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CK-1Chemical Kinetics Chemical kinetics is the study of time dependence of the change in the concentration of reactants and products. “The field of chemical.

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Presentation on theme: "CK-1Chemical Kinetics Chemical kinetics is the study of time dependence of the change in the concentration of reactants and products. “The field of chemical."— Presentation transcript:

1 CK-1Chemical Kinetics Chemical kinetics is the study of time dependence of the change in the concentration of reactants and products. “The field of chemical kinetics has not yet matured to a point where a set of unifying principles has been identified…There are many different theoretical models for describing how chemical reactions occur.” (M&S, 1137) Reactant  Product Generally, we want to understand the rate of reaction:

2 CK-2Recall from Slide CEq-2… ReactantsProducts For each constituent…

3 CK-3Rate of reaction, v(t) v(t), the rate of reaction, is defined as the rate of change in ξ (t) with time per unit volume Note all quantities are positive. What are units of v(t) ? 2NO(g) + O 2 (g)  2NO 2 (g) Examples: H 2 (g) + I 2 (g)  2HI (g)

4 CK-4The integrated rate law Rate laws must be determined experimentally and, generally, cannot be deduced from the balanced reaction!! The rate can be expressed as a function of the reactant concentrations. Most common function is of form: 2NO(g) + O 2 (g)  2NO 2 (g) General Example m th order in A n th order in B (m+n) th order overall The Order: Determined experimentally!

5 CK-5Finding rate laws experimentally Method of isolation Set up reaction so one reactant is in excess. Any change in rate will be due to changes in other reactant. Repeat for other reactant. Method of initial rates Measure concentration change as a function of time, ~ v(t), for a series of experimental conditions. (Conditions must include sets where the reactant A has the same initial concentration but B changes and vice versa). There are two common methods for determining rate laws: EX-CK2 where

6 CK-6 Units of k, rate constant Rate LawOrderUnits of k v = k 0 v =k[A] 1 v = k[A] 2 2 v = [A][B] 1 in [A], [B] 2 overall v = k[A] 1/2 1/2 concentration time (concentration) m (concentration) n

7 CK-7Rate laws can be complicated H 2 (g) + I 2 (g)  2HI (g) H 2 (g) + Br 2 (g)  2HBr (g) These rate laws suggest that these two reactions occur via different mechanisms (sets of individual steps). The first may be a elementary reaction (one step) whereas the latter is certainly a multistep process. We will soon explore how to obtain complicated rate laws from suggested mechanisms. EX-CK1

8 CK-8 Elementary Rxns vs. Complex Reactions Elementary Reactions Complex Reactions Reactants  Products Reactants  Intermediates  Products Mechanism of a complex reaction is a sequence of elementary reactions. Molecularity of Elementary Reactions: Unimolecular Bimolecular Termolecular

9 CK-9First order reactions The reaction:has rate law: Let’s integrate… Solution: First order reactions decay exponentially.

10 CK-10Ozone decays via first order kinetics k = 1.078 × 10 -5 s -1 at 300 K What is slope?

11 CK-11What happens as k increases? k = 0.0125 s -1 k = 0.0250 s -1 k = 0.0500 s -1 k = 0.1000 s -1

12 CK-12Half-life of a first order reaction Figure 28.3 The half-life, t 1/2, is the time it takes to fall to ½ of the starting concentration: At,

13 CK-13Other order reactions… Second order reaction: Second order rate: Integrated rate law: Zero order reaction: Zero order rate: Integrated rate law:

14 CK-14Pseudo-first order reactions You can “overload” the other reactants to determine the order with respect to one individual reactant (method of isolation). For, what happens if [ B ] >> [ A ]? Similar to S N 2 Lab

15 CK-15 Reversible reactions (small  r G ) ABAB k1k1 k -1 Assume first order, elementary rxn in both directions Rate: Conservation of Mass: Integrate:

16 CK-16At equilibrium ABAB k1k1 k -1 At equilibrium… What is the equilibrium constant for this reaction? The forward rate equals the reverse at equilibrium. In terms of rate constants?

