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Chapter 14: Kinetics aka Rates of Reactions

Chapter 14: Kinetics aka Rates of Reactions. Vs. The rate of a reaction can be increased by:. Increasing concentration. Increases the probability that the reactants will collide successfully. Increasing temperature.

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Chapter 14: Kinetics aka Rates of Reactions

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  1. Chapter 14: Kinetics aka Rates of Reactions Vs.

  2. The rate of a reaction can be increased by: • Increasing concentration • Increases the probability that the reactants will collide successfully • Increasing temperature • Increases the probability that the reactants will collide with energy > activation energy (Ea) • Decreasing particle size (increasing surface area of reactants) • The presence of a catalyst • A catalyst speeds up the rate of a reaction without being used up by the reaction • Enzymes are biological catalysts • Heterogeneous catalysts are in a different state of matter than the reactants • Homogeneous catalysts are in the same state of matter as the reactants

  3. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB

  4. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB Second premise: Reactants must have the correct orientation to form the products upon collision

  5. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB E < Ea Second premise: Reactants must have the correct orientation to form the products upon collision Third premise: Reactants must have sufficient energy for the collision to result in formation of products

  6. Collision Theory First premise: Reactants must collide in order to react and form products. A2 + B2 2 AB E > Ea Second premise: Reactants must have the correct orientation to form the products upon collision Third premise: Reactants must have sufficient energy for the collision to result in formation of products

  7. Ea Energy DHrxn Reaction progress Activated complex: an unstable transition state between reactants and products. It can either fall back down on the reactant side or go on to the product side.

  8. Ea Ea Energy DHrxn Reaction progress A catalyst speeds up the rate of a reaction by lowering the activation barrier

  9. Ea Energy DHrxn Reaction progress A catalyst speeds up the rate of a reaction by lowering the activation barrier

  10. The Ea for an endothermic reaction is: DHrxn + energy barrier Ea Energy DHrxn Reaction progress

  11. A B Reaction rate laws: This reaction is first order in [A] and first order overall Rate = k[A] • k = specific rate constant • depends upon temperature • unique for every reaction • Indicates the probability of a successful conversion of reactants to products Reaction order is given by the sum of the exponents on the molar concentrations in the rate law expression

  12. aA + bB products Double [A] rate doubles First order in [A] Double [B] rate quadruples Second order in [B] General form of rate law: Rate = k[A]m [B]n • The only time m=a and n=b is when the reaction proceeds in a single step, which is uncommon. • The reaction order must be found experimentally by comparing reactant concentrations. Rate = k[A][B]2

  13. Calculating rate constants • Determine the rate law. • Substitute in the reactant concentrations and measured rates. • Solve for k. Rate = k[A][B]2 Rate = 2[A][B]2

  14. Instantaneous rates vs. average rates: • Average rates are found using: D[A] Dt 0.150 M – 0.200 M Avg. rate = 8 s – 0 s Example: [NO]0 = 0.200 M and [NO] = 0.150 M after 8 s. What is the average rate? = -6.25 x 10-3 M/s • Instantaneous rates are found using rate laws. Example: What is the instantaneous rate if the reaction above has the following rate law: Rate = (-0.2 M-1s-1) [NO]2 Rate = (-0.2 M-1s-1) [0.150 M]2 = -4.5 x 10-3 M/s

  15. Average rates and stoichiometry: C + 2 D 3A + B +D[D] -D[A] -D[B] = = = +D[C] 2t 3t t t k = (M/s)/(M4) = s-1 M-3 The rate at which [A] decreases is 1/3 the rate at which [B] decreases. The rate at which [C] increases is half the rate at which [D] increases. Units of the specific rate constant: Ex: Determine the units for k if Rate = k[A]3[B] The rate of a reaction is always Molarity/time (usually seconds) The concentrations are always in Molarity M/s = k (M)3(M)

  16. Integrated Rate Laws • Rate laws allow you to determine the rate of the reaction if the concentrations are known. • Integrated rate laws allow you to determine the concentration of one of the reactants at a particular time. Rate law order Integrated Rate Law Rate = k[A] 1 ln[A]t = -kt + ln[A]0 Rate = k[A]2 2 Think about the implications of the y = mx + b form of the integrated rate laws…

  17. (1) 1 1 ln[A] [A] [A] ln[A] time time time (2) time Determine the reaction order from the following plots: 2nd order 1st order

