Reading, etc. Introduction to Chemical Kinetics. Energy profile for a reaction. The reverse direction... Reaction Rates.

Transcription

1 Reading, etc. Introduction to Chemical Kinetics CHEM 7 T. Hughbanks Kinetics, Chapter Read and understand section., but the material is mainly for your edification. Read Sections.7.8 are descriptive, read for general ideas. Energy profile for a reaction Energy reactants E rxn E a activated complex products Thermodynamic Quantity Reaction Coordinate Rate-determining Quantity The reverse direction... Energy products E rxn activated complex Rate-determining for reverse reaction E a (reverse) reactants Thermodynamic Quantity Reaction Coordinate Chemical Kinetics Reaction rates How fast? Reaction mechanisms How? Answers to these questions depend on the path taken from reactants to products. Reaction Rates αa + βb γc + δd Follow progress by measuring any one concentration: α Δ[A], β Δ[B], γ Δ[C], Δ[D] δ Rates of change related by coefficients from balanced equation.

2 rate = Δ[NO 2 ] = 2 Δ[NO] = 2 Δ[O 2 ] 2 NO 2 2 NO + O 2 concentration NO O 2 NO 2 Factors Which Influence Rates Identity & form of reactants, products H 2 + I 2 vs. H 2 + Br 2 solution vs. gas phase, etc. Concentrations of various species usually reactants somes products, other species Temperature usually, faster at higher T strong dependence Catalysts Concentration Effects: Rate Laws αa + βb Products Empirically, usually find that Rate = k [A] n [B] m n = order of reaction with respect to A m = order of reaction with respect to B n + m= overall order of reaction k = rate constant = k (T) Reaction Orders Order of a reaction can NOT be found by looking at a balanced equation! αa + βb Products Rate = k [A] n [B] m In general: α & n, β & m are not necessarily equal because this isn t an elementary step Reaction order can only be found by experiments BUT Examples 2 N 2 O 5 4NO 2 + O 2 rate = k [N 2 O 5 ] 2 NO 2 2NO + O 2 rate = k [NO 2 ] 2 CAN T predict these from equations! BUT More Examples rate = H 2 + I 2 2HI rate = k [H 2 ][I 2 ] H 2 + Br 2 2HBr k [H 2 ][Br 2 ] /2 + k [HBr][Br 2 ] -

3 Finding rate laws, rate constants Method of Initial Rates combine known amounts of reactants determine rate by measuring change in some concentration over a short repeat with different initial concentrations find experimental rate law Problem A + 2B products Expt. [A] [B] Initial Rate find rate law & rate constant, k (concentrations in M, rates in M/min) Answer to problem rate = 2 Δ[B] = Δ[A] = k [A] 2 = (.32)[A] 2 mol L min Integrated Rate Laws From initial concentrations & rate law, we can predict all concentrations at any t. Mathematically, this is an initial value problem often involving a simple differential equation. Simplest Case: First Order A 2P rate = d[a] = k [A] dt By doing some calculus (pp ), we can turn this into an equation relating concentration and : d[a ] = k d t [A] [A] ln [A] t = k t First Order Reactions ln [A] = k t [A] = e k t so = [A] e k t From this, see that a plot of ln[a] vs. t will be a line with a slope of -k.

4 Second Order, one reactant rate = d[a] = k [A] 2 dt d[a] = = k d t [A] 2 [A] Which has the solution: = k t or [A] t = k t + [A] If second order kinetics apply, a plot of /[A] vs. t will be a line with slope k. st vs. 2 nd Order Kinetics [A] remaining Time (s) Both cases shown with k =.693 st order st Order Test Plot: ln[a] vs. t 2 nd Order Test Plot: /[A] vs. t ln [A] -3-4 st order /[A] 6 4 st order (s) (s) 2 nd Order Test Plot: /[A] vs. t /[A] (s) A Real Example... 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Experiment at T = 338 K gives: (s) [N 2 O 5 ] (M) Find rate law?

5 Zero Order? First Order? y = -.x R 2 = [N 2 O 5 ] ln[n 2 O 5 ] y = -.48x R 2 =.9997 Second Order? Example... /[N 2 O 5 ] y =.7839x R 2 = Graphs show us that the reaction is first order, so: rate = k [N 2 O 5 ] We can also find k from slope of graph: ln [N 2 O 5 ] = ln [N 2 O 5 ] k t So slope is equal to k. Fit gives us: k =.48 s More from Example. (half life) At what will the [N 2 O 5 ] be equal to one-half of its original value? Use integrated rate law: ln [N 2 O 5 ] = ln [N 2 O 5 ] k t solve for t: k t = ln [N 2 O 5 ] ln [N 2 O 5 ] or t = k ln [N 2 O 5 ] [N 2 O 5 ] Example... (half-life) t = k ln [N O ] 2 5 [N 2 O 5 ] = k ln 2 =.693 k So with k =.48 s, we will get t = t /2 = 44 seconds This is called the half-life of the reaction. For a first order reaction, half-life is independent of initial amount.