Chemical Kinetics Read Chapter 5: p160-162 Problems: 5.81, 5.85, 5.87, 5.89 We will look at: 1. numerical descriptions of how fast reactions (rxns.) occur 2. the intermediates that form during a rxn (re. mechanism) 3. applying thermodynamics & the kinetic molecular theory to go from the descriptive learning to understanding. Our focus will be on gas and liquid phase (soln.) reactions. Introduction What is kinetics, and why should we care? 1. Kinetics examines how fast a rxn. proceeds. 2. When we really understand the details of kinetics we have more control over the rxn. (Medicine & industry) 3. Kinetic principles are the foundation of equilibrium 4. You will remain alive & healthy only as long as you can control the rates of chemical (and physical) processes in your body. Two examples at the extremes of reaction rates: 1st: Formation of iron oxide from iron metal: 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) 1. Formation of iron oxide is rusting. (photo from Google images) 2. It took my Datsun B-210 ~ 9 years to rust. 2 nd : A more rapid way to oxidize a car: http://www.youtube.com/watch?v=90tta_vzsf0 Any thoughts about the chemical reaction type in this video? A. Reaction Rates. Do speed analogy to understand rates in general. 1. Defining rate: change per unit time rate = concentration change time change Page1
Analogy Broadened: Distance traveled Time Speed or Rate 120 miles 2 hr 60 mi/hr Process Time Speed or Rate Frost 12 cupcakes 6 min 2 cupcakes frosted/min Based on this, how is speed/rate determined? Chemical process Time Speed or Rate A B In a 1 L solution In a 1 L solution 0.08 moles of B formed 10 s 0.08 moles B formed/10 s (0.08 moles/l = 0.08 M) or 0.008 moles B formed/s Because this is occurring in 1 L, the rate of increase in concentration of B for this reaction is 0.008 M/s 2. Example: H 2 (g) + Cl 2 (g) 2 HCl(g) How could you measure the rate? a. Measure the concentration of H 2, Cl 2, or HCl at various times as the reaction proceeds. b. If you measure one of the reactants, H 2 or Cl 2, would we see an increase or decrease over time? c. If you measured HCl, a product, would you see it increase or decrease over time? In the kinetics lab, did we measure a product or reactant of the reaction? Did it increase or decrease over time? 3. Another concern is how does one measure how much reactant or product is present? One of the most desirable ways is to see a visible change. 4. Reaction used in the Kinetics lab 6 H + + IO 3 + 8 I 3 I 3 + 3 H 2 O (which has color?) So you looked at the increase in concentration of the product, I 3 -, Page2
[I3 - ] (M) at different times by looking at its color. Let s look at some data from the kinetics lab. rate I 3 formation = [I 3 ] t Note: The bracket, [ ], means the concentration of whatever is inside the bracket (usually in M). Trial #5 Time (s) Abs [I 3 ] (M) 30 0.013 4.56E-07 45 0.028 9.82E-07 60 0.039 1.37E-06 75 0.050 1.75E-06 90 0.062 2.17E-06 105 0.073 2.56E-06 120 0.088 3.09E-06 135 0.099 3.47E-06 150 0.112 3.93E-06 Shown above are trial #5 data from Sp 05. You can get an approximate rate using the data from two time points, say 30 and 60 s. Try it. 5e-6 Trial #5 a) You will get a more reliable answer if you estimate y x for all of the points. b) You can do this by plotting the data [I 3 ] (y) vs. time (x) and using the trendline function, ie using the slope of the line. Remember, slope = rate!!! 4e-6 3e-6 2e-6 1e-6 0 20 40 60 80 100 120 140 160 time (s) The slope of the line is 2.85 x 10 8 M s & the R 2 value = 0.9988. What does the R 2 value tell you? Page3
Rate Laws & Reaction Order Now that we know how to measure rates, what use are they? A. We need to relate our numerical analysis of rxn. rate to a mental picture (model) of the atoms/molecules doing the rxn. Look at Boltzmann. www.google.com Boltzmann 3D How many collisions per 30 sec with: (428 K) 1. Five NO 2 molecules (MW ~ 46 amu) & two CO (MW ~ 28 amu) 2. Fifty NO 2 molecules & two CO molecules So what happened to the collision rate when more molecules were present in the same space? B. Consider the following general rxn.: a A + b B Products 1. You can write a general rate law for this reaction in the form: Rate = k [A] m [B] n k is the rate constant specific for this rxn., temp., etc. Note: a (coefficient in the reaction) is not necessarily = m & b is not necessarily = n Important: Changing [A] & [B] changes the rxn rate as shown by the Boltzmann demo. 2. What can you do with a rate law? a) Accurately predict what will occur. (Example: How fast will a reaction occur?) b) Get an understanding of the rxn mechanism (more below) C. Reaction order (example: Rate = k [A] 3 [B] 1 ) 1. Overall reaction order = sum of all exponents in the rate law. (ex: reaction order = ) 2. Rxn order w/ respect to specific reactant = exponent for that reactant. (example: reaction order with respect to A = & reaction order with respect to B = ) D. If you have a balanced chemical reaction., you still don t know the rate law. The rate law must be determined experimentally!!! Page4
