#### Transcript Topic 6 - uaschemistry

Topic 6 REACTION RATES IB Core Objective 6.1.1 Define the term rate of reaction. 6.1.1 Define the term rate of reaction. Different chemical reactions occur at different rates (ie. speeds) Rapid reactions, eg. The neutralisation of a strong acid by a strong base in aqueous solution Slow reactions, eg. Rusting of iron Rate of a chemical reaction is a measure of the speed at which products are formed, measured as the change in concentration divided by the change in time (units = mol dm-3s-1) IB Core Objective 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. Different changes can be measured vs. time to give a rate: Rate = Change/time Rate = Δ[Concentration]/ Δ[time] Rate = Δ[Pressure]/ Δ[time] Rate = Δ[Absorbance/ transmittance]/ Δ[time] Rate = Δ[pH]/ Δ[time] Reaction rate describes how fast a reaction will take place. 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. What factors will determine rate? Surface area (usually refers to solid) Concentration or pressure (if gas) Temperature Catalyst How is it found? Through experimentation only Rate of decreasing reactants −Δ[Concentration]/ Δ[time] Rate of increasing products Δ[Concentration]/ Δ[time] Notice the negative sign 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. pH changes (pH meter or acid/base titration) Volume, pressure or mass changes Conductivity (aq) (Conductivity meter or titration) Spectrometer or colorimeter for colour change 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. Data loggers and probes Conductivity Acid/base Temperature Titrations Acid/ base Concentration 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. TITRATION Removal of small samples from the reaction mixture at different times and then titrating the sample to determine the concentration of either one of the reactants or one of the products at this time. Results can then be used directly to generate a graph of concentration against time. Best for quite slow reactions 6.1.2 Describe suitable experimental procedures for . measuring rates of reactions Spectrometer Absorbance and transmittance Measures concentration Collecting gas 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. COLLECTION OF EVOLVED GAS Gas produced in the reaction is collected either in a gas syringe or in a graduated vessel over water. The volume of gas collected at different times can be recorded. This technique is limited to reactions that produce a gas (obviously!) PLUS if the gas is to be collected over water, the gas must not be water soluble. An alternative technique is to carry out the reaction in a vessel of fixed volume and monitor the increase in the gas pressure 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. COLLECTION OF EVOLVED GAS Example: Measuring the rate of reaction between a moderately reactive metal (such as zinc) and an acid (such as hydrochloric acid). Zn (s) + 2 H+ (aq) → Zn2+ (aq) + H2 (g) 6.1.2 Describe suitable experimental procedures for measuring rates of reactions. MEASUREMENT OF THE MASS OF REACTION MIXTURE The total mass of the reaction mixture will only vary if a gas is evolved. The gas should have a high molar mass (ie. not hydrogen) so there is a significant change in mass Example: measuring the rate of reaction between a metal carbonate (such as calcium carbonate, marble chips) and an acid (such as hydrochloric acid) by measuring the rate of mass loss resulting from the evolution of carbon dioxide: CaCO3 (s) + 2H+ (aq) → Ca2+ (aq) + H2O (l) + CO2 (g) IB Core Objective 6.1.3 Analyse data from rate experiments. Students should be familiar with graphs of changes in concentration, volume, and mass against time. 6.1.3 Analyse data from rate experiments. Numerical value will vary according to amount of the substance involved in the stoichiometric equation MnO4- (aq) + 8 H+ (aq) + 5 Fe2+ (aq) → Mn2+ (aq) 4 H2O (l) + 5 Fe3+ (aq) The rate of appearance of Fe3+ is five times as great as the rate at which MnO4- is consumed Rate = - ∆ [MnO4-] = 1 ∆ [Fe3+] ∆t 5 ∆t Units: mol dm-3 s-1 6.1.3 Analyse data from rate experiments. Or more simply for a reaction: a A → b B, then Rate = - 1 ∆ [A] = 1 ∆ [B] a ∆t b ∆t General Formula Any property that differs between the reactants and products can be used to measure the rate of the reaction 6.1.3 Analyse data from rate experiments. To find rate at an instant in time (instantaneous rate) we calculate the slope of a tangent on the experimentally obtained graph. Here, Δx is the change in time and Δy is the change in concentration. IB Core Objective How does kinetic 6.2.1 Describe the kinetic theory terms of the energyinand contact movement of particles whose differ average energy is for solids proportional to temperaturewhen in kelvins. compared to liquids and gases? As heat is supplied to a substance, the velocity (kinetic energy) of the particles will increase. When the velocity increases, so does the temperature. Therefore, the absoluteSolids temperature are fixed in kelvin is proportional to the average kinetic energy of all and only vibrate, the particles. so kinetic energy is limited IB Core Objective 6.2.2 Define the term activation energy, Ea. The minimum amount of energy required for reaction is the activation energy, Ea. 6.2.2 Define the term activation energy, Ea. • Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. IB Core Objective 6.2.3 Describe the collision theory Just because chemicals collide /interact does not mean they will react! Reaction rate depends on: Collision frequency Number of particles with E ≥ Ea. Appropriate collision geometry or orientation. IB Core Objective 6.2.4 Predict and explain, using collision theory, the qualitative effects of particle size, temperature, concentration and pressure on the rate of a reaction. 