Unit 6 – Kinetics AP CHemistry AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Collision Theory Key Ideas of Collision Theory: For a chemical reaction to occur, the reacting particles must collide. Not all collisions are successful (successful = collision results in a reaction)

What determines if a collision is successful? Particles need to have proper orientation (direction in which they collide)

What determines if a collision is successful? Particles need to have enough energy, or speed Activation Energy (EA) – Particles need to have a minimum amount of energy when they collide to have a successful collision EA needed to break existing bonds and form new bonds

Potential energy diagrams

Potential energy diagrams At all times, the matter has an amount of potential energy associated with it.

Potential energy diagrams Diagram shows: ΔH, EA

Nature of the reactants –Some reactant molecules react in a hurry, others react very slowly. Physical state- gasoline (l) vs. gasoline (g) ; K2SO4(s) + Ba(NO3)2(s) → no rxn; while both of these in the aqueous state will react. Heterogeneous reactants Homogeneous reactants

Factors that Affect Reaction rates Concentration of reactants –more molecules, more collisions. Temperature – heat >em up & speed >em up; the faster they move, the more likely they are to collide. An increase in temperature produces more successful collisions that are able to overcome the needed activation energy, therefore, a general increase in reaction rate with increasing temperature.

Factors that Affect Reaction rates Catalysts – accelerate chemical reactions but are not themselves transformed. Biological catalysts are proteins called enzymes. A catalyst is a substance that changes the rate of reaction by altering the reaction pathway. Most catalysts work by lowering the activation energy needed for the reaction to proceed, therefore, more collisions are successful and the reaction rate is increased. Remember! The catalyst is not part of the chemical reaction and is not used up during the reaction. (May be homogeneous or heterogeneous catalysts.)

Note: A catalyst lowers the activation energy barrier Note: A catalyst lowers the activation energy barrier. Therefore, the forward and reverse reactions are both accelerated to the same degree.

Factors that Affect Reaction rates Surface area of reactants – exposed surfaces affect speed. Except for substances in the gaseous state or solution, reactions occur at the boundary, or interface, between two phases. The greater surface area exposed, the greater chance of collisions between particles, hence, the reaction should proceed at a much faster rate. Ex. coal dust is very explosive as opposed to a piece of charcoal. Solutions are ultimate exposure! Adding an inert gas has NO EFFECT on the rate [or equilibrium] of the reaction.

Chemical reaction rates The speed of a reaction is expressed in terms of its “rate” = some measurable quantity is changing with time The rate of a chemical reaction is measured by the decrease in concentration of a reactant or an increase in concentration of a product in a unit of time: When writing rate expressions, they can be written in terms of reactant disappearance or product appearance Rate = change in concentration of a species time interval = ∆[𝑋] ∆𝑡

aA + bB  cC + dD Generalized rate of reaction: *The change in rate of a reactant is negative (because it’s disappearing) *The change in rate of a product is positive (because it’s appearing) 𝑅𝑎𝑡𝑒=− 1 𝑎 ∆ A ∆𝑡 =− 1 𝑏 ∆ B ∆𝑡 =+ 1 𝑐 ∆ C ∆𝑡 =+ 1 𝑑 ∆ D ∆𝑡

Average Rate of the Reaction – calculate using any time interval (Δt) You can choose your time interval range, but the longer the interval, the slower the rate. This is because the reaction slows down as it proceeds. 𝑅𝑎𝑡𝑒=− ∆ 𝐴 ∆𝑡 =− [𝐴] 𝑡 2 − [𝐴] 𝑡 1 𝑡 2 − 𝑡 1

What is the rate of the reaction with respect to A: 0 – 10 seconds? 1 – 20 seconds? 0 – 30 seconds?

H2O2 (aq) + 3I- (aq) + 2H+ (aq)  I3- (aq) + 2H2O (l) Consider this balanced chemical equation: H2O2 (aq) + 3I- (aq) + 2H+ (aq)  I3- (aq) + 2H2O (l) In the first 10.0 seconds of the reaction, the concentration of I- dropped from 1.000 M to 0.868 M. a) Calculate the average rate of this reaction in the time interval. 𝑅𝑎𝑡𝑒=− 1 3 ∆ I − ∆𝑡 =− 1 3 (0.868 𝑀 −1.000 𝑀) 10.0 𝑠 =𝟒.𝟒𝟎 × 𝟏𝟎 −𝟑 𝑴/𝒔

