CHEMICAL KINETICS H 2 S (g) + Zn 2+ (aq) ⇆ ZnS (s) + 2H + (aq) Chemical reactions can be viewed from different perspectives 4D-1 (of 21) STOICHIOMETRY Describes relationships based on conservation of atoms – predicts reaction quantities THERMODYNAMICS Describes energy and entropy changes – predicts if a reaction will occur KINETICS Describes how a reaction occurs – predicts the speed of a reaction THERMODYNAMICS KINETICS

CHEMICAL KINETICS 1 -Describes the speed of a chemical reaction and the factors that affect it 2 - Determines the mechanism of a chemical reaction at the molecular level COLLISION THEORY 1 - Molecules must collide to react 2 -Molecules must collide with sufficient energy to react (to break bonds) 3 - Molecules must collide in the proper orientation 4D-2 (of 21)

4D-3 (of 21)

Measured by how fast a reactant reacts away, or how fast a product is produced Rate =- Δ[reactant] _________________ Δt = - d[reactant] _________________ dt Rate =+ Δ[product] _________________ Δt = + d[product] _________________ dt REACTION RATES 4D-4 (of 21)

2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g)For the reaction 0.30 M/min- d[N 2 O 5 ] = _____________ dt Find the reaction rate with respect to NO 2 and O M N 2 O 5 _______________ min x 4 M NO 2 ____________ 2 M N 2 O 5 = 0.60 M NO 2 /min + d[NO 2 ] = ____________ dt 0.30 M N 2 O 5 _______________ min x 1 M O 2 ____________ 2 M N 2 O 5 = 0.15 M O 2 /min + d[O 2 ] = ____________ dt 4D-5 (of 21)

Fastest Reaction Rates Reaction rates are the slope of the tangent line (at any particular point) of a concentration vs. reaction time graph 0 Reaction Rate (equilibrium) M/min M/min M/min Concentration as a Function of Reaction Time 4D-6 (of 21)

Experimentally, rates of reactions are proportional to (1)Temperature (2)Concentration of reacting molecules RATE LAW – An algebraic expression that relates the rate of a reaction (how fast a reactant disappears or how fast a product appears) to the concentrations of the reactants and the temperature 4D-7 (of 21)

2N 2 O 5 (g) → 4NO 2 (g) + O 2 (g) Rate  T [N 2 O 5 ] Rate =k [N 2 O 5 ] SPECIFIC RATE CONSTANT (k) – The rate law constant that depends on temperature FIRST-ORDER REACTION – One in which the rate is proportional to the concentration of reactants to the first power 4D-8 (of 21)

2NO 2 (g) → 2NO (g) + O 2 (g) Rate =k [NO 2 ] 2 This is a SECOND-ORDER REACTION H 2 (g) + I 2 (g) → 2HI (g) Rate =k [H 2 ] [I 2 ] 1 st order in H 2 1 st order in I 2 2 nd order overall Rate laws can only be determined experimentally 4D-9 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [NO] [Cl 2 ] R = k [NO] x [Cl 2 ] y Choose 2 trials where [Cl 2 ] is constant = k [1.00] x [1.00] y ________________________________ = k [0.50] x [1.00] y 4 = 2 x  2 = x 4D-10 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [NO] [Cl 2 ] R = k [NO] x [Cl 2 ] y Choose 2 trials where [NO] is constant = k [0.50] 2 [1.00] y ________________________________ = k [0.50] 2 [0.50] y 2 = 2 y  1 = y 4D-11 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [NO] [Cl 2 ] R = k [NO] x [Cl 2 ] y R = k [NO] 2 [Cl 2 ] 1 The reaction is 2 nd order in NO, 1 st order in Cl 2, and 3 rd order overall 4D-12 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [HCl] [NO 2 ] R = k [HCl] x [NO 2 ] y Choose 2 trials where [NO 2 ] is constant = k [6.0] x [1.0] y ____________________________ = k [3.0] x [1.0] y 2 = 2 x  1 = x 4D-13 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [HCl] [NO 2 ] R = k [HCl] x [NO 2 ] y Choose 2 trials where [HCl] is constant = k [6.0] 1 [2.0] y ____________________________ = k [6.0] 1 [1.0] y 8 = 2 y  3 = y 4D-14 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [HCl] [NO 2 ] R = k [HCl] x [NO 2 ] y R = k [HCl] 1 [NO 2 ] 3 The reaction is 1 st order in HCl, 3 rd order in NO 2, and 4 th order overall 4D-15 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [HCl] [NO 2 ] R = k [HCl] x [NO 2 ] y R = k [HCl] 1 [NO 2 ] 3 Find the value of k, with its units R = k ________________ [HCl] 1 [NO 2 ] 3 = M/s ____________________ (3.0 M) 1 (1.0 M) 3 = M -3 s -1 4D-16 (of 21)