  18. After 1 half life (t1/2): [A]t = ½[A]0 Determining half-life from integrated rate law: • A half-life is the amount of time that is needed for ½ of the material to react. 1st order reactions: ln[A]t = -kt + ln[A]0 ln(½[A]0) = -kt1/2 + ln[A]0 ln(½) + ln[A]0 - ln[A]0 = -kt1/2 t1/2 = 0.693/k Note that the half-life is independent of concentration for 1st order reactions

  19. Second order reactions: At t1/2, [A]t = ½ [A]0 Note that the half-life increases with decreasing concentration.

  20. Temperature and Reaction Rates • Recall that the reactants have to have an energy > Ea for the reaction to be successful. • As a rough estimate: for every 10C temperature change, the reaction rate doubles. • Because temperature is a measure of the AVERAGE kinetic energy of the particles in the system, only a fraction of the particles present will have and energy > Ea. This fraction can be found using: Where Ea is in Joules, R = 8.314 J/mol·K and temperature is in Kelvin.

  21. The Arrhenius Equation: allows you to calculate Ea for a reaction from its specific rate constant • A is the frequency factor and is related to the number of collisions that will have the correct orientation. • A is essentially independent of temperature Ea can be calculated by determining k for the same reaction at several different temperatures by plotting lnk vs. 1/T Q: How would you determine Ea and A from an lnk vs. 1/T plot? A: The slope = -Ea/R and the intercept is lnA.

  22. Reaction Mechanisms: the steps through which a multistep chemical reaction occurs. Step 1: (fast) (CH3)3AuPH3 (CH3)3Au + PH3 Step 2: (slow) Step 3: (fast) (CH3)3Au C2H6 + (CH3)Au k1 (CH3)Au + PH3 (CH3)AuPH3 (CH3)3AuPH3 C2H6 + (CH3)AuPH3 k-1 14.88 In a hydrocarbon solution, the gold compound (CH3)3AuPH3 decomposes into ethane (C2H6) and a different gold compound, (CH3)AuPH3. The following mechanism has been proposed for this decomposition: k2 rds k3 Overall rxn: • Reactants and products in the elementary steps that cancel out and thus do not appear in the overall reaction are reaction intermediates.

  23. Molecularity: the number of molecules that must collide for an elementary reaction step to proceed Step 1: (fast) k2 (CH3)3AuPH3 (CH3)3Au + PH3 Step 2: (slow) Step 3: (fast) (CH3)3Au C2H6 + (CH3)Au k1 (CH3)Au + PH3 (CH3)AuPH3 k-1 # of molecules Molecularity 1 unimolecular 2 bimolecular 3 termolecular (very rare) k3 What is the molecularity of each step in the reaction mechanism? Step 1: unimolecular Step 2: unimolecular Step 3: bimolecular

  24. Rate laws and elementary steps: Recall that the coefficients of the reactants in single step reactions (one elementary step) are the reaction order for that reactant. Step 1: (fast) k2 (CH3)3AuPH3 (CH3)3Au + PH3 Step 2: (slow) Step 3: (fast) (CH3)3Au C2H6 + (CH3)Au k1 (CH3)Au + PH3 (CH3)AuPH3 k-1 k3 What is the rate law for each elementary step? Reaction order Step 1: rate = k1[(CH3)3AuPH3] 1st rate = k2[(CH3)3Au] Step 2: 1st Step 3: rate = k3[(CH3)Au][PH3] 1st in each reactant 2nd order overall

  25. Step 1: (fast) k3 k2 (CH3)3AuPH3 (CH3)3Au + PH3 Step 2: (slow) Step 3: (fast) (CH3)3Au C2H6 + (CH3)Au k1 (CH3)Au + PH3 (CH3)AuPH3 k-1 What is the rate law for this reaction? • Step 2 is rds, so rate = k2[(CH3)3Au] • Because the [(CH3)3Au] is unknown (it’s an intermediate), have to use the equilibrium in Step 1 to solve for [(CH3)3Au]. (Note that [(CH3)3Au]=[PH3]) [(CH3)3Au]2=K1[(CH3)3AuPH3] [(CH3)3Au]=(K1[(CH3)3AuPH3])1/2 Substituting in for [(CH3)3Au] rate = k2(K1[(CH3)3AuPH3])1/2

  26. Reaction energy diagrams for multistep reactions • Based on the following reaction profile, how many intermediates are formed in the reaction A  D? • How many transition states are there? • Which step is the fastest? • Is the reaction A  D exothermic or endothermic? 2 (B and C) 3 C  D endothermic B  C (slow) C  D (fast) A  B (fast)

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