Quiz #4 Next Week (Feb 12-16) Covers sections During lectures next week, we will complete ALL OF Chapter 15 and parts of Chapter 16! The Second Midquarter Exam will cover Chapters 14, 15, and

Week six, a continuation of Chapter 14 Chemical Kinetics 14.1Factors that Affect Reaction Rates 14.2Reaction Rates 14.3Concentration and Rate 14.4The Change of Concentration with Time 14.5Temperature and Rate The Collision Model Activation Energy The Orientation Factor The Arrhenius Equation and Activation Energies 14.5Reaction Mechanisms Elementary Steps; Multistep Mechanisms Rate Laws for Elementary Steps Rate Laws for Multistep Mechanisms 14.7Catalysis Homogeneous and Heterogeneous Catalysis Enzymes

Note the DRAMATIC effect of temperature on k

Activation Energy Consider the rearrangement of acetonitrile: –In H 3 C-N  C, the C-N  C bond bends until the C-N bond breaks and the N  C portion is perpendicular to the H 3 C portion. This structure is called the activated complex or transition state. –The energy required for the above twist and break is the activation energy, E a. –Once the C-N bond is broken, the N  C portion can continue to rotate forming a C-C  N bond.

Activation Energy The change in energy for the reaction is the difference in energy between CH 3 NC and CH 3 CN. The activation energy is the difference in energy between reactants, CH 3 NC and transition state. The rate depends on E a. Notice that if a forward reaction is exothermic (CH 3 NC  CH 3 CN), then the reverse reaction is endothermic (CH 3 CN  CH 3 NC).

Activation Energy Consider the reaction between Cl and NOCl: –If the Cl collides with the Cl of NOCl then the products are Cl 2 and NO. –If the Cl collided with the O of NOCl then no products are formed. We need to quantify this effect.

Activation Energy

The Arrhenius Equation Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: –k is the rate constant, E a is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. –A is called the frequency factor, and is a measure of the probability of a favorable collision. –Both A and E a are specific to a given reaction.

The Arrhenius Equation If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: If we do not have a lot of data, then we can use Look Familiar???

Sample Exercise Temp. / o Ck / (s -1 ) x x x x10 -5

Sample Exercise Temp. ( o C)T (K)1/T (K -1 )k (s -1 )ln k x x x x10 -5

Sample Exercise Temp. ( o C)T (K)1/T (K -1 x 10 3 )k (s -1 )ln k x x x x10 -5

Sample Exercise Temp. ( o C)T (K)1/T (K -1 x 10 3 )k (s -1 )ln k x x x x

Fig For CH 3 NC  CH 3 CN reaction

From the graph we find the slope = -1.9 x 10 4 K But this is also equal to - E a /R Or E a = -(slope)(R) = - ( - 1.9x10 4 K)(8.314 J mol -1 K -1 )(1 kJ / 1000 J) = 1.6 x 10 2 kJ/mol or 160 kJ/mol

We can now use these results to calculate the rate constant at any temperature. To calculate k 1 for a temperature of K, make substitutions for all other parameters: E a = 1.6 x 10 2 kJ/mol k 2 = 2.52 x s -1 T 1 = K T 2 = K and

Activation Energy-orientation factor

Note the “A” term.

Note that termolecular reactions are extremely unlikely !

Multistep Mechanisms and Rate Laws Overall: NO 2 (g) + CO (g)  NO (g) + CO 2 (g) with an observed rate law of Rate = k[NO 2 ] 2 It appears that at temperatures below 225 o C, the reaction proceeds via two elementary steps: NO 2 + NO 2  NO 3 + NO (1) NO 3 + CO  NO 2 + CO 2 (2) Which yields two rate expressions: Rate 1 = k 1 [NO 2 ] 2 Rate 2 = k 2 [NO 3 ][CO] Important requirements: (a)Multi steps must add up to yield overall reaction. (b)may involve reactive intermediates (different from activated complexes). (c)One of these may be the ‘rate-determining’ step—the slower one! In this case given above, step (1) is the rate-determining step and step (2) is a faster step.

Another example: 2 NO 2 + F 2  2 NO 2 F observed rate = k[NO 2 ][F 2 ] Proposed mechanism: NO 2 + F 2  NO 2 F + F (1) slow F + NO 2  NO 2 F (2) fast Sum gives overall reaction, and rate = rate 1 = k 1 [NO 2 ][F 2 ]

Another example: overall 2 NO + Br 2  2 NOBr with an observed rate law of rate = k[NO] 2 [Br 2 ] The proposed mechanism is: NO + Br 2 == NOBr 2 (1) (fast) NOBr 2 + NO  2 NOBr(2) (slow) Our earlier guideline would suggest we use the slow step to determine the rate law, giving rate = k 2 [NOBr 2 ][NO] but this presents an immediate problem, since we don’t know what experimental quantities to put in for [NOBr 2 ] ! The solution comes from an analysis of the reversible fast reaction (1). rate forward = k 1 [NO][Br 2 ] and rate reverse = k -1 [NOBr 2 ] but these exist in a fast, dynamic equilibrium where rate forward = rate reverse

And this gives us the relationship rate forward = k 1 [NO][Br 2 ] = k -1 [NOBr 2 ] = rate reverse and

Quiz #4 Next Week (Feb 12-16) Covers sections During lectures next week, we will complete ALL OF Chapter 15 and parts of Chapter 16! The Second Midquarter Exam will cover Chapters 14, 15, and

Somewhat more complicated: 2 O 3  3 O 2 obs rate = k[O 3 ] 2 [O 2 ] -1 Proposed mechanism: O 3 = O 2 + O(1) fast, reversible O + O 3  2 O 2 (2) slow tells us rate = k 2 [O][O 3 ] now what??? assume k 1 [O 3 ] = k -1 [O 2 ][O] so that

Cl 2 + CHCl 3  HCl + CCl 4 rate = k[Cl 2 ] 1/2 [CHCl 3 ] Proposed mechanism: Cl 2 = 2 Cl(1) fast, reversible Cl + CHCl 3  HCl + CCl 3 (2) slow CCl 3 + Cl  CCl 4 (3) fast Evaluate the rate constant using this mechanism.