17 CK-17 Temperature Dependence of k Svante Arrhenius Winner of the 3 rd Nobel Prize in Chemistry The rate constant can vary in different ways with T. Differential form of the Arrhenius Equation:

18 CK-18Arrhenius Parameters E a is the activation energy. This is the energy required to get over a barrier (at the activated or transition state) between the reactants and products. E a has units of energy and is T independent. Integrated forms of Arrhenius equation: Activated (or transition) state A is the pre-exponential or Arrhenius factor and is T dependent. A is a measure of rate at which collisions occur (and takes lots of things into acct such as orientation, molecular size, number of molecules per volume, molecular velocity, etc). 2HI(g)→I 2 (g) + H 2 (g)

19 CK-19Transition-State Theory AB ‡ is the transition state (or activated complex.) Transition state theory assumes that the transition state and reactants are in equilibrium with each other, and uses concepts from chemical equilibrium and statistical mechanics to find kinetic info such as rate constants! ‡ ‡ Eyring Equation (key to transition-state theory) ‡ From CEq: So… ‡‡ ‡ Change in Gibbs energy from reactants to TS Entropy of activation Enthalpy of activation

20 CK-20 Relating E a to thermodynamics! Arrhenius Equation: Differentiate wrt T: or From Eyring Equation: van’t Hoff Equation (for K c ): Putting it all together… or Necessary Pieces… ‡ ‡ ‡

21 CK-21 What about A, the pre-exponential? and so A  A ‡  Products A+B  AB ‡  Products Unimolecular Gas Phase Reaction Bimolecular Gas Phase Reaction so What is A ? ‡ ‡‡‡ ‡ ‡ ‡ ‡ ‡ ‡ Same for reactions in solution ‡‡

22 CK-22Transition State Theory and NMR Lab In the NMR/N,N-DMA Paper, Gasparro et al. found an activation energy of 70.3 kJ/mol and a pre-factor of 1.87 × 10 10 s -1. Using these values, and a temperature of 298 K, find… ‡‡‡ Why is TST important? 1.Provides details of a reaction on the molecular scale. 2.Connects quantum mechanics and kinetics. 3.Currently used for many computational studies on reaction rates. EX-CK3

23 CK-23 Chapter 29: Reaction Mechanisms

24 CK-24Always remember…. One can never prove a reaction mechanism, although evidence may disprove a mechanism. Verifying proposed mechanisms requires extensive experimental verification of each proposed step!

25 CK-25Let’s examine a reaction … Reaction could progress in multiple ways… How can we distinguish? Case 1: One elementary stepCase 2: Two step reaction EX-CK4

26 CK-26Now let’s focus on the intermediate… How do k 1 and k 2 relate in case a? in case b? (a) I forms quickly but decays slowly… k 1 is fast relative to k 2. (b) I builds up to a constant, nearly negligible, concentration until near end of reaction. … k 1 is slow relative to k 2. Steady state approximation... Valid only if k2 > k1. EX-CK5

27 CK-27 Rate Laws do not yield unique mechanisms EX-CK6 An empirically determined rate law does not imply a unique reaction mechanism! Consider reaction: Experimentally, it was determined that the rate is given by: Researchers proposed two possible mechanisms. They need to determine if one of them is correct. So how would researchers distinguish between the mechanisms?

28 CK-28Remember the Chain Rxn from CK-6? H 2 (g) + Br 2 (g)  2HBr (g) Proposed Mechanism Initiation: Propagation: Inhibition: Termination: EX-CK7

29 CK-29What about the solution kinetics lab?! Now that we’ve explored reaction mechanisms and rate laws, let’s try to derive the rate laws from the solution kinetics lab… EX-CK8

30 CK-30Catalysis Catalyst: A substance that participates in the chemical reaction but is not consumed. Provides a new mechanism for reaction and can cause reaction to occur faster. In an experiment involving a catalyst, there are two competing reactions: If both reactions are elementary, overall rate is given by: EX-CK9

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