Find the rate law given the following experimental data Initial Rate (M/s) [HCl] [NO 2 ] R = k [HCl] x [NO 2 ] y R = k [HCl] 1 [NO 2 ] 3 Calculate the rate of the reaction when [HCl] = 1.0 M and [NO 2 ] = 3.0 M R = k [HCl] 1 [NO 2 ] 3 = ( M -3 s -1 ) (1.0 M) 1 (3.0 M) 3 = 0.36 Ms -1 4D-17 (of 21)

Find the rate law given the following experimental data Initial Rate (M/min) [F 2 ] [Cl 2 ] R = k [F 2 ] x [Cl 2 ] y Choose 2 trials where [Cl 2 ] is constant 0.18 = k [0.30] x [0.20] y _____________________________ 0.09 = k [0.15] x [0.20] y 2 = 2 x  1 = x 4D-18 (of 21)

Find the rate law given the following experimental data Initial Rate (M/min) [F 2 ] [Cl 2 ] R = k [F 2 ] x [Cl 2 ] y Choose 2 trials where [F 2 ] is constant 1.08 = k [0.60] 1 [0.60] y _____________________________ 0.18 = k [0.30] 1 [0.20] y 6 = (2) 3 y  1 = y ??????? 3 = 3 y 4D-19 (of 21)

Find the rate law given the following experimental data Initial Rate (M/min) [F 2 ] [Cl 2 ] R = k [F 2 ] x [Cl 2 ] y R = k [F 2 ] 1 [Cl 2 ] 1 Find the value of k, with its units R = k _____________ [F 2 ] 1 [Cl 2 ] 1 = 0.18 M/min ________________________ (0.30 M) 1 (0.20 M) 1 = 3.0 M -1 min -1 4D-20 (of 21)

Find the rate law given the following experimental data Initial Rate (M/min) [F 2 ] [Cl 2 ] R = k [F 2 ] x [Cl 2 ] y R = k [F 2 ] 1 [Cl 2 ] 1 Calculate the rate of the reaction when [F 2 ] = 0.20 M and [Cl 2 ] = 0.40 M R = k [F 2 ] 1 [Cl 2 ] 1 = (3.0 M -1 min -1 ) (0.20 M) 1 (0.40 M) 1 = 0.24 Mmin -1 4D-21 (of 21)

2N 2 O (g) → 2N 2 (g) + O 2 (g) Rate =k [N 2 O ] 0 This is a ZERO-ORDER REACTION Rate =kThe rate law for this reaction is:, or 0º 4E-1 (of 18) -d[N 2 O] = k _________ dt d[N 2 O] = – k dt ∫ 0 t ∫ 0 t [N 2 O] t – [N 2 O] 0 = – kt – (– k0) [N 2 O] t – [N 2 O] 0 = – kt [N 2 O] t = [N 2 O] 0 – kt

[N 2 O] t = -kt + [N 2 O] 0 To plot a linear graph:y = mx + b y= [N 2 O] t m= -k x= t b= [N 2 O] 0 4E-2 (of 18) For all reactions with zero-order kinetics: [X] t = -kt + [X] o a plot of [reactant] t vs. t will yield a line

N 2 O 5 (g) → N 2 O (g) + 2O 2 (g) This is a FIRST-ORDER REACTION Rate =k[N 2 O 5 ] 1 The rate law for this reaction is:, or 1º: 4E-3 (of 18) -d[N 2 O 5 ] = k[N 2 O 5 ] _________ dt d[N 2 O 5 ] = – k dt ________ [N 2 O 5 ] ∫ 0 t ∫ 0 t ln[N 2 O 5 ] t – ln[N 2 O 5 ] 0 = – kt – (– k0) ln[N 2 O 5 ] t – ln[N 2 O 5 ] 0 = – kt ln[N 2 O 5 ] t = ln[N 2 O 5 ] 0 – kt

To plot a linear graph: y= ln[N 2 O 5 ] t m= -k x= t b= ln[N 2 O 5 ] o ln[N 2 O 5 ] t = -kt + ln[N 2 O 5 ] o 4E-4 (of 18) For all reactions with first-order kinetics: ln[X] t = -kt + ln[X] o a plot of ln[reactant] t vs. t will yield a line

