Presentation on theme: "Chapter 14 Chemical Kinetics. Study of rxn rates (how fast rxn progresses) –Measured in [X]/ t Also deals with reaction mechanisms –Steps rxn goes."— Presentation transcript:
Chapter 14 Chemical Kinetics
Study of rxn rates (how fast rxn progresses) –Measured in [X]/ t Also deals with reaction mechanisms –Steps rxn goes through to move from reactants to products
Factors That Influence Reaction Rate Under a specific set of conditions, every reaction has its own characteristic rate, which depends upon the chemical nature of the reactants. Four factors can be controlled during the reaction: 1.Concentration - molecules must collide to react; 2.Physical state - molecules must mix to collide; 3.Temperature - molecules must collide with enough energy to react; 4.The use of a catalyst.
Expressing Rxn Rates Rates are expressed in some unit quantity per time –SI units for rates of distance (speed) are in m/s –Chemical rxn rate units are M/s Consider the rxn A→B – Rxn rates given can be initial, instantaneous, or average
Expressing Rate aA + bB cC + dD rate = 1 a -= - [A] tt 1 b [B] tt 1 c [C] tt = + 1 d [D] tt = + The numerical value of the rate depends upon the substance that serves as the reference. The rest is relative to the balanced chemical equation.
Practice PROBLEM:Determine the average rxn rate over the course of the entire experiment for the data listed in Table 16.1 PROBLEM:(a) Balance the following equation and express the reaction rate in terms of the change in concentration with time for each substance: NO (g) + O 2(g) → N 2 O 3(g). (b) How fast is [O 2 ] decreasing when [NO] is decreasing at a rate of 1.60x10 -4 M/s?
The Rate Law The rate law governs the progress of a given rxn. For a general rxn: aA + bB → cC + dD –The rate law is given by the equation below: –Rate = k[A] x [B] y, k – rate constant; x & y are rxn orders wrt A & B All components of a rxn’s rate law must be determined experimentally –Measure physical quantity that you can relate to the concentration of a reactant a specific instant (initial rate method) or over time (integrated rate law)
Determining Rate Laws The Initial rate method –Used to determine rxn orders experimentally –Measure initial rate of different reactant concentrations –Data is listed in a table –Ratio data, in general rate law form, from 2 lines in the table to determine order of each reactant in rxn Choose lines where conc. reactant in question changes and conc. of all other reactants stays the same
Practice Experiment Initial Reactant Concentrations (mol/L) Initial Rate (mol/L*s) O2O2 NO 1.10x x x x x x x x x x x x x x x NO( g ) + O 2 ( g ) 2NO 2 ( g ) 1.Determine the general rate law for the rxn. 2.Calculate the rate constant for experiment 2.
The Rate Constant Specific for a particular rxn at a particular temperature, within experimental error Units for k tell you the overall rxn order –Remember units for rate are M/time –Units for [A] x are M x –For a rxn with an overall order R, the unit for k can be found by
The Integrated Rate Law Can be used for 2 reasons 1.Determine reactant concentration after an elapsed time--- must know order of reactant, rate constant, correct formula 2.Determine rxn order for a specific reactant--- must graph different quantities vs. time and see which gives most linear plot Can only be used for 0, 1 st, and 2 nd order rates
Integrated Rate Law Formulas & Plots
Practice PROBLEM: At 25 0 C, hydrogen iodide (HI) breaks down ver slowly to hydrogen and iodine: rate = k[HI] 2.The rate 25 0 C is 2.4x L/mol s. It mol of HI is place in a 1.00L container, how long will it take for the concentration of HI to reach M (10% reacted)? Determine the rxn order for N 2 O 5 using the graphical data given PROBLEM:
Reaction Half-Life Time required for reactant concentration to reach ½ its initial value
The Arrhenius Equation Describes the relationship between temperature and rxn rate Higher T Larger k Increased/faster rate Smaller E a Larger k Increased/faster rate Lower E a (or T) Smaller k Decreased/slower rate A is related to both the collision frequency an orientation probability factor (dependent on structural complexity) where k is the kinetic rate constant at T E a is the activation energy R is the energy gas constant T is the Kelvin temperature A is the collision frequency factor
Activation Energy Energy that must be overcome for reactants to form products –All rxns regardless of initial and final energies have E a > 0 –Some bonds must break and new bonds must form –Reactant molecules gain this energy through collisions with one another –Increasing temperature increases rate as # collisions and energy of collisions increase
Reaction Energy Diagrams Used to depict changes reactant molecules undergo to form products
Reaction Mechanism Sequence of single rxn steps that sum to the overall rxn It is impossible to prove rxn mechanism experimentally Rxn energy diagrams can elucidate # steps in a mechanism Steps in a mechanism for an overall rxn are elementary steps in which the coefficients of each reactant denote the reaction rate order wrt the reactant The sum of all reactant coefficients in an elementary step denote the molecularity of the step The higher the molecularity of an elementary step, the slower its rate
2NO 2(g) + F 2(g) → 2NO 2 F (g) Experimental rate law determined: –rate = k[NO 2 ][F 2 ] Accepted mechanism: –NO 2(g) + F 2(g) → NO 2 F (g) + F (g) –NO 2(g) + F (g) → NO 2 F (g)
Correlating Mechanism w/ Rate Law For a mechanism to be reasonable, its elementary steps must meet 3 criteria: 1.Elementary steps must add up to overall balanced eqtn 2.Elementary steps must be physically reasonable (usu. bi- or lower molecularity) 3.Mechanism must correlate with the rate law Overall rate law is usually equivalent to the slowest step’s (the rate limiting step, RLS) rate law RLS can be picked out in a rxn energy diagram and predicted in a mechanism
Practice The rxn and rate law for the decomposition of dinitrogen pentoxide are 2N 2 O 5(g) → 4NO 2(g) + O 2(g) rate = k[N 2 O 5 ] 2 and the rxn energy diagram is given above. Which of the following mechanisms is most likely? A.One-step collisionC. 2N 2 O 5(g) → N 4 O 10(g) [fast] N 4 O 10(g) → 4NO 2(g) + O 2(g) [slow] B.2N 2 O 5(g) → N 4 O 10(g) [slow] N 4 O 10(g) → 4NO 2(g) + O 2(g) [fast]
Catalysis Increasing rxn rate by adding a catalyst Catalyst function: –Lowers E a increases k without being consumed or changing product amount Usually lowers E a by providing a different mechanism