6.2.4 Surface Area By increasing the surface area we increase the contact area. Collision rate will increase. Marble Marble chip 6.2.4 Concentration & Pressure Increasing the concentration or pressure will increase opportunity for molecules to react Collision rate will increase Pressure only affects gases 6.2.1 and 6.2.4 Temperature and Rate • Temperature is directly related to kinetic energy or particle speed. • Faster particles increases probability of molecule interaction • Will increase # of particles with sufficient energy to over come activation energy. Orientation IB Core Objective 6.2.6 Describe the effect of a catalyst on a chemical reaction. What do you know about catalysts? How would you define one in your own terms? Have you heard the expression “catalyst for change”? Catalysts 6.2.6 Catalysts provide are just an a facilitator alternate reaction pathway. Theythat are takes not used One lessup energy. Similar to a tunnel a mountain side, whether you go up through They lower the activation energy allowing forand more over, or go still end up atrequirement the same place. particles tothrough have theyou correct energy Analogy Catalysts are like a dating service They bring compatible people together They are not involved in what happens after the pairs are together They can be reused again and again. This is not to say that people will not get together on their own, however it lowers the energy required to find a match 6.2.6 Describe the effect of a catalyst on a chemical reaction. One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break. Draw an Energy Diagram for a catalyzed reaction IB Core Objectives 6.2.5 Sketch and explain qualitatively the Maxwell-Boltzmann energy distribution curve for a fixed amount of gas at different temperatures and its consequences for changes in reaction rate. Students should be able to explain why the area under the curve is constant and does not change with temperature. Maxwell–Boltzmann Distributions 6.2.5 Activation Energy Maxwell–Boltzmann Distributions 6.2.5 As the temperature increases, the curve flattens and broadens. At higher This system has a fixed number of particles temperatures, a larger number of molecules has higher energy. Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. • As a result, the reaction rate increases. Be sure to draw the higher temp. Curve shorter and wider than the original. IB Core Objective 6.2.7 Sketch and explain Maxwell-Boltzmann curves for reactions with and without catalysts. Catalysts and Activation Energy 6.2.7 Is this a MaxwellBoltzmann distribution curve? Why or why not? Sketch what a MaxwellBoltzmann curve would look like. 0. 693 1 [lnA12] 1 t1 / 2 k t0 [[A A]] t k2k [kA] 0 [ kA] 0 HL Rate Law Return Order in [A] Rate Law Integrated Form, y = mx + b Straight Line Plot Half-Life t1/2 zeroth order (n = 0) rate = k [A]o= k [A]t = - k t +[A]o [A]t vs. t (slope = - k) t½ = [A]0/2k ln[A]t vs. t t½ = ln 2/k first order (n = 1) second order (n = 2) rate = k [A]1 rate = k [A]2 ln[A]t = - k t + ln[A]o 1/[A]t = kt + 1/[A]o (slope = - k) 1/[A]t vs. t (slope = k) t½ = 1/(k[A]o) HL Objective 16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant. 16.1.1 Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant. Rate Rxn = Δ[A]/ Δt ‘k’ represents a constant and does not change EXCEPT •Temperature!! •Particle size (solids) Rate = k[A]m [B]n k = rate constant A and B are reactants. m and n represent the order of reaction for each reactant. m + n = overall order of reaction HL Objective 16.1.2 Deduce the rate expression for a reaction from experimental data. 16.1.2 Deduce the rate expression for a reaction from experimental data. 1) If concentration of [A] is doubled yet no change in rate, than order = zero. [A]0 2) If doubling conc. of [A] = a doubling of rate, than order = one. [A]1 3) If doubling conc. of [A] = a quadrupling of rate, than order = 2. [A]2 16.1.2 Deduce the rate expression for a reaction from experimental data. For the reaction A + 2B → C Expt. Number 1. 2. 3. 4. [A] mol dm-3 [B] mol dm-3 Rate of formation of C mol dm-3 s-1 1 0.10 0.05 0.02 x 102 2 0.10 0.10 0.04 x 102 3 0.05 0.10 0.01 x 102 4 0.10 0.20 0.08 x 102 Deduce the order of reaction in respect to reactant A Deduce the order of reaction in respect to reactant B State what the overall order of the reaction is. Deduce the rate expression for this reaction A: 1. 2nd order. 2. 1st order 3. 2+1=3 3rd order 4. rate = k[A]2[B] IB HL Objective 16.1.3 Solve problems involving the rate expression. 16.1.3 Solve problems involving the rate expression. From the previous problem, rate = k[A]2[B]. To find the rate constant, plug in numbers from the results: Expt. Number [A] mol dm-3 [B] mol dm-3 Rate of formation of C mol dm-3 s-1 1 0.10 0.05 0.02 x 102 2 0.10 0.10 0.04 x 102 3 0.05 0.10 0.01 x 102 4 0.10 0.20 0.08 x 102 A: k= 4 x 103 mol-2 dm6 s-1 Notice the units! 