must be found experimentally and cannot be found from looking at the equation (i.e.m and n are nothing to do with x and y) - so the rate equation can only be found by experiment. Rate equations provide a lot of information about the mechanism for the reaction. In some reactions, a catalyst appears in the rate equation (or even one of the products if the product is acatalyst for that reaction – autocatalysis). k is the rate constant for a reaction (k is different for different reactions). The only thing that affects the value of k is temperature (k increases with temperature). The units of k depend on the order of the reaction (note rate is usually in mol dm s , concentrations are in-3mol dm ).-3 www.CHEMSHEETS.co.uk23-Feb-2016-1Chemsheets A2 1001Page 6
TASK 2 – What orders mean1 Consider this reaction:A B C DThe rate equation is:rate k [B] [C]2 Consider this reaction:The rate equation is:P Q R2rate k [P] [T]Complete the table to show how changing the concentrations affects the rate.2Complete the table to show how changing the concentrations affects the rate.Initial rate-3 -1(mol dm s )Change in concentrationof reagents6.0[P] x 25.0[Q] x 510.0[R] x 3Initial rate-3 -1(mol dm s )Change in concentrationof reagentsEffect on rateNew initial rate-3 -1(mol dm s )2.5[A] x 3None2.50.75[B] x 40.80[T] x 412[C] x 108.0[P] 30.50[D] x 512.5[Q] 20.25[A] 460[R] 52.8[B] 1050[T] 103.5[C] 312[P] x 2, [Q] x 210[P] x 2, [T] x 30.80[D] 240[Q] x 2, [T] 310.3[A] x 2, [B] x 225[R] x 3, [T] 26.5[B] x 2, [C] x 310[P] x 4, [Q] 2, [T] x 212.5[A] x 2, [B] 320[P] x 2, [Q] x 10, [T] 1.54.8[B] x 3, [C] 230[P] x 3, [Q] x 10, [T] 1012.5[A] x 6, [B] 4, [C] x 25[P] 2, [Q] 2, [T] 3[A] x 2, [B] x 10, [C] 1.512[P] x 2, [Q] 10, [T] x 52.916[P] 3, [Q] 2.5, [T] x 315.5[B] x 3, [C] x 10, [D] 108[P] x 2.5, [Q] 4, [T] 2 www.CHEMSHEETS.co.ukT acts as catalyst23-Feb-2016Effect on rateNew initial rate-3 -1(mol dm s )Chemsheets A2 1001Page 7
TASK 3 – Finding the units of the rate constantWork out the units for the rate constant in each of the following examples.Rate equation1)rate k [A]2)rate k [C] [H]3)rate k [S]4)rate k [J] [H]5)rate k [T]6)rate k [S] [E] [G]7)rate k [D] [C]8)rate k [A] [B] [C]9)rate k [J] [G]10)rate k [H ] [Br ]Rearrange to give kWorkingk rate[A]k (mol dm ) s-3(mol dm )-3Units for k-1s-122222 www.CHEMSHEETS.co.uk-23-Feb-2016Chemsheets A2 1001Page 8
SECTION 4 – Find orders using initial rates dataThe order with respect to each reagent, and so the rate equation, can be found by doing a series of experimentswhere the initial concentration of each reagent is changed to see how it affects the initial rate.Example 11Initial [R]-3(mol dm )1.0Initial [S]-3(mol dm )1.0Initial rate-3 -1(mol dm s )0.20022.01.00.40031.02.00.200ExperimentImagine a reaction where R reacts with S to makesome products:R S productsIf we compare experiments 1 and 2: The concentration of R has been doubled, but the concentration of S remains the same. The rate has doubled. Therefore the effect of doubling the concentration of R is to double the rate, and so therefore the reaction isfirst order with respect to the concentration of RIf we compare experiments 1 and 3: The concentration of S has been doubled, but the concentration of R remains the same. The rate has stayedthe same. Therefore doubling the concentration of S has no effect on the rate, and so therefore the reaction is zero orderwith respect to the concentration of STherefore, the rate equation is:rate k[R]We can then work out the rate constant using any of the experiments:Using experiment 2:k rate 0.400 0.200 s[R]2.0Example 2 rate 0.200 [R]-14Initial [T]-3(mol dm )1.0Initial [U]-3(mol dm )1.0Initial rate-3 -1(mol dm s )0.50050.51.00.25062.02.04.000ExperimentImagine a reaction where T reacts with U to makesome products:T U productsIf we compare experiments 4 and 5: The concentration of T has been halved, but the concentration of U remains the same. The rate has halved. Therefore the effect of halving the concentration of T is to halve the rate, and so therefore the reaction is firstorder with respect to the concentration of UThere is no pair of experiments where only the concentration of U has been changed, so we will have to use thefact that we have just worked out that the reaction is first order with respect to the concentration of U to do this.If we compare experiments 4 and 6: The concentration of T has been doubled, and this on its own would double the rate to 1.000 mol dm s .-3-1 The concentration of U has also been doubled, and this has effectively increased the rate from 1.000 to 4.000-3 -1mol dm s – in other words it has made it 4 times bigger. Therefore the effect of doubling the concentration of U is to make the rate 4 times faster, and so therefore thereaction is second order with respect to the concentration of UTherefore, the rate equation is:rate k[T][U]2We can then work out the rate constant using any of the experiments:Using experiment 6: www.CHEMSHEETS.co.ukk -26rate 4.00 0.500 mol dm s22[T][U]2.0 x 2.023-Feb-2016-1 rate 0.500 [T][U]2Chemsheets A2 1001Page 9
TASK 4 – Finding rate equations using initial rates datae) P reacts with Q in the presence of an acid catalyst.1Deduce the rate equation for each of the following reactions.1Initial [A]-3(mol dm )3Initial [B]-3(mol dm )1Initial rate-3 -1(mol dm s )102614033210Experimentb) D reacts with E to form F and G.Initial [Q]-3(mol dm )Initial [H ]-3(mol dm 20.080.132Initial [D]-3(mol dm )Initial [E]-3(mol dm )Initial rate-3 -1(mol dm s )40.10.1550.40.12060.80.28017Initial [X]-3(mol dm )0.05Initial rate-3 -1(mol dm s )0.016180.050.050.100190.100.150.400g) M reacts with N to make O.c) J reacts to form K and L in the presence of an acid catalyst.Experiment7Initial [J]-3(mol dm )0.2 Initial [H ]-3(mol dm )0.5J K L80.20.25390.11.024M N 2O20Initial [M]-3(mol dm )0.1Initial [N]-3(mol dm )0.2Initial rate-3 -1(mol dm s )0.15210.30.21.35220.40.32.40ExperimentInitial rate-3 -1(mol dm s )12h) HI dissociates to form hydrogen and iodine:d) S reacts with T to form U.S T 2UExperiment-310Initial [S]-3(mol dm )0.005Initial [T]-3(mol dm )0.2Initial rate-3 -1(mol dm s )300110.0010.3450120.0100.2300Experiment www.CHEMSHEETS.co.ukInitial [HI] (mol dm )-3-1Initial rate (mol dm s )223-Feb-2016)Initial [W]-3(mol dm )0.02ExperimentExperiment12W X Y Zf) W reacts with X to form Y and Z.D 2E F GInitial rate-3 (mol dm s Initial [P]-3(mol dm )ExperimentA B Ca) A reacts with B to form C.P Q 2R2 HI(g) H2(g) I2(g)232425261.643.284.926.560.411.643.696.56For each of the rate equations you deduced in question 1, calculate therate constant and deduce its units.Chemsheets A2 1001Page 10
TASK 5 – Using initial rates data1 G reacts with H:G 2 H productsComplete the table with initial rates data.rate k[G][H]ExperimentInitial [G](mol dm-3)Initial [H](mol dm-3)Initial rate(mol dm-3 00050.1500.10060.8000.80070.3001.20080.2500.4002 J reacts with K:J K productsComplete the table with initial rates data.rate k[J]Initial [J](mol dm-3)Initial [K](mol dm-3)Initial rate(mol dm-3 00.1800.100rate k[L][M]Initial [M](mol dm-3)Initial rate(mol dm-3 s-1)140.1000.1000.0500150.1000.300160.100 www.CHEMSHEETS.co.uk0.200Initial [A](mol dm-3)Initial [B](mol dm-3)Initial rate(mol dm-3 