H2O2 (aq) + 3I- (aq) + 2H+ (aq)  I3- (aq) + 2H2O (l) Consider this balanced chemical equation: H2O2 (aq) + 3I- (aq) + 2H+ (aq)  I3- (aq) + 2H2O (l) In the first 10.0 seconds of the reaction, the concentration of I- dropped from 1.000 M to 0.868 M. b) Determine the rate of change in the concentration of H+ (that is, Δ[H+]/Δt). 𝑅𝑎𝑡𝑒=− 1 2 ∆ H + ∆𝑡 𝑅𝑎𝑡𝑒=− 1 2 ∆ H + ∆𝑡 ∆ H + ∆𝑡 =−2(𝑅𝑎𝑡𝑒) =−2(𝟒.𝟒𝟎 × 𝟏𝟎 −𝟑 𝑴/𝒔) =− 𝟖.𝟖𝟎 × 𝟏𝟎 −𝟑 𝑴/𝒔

H2O2 (aq) + 3I- (aq) + 2H+ (aq)  I3- (aq) + 2H2O (l) Consider this balanced chemical equation: H2O2 (aq) + 3I- (aq) + 2H+ (aq)  I3- (aq) + 2H2O (l) For the reaction shown above (example 1), predict the rate of change in concentration of a) H2O2 (Δ[H2O2]/Δt) b) I3- (Δ[I3-]/Δt) 𝑅𝑎𝑡𝑒=− 1 1 ∆ H2O2 ∆𝑡 =− 𝟒.𝟒𝟎 × 𝟏𝟎 −𝟑 𝑴/𝒔 𝑅𝑎𝑡𝑒=+ 1 1 ∆ I 3 − ∆𝑡 =𝟒.𝟒𝟎 × 𝟏𝟎 −𝟑 𝑴/𝒔

Group Practice: Expressing Reaction Rates In groups, express the rate of reaction in terms of the change in concentration for each of the reactants and products, When the first reactant is decreasing at a rate of 0.100 M/s, how fast are the other reactants decreasing? How fast are the products increasing?

Group Practice: Expressing Reaction Rates 2 NO2(g) + F2(g)  2 NO2F(g) CH3Cl(g) + 2Cl2(g)  CCl4(g) + 3HCl(g) 2H2O2(aq)  2H2O(l) + O2(g) NO2(g)  NO(g) + ½ O2(g) Cl2(g) + 3F2(g)  2ClF3(g)

Warm-up #1 For the reaction A + 2B  C under a given set of conditions, the initial rate is 0.100 M/s. What is Δ[B]/Δt under the same conditions? -0.0500 M/s -1.000 M/s -0.200 M/s C (-0.200 M/s) AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #2 Dinitrogen monoxide decomposes into nitrogen and oxygen when heated. The initial rate of the reaction is 0.022 M/s. What is the initial rate of change of the concentration of N2O (that is, Δ[N2O]/Δt)? 2 N2O (g)  2 N2 (g) + O2 (g) -0.022 M/s -0.011 M/s -0.044 M/s +0.022 M/s C (-0.044 M/s) AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #3 A burning splint will burn more vigorously in pure oxygen than in air because oxygen is a reactant in combustion and concentration of oxygen is higher in pure oxygen than it is in air. oxygen is a catalyst for combustion. oxygen is a product of combustion. nitrogen is a product of combustion and the system reaches equilibrium at a lower temperature. nitrogen is a reactant in combustion and its low concentration in pure oxygen catalyzes the combustion.

Warm-up #4 As the temperature of a reaction is increased, the rate of the reaction increases because the reactant molecules collide less frequently reactant molecules collide more frequently and with greater energy per collision activation energy is lowered reactant molecules collide less frequently and with greater energy per collision reactant molecules collide more frequently with less energy per collision B) reactant molecules collide more frequently and with greater energy per collision AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #5 5. Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2  4NO2 + 6H2O All of the above are valid expressions of the reaction rate. E) All of the above are valid expressions of the reaction rate. AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

The Rate law Rate=𝒌 [A] 𝒏 The rate of a reaction often depends on the concentration of one or more of the reactants. A  Products We can express the relationship between the rate of the reaction and the concentration of the react: Rate=𝒌 [A] 𝒏 k = rate constant (“proportionality constant”) n = reaction order

Reaction order Rate=𝒌 [A] 𝒏 Reaction order (n) – determines how the rate depends on the concentration n = Order How rate is related to [A] n = 0 Zero order Rate is independent of A Rate = k[A]0 = k(1) n = 1 First order Rate is direction proportional to [A] Rate = k[A]1 = k[A] n = 2 Second order Rate is proportional to [A]2 Rate = k[A]2

Reactant concentration as a function of time for different reaction orders INSERT FIGURE OF REACTANT CONC AS A FUNCTION OF TIME FOR DIFFERENT REACTION ORDERS Like Fig. 13.5 AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Determining the order of a reaction The order of a reaction can be determined ONLY by experiment! Method of initial rates – initial rates measured for different initial reactant concentrations Rate measured for a short period of time at beginning of the experiment Use to determine the effect of concentration for each reactant on the rate.

