1 / 87

930 likes | 1.67k Views

Vanessa N. Prasad-Permaul CHM 1046 Valencia Community College. Chapter 13: rates of reaction. Introduction. Chemical kinetics is the study of reaction rates 2. For a chemical reaction to be useful it must occur at a reasonable rate

Download Presentation
## Chapter 13: rates of reaction

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**Vanessa N. Prasad-Permaul**CHM 1046 Valencia Community College Chapter 13: rates of reaction**Introduction**• Chemical kinetics is the study of reaction rates 2. For a chemical reaction to be useful it must occur at a reasonable rate 3. It is important to be able to control the rate of reaction 4. Factors that influence rate a) Concentration of reactants (molarity) b) Nature of reaction, process by which the reaction takes place c) Temperature d) Reaction mechanism (rate determining step) e) Catalyst**Reaction rate**Reaction rate- Positive quantity that expresses how the conc. of a reactant or product changes with time 2N2O5(g) 4NO2(g) + O2(g) 1. [ ] = concentration, molarity, mole/L 2. [N2O5] decreases with time 3. [NO2] increases with time 4. [O2] increases with time 5. Because of coefficients the concentration of reactant and products does not change at the same rate 6. When 2 moles of N2O5 is decomposed, 4 moles of NO2 and 1 mole of O2 is produced**2N2O5(g) 4NO2(g) + O2(g)**-[N2O5] = [NO2] = [O2] 2 ½ -[N2O5] because it’s concentration decreases, other positive because they increase Rate of reaction can be defined by dividing by the change in time, t rate = - [N2O5] = [NO2] = [O2] t 2t ½ t**Reaction Rates**Generic Formula: aA + bB cC + dD rate = - [A] = - [B] = [D] = [C] at bt dt ct**RELATING THE DIFFERENT WAYS OF EXPRESSING REACTION RATES**EXAMPLE 13.1: CONSIDER THE REACTION OF NO2 WITH F2 TO GIVE NO2F 2NO2(g) + F2(g) 2NO2F(g) HOW IS THE RATE OF FORMATION RELATED TO THE RATE OF REACTION OF F2? NO2F = [NO2F] F2 = -[F2] t t [NO2F] = -[F2] 2t t RATE OF FORMATION RATE OF REACTION**RELATING THE DIFFERENT WAYS OF EXPRESSING REACTION RATES**EXERCISE 13.1: CONSIDER THE REACTION OF NO2 WITH F2 TO GIVE NO2F 2NO2(g) + F2(g) 2NO2F(g) HOW IS THE RATE OF FORMATION RELATED TO THE RATE OF REACTION OF NO2?**Calculating the average reaction rate**• EXAMPLE 13.2: CALCULATE THE AVERAGE RATE OF DECOMPOSITION OF N2O5 • N2O5(g) 2NO2(g) + 1/2O2(g) • N2O5 = -[N2O5] • t • (0.93x10-2M – 1.24x10-2M) = -0.31x10-2M = 5.2x10-6M/s • 1200s – 600s 600s RATE OF DECOMPOSITION**Calculating the average reaction rate**EXERCISE 13.2: IODIDE ION IS OXIDIZED BY HYPOCHLORITE ION IN BASIC SOLUTION. CALCULATE THE AVERAGE RATE OF REACTION OF I- DURING THIS TIME INTERVAL: I-(aq) + ClO-(aq) Cl-(aq) + IO-(aq)**Reaction Rate and Concentration**The higher the conc. of starting reactant the more rapidly a reaction takes place 1. Reactions occur as the result of collisions between reactant molecules 2. The higher the concentration of molecules, the greater the # of collisions in unit time and a faster reaction 3. As the reactants are consumed the concentration decreases, collisions decrease, reaction rate decreases 4. Reaction rate decreases with time and