For the 1º decomposition of N 2 O 5, k = s -1. Calculate the molarity of N 2 O 5 remaining from 10.0 M N 2 O 5 after 100. seconds [N 2 O 5 ] t = [N 2 O 5 ] o e -kt = (10.0 M) e –( s -1 )(100. s) = 2.89 M 4E-5 (of 18) This is the same as radioactive decay, which follows first order kinetics First order kinetics can also be written as: [X] t = [X] o e -kt

2HI (g) → H 2 (g) + I 2 (g) This is a SECOND-ORDER REACTION Rate =k[HI ] 2 The rate law for this reaction is:, or 2º 4E-6 (of 18) -d[HI] = k[HI] 2 _______ dt d[HI] = – k dt ______ [HI] 2 ∫ 0 t ∫ 0 t –1 – –1 = – kt _____ _____ [HI] t [HI] 0 –1 = –1 – kt _____ _____ [HI] t [HI] 0

1 = kt + 1 ____ _____ [HI] t [HI] o To plot a linear graph: y= 1/[HI] t m= k x= t b= 1/[HI] o 4E-7 (of 18) For all reactions with second-order kinetics: 1 = kt + 1 ____ ____ [X] t [X] o a plot of 1/[reactant] t vs. t will yield a line

Order Rate Law R = k R = k [X] 1 R = k [X] 2 Equality [X] t = -kt + [X] o ln [X] t = -kt + ln [X] o 1 = kt + 1 ____ [X] t [X] o Linear Plot [X] vs. t ln [X] vs. t 1vs. t ____ [X] 4E-8 (of 18)

Find the rate law for the reaction A → B given the following: [A] (M): Time (s): R = k [A] X 4E-9 (of 18)

Find the rate law for the reaction A → B given the following: [A] (M): Time (s): Test for 0º Linear plot would be [A] vs. t If linear, the slope calculated with any 2 points will be constant M – M ____________________________ 0.00 s – 5.00 s = Ms M – M ____________________________ 5.00 s – s = Ms -1 Slopes are not constant,  not 0º 4E-10 (of 18)

Find the rate law for the reaction A → B given the following: [A] (M): Time (s): Test for 1º Linear plot would be ln[A] vs. t ln ( M) – ln ( M) ____________________________________ 0.00 s – 5.00 s = s -1 ln ( M) – ln ( M) ____________________________________ 5.00 s – s = s -1 Slopes are not constant,  not 1º 4E-11 (of 18)

Find the rate law for the reaction A → B given the following: [A] (M): Time (s): Test for 2º Linear plot would be 1/[A] vs. t (1/ M) – (1/ M) ____________________________________ 0.00 s – 5.00 s = M -1 s -1 (1/ M) – (1/ M) ____________________________________ 5.00 s – s = M -1 s -1 Slopes are constant,  the reaction is 2º R = k [A] 2 4E-12 (of 18)

REACTION MECHANISMS Chemical reactions occur as a specific series of collisions and each collision is considered a STEP Each step has a MOLECULARITY a)If 1 molecule decomposes, the step is UNIMOLECULAR b)If 2 molecules collide to react, the step is BIMOLECULAR c)If 3 molecules collide to react, the step is TRIMOLECULAR (rare) 4E-13 (of 18) ELEMENTARY REACTION – A reaction the occurs in only one step (or one collision)

REACTION MECHANISM – The series of steps that yield the balanced chemical reaction RATE DETERMINING STEP (or RATE LIMITING STEP) – The slowest step in the reaction mechanism The reactants in a reaction’s rate law are the reactants in the rate determining step 4E-14 (of 18)

2NO (g) + 2H 2 (g) → N 2 (g) + 2H 2 O (g) This reaction occurs via a 3 step mechanism: (1) 2NO ⇆ N 2 O 2 (fast equilibrium) (2)N 2 O 2 + H 2 → N 2 O + H 2 O(slow) (3) N 2 O + H 2 → N 2 + H 2 O(fast) R = k 2 [N 2 O 2 ] [H 2 ] This is not a reactant in the reaction, it is a REACTION INTERMEDIATE K1K1 k2k2 k3k3  [N 2 O 2 ] must be substituted out 4E-15 (of 18)

2NO (g) + 2H 2 (g) → N 2 (g) + 2H 2 O (g) This reaction occurs via a 3 step mechanism: (1) 2NO ⇆ N 2 O 2 (fast equilibrium) (2)N 2 O 2 + H 2 → N 2 O + H 2 O(slow) (3) N 2 O + H 2 → N 2 + H 2 O(fast) R = k 2 [N 2 O 2 ] [H 2 ] K 1 = [N 2 O 2 ] ________ [NO] 2 K1K1 k2k2 k3k3 R = k 2 K 1 [NO] 2 [H 2 ] K 1 [NO] 2 = [N 2 O 2 ] R = k [NO] 2 [H 2 ] 4E-16 (of 18)

