Mullis 1

2 Kinetics Concept of rate of reaction Use of differential rate laws to determine order of reaction and rate constant from experimental data Effect of temperature change on rates Energy of activation; the role of catalysts The relationship between the rate- determining step and a mechanism

Mullis 3 Rate of Reaction Rate = Δ[concentration] or d [product] Δ time dt Rate of appearance of a product = rate of disappearance of a reactant Rate of change for any species is inversely proportional to its coefficient in a balanced equation.

Mullis 4 Rate of Reaction Assumes nonreversible forward reaction Rate of change for any species is inversely proportional to its coefficient in a balanced equation. 2N 2 O 5  4NO 2 + O 2 Rate of reaction = -Δ[N 2 O 5 ] = Δ[NO 2 ] = Δ[O 2 ] 2 Δt 4 Δt Δt where [x] is concentration of x (M) and t is time (s)

Mullis 5 Reaction of phenolphthalein in excess base Use the data in the table to calculate the rate at which phenolphthalein reacts with the OH- ion during each of the following periods: (a) During the first time interval, when the phenolphthalein concentration falls from M to M. (b) During the second interval, when the concentration falls from M to M. (c) During the third interval, when the concentration falls from M to M. Conc. (M)Time (s)

Mullis 6 Reactant Concentration by Time

Mullis 7 Finding k given time and concentration Create a graph with time on x-axis. Plot each vs. time to determine the graph that gives the best line: – [A] – ln[A] – 1/[A] – (Use LinReg and find the r value closest to 1) – k is detemined by the slope of best line (“a” in the linear regression equation on TI-83) – 1 st order (ln[A] vs. t): k is –slope – 2 nd order (1/[A] vs t: k is slope)

Mullis 8 Rate Law Expression As concentrations of reactants change at constant temperature, the rate of reaction changes. According to this expression. Rate = k[A] x [B] y … Where k is an experimentally determined rate constant, [ ] is concentration of product and x and y are orders related to the concentration of A and B, respectively. These are determined by looking at measured rate values to determine the order of the reaction.

Mullis 9 Finding Order of a Reactant - Example 2ClO 2 + 2OH -  ClO ClO H 2 O Start with a table of experimental values: To find effect of [OH - ] compare change in rate to change in concentration. When [OH - ] doubles, rate doubles. Order is the power: 2 x = 2. x is 1. This is 1 st order for [OH - ]. [ClO 2 ] (M)[OH - ] (M)Rate (mol/L-s) x x x x

Mullis 10 Finding Order of a Reactant - Example 2ClO 2 + 2OH -  ClO ClO H 2 O Start with a table of experimental values: To find effect of [ClO 2 ] compare change in rate to change in concentration. When [ClO 2 ] triples, rate increases 9 times. Order is the power: 3 y = 9. y is 2. This is 2nd order for [ClO 2 ]. [ClO 2 ] (M)[OH - ] (M)Rate (mol/L-s -1 ) x x x x 9x

Mullis 11 Finding Order of a Reactant - Example 2ClO 2 + 2OH -  ClO ClO H 2 O Can use algebraic method instead. This is useful when there are not constant concentrations of one or more reactants. This example assumes you found that reaction is first order for [OH - ] x =k(0.010) x (.030) x = k (0.030) x (.060) =.333 x (.5) For [ClO 2 ]x, x = 2 [ClO 2 ] (M)[OH - ] (M)Rate (mol/L-s -1 ) x x x10 -2

Mullis 12 Rate Law: 2ClO 2 + 2OH -  ClO ClO H 2 O Rate = k[ClO 2 ] 2 [OH - ] To find k, substitute in any one set of experimental data from the table. For example, using the first row: k = rate/[ClO 2 ] 2 [OH - ] k = 6.00x10 -4 Ms -1 = 200 M -2 s -1 [0.010M] 2 [0.030M] Overall reaction order is 2+1=3. Note units of k.

Mullis 13 Determining k Given Overall Reaction Order Rate(M/s) = k[A] x x = overall order of reaction [A] = the reactant concentration (M) Overall reaction order ExampleUnits of k 1Rate=k[A](M/s)/M = s -1 2Rate=k[A] 2 (M/s)/M 2 = M -1 s -1 3Rate=k[A] 3 (M/s)/M 3 = M -2 s Rate=k[A] 1.5 (M/s)/M 1.5 = M -0.5 s -1

Determine Reaction Order Mullis 14 From the following experimental data, determine the order of the reaction and the rate constant for the reaction: C 4 H 8 (g) → 2C 2 H 4 (g) The reaction was carried out in a constant volume container at 532 °C. Time (s)P (mm Hg)

Another Reaction Order Example Mullis 15 The following data were collected during a study of its decomposition at a certain temperature: [H 2 O 2 ] (M)Time (s)

Mullis 16 Integrated Rate Law Use when time is given or requested Relates concentration and time to rate 1 st order: ln[A] = -kt + ln[A] 0 or [A]=[A] 0 e -kT 2nd order: 1__ = kt + 1__ [B][B] 0 Wow! y= mx + b Both equations can be used with linear regression to solve for slope, or k.

