Outline:1/31/07 n n Turn in Research Symposium Seminar reports – to me n n Exam 1 – two weeks from Friday… n Today: Start Chapter 15: Kinetics Kinetics & Reaction mechanisms

Chapters 6 and 14 introduced Thermodynamics: heat, work, energy, 1 st, 2 nd laws, state vs. path variables, spontaneity, etc. as related to chemical reactions…. Chapter 15 introduces: l the rate of reactions (kinetics) l the mechanisms of reactions These two concepts are closely related on a molecular level!

Is the rate of a reaction important? e.g. airbags…. Both rate of reaction and mechanism are vital to understanding this problem! Is the exact mechanism important? e.g. Ozone destruction (i) O 3 + Cl  O 2 + ClO (ii) O + ClO  O 2 + Cl O 3 + O  2 O 2

CFC + ultraviolet light  free Cl atoms h

Speed of a reaction is measured by the change in concentration with time. For a reaction A  B Reaction Rates

Consider: C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Most useful units for this rate = molarity/time. (Since volume is constant, molarity and moles are directly proportional.) Reaction Rates

 The average rate decreases with time?

 How do we get a useful number?  Plot [C 4 H 9 Cl] versus time:  The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve.  Instantaneous rate is different from average rate. Reaction Rates

All reaction “rates” slow down when viewed this way…

n What is the “rate” of a reaction? = the number of reactions/unit time = the number of reactions/unit time Why does the reaction slow down? n Go back to a molecular picture…

Answer: Per molecule it doesn’t !!! 6/12 = 50%3/6 = 50% (For some reactions…)

In general: rates decrease as concentrations decrease... there are stoichiometric factors… For a reaction (in general): aA + bB  cC + dD Reaction Rates (if rate is not specified for a particular substance!)

For example: NH 4 + (aq) + NO 2  (aq)  N 2(g) + 2H 2 O (l) Concentration and Rate

For the reaction NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) –as [NH 4 + ] doubles, the rate doubles... –as [NO 2 - ] doubles, the rate doubles... –rate  [NH 4 + ][NO 2 - ]. Rate law: The constant k is the rate constant. Concentration and Rate

Rate = k [A] x [B] y where: k is the rate constant [A],[B] are the concentrations of A,B x,y exponents are the reaction “order” Rate Laws

For a general reaction with rate law the reaction is mth order in reactant 1 and nth order in reactant 2. The overall order of reaction is m + n A reaction can be zeroth order if m, n, are zero. Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry.

A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. A reaction is first order if doubling the concentration causes the rate to double. A reaction is nth order if doubling the concentration causes an 2 n increase in rate. Note that the rate constant (k) does not depend on concentration.

Examples of reaction order: n n 1 st order: x = 1 or y = 1 e.g. Rate = k [A] n n 2 nd order: x = 2 or y = 2 or (x = 1, y = 1) e.g. Rate = k[A] 2 or Rate = k[A][B] n n 3 rd order: x = 3 or (x = 2, y = 1) etc. e.g. Rate = k[A] 3 or Rate = k[A] 2 [B]

Determining the Reaction Rate: Two proposed mechanisms for 2 NO 2  2 NO + O 2 A) step 1: NO 2  NO + O (slow) step 2: NO 2 + O  NO + O 2 (fast) B) step 1: 2 NO 2  NO 3 + NO (slow) step 2: NO 3  NO + O 2 (fast) Which is correct???

Determining the Reaction Rate: Two proposed mechanisms for 2 NO 2  2 NO + O 2 A) step 1: NO 2  NO + O (slow) step 2: NO 2 + O  NO + O 2 (fast) B) step 1: 2 NO 2  NO 3 + NO (slow) step 2: NO 3  NO + O 2 (fast) Unimolecular, so Rate = k[NO 2 ]Bimolecular, so Rate = k[NO 2 ] 2

Kinetics tells us about the mechanism! M/s Rate = k [NO 2 ] x 0.20 M/s = k [4.1] x 0.08 M/s = k [2.5] x 2.5 = [1.6] x  x = 2

B) step 1: 2 NO 2  NO 3 + NO (slow) step 2: NO 3  NO + O 2 (fast) Bimolecular, so Rate = k[NO 2 ] 2 Determining the Reaction Rate: Find the rate limiting step and use the reactant(s) and coefficient(s) in the rate law.

Rate = k [NO 2 ] 2 Rate-determining step

The rate equation cannot be predicted, it can only be measured empirically. Bottom line: Rate Law is related to the mechanism of the rate-determining step!

The rate equation cannot be predicted, it can only be measured empirically. n n Calculate k from initial rates n n Use the integrated form of the rate eqn. to solve for concentration (Section 15.4) There are two forms to know: First order: ln[A] = ln[A] o  k t Second order: 1/[A] = 1/[A] o  k t

k Can use data to find k and reaction order How do you find the reaction order?

Plot both…. ln[A] = ln[A] o  k t 1/[A] = 1/[A] o  k t Only one will be truly linear…. Rate = k [NO 2 ] 2 slope