eventually = 0, all reactants consumed 5. Instantaneous rate, rate at a particular time 6. Initial rate at t = 0**Rate Expression and Rate Constant**Rate expression / rate law: rate = k[A] where k = rate constant, varies w/ nature and temp. [A] = concentration of A Rate law: an equation that related the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers**Order of rxn involving a single reactant**General equation rate expression A products rate = k[A]m where m=order of reaction m=0 zero order m=1 first order m=2 second order m, can’t be deduced from the coefficient of the balanced equation. Must be determined experimentally!**Order of rxn involving a single reactant**Rate of decomposition of species A measured at 2 different conc., 1 & 2 rate2 = k[A2]m rate1 = k[A1]m By dividing we can solve for m, to find the order of the reaction Rate2 = [A2]m Rate1 [A1]m(Rate2/Rate1) = ([A2]/[A1])m**Determining the order of reaction from rate law**EXAMPLE 13.3: BROMIDE ION IS OXIDIZED BY BROMATE ION IN ACIDIC SOLUTION 5Br -(aq) +BrO3-(aq) + 6H+(aq) 3Br2(aq) + 3H2O(l) Rate = k[Br -][BrO3-][H+]2 WHAT IS THE ORDER OF REACTION WITH RESPECT TO EACH REACTANT SPECIES? WHAT IS THE OVERALL ORDER OF REACTION? Br - IS FIRST ORDER BrO3- IS FIRST ORDER H+ IS SECOND ORDER THE REACTION IS FOURTH ORDER OVERALL (1+1+2)=4**Determining the order of reaction from rate law**EXERCISE 13.3: WHAT ARE THE REACTIO ORDERD WITH RESPECT TO EACH REACTANT SPECIES FOR THE FOLLOWING REACTION? WHAT IS THE OVERALL REACTION? NO2(g) + CO(g) NO(g) + CO2(g) RATE = k[NO2]2**Example 1**CH3CHO(g) CH4(g) + CO(g) Using the given data determine the reaction order**Example 1**Rate2 = [A2]m Rate1 [A1]m 4 = 2m log 4 m = 2 log 2 second order rate = k[CH3CHO]2 once the order of the rxn is known, rate constant can be calculated, let’s calculate the rate constant, k**Example 1**rate = k[CH3CHO]2 rate = .085 mol/L s conc = .10 mol/L k = rate = 0.085 mol/L s = 8.5 L/mol s [CH3CHO]2 (0.10 mol/L)2 now we can calc. the rate at any concentration, let’s try .55 M**Example 1**rate = k[CH3CHO]2 rate = 8.5 L/mol s [.55]2 = 2.6 mol/ L s Rate when [CH3CHO] = .55 M is 2.6 mol/L s**Order of rxn with more than 1 reactant**aA + bB prod rate exp: rate = k [A]m [B]n m = order of rxn with respect to A n = order of rxn with respect to B Overall order of the rxn is the sum, m + n**Key**• When more than 1 reactant is involved the order can be determined by holding the concentration of 1 reactant constant while varying the other reactant. From the measured rates you can calculate the order of the rxn with respect to the varying reactant**Example 2**(CH3)3CBr + OH-(CH3)3COH + Br- Find the order of the reaction with respect to both reactants, write the rate expression, and find the overall order of the reaction**Reactant concentration and time**Rate expression rate = k[A] Shows how the rate of decomposition of A changes with concentration More important to know the relation between concentration and time Using calculus: Integrated rate equations relating react conc. to time**For the following rate law, what is the overall order**of the reaction? Rate = k [A]2 [B] • 