How can we do that? Experimental Determination of a Rate Law One method is to measure initial rate at different [reactant]. (The bracket here means concentration of reactant in mol/l.) Then see what exponent (remember m and n?) fits the data. Rxn.: 6 H + + IO 3 + 8 I 3 I 3 + 3 H 2 O Rate law (general form): rate = k [IO 3 ] m [I ] n Aside: You may be wondering what happened to the [H + ] term. Two points with respect to that: There are ways to set up the reaction conditions such that we make that term disappear (actually combine with k). A. To get m, you will compare experimental trials. Here are some data from Sp 05 (the [I ] was the same in trials #1-3 & #4-6, and the [IO 3 ] was twice as high in trials #4-6 as #1-3): Relative Concentrations Trials I - - IO 3 Rate (M/s) #1-3 1 1 1.35 x 10-8 #4-6 1 2 2.70 x 10-8 #7-9 2 1 ------------ rate for #1-3 = 1.35 x 10 8 M/s rate for #4-6 = 2.70 x 10 8 M/s Make a ratio of the rate law for trials 4-6 divided by the rate law for trials 1-3. rate 4-6 = k [IO 3 ] m 4-6 [I ] n rate 1-3 = k [IO 3 ] m 1-3 [I ] n Then fill in the appropriate concentrations and rate values 2.70 x 10 8 M/s = k [2] m [I ] n Can you see that the k 1.35 x 10 8 M/s k [1] m [I ] n and [I ] n cancel? Page5
2.70 x 10-8 M/s 1.35 x 10-8 M/s = 2 m 1 m 2 = 2 m What must m equal, for the equation to be true? 1 1 m If you can t see what m equals, just try some integer values. Hopefully, you will find that m = That is because as the concentration doubled, the rate also doubled. If the rate quadruples (increases by a factor of 4), when the concentration doubles, then m = If the rate goes up by a factor of 8, when the concentration doubles, then m = The logic is based on: 2 1 = 2, (Note: 2 2 2 = 4, 2 3 = 8, etc. 1 = 2 1. That is, you can factor out the m.) B. Following this logic, you could also determine n. What reactions should we compare to accomplish this? Trials. # Note: In this case we want [IO 3 ] to be constant. and #. C. Now that we know m = and n =, do we have enough information to determine k? Rearrange the rate law to solve for k in the space at right: Reaction Mechanisms What is really happening during the rxn? A reaction mechanism describes the actual changes that occur during a chemical rxn. A. Let s look at a mechanism for NO 2 (g) + CO (g) NO (g) + CO 2 (g) 1. Experiments have suggested a two step process: 1 st step: NO 2 (g) + NO 2 (g) NO(g) + NO 3 (g) 2 nd step: NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) Page6
2. While some of the reaction components are unstable, one component of elementary rxns. is very unstable. That is NO 3. A component like NO 3 that exists transiently (for a really short time) in the mechanism is a reaction intermediate. 3. Notice that the sum of the individual elementary steps must add up to the overall, balanced molecular equation: 1 st step: NO 2 (g) + NO 2 (g) NO(g) + NO 3 (g) 2 nd step: NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) NO 2 (g) + NO 2 (g) + NO 3 (g) + CO(g) NO(g) + NO 3 (g) + NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g) NO(g) + CO 2 (g) Terms occurring on both reactant & products sides of the sum are shown in bold. When these terms are dropped, the equation must match the original, balanced overall rxn. How does a mechanism relate to a rate law? The balanced molecular equation for a rxn describes the reactant & product ratios (stoichiometry), not necessarily the mechanism. However, the rate law for the rate limiting step in a mechanism defines the rate law for the overall rxn. A. The rate limiting step in any process is the slowest step. (Think about waiting in line.) Have you ever worked on/or seen an assembly line? B. When the overall rxn occurs in a single step, then the rate law is determined by the molecular equation. Such a reaction is: CH 3 Br(aq) + OH (aq) CH 3 OH(aq) + Br (aq) So: Rate = Δ CH 3 OH Δt = k [CH 3 Br] [OH ] C. When a rxn. occurs with more than one step in the reaction mechanism, usually one of the steps is much slower than the others. This is the rate limiting step, and it determines the mechanism 1. Let s go back and look at the rxn. of NO 2 with CO: Balanced rxn.: NO 2 (g) + CO (g) NO (g) + CO 2 (g) Page7
