Chemistry 232 Chemical Kinetics

The Connection Between Chemical Reactions and Time Discussed the energetics of chemical reactions. 2 A  B combination reaction. How long it will take for this reaction to occur?

Chemical Kinetics Chemical kinetics - speed or rate at which a reaction occurs How are rates of reactions affected by Reactant concentration? Temperature? Reactant states? Catalysts?

The Instantaneous Reaction Rate Consider the following reaction A + B  C Define the instantaneous rate of consumption of reactant A, A

Reaction Rates and Reaction Stoichiometry Look at the reaction O3(g) + NO(g) ® NO2(g) + O2(g)

Another Example 2 NOCl (g)  2 NO + 1 Cl2 (g) WHY? 2 moles of NOCl disappear for every 1 mole Cl2 formed.

The General Case a dt b dt c dt d dt a A + b B ® c C + d D rate = -1 d[A] = -1 d[B] = +1 [C] = +1 [D] a dt b dt c dt d dt Why do we define our rate in this way? removes ambiguity in the measurement of reaction rates in that we now obtain a single rate for the entire equation, not just for the change in a single reactant or product.

Alternative Definition of the Rate Rate of conversion related to the advancement of the reaction, . V = solution volume

An Example Examine the following reaction 2 N2O5 (g)  4 NO2 (g) + O2 (g) N2O5 NO2 O2 Initial nø Change -2 +4 + Final nø - 2 4 

The N2O5 Decomposition Note – constant volume system

The Rate Law Relates rate of the reaction to the reactant concentrations and rate constant For a general reaction a A + b B + c C ® d D + e E

Rate Laws (Cont’d) The only way that we can determine the superscripts (x, y, and z) for a non-elementary chemical reaction is by experimentation. Use the isolation method (see first year textbooks).

 x + y + z = reaction order For a general reaction  x + y + z = reaction order e.g. X = 1; Y = 1; Z = 0 2nd order reaction (x + y + z = 2) X = 0; Y = 0; Z = 1 (1st order reaction) X = 2; Y = 0; Z = 0 (2nd order)

Integrated Rate Laws The rate law gives us information about how the concentration of the reactant varies with time How much reactant remains after specified period of time? Use the integrated rate laws.

Rate = v = - d[A]/d t = k[A] First Order Reaction A  product Rate = v = - d[A]/d t = k[A] How does the concentration of the reactant depend on time? k has units of s-1

The Half-life of a First Order Reaction For a first order reaction, the half-life t1/2 is calculated as follows.

Radioactive Decay Radioactive Samples decay according to first order kinetics. This is the half-life of samples containing e.g. 14C , 239Pu, 99Tc. Example

A + B ® products v = k[A][B] Second Order Reaction A ® products v = k[A]2 A + B ® products v = k[A][B] Case 1 is 2nd order in A Case 2 is 1st order in A and B and 2nd order overall

The Dependence of Concentration on Time For a second order process where v = k[A]2

Half-life for This Second Order Reaction. [A] at t = t½ = ½ [A]0

Other Second Order Reactions Examine the Case 2 from above A + B ® products v = k[A][B]

A Pseudo-first Order Reaction Example hydration of methyl iodide CH3I(aq) + H2O(l)  CH3OH(aq) + H+(aq) + I- (aq) Rate = k [CH3I] [H2O] Since we carry out the reaction in aqueous solution [H2O] >>>> [CH3I] \ [H2O] doesn’t change by a lot

Pseudo-first Order (cont’d) Since the concentration of H2O is essentially constant v = k [CH3I] [constant] = k` [CH3I] where k` = k [H2O] The reaction is pseudo first order since it appears to be first order, but it is actually a second order process.

Reactions Approaching Equilibrium Examine the concentration profiles for the following simple process. A  B

Approaching Equilibrium Calculate the amounts of A and B at equilibrium.

The Equilibrium Condition At equilibrium, vA,eq = vB,eq.

Temperature Dependence of Reaction Rates Reaction rates generally increase with increasing temperature. Arrhenius Equation A = pre-exponential factor Ea = the activation energy

Rate Laws for Multistep Processes Chemical reactions generally proceed via a large number of elementary steps - the reaction mechanism The slowest elementary step Þ the rate determining step (rds).

Elementary steps and the Molecularity Kinetics of the elementary step depends only on the number of reactant molecules in that step! Molecularity  the number of reactant molecules that participate in elementary steps

The Kinetics of Elementary Steps For the elementary step unimolecular step For elementary steps involving more than one reactant a bimolecular step

For the step a termolecular (three molecule) step. Termolecular (and higher) steps are not that common in reaction mechanisms.

The Steady-State Approximation Examine the following simple reaction mechanism Rate of product formation, vp,is proportional to the concentration of an intermediate.

What Is an Intermediate? B is an intermediate in the above reqction sequence. A species formed in one elementary step of a reaction mechanism and consumed in one or more later steps. Intermediates – generally small, indeterminate concentrations.

Applying the Steady State Approximation (SSA) Look for the intermediate in the mechanism. Step 1 – B is produced. Reverse of Step 1 – B is consumed. Step 2 – B is consumed.

The SSA (Cont’d) The SSA applied to the intermediate B.

SSA – The Final Step Substitute the expression for the concentration of B into the rate law vp.

The Michaelis-Menten Mechanism Enzyme kinetics – use the SSA to examine the kinetics of this mechanism. ES – the enzyme-substrate complex.

Applying the SSA to the Mechanism Note that the formation of the product depends directly on the [ES] What is the net rate of formation of [ES]?

ES – The Intermediate Apply the SSA to the equation for d[ES]/dt = 0

Working Out the Details Let [E]o = [E] + [ES] Complex concentration Initial enzyme concentration Free enzyme concentration Note that [E] = [E]o - [ES]

The Final Equation Substituting into the rate law vp.

The Michaelis Constant and the Turnover Number The Michaelis Constant is defined as The rate constant for product formation, k2, is the turnover number for the catalyst. Ratio of k2 / KM – indication of catalytic efficiency.

The Maximum Velocity As [S]o gets very large. Note – Vmax is the maximum velocity for the reaction. The limiting value of the reaction rate high initial substrate concentrations.

Lineweaver-Burk Equation Plot the inverse of the reaction rate vs. the inverse of the initial substrate concentration.

Lindemann-Hinshelwood Mechanism An Early attempt to explain the kinetics of complex reactions. Mechanism Rate Laws

The ‘Activated’ Intermediate Formation of the product depends directly on the [A*]. Apply the SSA to the net rate of formation of the intermediate [A*]

Is That Your ‘Final Answer’? Substituting and rearranging

The ‘Apparent Rate Constant’ Depends on Pressure The rate laws for the Lindemann-Hinshelwood Mechanism are pressure dependent. High Pressure Case Low Pressure Case

The Pressure Dependence of k’ In the Lindemann-Hinshelwoood Mechanism, the rate constant is pressure dependent.