A  Products In an experiment, the initial rate is measured at several different initial concentrations with the following results: [A] (M) Initial Rate (M/s) 0.10 0.015 0.20 0.030 0.40 0.060 Doubling [A] resulted in a doubled rate. Quadrupling [A] resulted in a quadrupled rate (x4). ***Rate is directly proportional to initial concentration. Therefore, “first order in A”

Solving for k Rate=𝑘 [A] 1 𝑘= rate [A] = 0.015 𝑀/𝑠 0.10 𝑀 =0.15 𝑠 −1 We can determine the value of the rate constant, k, by solving the rate law for k and substituting the concentration and the initial rate from any one of the three measurements: Rate=𝑘 [A] 1 𝑘= rate [A] = 0.015 𝑀/𝑠 0.10 𝑀 =0.15 𝑠 −1

Zero order (n=0) [A] (M) Initial Rate (M/s) 0.10 0.015 0.20 0.40 Zero order reaction, the initial rate is independent of the reactant concentration – the rate is the same at all measured initial concentrations!

Second order (n=2) [A] (M) Initial Rate (M/s) 0.10 0.015 0.20 0.40 0.060 0.40 0.240 Second order reaction, the initial rate quadruples for a doubling of the reactant concentration – the relationship is quadradic!

Determine Rate Order When it’s not as obvious to see how the rate changes, you can substitute any two initial concentrations and the corresponding initial rates into a ratio of the rate laws: 𝟎.𝟐𝟒𝟎 𝑴/𝒔 𝟎.𝟎𝟔𝟎 𝑴/𝒔 = 𝒌 (𝟎.𝟒𝟎 𝑴) 𝒏 𝒌 (𝟎.𝟐𝟎 𝑴) 𝒏 𝟒.𝟎= 𝟎.𝟒𝟎 𝟎.𝟐𝟎 𝒏 𝒓𝒂𝒕𝒆 𝟐 𝒓𝒂𝒕𝒆 𝟏 = 𝒌 [𝑨] 𝟐 𝒏 𝒌 [𝑨] 𝟏 𝒏 log 𝟒.𝟎=log (𝟐 𝒏 ) log 𝟒.𝟎=𝒏 log (𝟐) 𝒏= log 𝟒 log 𝟐 =𝟐

Another way to Determine Rate Order Because k is in both rate laws, and it is the same value for k (it’s the same reaction just different concentrations), simplify the ratio of rate laws: 𝟎.𝟐𝟒𝟎 𝑴/𝒔 𝟎.𝟎𝟔𝟎 𝑴/𝒔 = 𝒌 (𝟎.𝟒𝟎 𝑴) 𝒏 𝒌 (𝟎.𝟐𝟎 𝑴) 𝒏 𝟒.𝟎= 𝟎.𝟒𝟎 𝟎.𝟐𝟎 𝒏 𝒓𝒂𝒕𝒆 𝟐 𝒓𝒂𝒕𝒆 𝟏 = 𝒌 [A] 𝟐 𝒏 𝒌 [A] 𝟏 𝒏 log 𝟒.𝟎=log (𝟐 𝒏 ) log 𝟒.𝟎=𝒏 log (𝟐) 𝒓𝒂𝒕𝒆 𝟐 𝒓𝒂𝒕𝒆 𝟏 = [A] 𝟐 [A] 𝟏 𝒏 𝒏= log 𝟒 log 𝟐 =𝟐

Rate Constants Rate Constant (k) – depends on the reaction Zero order: Experimentally determined A constant value that lets us mathematically compare RATE to INITIAL CONCENTRATION Units are different for each reaction order: Zero order, k = M/s (M· s-1) First order, k = 1/s (s-1) Second order, k = M-1· s-1 If you forget which is which, just look at the units required in the rate law to so that Rate = M/s Zero order: Rate=𝑘 First order: Rate=𝑘[A] Second order: Rate=𝑘 [A] 2