CHCl 3 (g) + Cl 2 (g) → CCl 4 (g) + HCl (g) This reaction occurs via a 3 step mechanism: (1) Cl 2 ⇆ 2Cl(fast equilibrium) (2)CHCl 3 + Cl → CCl 3 + HCl(slow) (3) CCl 3 + Cl → CCl 4 (fast) R = k 2 [CHCl 3 ] [Cl] K1K1 k2k2 k3k3 K 1 = [Cl] 2 ______ [Cl 2 ] R = k 2 [CHCl 3 ] K 1 ½ [Cl 2 ] ½ K 1 [Cl 2 ] = [Cl] 2 R = k [CHCl 3 ] [Cl 2 ] ½ K 1 ½ [Cl 2 ] ½ = [Cl] 4E-17 (of 18)

H 2 (g) + l 2 (g) → 2HI (g) This reaction occurs via a 3 step mechanism: (1) l 2 ⇆ 2l(fast equilibrium) (2) H 2 + l ⇆ H 2 l(fast equilibrium) (3) H 2 I + l → 2HI(slow) R = k 3 [H 2 I] [l] K1K1 K2K2 k3k3 K 2 = [H 2 I] _______ [H 2 ][I] K 2 [H 2 ] [I] = [H 2 I] R = k 3 K 2 [H 2 ] [l] [I] K 1 = [I] 2 ____ [I 2 ] K 1 [I 2 ] = [I] 2 R = k 3 K 2 [H 2 ] K 1 [I 2 ] R = k [H 2 ] [I 2 ] 4E-18 (of 18)

ACTIVATION ENERGY (E a ) – The minimum energy needed by the reacting molecules for an effective collision Reactants Products ΔHΔH EaEa 4F-1 (of 12)

On the microscopic level, rates of reactions depend on: (1)The collision frequency of the reacting molecules (2)The fraction of collisions that have the proper orientation (3)The fraction of collisions that have the activation energy On the macroscopic level, rates of reactions depend on: (1)Concentrations of reactants (2)Temperature CALCULATING THE ACTIVATION ENERGY 4F-2 (of 12)

1889 SVANTE ARRHENIUS Proposed that the specific rate constant, k, is the product of the collision frequency, the fraction of collisions with the proper orientation, and the fraction of collisions with the activation energy k= zpe -E a /RT z= collision frequency p= fraction of collisions with the proper orientation e -E a /RT = fraction of the collisions with the activation energy A is the PRE-EXPONENTIAL FACTOR E a is the ARRHENIUS ACTIVATION ENERGY k= Ae -E a /RT 4F-3 (of 12)

The activation energy can be determined graphically by knowing specific rate constants at different temperatures k= Ae -E a /RT ln k= ln A - E a ____ RT ln k= -E a + ln A _____ RT 4F-4 (of 12)

ln k= -E a 1 + ln A _____ ___ R T The activation energy can be determined graphically by knowing specific rate constants at different temperatures y= ln k m= -E a /R x = 1/T b= ln A 4F-5 (of 12)

ln k= -E a 1 + ln A _____ ___ R T The activation energy can be determined graphically by knowing specific rate constants at different temperatures 4F-6 (of 12) -E a = m _____ R E a = -Rm E a = -(8.314 J/K)( K)= 28,400 J

Reactants Products ΔHΔH EaEa ΔHΔH EaEa Exothermic Negative ΔH Endothermic Positive ΔH 4F-7 (of 12) Reactants Products REACTION ENERGY PROFILES

Reactants Products Enthalpy Change (ΔH) Forward Reaction E a (E a-for ) Reverse Reaction E a (E a-rev ) E a-for – E a-rev = ΔH 4F-8 (of 12)

Reactants Products At the highest point of the graph, the reactants go through a high energy state called the TRANSITION STATE At the transition state, the colliding molecules form a single unit called the ACTIVATED COMPLEX ( ǂ ) ǂ 4F-9 (of 12)

The Activated Complex A single unit in which old bonds are breaking and new bonds are forming 4F-10 (of 12)

CATALYSIS Increases the rate of a reaction by allowing the reaction to take place via a different pathway with a lower activation energy EaEa EaEa A catalyst brings a reaction to equilibrium faster, it does not change the equilibrium concentrations 4F-11 (of 12)

N 2 (g) + 3H 2 (g) → 2NH 3 (g) Metal surfaces can act as catalysts for gaseous reactions 4F-12 (of 12)