Catalysis A catalyst changes the rate of a chemical reaction without being consumed. Homogeneous catalysis: catalyst and reaction are in the same, single phase. Heterogeneous catalysis: catalyst and reaction are in different phases. Often the catalyst is a solid in contact with gaseous or liquid reactions. Enzymes: In living systems, usually a large molecule which catalyzes a specific reaction, an enzyme-substrate complex.

Homogeneous: catalyst and reaction are in same phase: Hydrogen peroxide decomposes very slowly: 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g) In the presence of the bromide ion, the decomposition occurs rapidly: –2Br - (aq) + H 2 O 2 (aq) + 2H + (aq)  Br 2 (aq) + 2H 2 O(l). –Br 2 (aq) is brown. –Br 2 (aq) + H 2 O 2 (aq)  2Br - (aq) + 2H + (aq) + O 2 (g). –Br - is a catalyst because it can be recovered at the end of the reaction and it makes the reaction rate faster. Generally, catalysts operate by lowering the activation energy for a reaction.

Catalysts can operate by increasing the number of effective collisions. That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing E a. A catalyst may add intermediates to the reaction. Example: In the presence of Br -, Br 2 (aq) is generated as an intermediate in the decomposition of H 2 O 2. When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction.

Heterogeneous Catalysis The catalyst is in a different phase than the reactants and products. Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars). Most industrial catalysts are heterogeneous. First step is adsorption (the binding of reactant molecules to the catalyst surface). Adsorbed species (atoms or ions) may have increased reactivity, but they are always easily available. Reactant molecules are also adsorbed onto the catalyst surface and may migrate to active sites.

–Consider the hydrogenation of ethylene: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g),  H o = -137 kJ/mol. –The reaction is slow in the absence of a catalyst. –In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature. –First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface. –The H-H bond breaks and the H atoms migrate about the metal surface. –When an H atom collides with an ethylene molecule on the surface, the C-C  bond breaks and a C-H  bond forms. –When C 2 H 6 forms it desorbs from the surface. –When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered.

ENZYMES Enzymes are biological catalysts. Most enzymes are protein molecules with large molecular masses (10,000 to 10 6 ). Enzymes have very specific shapes. Most enzymes catalyze very specific reactions. Substrates are the reactants that undergo reaction at the active site of an enzyme. A substrate locks into an enzyme and a fast reaction occurs. The products then move away from the enzyme.

Considerable research is currently underway to modify enzymes to prevent undesirable reactions and/or to prepare new products.

Only substrates that fit into the enzyme lock can be involved in the reaction. If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors). The number of events (turnover number) catalyzed is large for enzymes ( per second).

Nitrogenase in legumes converts N 2 to NH 3. ‘Fixation’ of N 2 ’ converts it to compounds useful to plants

Consider the reaction 2 N 2 O 5  4 NO 2 + O 2 Calculate E a from the following data: k/s -1 T/ o C 2.0 x x x x x

Formic acid alone, in the gas phase. Formic acid in presence of ZnO. Sample Problem, page :

The decomposition of formic acid shown on the previous slide is given by HCOOH (g )  CO 2 (g) + H 2 (g) It has been found to be first order at 838 K. see pp a)Estimate the half-life and first-order rate constant for the decomposition of pure formic acid and formic acid in the presence of ZnO. b)What is the effect of ZnO? c)Suppose we express the concentration of formic acid in mol/L. What effect would that have on the rate constant? d)The pressure of formic acid at the beginning of the reaction can be read from the graph. Assume constant T, ideal gas behavior, and a reaction volume of 436 cm 3. How many moles of gas are in the container at the end of the reaction? e)The standard heat of formation of formic acid vapor is ΔH o f = kJ/mol. Calculate ΔH o (see Sec. 5.7) for the overall reaction. Assume the activation energy, E a, is 184 kJ/mol, sketch an approximate energy profile for the reaction, and label E a, ΔH o f, and the transition state.

Chapter 15Chemical Equilibrium 15.1The Concept of Equilibrium 15.2The Equilibrium Constant The Magnitude of Equilibrium Constants The Direction of the Chemical Equation and K eq Other ways to Manipulate Chemical Equations and K eq Units of Equilibrium Constants– K p and K c 15.3Heterogeneous Equilibria 15.4Calculating Equilibrium Constants 15.5Applications of Equilibrium Constants Predicting the Direction of Reaction Calculation of Equilibrium Concentrations 15.6Le Châtelier’s Principle Change in Reactant or Product Concentrations Effects of Volume and Pressure Changes Effect of Temperature Changes The Effect of Catalysts

The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ]

The Equilibrium Constant Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2 What’s wrong (or not always correct) about this last statement?

The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = Always correct NOT always correct.

The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

The Equilibrium Constant To generalize this expression, consider the reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD