5.111 Lecture Summary #30 Monday, November 24, 2014

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1 5.111 Lecture Summary #30 Monday, November 24, 2014 Reading for Today: in 5 th ed and in 4 th ed. Reading for Lecture #31: 14.6, 17.7 in 5 th ed and 13.6, 17.7 in 4 th ed. Topic: Introduction to Kinetics I. Rates of Chemical Reactions II. Rate Expressions and Rate Laws Kinetics Versus Thermodynamics When considering a chemical reaction, one must ask whether the reaction will go forward spontaneously (thermodynamics), and the reaction will go (kinetics). Stable/unstable refers to ( Labile/nonlabile (inert) refers to the tendency to decompose) at which this tendency is realized Rate is important. A chemical kinetics experiment measures the rate at which the concentration of a substance taking part in a chemical reaction changes with time. Factors affecting rates of chemical reactions Let s consider the oscillating clock reaction To understand this reaction, one must consider thermodynamics, chemical equilibrium, acidbase, oxidationreaction, kinetics, and the influence of oxidation and liganded state to color. The overall reaction is: IO H 2 O 2 + CH 2 (CO 2 H) 2 + H + Its mechanism involves multiple steps, including: ICH(CO 2 H) O H 2 O (a) IO 3 + I + 2 H 2 O H + 2 O H 2 O + I 2 (spontaneous when [I 2 ] is low) (b) I 2 + CH 2 (CO 2 H) 2 ICH(CO 2 H) 2 + H + + I (spontaneous when [I 2 ] is high) Reaction (a): addition of IO 3 and I to hydrogen peroxide (H 2 O 2 ) under acidic conditions, turns a clear solution to amber (I is clear and I 2 is amber). Reaction (b): addition of I 2 (I 2 is amber) to malonic acid (CH 2 (CO 2 H) 2 ), generates a complex that is blue. Thus, the color of I depends on both oxidation and liganded state. Let s think about the oxidationreduction processes in Reaction (a): I in IO 3 is being to I2; I is being to I2; O in H 2 O 2 is being to O2; O in H 2 O 2 is being to H 2 O. With a large (+) E, H 2 O 2 is The reaction rate is also sensitive to temperature. 1

2 I. Rates of Chemical Reactions Measuring average reaction rates Consider: NO 2 (g) + CO (g) NO (g) + CO 2 (g) [NO] (M) Can monitor the changes in concentration of NO average rate = average rate = change in concentration change in time time (sec) average rate = M = sec average rate depends on time interval chosen Measuring instantaneous reaction rates Consider: NO 2 (g) + CO (g) NO (g) + CO 2 (g) Instantaneous rate = limit t 0 [NO] t + t [NO] t = d[no] t [NO] (M) As t approaches 0, the rate becomes the slope of the line tangent to the curve at time t. Instantaneous rate at t=150 sec is M = 7.7 x 10 5 M s sec Initial rate = Instantaneous rate at time equals sec time (sec) 2

3 Rate expressions Consider again: NO 2 (g) + CO (g) NO (g) + CO 2 (g) Can monitor NO or CO 2 increase or NO 2 or CO decrease rate = d[no 2 ] = d[co] = = Assuming no intermediate species and/or that the concentration of intermediates is independent of time Generally aa + bb cc + dd rate = 1 d[a] = 1 d[b] = 1 d[c] = 1 d[d] a b c d Example 2HI (g) H 2 (g) + I 2 (g) rate = = = II. Rate Laws The rate law is the relationship between the rate and the concentration, which are related by a proportionality constant,,call ed the. aa + bb cc + dd rate = k [A] m [B] n... m and n are order of reaction in A and B, respectively k is the rate constant Truths about rate laws (1) Rate law is a result of experimental observation. You CANNOT look at the stoichiometry of the reaction and predict the rate law (unless the reaction is an elementary reaction we will come back to this later). (2) The rate law is not limited to reactants. It can have a product terms, i.e., rate =k[a] m [B] n [C] c 3