Mullis 17 Half life for 1 st vs 2 nd Order Reactions 1 st order:t 1/2 = k 2 nd order: t 1/2 = 1__ k[A] 0

Mullis 18 OrderRate LawConcentration-Time Equation Half-life Equation Graphical AnalysisGraph 1Rate = k[A]ln = – ktt ½ = ln[A] vs. time slope = –k ln[A] t 2Rate = k[A] 2 = kt +t ½ = vs. time slope = k t 0Rate = k[A] 0 [A] = –kt + [A] 0 t ½ = [A] vs. t slope = –k [A] t

Mullis 19 The Arrhenius Equation and Finding E a k=Ae -Ea/RT Where A is the frequency factor – Related to frequency of collisions and favorable orientation of collisions E a is activation energy in J R = J/mol-K T is temp in K k is the rate constant

Mullis 20 Using the Arrhenius equation As E a increases, rate decreases. – Fewer molecules have the needed energy to react. As temp increases, rate increases – More collisions occur and increased kinetic energy means more collisions have enough energy to react. – Mathematically, T is in denominator of the power –E a /T. ln k = -E a + lnA T

Mullis 21 Activation Energy: Energy vs. Reaction Progress E a is lowered with the addition of a catalyst. Peak is where collisions of reactants have achieved enough energy to react The arrangement of atoms at the peak is activated complex or the transition state.

Mullis 22 Temperature effects on a reaction Two factors account for increased rate of reaction. 1. Energy factor: When enough energy is in collision for formation of activation complex, bonds break to begin reaction. With higher temp, more collisions have this energy. 2. Frequency of collision: Particles move faster and collide more frequently with higher temp, increasing chance of reaction. Increasing temperature increases the rate of a reaction more if the reaction is endothermic to start with. K increases according to k=Ae -Ea/RT

Mullis 23 Finding E a given info at two temperatures ln k 1 = E a [1_ - 1_ ] k 2 RT 2 T 1 Similar to vapor pressure equation, Clausius-Clapeyron Equation: ln P vapT1 = ΔH vap [ 1_ - 1_ ] P vapT2 R T 2 T 1 In both cases, use R = J/K-mol and be sure E a or ΔH vap are in J, not kJ.

Mullis 24 Mechanisms: Multistep Reactions The following reaction occurs in a single step. CH 3 Br (aq) + OH - (aq)  CH 3 OH (aq) + Br - (aq) – Rate = k(CH 3 Br)(OH - ) This one occurs in several steps: (CH 3 ) 3 CBr(aq) + OH - (aq)  (CH 3 ) 3 COH (aq) + Br - (aq) 1. (CH 3 ) 3 CBr  (CH 3 ) 3 C + + Br - Slow step 2. (CH 3 ) 3 C + + H 2 O  (CH 3 )COH 2+ Fast step 3. (CH 3 ) 3 COH 2+ + OH -  (CH 3 ) 3 COH + H 2 O Fast step – Rate = k((CH 3 ) 3 CBr)

Mullis 25 Mechanisms: Multistep Reactions The overall rate of reaction is more or less equal to the rate of the slowest step. (rate-limiting step) If only one reagent is involved in the rate-limiting step, the overall rate of reaction is proportional to the concentration of only this reagent. Ex. For the reaction with Rate = k((CH 3 ) 3 CBr) Although the reaction consumes both (CH 3 ) 3 CBr and OH-, the rate of the reaction is only proportional to the concentration of (CH 3 ) 3 CBr. The rate laws for chemical reactions can be explained by the following general rules. – The rate of any step in a reaction is directly proportional to the concentrations of the reagents consumed in that step. – The overall rate law for a reaction is determined by the sequence of steps, or the mechanism, by which the reactants are converted into the products of the reaction. – The overall rate law for a reaction is dominated by the rate law for the slowest step in the reaction.

Mullis 26 Substituting for an Intermediate Step 1 : N 2 H 2 O 2  N 2 HO 2- + H + fast equilibrium Step 2: N 2 HO 2-  N 2 O + OH - slow Step 3: H + + OH -  H 2 Ofast Requirement: A fast equilibrium prior to the rate determining (slow) step that contains the intermediate for which you wish to substitute. 1. N 2 H 2 O 2  N 2 HO 2- + H + fast equilibrium 2. For the fast equilibrium, write the rate law (leaving out the k and R) for the reactants and set it equal to the rate law for the products. This can be done because in an equilibrium reaction the forward rate must be equal to the reverse rate. [N 2 H 2 O 2 ] = [N 2 HO 2- ] [H + ] 3. Algebraically solve for the intermediate, N 2 HO 2- [N 2 H 2 O 2 ] / [H + ] = [N 2 HO 2- ] 4. Algebraically substitute into the rate law for N 2 HO 2- Rate law with intermediate is: R = k [N 2 HO 2- ], so R = k [N 2 H 2 O 2 ] / [H + ]