Rate Constants Zero order: Rate=𝑘 First order: Rate=𝑘(𝑀) Second order: Units are different for each reaction order: Zero order, k = M/s (M· s-1) First order, k = 1/s (s-1) Second order, k = M-1· s-1 If you forget which is which, just look at the units required in the rate law to so that Rate = M/s Zero order: Rate=𝑘 First order: Rate=𝑘(𝑀) Second order: Rate=𝑘 (𝑀 2 ) Rate= (𝑴∙𝒔 −𝟏 ) Rate= (𝒔 −𝟏 )(𝑴) Rate= (𝑴 −𝟏 ∙ 𝒔 −𝟏 ) (𝑴 𝟐 )

This reaction is first order in N2O5: N2O5(g)  NO3(g) + NO2(g) The rate constant for the reaction at a certain temperature is 0.053/s. Calculate the rate for the reaction when [N2O5] = 0.055 M. What would the rate of the reaction be at the concentration indicated in part a if the reaction were second order? Zero order?

What is the order of the reaction? Consider the data showing the initial rate of a reaction (A  Products) at different concentrations of A. What is the order of the reaction? Write a rate law for the reaction including the value of the rate constant, k. [A] (M) Initial Rate (M/s) 0.100 0.053 0.200 0.210 0.300 0.473

Consider the data showing the initial rate of a reaction (A  Products) at different concentrations of A. What is the order of the reaction? Write a rate law for the reaction including the value of the rate constant, k. [A] (M) Initial Rate (M/s) 0.100 0.053 0.200 0.210 0.300 0.473 2nd order 𝒓𝒂𝒕𝒆 𝟐 𝒓𝒂𝒕𝒆 𝟏 = [A] 𝟐 [A] 𝟏 𝒏 Rate=𝟓.𝟐𝟓 𝑴 −𝟏 𝒔 −𝟏 [A] 𝟐 𝑘= rate [A]

Reaction order for multiple reactants aA + bB  cC + dD For the generic reaction: Each reactant has its own reaction order. The overall order is the sum of the exponents: overall order = m + n 𝑹𝒂𝒕𝒆=𝒌 [A] 𝒎 [B] 𝒏

[NO2] (M) [CO] (M) Initial Rate (M/s) 0.10 0.20 0.40 Determine the order experimentally! Use method of initial rates (like before). When looking at reaction order for A, choose two experiments where [A] changes but all other concentrations stay constant. Do likewise for B (choose where [B] changes, holding [A] constant). [NO2] (M) [CO] (M) Initial Rate (M/s) 0.10 0.0021 0.20 0.0082 0.0083 0.40 0.033 When looking for reaction order of NO2…

[NO2] (M) [CO] (M) Initial Rate (M/s) 0.10 0.20 0.40 Determine the order experimentally! Use method of initial rates (like before). When looking at reaction order for A, choose two experiments where [A] changes but all other concentrations stay constant. Do likewise for B (choose where [B] changes, holding [A] constant). [NO2] (M) [CO] (M) Initial Rate (M/s) 0.10 0.0021 0.20 0.0082 0.0083 0.40 0.033 When looking for reaction order of CO…

NO2(g) + CO(g) --> NO(g) + CO2(g) Consider the reaction between nitrogen dioxide and carbon monoxide: NO2(g) + CO(g) --> NO(g) + CO2(g) (a) the rate law for the reaction (b) the rate constant (k) for the reaction. [NO2] (M) [CO] (M) Initial Rate (M/s) 0.10 0.0021 0.20 0.0082 0.0083 0.40 0.033

8 (doubled, then quadrupled) A reaction in which A, B, and C reacto to form products is first order in A, second order in B, and zero order in C. Write a rate law for the reaction.  What is the overall order of the reaction? By what factor does the reaction rate change... if [A] is doubled ? if [B] is doubled? if [C] is doubled? By what factor does the reaction rate change if the concentration of all three reactants are doubled? Rate = k[A][B]2 3rd order 2 (doubled) 4 (quadrupled) 1 (no change) 8 (doubled, then quadrupled)