4 (3) For rate = k[a] m [B] n, m is the order of reaction in A, n is the order of reaction in B. m and n can be integers, fractions, negative or positive. m = 0 Double concentration/ m = ½ Double concentration/ m = 1 First order k[a] Double concentration/ m = 2 Second order k[a] 2 Double concentration/ Triple concentration/ m = 1 Double concentration/ m = 1/2 Double concentration/ (4) The overall reaction order is the sum of the exponents in the rate law. For rate = k[a] 2 [B], the overall reaction order is order. order in A and order in B (5) The units for k vary. Determine units for k by considering units for rate and for concentration. Integrated Rate Laws Measuring initial rates can be difficult because it involves determining in concentrations that occur during short intervals in time. changes An alternative is to use the integrated rate law, which expresses concentrations directly as a function of time. 4

5 Integrated firstorder rate law First Order A B rate = d[a] = k[a] separate concentration and time terms 1 d[a] = k [A] [A] t 1 [A] d[a] = k o t ln [A] t ln = kt or ln [A] t = kt + ln Equation for straight line ln [A] t = kt [A] t [A] t = = e kt e kt Integrated 1 st order rate law Let s plot ln [A t ] versus time intercept = ln [A] t slope = ln [A] t = kt + ln y = mx + b time (sec) Rate constants can be determined from experiment by plotting data in this manner. 5

6 Firstorder Halflife Halflife is the time it takes for the original concentration to be reduced by hal f ( ). From above ln [A] t = kt ln 2 = kt 1/2 ln 1/2 = kt 1/ = kt 1/2 t 1/2 = First order half life depend on concentration. Half life depends on k, and k depends on the material in question. For the same material does it take longer to go from 1 ton to a ½ ton or 1 gram to a ½ gram?. k conc t 1/2 2nd 3rd half halflife life 6

7 Equation Sheet Exam 4 c = x 10 8 m/s ΔG = ΔH TΔS h = x J s ΔG = ΔG + RT ln Q 23 NA = x 10 mol 1 ΔG = RT ln K R = J/(K mol) ΔG = RT ln Q/K 1 ev = x J ln (K 2 /K 1 ) = (ΔH /R)(1/T 2 1/T 1 ) Kw = 1.00 x at 25.0 C ph pk a log (HA/A ) = ph + poh at 25.0 C + 1 I (Faraday's constant) = 96,485 C mol ph= log [H3O ] poh= log [OH ] Electromagnetic Spectrum: Violet ~ nm K w = K a K b pk = log K Blue ~ nm Green ~ nm Q = It Yellow ~ nm Orange ~ nm ΔG cell = (n)(i) ΔE cell Red ~ nm Complementary Colors: red/green, ΔE (cell) = E (cathode) E (anode) blue/orange, yellow/violet ΔE = E (reduction) E (oxidation) I < Br < Cl (weak field ligands) <F < OH < H 2 O (intermediate) ΔE cell = E cell (RT/nI)lnQ 9 <NH 3 < CO < CN (strong field ligands) RT/I = V at 25.0 C 1 Coulomb Volt = 1 Joule I/RT = V 1 at 25.0 C 1 Bq = 1 nuclei/sec 1A = 1C/s ln = log 1 J = 1 kgm 2 s 2 x = [b ± (b 2 4ac) 1/2 ]/2a ax 2 + bx + c = 0 E = hν = hc/λ 1W = 1 J/s ΔE cell = E cell [( V)(lnQ)/n] at 25.0 C ΔE cell = E cell [( V)(log Q)/n] at 25.0 C ln K = (ni/rt) ΔE A = A oe kt N = No e kt A = kn c = νλ [A] = [A] o e kt t ½ = ln2 / k 1/[A] = 1/[A] o + kt t ½ = 1 / k[a] o 7

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