1 • 2 • 3 • 4**Zero Order**Zero order: A Products rate = k [A]0 – [A] = kt t½ = [A]0 2k m = 0: zero order - rate is independent of the concentration of reactant. Doubling the concentration has no effect on rate. RATE EXPRESSION HALF LIFE CONC-TIME RELATION**First Order**First Order: A Products rate = k[A] ln [A]o/[A] = kt t½ = .693/k [A]o = original conc. of A [A]= Conc. of A at time, t k = first order rate constant ln = natural logarithm m = 1: first order - rate is directly proportional to the concentration of the reactant. Doubling the concentration increases the rate by a factor of 2. RATE EXPRESSION HALF LIFE CONC-TIME RELATION**Second Order**Second order: A Products rate = k[A]2 1 – 1 = kt t½ = 1 [A] [A]0 k[A]0 m = 2: second order - the rate is to the square of the concentration of the reactant. Doubling the concentration increases the rate by a factor of 4. RATE EXPRESSION HALF LIFE CONC-TIME RELATION**Determining THE RATE LAW FROM INITIAL REACTIONS**• EXAMPLE 13.4: IODIDE ION IS OXIDIZED IN ACIDIC SOLUTION TO TRIIODIDE ION I3-, BY HYDROGEN PEROXIDE • H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + H2O(l) • OBTAIN THE REACTION ORDER WITH RESPECT TO EACH REACTANT • FIND THE RATE CONSTANT**Determining THE RATE LAW FROM INITIAL REACTIONS**• Rate = k[H2O2]m [I-]n [H+]p • (Rate) 2 = k[H2O2]m [I-]n [H+]p • (Rate) 1 = k[H2O2]m [I-]n [H+]p • Rate constant cancels • 2.30 x 10-6 = [0.020]m [0.010]n [0.00050]p • 1.15 x 10-6 = [0.010]m [0.010]n [0.00050]p • = 2m • m = 1 • doubling the H2O2 concentration doubles the rate • Rate = k[H2O2] [I-] • Reaction order = 1, 1, 0**Determining THE RATE LAW FROM INITIAL REACTIONS**Rate = k[H2O2] [I-] 1.15 x 10-6 mol = k x 0.010 mol x 0.010 mol L.s L L k = 1.15 x 10-6 mol L.s 0.010 mol x 0.010 mol L L k = 1.2 x 10-2 L/(mol.s)**Determining THE RATE LAW FROM INITIAL REACTIONS**EXERCISE 13.4: THE INITIAL-RATE METHODWAS APPLIED TO THE DECOMPOSITION OF NITROGEN DIOXIDE NO2 (aq) 2NO (g) + O2(g) FIND THE RATE LAW AND THE VALUE OF THE RATE CONSTANT WITH RESPECT TO O2 FORMATION**Example 3**The following data was obtained for the gas-phase decomp. of HI Is this reaction zero, first, or second order in HI? Hint: Graph each Conc. Vs. time corresponding to correct [A], ln [A], or 1/[A]**[HI] vs. time not linear so it is not zero order**[A] vs. TIME FOR ZERO ORDER REACTION**ln [HI] vs. time not linear so it is not first order**ln[A] vs. TIME FOR 1ST ORDER REACTION**1/[HI] vs. time is linear so it is second order**1/ln[A] vs. TIME FOR 2ND ORDER REACTION**USING AN INTEGRATED RATE LAW**EXAMPLE 13.5: HOW LONG WOULD IT TAKE FOR THE CONC. OF N2O5 TO DECREASE TO 1.00 x 10-2 mol/L FROM THE INITIAL VALUE? kt = ln [A]0 [A]t 4.80 x 10-4/s (t) = ln [1.65 x 10-2 mol/L] [1.00 x 10-2 mol/L] 4.80 x 10-4/s (t) = 0.500775 t = 0.500775 4.80 x 10-4/s t = 1043.28125s * 1min 60sec t = 17.4min**USING AN INTEGRATED RATE LAW**EXERCISE 13.5: A) WHAT WOULD THE CONCENTRATION BE OF DINITROGEN PENTOXIDE IN THE EXPERIMENT DESCRIBED IN EXAMPLE 13.5 AFTER 6.00 x 102 s? B) HOW LONG WOULD IT TAKE FOR THE CONCENTRATION OF N2O5 TO DECREASE TO 10% OF ITS INITIAL VALUE?