16.1.3 Solve problems involving the rate expression. Units for rate constant Zero order overall: mol dm-3 s-1 First order overall: s-1 Second order overall: mol-1 dm3 s-1 Third order overall: mol-2 dm6 s-1 IB HL Objective 16.1.4 Sketch, identify and analyse graphical representation for zero-, first- and secondorder reactions. Note: Students should be familiar with both concentration-time and rate-concentration graphs. 16.1.4 Sketch, identify and analyse graphical representation for zero-, first- and second-order reactions. Concentration Order of Reaction Zero order reaction doubled. The rate: Does not change Triples (x3) x9 Rate Concentration st order reaction 1 Does not change Zero 2nd order reaction Doubles (x2) 1st Quadruples (x4) 2nd Concentration tripled. The rate: Time Time 16.1.4 Sketch, identify and analyse graphical representation for zero-, first- and second-order reactions. Zero order = linear. Decrease in concentration does not affect the rate of reaction. First order = exponential decay (same as radioactive decay). (t1/2) Second order = parabolic, because it depends on the square of the concentration. Other Useful Information FOR 1ST ORDER REACTIONS ONLY Half life: The time it takes for half a substance to decay/ disappear t1/2 = ln(2)/k Concentration / Pressure Calculate the half life for the previous question Half life IB HL Objective 16.2.1 Explain that reactions can occur by more than one step and that the slowest step determines the rate of reaction (ratedetermining step). 16.2.1 Explain that reactions can occur by more than one step and that the slowest step determines the rate of reaction (ratedetermining step). Several reactions may have simple equations, but sometimes there are several intermediate steps which occur to get to the final product(s). These various intermediate steps can occur at different rates. The slowest step is the rate-determining step. 16.2.1 Explain that reactions can occur by more than one step and that the slowest step determines the rate of reaction (ratedetermining step). Unimolecular: One species breaks up or rearranges to form products. Bimolecular: Two species collide and interact to form the product. (Doubling the concentration of either will double the collision rate. What would be the overall order for this?) Molecularity Reaction Rate Law Uni A Product Rate = k[A] Bi A +B Product Rate = k[A][B] Bi A + A Product Rate = k[A]2 IB HL Objective 16.2.2 Describe the relationship between reaction mechanism, order of reaction and rate-determining step. Mechanism Must accurately represent the original stoichiometric equation 2XY2 2XY + Y2 Possible Mechanism XY + XY X Y X Y X + 2Y X + Y 2XY 2 2 2 2 4 2 2 2 4 2 Determining the Slowest Step 1) Determine the overall reaction equation 2) Check to see if adding the reactions = the overall stoichiometric equation 3) Check for consistency with regards to the rate equation and the rate determining step Does the proposed rate equation = the overall reaction equation? Question Through experimentation it was found that overall rate equation for: NO2(g) + CO(g) NO(g) + CO2(g) Rate = k[NO2]2 Found to be zero order The proposed reaction mechanism is: (Slow) NO2 + NO2 N2O4 (Fast) N2O4 + CO NO + CO2 + NO2 NO2(g) + CO(g) NO(g) + CO2(g) Does the sum match the original stoichiometric equation? Question Continued The proposed reaction mechanism is: (Slow) NO2 + NO2 N2O4 (Fast) N2O4 + CO NO + CO2 + NO2 NO2(g) + CO(g) NO(g) + CO2(g) Does the RDS rate eqn = the overall rate eqn? RDS rate = k[NO2][NO2] k[NO2]2 Appears to be consistent with the original rate equation Question 2 2NO(g) + 2H2(g) → N2(g) + 2H2O(g) rate = k[H2][NO]2 State and explain whether this mechanism agrees with the experimental rate expression in H2 + NO X X + NO Y + H2O Y + H2 N2 + H2O fast step slow step fast step NO is part of two steps, therefore must be considered in the rate expression. Question 3 2H2O2 Br- cat. 2H2O + O2 Which of the following mechanisms are correct? Mech. 1 H2O2 2OH 2OH + Br- BrO- + 2H2O2 2H2O2 BrO- 2H2O + O2 + BrMech. 2 H2O2 + Br- H2O + BrOBrO- + H2O2 H2O + O2 + BrA: Mech 2 IB HL Objective 16.3.1 Describe qualitatively the relationship between the rate constant (k) and temperature (T). 16.3.2 Determine activation energy (Ea) values from the Arrhenius equation by a graphical method. 16.3.1 Describe qualitatively the relationship between the rate constant (k) and temperature (T). The Arrhenius Equation is k = Ae(-Ea/RT). This equation can be found in your data booklet. A is the Arrhenius constant. It is dependent on collision rate and steric factors (geometry of the colliding particles). This expression indicates that the rate constant k depends exponentially on temperature. Temperature has a large effect on the reaction rate. Arrhenius Equation (16.3) The relationship between rate constant and temperature. The equation of a straight line k = Ae(-Ea/RT) can be determined by taking the natural log. Activation Energy can be found by varying the temperature of an experiment lnk = lnA –(Ea/R)1/T Arrhenius Equation (16.3) Graphing ln k VS 1/T will give a straight line Slope will = -Ea/R ln k Extrapolate to find: ln [A] 1/T