B. Activation Energy: Ea

Transcription

1 B. Activation Energy: Ea a) Example reaction: the burning of charcoal in the BBQ C (s) + O 2(g) CO 2 (remember, burning is VERY exothermic) Question: Will charcoal in your BBQ spontaneously catch fire? You need to add a flame (and usually some lighter fluid). This provides the Question: Do you need to continue to add more energy to keep the reaction going? The energy released from the initial portion of the exothermic reaction provides the Ea for further reactants to become products. Activation Energy: This is the energy required to form the transition state complex from the reactants. This is an energy barrier that must be climbed over for the reaction to proceed. for an reaction For an reaction Kinetics - 25

2 C. Arrhenius Equation What It Tells Us The Arrhenius Equation gives the relationship between and. Rxn Rate Constant = k = A e -Ea/RT Arrhenius Equation Ea = activation energy in J/mol R = universal gas constant = A = Arrhenius pre-exponential (frequency) factor (very large #) This accounts for the frequency of collisions that have the correct to react. e -Ea/RT accounts for the fraction of collisions that have with enough to react The reaction rate increases exponentially as the temperature increases. This equation can also be linearized: Ea 1 ln k ln A R T y = b + (m) x Arrhenius Plots We can obtain a line by plotting vs. To obtain this graph: Do experiments at different. Determine k at each temp., take its natural log, and plot vs 1/T. These plots are used to determine the value of Ea (multiply the slope of the graph by to get Ea.) Kinetics - 26

3 D. Alternate (2-point / Slope) Version of Arrhenius Equation The activation energy can be related to the rate constants at just two temperatures using another form of the equation: ln k 2 E a 1 1 k 1 R T 2 T 1 For lab: Don t use 2 point slope form. Use best-fit line!!! LP # 9. The Ea for the reaction: 2 ClO 2 F (g) 2 ClOF (g) + O 2(g) is 186 kj/mol. If the value of k is s -1 at 322 o C, what is the value of k at 50 o C? SOLUTION: T1 = T2 = Ea = Kinetics - 27

4 Reaction Mechanisms Definition: The sequence of molecular events, or elementary reaction steps that defines the pathway from reactants to products. A. Elementary Reaction Steps 1. Definition: A molecular event ; (most often times it is a collision) 2. It is a single step in a reaction mechanism. Molecularity Steps are classified on the basis of their molecularity, or the number of particles involved in the collision. reaction - elementary reaction step that involves a single reactant molecule reaction - elementary reaction step that involve two atoms or molecules reaction - elementary reaction step that involve three atoms or molecules; rare An Is a reaction only having a single elementary reaction step. B. Multiple Step Reactions Multi-step rxns will have more than one hump in the reaction coordinate diagram. A species that is formed in one step of a reaction mechanism and consumed in a subsequent step. a. do not appear in the net equation for the overall reaction b. presence is only noticed in the elementary steps Kinetics - 28

5 C. An Overall Reaction It describes reaction stoichiometry. (The types of reactions we are used to writing.) It is the sum of all of the Elementary Reaction Steps Experimentally observed rate law for an overall reaction depends on the reaction mechanism. It will correspond to the rate determining elementary step. (This is the step with the largest.) (This is the step) Since elementary reactions have only one step, we can write the rate law directly from the reaction equation. LP# 10. Write the rate law for the following elementary reaction: O 3 + NO O 2 + NO 2 D. Rate Law from Reaction Mechanism For elementary reaction steps, reaction law coefficients are equal to the rate law exponents. For multi-step reaction, the rate law is based on the slowest step. Mechanism with a Slow Initial Step If the first step is slow, then the overall reaction will proceed at the rate of this slow initial step. The subsequent steps can only proceed as the first one provides products for it. Step Molecularity Example: 2NO(g) + H2(g) N2O(g) + H2O(g) Slow N2O(g) + H2(g) N2(g) + H2O(g) Fast 2NO(g) + 2H2(g) N2(g) + 2H2O(g) Overall Rate Law: The rate law will only include species that are part of the overall reaction. Kinetics - 29

6 Mechanism with a Fast Initial Step The rate law is based on a step other than the first step. The rate law will include species that are reaction that do not appear in the overall reaction. Through a series of substitutions, these reaction intermediate factors can be replaced in the rate law with factors of species that do appear in the overall reaction. Note: Although a proper rate law should not include any intermediates, for the purpose of this course, we will not spend the time on the process of substitutions. You may include intermediates in rate laws that you use for this course. E. Procedure used for establishing a reaction mechanism. 1. Determine the overall rate law experimentally. 2. Devise a series of elementary steps. 3. Predict the rate law based on the reaction mechanism. 4. If observed and predicted rate laws agree, the proposed mechanism is a probable pathway for the reaction. 5. Easy to disprove a mechanism; impossible to "prove" a mechanism. VIII. Catalysis A. How Do Catalysts Work? They work (generally) by OR Occasionally, they can make collisions more effective. Remember k = A e -Ea/RT Lowering Ea: Kinetics - 30

7 Example: 2 H 2 O 2(aq) 2 H 2 O (l) + O 2(g) + energy Reaction coordinate diagrams w/ and w/out the catalyst The two humps show that there are 2 steps in the reaction mechanism w/ catalyst. The # humps = # of steps in the reaction mechanism Hump sizes may be different. The larger hump is the slower, rate-determining step. Catalysts decrease Activation Energy by allowing access to a lower energy reaction mechanism. Hypothesized catalyzed Reaction Mechanism Step #1 H2O2 + I - H2O + IO - Step #2 H2O2 + IO - H2O + O2 + I - Overall 2 H2O2 2 H2O + O2 is a reaction intermediate is the catalyst It participates in forming the intermediate It is returned to solution. (It is not used up.) If there are 2 humps, there must be 2 transition states. Question: If this mechanism is correct, what would the rate law be? Which step is rate limiting? Ans: Step #1 is slow because it has the higher Ea hump Kinetics - 31

8 B. How do Changes in Ea Affect the Rate? The lower the value of Ea, the more molecules have enough energy to react. LP# 11.:Let s assume that the only change is to the activation energy. By what factor does the reaction rate change if we lower the activation energy from 75kJ/mol to 57kJ/mol (assuming T = 25 C)? Note: Both fractions of molecules w/ enough energy to react are very low. The bigger the value of Ea, the lower the fraction. When we lowered the value of Ea by only 25%, the reaction rate went up by a factor of about This means that there are almost 1450 times more particles with enough energy to react. Kinetics - 32