**RELATING THE HALF-LIFE OF A HALF-REACTION TO THE RATE**CONSTANT • EXAMPLE 13.6: SULFURYL CHLORIDE, SOC2Cl2, IS A COLORLESS CORROSIVE LIQUID WHOSE VAPOR DECOMPOSES IN A FIRST-ORDER REACTION TO SULFUR DIOXIDE AND CHLORINE. • SO2Cl2(g) SO2(g) + Cl2(g) • AT 320oC, THE RATE CONSTANT IS 2.20 x 10-5/s. WHAT IS THE HALF-LIFE OF SO2Cl2 VAPOR AT THIS TEMPERATURE? • HOW LONG (IN HOURS) WOULD IT TAKE FOR 50.0% OF • THE SO2Cl2 TO DECOMPOSE? • B) HOW LONG WOULD IT FOR 75.0% OF THE SO2Cl2?**RELATING THE HALF-LIFE OF A HALF-REACTION TO THE RATE**CONSTANT EXAMPLE 13.6: t½ = 0.693/k t½ = 0.693 2.20 x 10-5/s t½ = 3.15 x 104 s The half-life is the time required for 50.0% of the SO2Cl2 to decompose. This is 8.75 hours 8.75 hrs x 2 = 17.5 hrs kt = ln [A]0 [A]t**RELATING THE HALF-LIFE OF A HALF-REACTION TO THE RATE**CONSTANT • EXERCISE 13.6: THE ISOMERIZATION OF CYCLOPROPANE TO PROPYLENE IS FIRST ORDER IN CYCLOPROPANE AND OVERALL. AT 1000oC THE RATE CONSTANT IS 9.2/s • A) WHAT IS THE HALF-LIFE? • HOW LONG WOULD IT TAKE FOR THE CONCENTRATION • OF CYCLOPROPANE TO DECREASE TO 50% OF ITS INITIAL VALUE? • C) TO 25% OF ITS INITIAL VALUE?**Activation Energy**Activation Energy: Ea (kJ) For every reaction there is a certain minimum energy that molecules must possess for collisions to be effective. 1. Positive quantity (Ea>0) 2. Depends only upon the nature of reaction 3. Fast reaction = small Ea 4. Is independent of temp and concentration**For the following reaction:**A + B C If the concentration of A is doubled, and B is constant, the rate doubles. What is the order of the reaction with respect to A? • 0 • 1 • 2**Reaction Rate and Temp**Reaction Rate and Temp • As temp increases rate increases, Kinetic Energy increases, and successful collisions increase 2. General rule for every 10°C inc. in temp, rate doubles**If I increase the temperature of a reaction from**110 K to 120 K, what happens to the rate of the reaction? • Stay the same • Doubles • Triples**The Arrhenius Equation**The Arrhenius Equation f = e-Ea/RT f = fraction of molecules having an En. equal to or greater than Ea R = gas constant A = constant T = temp in K ln k = ln A –Ea/RT plot of ln k Vs. 1/T linear slope = -Ea/R Two-point equation relating k & T ln k2 = Ea [1/T1 – 1/T2] k1 R**USING THE ARRHENIUS EQUATION**EXAMPLE 13.7: THE RATE CONSTANT FOR THE FORMATION OF HYDROGEN IODIDE FROM THE ELEMENTS IS 2.7 x 10-4L/(mol.s) @600 K AND 3.5 x 10-3L/(mol.s) AT 650 K. FIND THE ACTIVATION ENERGY Ea ln k2 = Ea [1/T1 – 1/T2] k1 R ln 3.5 x 10-3L/(mol.s) = Ea1 - 1 2.7 x 10-4L/(mol.s) 8.314 J/(mol.K) 600K 650K Ea = 2.56 * (8.314 J/mol ) 1.28 x 10-4 Ea = 1.66 x 105 J/mol or 166kJ**USING THE ARRHENIUS EQUATION**EXERCISE 13.7: ACETALDEHYDE DECOMPOSES WHEN HEATED. THE RATE OF DECOMPOSITION IS 1.05 x 10-3 L/mol.s AT 759 K AND 2.14 x 10-2 L/mol.s AT 836 K. CH3CHO(g) CH4(g) + CO(g) WHAT IS ACTIVATION ENERGY FOR THIS DECOMPOSITION? WHAT IS THE RATE CONSTANT AT 865 K?**Example 4**For a certain rxn the rate constant doubles when the temp increases from 15 to 25°C. • Calc. The activation energy, Ea b) Calc. the rate constant at 100°C, taking k at 25°C to be 1.2 x 10-2 L/mol s

More Related