Proposed mechanism: k 1 1 st step: NO 2 (g) + NO 2 (g) NO(g) + NO 3 (g) k 2 2 nd step: NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) The 1 st step is the slow step in this sequence. By slow, I mean slow relative to the 2 nd step. That is, k 1 << k 2. This means that the rate of the overall process can be no faster than the rate of step 1. Can we write a rate law for the 1 st elementary rxn.? Rate = -Δ NO 2 Δt = k 1 [NO 2 ] 2 The experimentally determined rate law rxn does turn out to be: Rate = -Δ NO 2 = k 1 [NO 2 ] 2 Δt Therefore our mechanism predicted the correct rate law. There is a mutually reinforcing relationship between the mechanism and the rate law: The rate law helps to confirm the mechanism, and the mechanism gives us insight into the rate law. Reaction Rates and the Effect of Temperature A. Some of the thoughts behind this part of kinetics started with analysis of collision rates in the gas phase. At 20 C and 1 atm, each gas molecule has about 10 9 collisions/sec. Imagine a reaction occurring with each collision. BOOM!!! This does seem consistent with the rates of explosive chemical rxns, It is not consistent with the rates of most chemical rxns. So, for most rxns, only a small % of the collisions result in a rxn. These productive collisions have appropriate energy & orientation. B. Energy input (activation energy) is required for a reaction to occur. 1. Consider the rxn.: O(g) + HCl(g) OH(g) + Cl(g) Remember: A reaction can only occur between two atoms/ molecules if they collide with one another. In this rxn, the H Cl bond must be broken (requires energy), and an O H bond must be formed.(think about overlap of electron clouds.) O H Cl O H Cl O H + Cl Page8
2. The intermediates in a rxn (O H Cl, in this case) are usually very unstable & high in potential energy. Terms: a) activation energy (E a ) is the energy hill that must be achieved before a reaction can proceed. b) transition state (synonym: activated complex) is the highest energy point, or the top of the energy hill. (Try to relate the process of going from reactants to the transition state climbing up a mountain.) G reaction progress C. Reaction rates tend to increase with increasing temperature. Why is this so? The fraction of collisions with enough energy to reach E a is related to the Boltzmann Distribution. Assume for a given reaction we need a velocity greater than 1000 m/s to reach E a. a) Look at Boltzmann at 302 K for O 2 (MW = 32). What % of the O 2 molecules has a velocity greater than 1000 m/s? b) Now increase the temperature to around 600 K. Does the fraction with a velocity greater than 1000 m/s increase? This is why increasing T increases the rxn rate, by increasing k. D. Appropriate orientation is required for a productive collision. Go back to our previous example of a gas phase rxn.: O(g) + HCl(g) OH(g) + Cl(g) Page9
What if O collides with the Cl side of the HCl molecule, instead of the H side H Cl O N.R. No H to O bonding occurs. Therefore, many collisions do not result in a reaction. E. Why do people investigate this stuff? 1. If you understand a reaction, you can modify it to change the rate, improve efficiency, and so on. 2. Sometimes scientists look for molecules that can act as catalysts to speed up reactions. 3. Enzymes are catalysts that increase the rate of reactions in living systems. (What enzyme did we discuss on the first day of class?). Sometimes scientists look for inhibitors of an enzyme to reduce the rate of a reaction. Catalysis How we make life go. A. Catalysts work by lowering the activation energy. B. They do this by creating a different reaction pathway. Analogy: Take a 600 pound cow from the 1 st floor of the Smith Building to the 4 th floor. Think of two different pathways. 1. 2. 3. Another silly way? Which pathway has the lowest energy? C. Do catalysts speed up the rate of the reverse rxn too? Look at the reaction coordinate diagram. G reaction progress What components of the diagram relate to kinetics? Page10