s-1)1911220128212216D E productsExperimentInitial [D](mol dm-3)Initial [E](mol dm-3)Initial rate(mol dm-3 s-1)22110.2023210.2024440.80P(g) Q(g) productsExperimentInitial [P](mol dm-3)Initial [Q](mol dm-3)Initial rate(mol dm-3 80.40.20.01598 Initial [L](mol dm-3)18Experimentc) P reacts with Q to form various products.2Experimentd) The reaction between iodine and propanone is catalysed by H (aq) ions. I2(aq) CH3COCH3(aq) CH3COCH2I(aq) H (aq) I (aq)0.15000.200A B Cb) D reacts with E to form various products.0.1003 L reacts with M:2L 2M N 2PComplete the table with initial rates data.17a) A reacts with B to form C.2Experiment134 In each of the following questions, use the initial rate data to find: the rate equation the rate constant k (including its units) at the experiment’s l [H ](mol dm-3)Initial [CH3COCH3](mol dm-3)Initial [I2](mol dm-3)Initial rate(mol dm-3 s-1)290.56.00.029.0 x 10-5300.58.00.041.2 x 10-4311.06.00.021.8 x 10-4321.08.00.022.4 x 10-4Chemsheets A2 1001Page 11
SECTION 5 – Rate determining step The mechanism for a reaction may only involve one step, or could actually involve a series of steps. If a reaction mechanism has a series of steps, then the overall rate is dependent on the rate of slowest step. The slowest step is called the rate determining step (RDS). In any mechanism, species that are only involved in steps after the RDS do not appear in the rate equation. However,species that appear in the steps up to and including the RDS are in the rate equation. The rate equation gives us clues about the mechanism and which is the rate determining step. We find the rate equation by experiment and then use the rate equation to deduce possible mechanisms (we do not use themechanism to work out the rate equation!). A substance that acts as a catalyst will appear in the rate equation but not in the overall chemical equation for the reaction.E.g. 1Overall equationRate equationPossible mechanism that fits rate equationA 2B C Drate k[A][B]Step 1A B Pslow, RDSStep 2P B C DfastUp to and including the RDS: one particle of A and oneparticle of B are involved.E.g. 2Overall equationA 2B D ERate equationrate k[B]2Possible mechanism that fits rate equationStep 12B B2slow, RDSStep 2B2 A D EfastUp to and including the RDS: two particles of B are involved.E.g. 3Overall equationRate equationA 2B D E2rate k[A][B]Possible mechanism that fits rate equationStep 12B B2fastStep 2B2 A D Eslow, RDSUp to and including the RDS: one particle of A and twoparticles of B are involved.E.g. 4Overall equationA 2B D E H acts as a catalystRate equation rate k[A][H ]Possible mechanism that fits rate equationStep 1A H AHStep 2AH B AB HStep 3AB B D E slow, RDS fastfastUp to and including the RDS: one particle of A and one particle of H are involved. Note that H is a catalyst as it appears in the rate equationbut not the overall reaction equation. www.CHEMSHEETS.co.uk23-Feb-2016Chemsheets A2 1001Page 12
TASK 6 – Rate determining stepIn each of the following reactions, deduce which step is the rate determining step.1236Overall reaction:A 2B DRate equation:rate k [B]Mechanism:step 12B Cstep 2C A DOverall reaction:In each of the following reactions, state which of the proposed mechanisms could be apossible mechanism and which could not be a correct mechanism. Do not assume thatthere is one answer in each case!2E 2F GOverall reaction:P 2Q RRate equation:rate k [P]Possible mechanism 1:step 1 (rds)P Q PQstep 2PQ Q Rstep 12 Q Q2step 2 (rds)Q2 P RPossible mechanism 2:2Rate equation:rate k [E] [F]Mechanism:step 1E F EFstep 2EF F G7Overall reaction:J K LRate equation:rate k [J] [H ]Mechanism:step 1J H JHstep 2JH K L H Overall reaction:X Y Z BRate equation:rate k [X] [Y]Possible mechanism 1:step 1X A Bstep 2 (rds)A Y Zstep 1 (rds)X Q P Bstep 2Y P A Q Possible mechanism 2:45Overall reaction:M2 2 N 2 MNRate equation:rate k [M2]Mechanism:step 1M2 2 Mstep 2M N MNOverall reaction:2A C DRate equation:rate k [A] [C]Mechanism:step 12A Bstep 2B C D www.CHEMSHEETS.co.uk8Overall reaction:2C D FRate equation:rate k [C] [D]Possible mechanism 1:step 1C D Estep 2 (rds)C E Fstep 12 C C2step 2 (rds)C2 D F2Possible mechanism 2:23-Feb-20162Chemsheets A2 1001Page 13
SECTION 6 – The Arrhenius EquationThe Arrhenius equation shows the link between the rate constant, activation energy and temperature.Ealn k ln A RTEa RTk AekAEaRT rate constant (units vary depending on order of reaction) Arrhenius constant (same units as k) (also called pre-exponential or frequency factor)-1 activation energy (J mol )-1 -1 gas constant (8.31 J mol K ) temperature (K)[Note that e is a very special number in maths. It can be used on a calculator by using the SHIFT ln buttons.]We can do various calculations using the equation, including some to show how changes in temperature andthe addition of a catalyst affects the rate constant (and remember – the bigger the rate constant, the faster thereaction).Example 1 – Finding the activation energy for a reactionln k ln A -1Calculate the activation energy for a reaction in kJ mol given5-13 -1the Arrhenius constant (A) 5.25 x 10 mol dm s and the-4-13 -1rate constant (k) 3.36 x 10 mol dm s at 298 K.EaRTEa ln A ln kRTEa RT (ln A ln k)Ea (8.31 x 298) (13.17 8.00)Ea 52425 J
www.CHEMSHEETS.co.uk 23-Feb-2016 Chemsheets A2 1001 Page 3 As the temperature changes, the average energy of the particles increases and so the
www.CHEMSHEETS.co.uk 14-Jul-12 Chemsheets A2 009 7 TASK 3 – pH of strong bases 1) Calculate the pH of the following solutions. a) 0.15 mol dm-3 KOH b) 0.05 mol .
www.CHEMSHEETS.co.uk 10-Mar-2016 Chemsheets A2 1014 Page 7 Example 2 Ethanol has the formula C 2 H 5 OH and is used as a fuel (e.g. for cars in Brazil). It burns .
www.CHEMSHEETS.co.uk 10-Jul-12 Chemsheets AS 029 3 CALORIMETRY The enthalpy change for a reaction can be found by measuring the temperature change in a reaction.File Size: 1MB
a) [Co(H 2O) 6] 2 Co 2 e) [FeO 4] 2– Fe 6 i) K 2[CoCl 4] Co 2 b) [CrCl 6] 3– Cr 3 f) [Mn(CN) 6] 4– Mn 2 j) K 3[AuF 6] Au 3 c) [Co(NH 3) 6]Cl 2 Co 2 g) [Ni(CO) 4] Ni 0 k) (NH 4) 2[IrCl 6] Ir 4 d) [Co(NH 3) 5Cl]Cl 2 Co 3 h) [Ni(EDTA)] 2– Ni 2 l) Na[Mn(CO) 5] Mn -1 TASK 2 – Writing half equations a) 2I– I 2 2e – b .
metal carbonate metal oxide carbon dioxide (on heating) CaCO 3 CaO CO 2 TASK 4 – WRITING BALANCED EQUATIONS 1) Balance the following equations.
www.CHEMSHEETS.co.uk 16-July-2016 Chemsheets A2 1078 b) By reduction of nitrile compounds This is the commonest way as it only forms one amine rather than a mixture.
Jan 30 Last day of Ukulele Kids session 1 Jan 30 Dental Screening (K & Gr. 2) Jan 30 Girls Basketball Tournament Feb. 6 Pasta Night Feb 13 School Valentine’s Day celebrations Feb 14 PA Day Feb 17 Family Day Feb 19 Confirmation Retreat Gr. 7 Feb 22 Gr. 7 Confirmation Feb 25 Shrove Tuesday/ Pancake Tuesday Feb 26 Ash Wednesday
ASTM C167 Standard test methods for thickness and density of blanket or batt thermal insulations ASTM C518 Standard test method for steady-state thermal transmission properties by means of the heat flow meter apparatus . TL-205 HOME INNOVATION RESEARCH LABS Page 6 of 6. ASTM C653 Standard guide for determination of the thermal resistance of low-density blanket-type mineral fiber insulation .