Warm-up #1 For the following reaction: NO2(g) + CO(g) → NO(g) + CO2(g), the rate law is: Rate = k[NO2]2. If a small amount of gaseous carbon monoxide (CO) is added to a reaction mixture that was 0.10 molar in NO2 and 0.20 molar in CO, which of the following statements is true? Both k and the reaction rate remain the same. Both k and the reaction rate increase. Both k and the reaction rate decrease. Only k increases, the reaction rate remains the same. Only the reaction rate increases; k remains the same. A—The value of k remains the same unless the temperature is changed or a catalyst is added. Only materials that appear in the rate law, in this case NO2, will affect the rate. Adding NO2 would increase the rate, and removing NO2 would decrease the rate. CO has no effect on the rate AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

I- temperature II- concentration Warm-up #2 Changes in which of the factors affect both rate and rate constant? I- temperature II- concentration I only II only Both I and II Neither I or II I only (Temperature only) AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #3 A reaction was found to be second order in carbon monoxide concentration. The rate of the reaction __________ if the [CO] is doubled, with everything else kept the same. doubles remains unchanged triples increases by a factor of 4 is reduced by a factor of 2 D) increases by a factor of 4 AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #4 The overall order of a reaction is 2. The units of the rate constant for the reaction are __________. M/s M-1s-1 1/s 1/M s/M2 B) M-1s-1 AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #5 If the rate law for the reaction, 2A + 3B  products is first order in A and second order in B, then the rate law is R = k[A][B] k[A]2[B]3 k[A][B]2 k[A]2[B] k[A]2[B]2 C) k[A][B]2 AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #6 The rate law for a reaction is rate = k[A][B]2 Which one of the following statements is false? The reaction is first order in A. The reaction is second order in B. The reaction is second order overall. k is the reaction rate constant If [B] is doubled, the reaction rate will increase by a factor of 4. C) The reaction is second order overall. AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Integrated rate law (dependence of concentration on time) Integrated rate law – a relationship between the concentrations of reactants and time Different for each reaction order Choose the order that gives a STRAIGHT LINE when plotted Rate = slope (or – slope) y = mx + b

𝐥𝐧 A 𝒕 =−𝒌𝒕+ln [A] 𝟎 𝐥𝐧 A 𝒕 A 𝟎 =−𝒌𝒕 [A]t = concentration of A at any time, t [A]0 = initial concentration of A (time zero) k = rate constant FIRST ORDER: or 𝐥𝐧 A 𝒕 =−𝒌𝒕+ln [A] 𝟎 𝐥𝐧 A 𝒕 A 𝟎 =−𝒌𝒕

SECOND ORDER: 𝟏 A 𝒕 = 𝟏 A 𝟎 +𝒌𝒕

ZERO ORDER: A 𝒕 =−𝒌𝒕+ A 𝟎

Example #1 Indicate the order of reaction consistent with each observation: A plot of the concentration of the reactant versus time yields a straight line. A plot of the inverse of the concentration versus time yields a straight line. A plot of the natural log of the concentration of the reactant versus time yields a straight line.

Example #2 Time (s) [AB] (M) First order   Second Order 0.950 50 0.459 100 0.302 150 0.225 200 0.180 250 0.149 300 0.128 350 0.112 400 0.0994 450 0.0894 500 0.0812

HALF LIFE Half-life (t1/2) – the time required for the concentration of a reactant to decrease by one-half of its initial value EX: If a reaction has a half-life of 100 seconds, and if the initial concentration of the reactant is 1.0 M, the concentration will fall to 0.50 M in 100 seconds. The half-life expression is different for each reaction order!

Half-life: First order Molecular iodine dissociates at 625 K with a first-order rate constant of 0.271 s-1. What is the half-life of this reaction? t1/2 is independent of initial concentrations! 𝒕 𝟏/𝟐 = 𝟎.𝟔𝟗𝟑 𝒌 𝑡 1/2 = 0.693 0.271 𝑠 −1 =2.56 𝑠

Half-life: SECOND order t1/2 depends on initial concentration Half-life gets longer as concentration decreases SECOND ORDER: The decomposition of XY is second order in XY and has a rate constant of 7.02 x 10-3 M-1s-1 at a certain temperature. What is the half-life for this reaction at an initial concentration of 0.100 M?  How long will it take for the concentration of XY to decrease to 12.5% of its initial concentration when the initial concentration is 0.100 M? 𝒕 𝟏/𝟐 = 𝟏 𝒌 [A] 𝟎

Half-life: ZERO-order t1/2 depends on initial concentration Half-life gets shorter as concentration decreases ZERO ORDER: 𝒕 𝟏/𝟐 = [A] 𝟎 𝟐𝒌

