A. Reaction Rates. Do speed analogy to understand rates in general. miles traveled. kilometers traveled

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1 Chemical Kinetics On-line: 1. Chap Chapter sections (pp ) Introduction What is kinetics, and why should we care? 1. Kinetics examines how fast a rxn. proceeds. 2. 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 O2(g) 2 Fe2O3(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: Any thoughts about the chemical reaction type in this video? A. Reaction Rates. Do speed analogy to understand rates in general. miles traveled 0. Think about driving. Your speed is how many or (mph). If we were in Canada, it would be kilometers traveled hour hour In both cases it is the distance traveled divided by the time. How do you calculate the distance traveled? Let say the car odometer read 1557 mile at the beginning of your trip and 1598 mile at the end of your trip. The change in distance If it took 0.70 hours, your speed 1. Defining rate: change per unit time rate concentration change time change Page1

2 Chemical process A B Time 0.08 moles of B formed 10 s 0.08 moles B formed 10 sec Speed or Rate or moles B formed 1 sec Say this was in 1 Liter so (0.08 moles/l 0.08 M) 0.08 M B formed 10 s 0.08 M B formed 10 sec or M B formed 1 sec So the rate of increase in concentration of B for this reaction is M/s 2. Example: H2(g) + Cl2(g) 2 HCl(g) How could you measure the rate? a. Measure the concentration of H2, Cl2, or HCl at various times as the reaction proceeds. b. If you measure one of the reactants, H2 or Cl2, 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 + + IO3 + 8 I 3 I3 + 3 H2O (which has color?) So you looked at the increase in concentration of the product, I3 -, at different times by looking at its color. Let s look at some data from the kinetics lab. rate I3 formation Δ [I3 ] Δ t Note: The bracket, [ ], means the concentration of whatever is inside the bracket (usually in M). Page2

3 [I3 - ] (M) Trial #5 Time (s) Abs [I3 ] (M) E E E E E E E E E-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. concentration time conc. final - conc. initial time final - time initial 3.09 x x 10 6 M s Now lets do it again with the data from 30 and 60 s. Change in concentration change in time Trial #5 But a) You will get a more reliable answer if you estimate Δy for all Δx of the points. 5e-6 4e-6 3e-6 b) You can do this by plotting the data [I3 ] (M) on the y-axis and time (s) on the x-axis, and using the trendline function, ie using the slope of the line. Remember, slope rate!!! The slope of the line is 2.85 x 10 8 M s & the R 2 value What does the R 2 value tell you? 2e-6 1e time (s) Page3

4 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. Boltzmann 3D How many collisions per 30 sec with: (428 K) 1. Five NO2 molecules (MW ~ 46 amu) & two CO (MW ~ 28 amu) 2. Fifty NO2 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 the concentration of reactants changes the reaction 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 C. Reaction order (example: Rate k [A] 2 [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. Page4

5 The rate law must be determined experimentally. 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 + + IO3 + 8 I 3 I3 + 3 H2O Rate law (general form): rate k [IO3 ] 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 [IO3 ] was twice as high in trials #4-6 as #1-3): Relative Concentrations Rate Trials I - - IO 3 (M/s) # x 10-8 # x 10-8 # rate for # x 10 8 M/s rate for # 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. rate4-6 k [IO3 ]4-6 m [I ] n rate1-3 k [IO3 ]1-3 m [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? 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. Page5

6 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, 2 2 4, 2 3 8, etc. (Note: 2m 1 m ( 2 1 )m. 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. # and #. Note: In this case we want [IO3 ] to be constant. 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 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. In the rate law, these energy and orientation aspects are collected into the rate constant, k. B. Energy input (activation energy) is required for a reaction to occur. 1. Consider the rxn.: O(g) + HCl(g) OH(g) + Cl(g) Page6

7 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 intermediate structure 2. The intermediates in a rxn (O H Cl, in this case) are usually very unstable & high in potential energy. Terms: a) activation energy (Ea) 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.) 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? Page7

8 Initial temp Temp has been raised This is why increasing T increases the rxn rate. Increasing T increases k. Larger k means a faster rate. Rate k [A] m [B] n 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) 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 Page8

9 3. Enzymes are catalysts that increase the rate of reactions in living systems.. Sometimes scientists look for 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 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. What components of the diagram relate to kinetics? Page9