Warm-up How many half lives have passed [A]nh = (1/2)n[A]0

Warm-up #1 A reaction follows the rate law: Rate = k[A]2. Which of the following plots will give a straight line? 1/[A] versus 1/time [A]2 versus time 1/[A] versus time ln[A] versus time [A] versus time C—The "2" exponent means this is a second-order rate law. Second-order rate laws give a straight-line plot for 1/[A] versus t. AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #2 For a first-order reaction, a plot of __________ versus __________ is linear. ln [A]t, 1/t ln [A]t, t 1/[A]t, t [A]t, t t, 1/[A]t B) ln [A]t, t AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #3 The reaction 2NOBr(g) → 2NO(g) + Br2(g) is a second-order reaction with a rate constant of 0.50 M-1s-1at 11ᵒC. If the initial concentration of NOBr is 1.0 M, the concentration of NOBr after 10.0 seconds is __________. 0.0500 M 0.0250 M 0.0312 M 0.5000 M 0.0625 M C) 0.0312 M AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #4 The following reaction is second order in [A] and the rate constant is 0.039 M-1s-1 A → B The concentration of A was 0.30 M at 23 s. The initial concentration of A was __________ M. 2.4 0.27 0.41 3.7 1.2 x 10-2 C) 0.41 AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Warm-up #5 The reaction A → B is first order in [A]. Consider the following data. The half-life of this reaction is __________ s. 0.97 7.1 5.0 3.0 0.14 C) 5.0 AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Integrated Rate Laws 𝟏 A 𝒕 = 𝟏 A 𝟎 +𝒌𝒕 𝐥𝐧 A 𝒕 =−𝒌𝒕+ln [A] 𝟎

Maxwell boltzman diagram

Describe what is happening here..

MC Questions 2. The specific rate constant, k, for radioactive beryllium–11 is 0.049 s–1. What mass of a 0.500 mg sample of beryllium–11 remains after 28 seconds? 0.250 mg 0.125 mg 0.0625 mg 0.375 mg 0.500 mg B—The half-life is 0.693/k = 0.693/0.049 s–1 = 14 s. The time given, 28 s, represents two half-lives. The first half-life uses one-half of the isotope, and the second half-life uses one-half of the remaining material, so only one-fourth of the original material remains. AP Chemistry Unit 9 KINETICS (Unit 9 slides.pptx)

Reaction Mechanisms Most chemical reactions do not occur in a single step, but over the course of several steps. A chemical equation usually represents the overall reaction (not the series of individual steps). For example, the reaction H2(g) + 2 ICl(g)  2 HCl(g) + I2(g), it actually occurs in TWO steps: Step 1 H2(g) + ICl(g)  HI(g) + HCl(g) Step 2 HI(g) + ICl(g)  HCl(g) + I2(g) Overall H2(g) + 2 ICl(g)  2 HCl(g) + I2(g)

Molecularity Elementary steps are characterized by their molecularity (the number of reactant molecules that collide together in an elementary step). UNIMOLECULAR: BIMOLECULAR: TERMOLECULAR: A  Products A + A  Products A + B  Products A + B + C  Products

Each step in a reaction mechanism is called an elementary step. Cannot be broken down into simpler steps Occur as written For a valid reaction mechanism, the individual steps must add to the overall reaction. Reaction Intermediate – forms in one elementary step, and is consumed in another Step 1 H2(g) + ICl(g)  HI(g) + HCl(g) Step 2 HI(g) + ICl(g)  HCl(g) + I2(g)

Rate Laws for Elementary Steps The rate law for a general reaction has to be determined by experiment. The rate law for elementary steps can be determined just from their balanced chemical equations. Elementary Step Molecularity Rate Law A  Products  1 Rate = k[A] A + A  Products  2 Rate = k[A]2 A + B  Products Rate = k[A][B] A + A + A  Products  3 Rate = k[A]3 A + A + B  Products Rate = k[A]2[B]

Rate Determining Step One of the steps will be significantly slower than the others. The slowest step limits (or determines) the overall rate of reaction (or how fast it can proceed) The RDS determines the rate law for the overall reaction! NO2 + NO2  NO3 + NO (slow) NO3 + CO  NO2 + CO2 (fast) NO2 + CO  NO + CO2 Overall

Energy Diagram for a Two-Step Mechanism: Because Ea for Step 1 > Ea for step 2, Step 1 has the smaller rate constant and is rate limiting. Step 1 has higher activation energy Step 1 has smaller rate constant Step 1 determines overall rate