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Chemical Kinetics

Chemical Kinetics. Unit 11. Chemical Kinetics. Chemical equations do not give us information on how fast a reaction goes from reactants to products. KINETICS : the study of reaction rates and their relation to the way the reaction proceeds, i.e. its mechanism

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Chemical Kinetics

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  1. Chemical Kinetics Unit 11

  2. Chemical Kinetics • Chemical equations do not give us information on how fast a reaction goes from reactants to products. • KINETICS: the study of reaction rates and their relation to the way the reaction proceeds, i.e. its mechanism • We can use thermodynamics to tell if a reaction is product – or reactant – favored • Only kinetics will tell us how fast the reaction happens!

  3. Rate of Reaction • A rate is any change per interval of time. • Example: speed (distance/time) is a rate! • Reaction rate = change in concentration of a reactant or product with time

  4. Expressing a Rate For the reaction A  P = Appearance of product Disappearance of reactant

  5. Reaction Conditions & Rates Collision Theory of Reactants • Reactions occur when molecules collide to exchange or rearrange atoms • Effective collisions occur when molecules have correct energy and orientation

  6. Factors Affecting Rates • Concentrations (and physical state of reactants and products) • Temperature • Catalysts • Catalysts are substances that speed up a reaction but are unchanged by the reaction

  7. Effect of Concentration on Reaction Rate To propose a reaction mechanism, we study the reaction rate and its concentration dependence.

  8. Rate Laws or Rate Expressions The rate law for a chemical reaction relates the rate of reaction to the concentration of reactants. For aA + bB  cC + dD The rate law is: Rate = k[A]m[B]n • The exponents in a rate law must be determined by experiment. • They are NOT derived from the stoichiometry coefficients in an overall chemical equation.

  9. Rate Laws & Orders of Reactions Rate Law for a reaction: Rate = k[A]m[B]n[C]p The exponents m, n, and p • Are the reaction order • Can be 0, 1, 2, or fractions (may be other whole numbers in fictional examples) • Must be determined by experiment Overall Order = sum of m, n, and p

  10. Interpreting Rate Laws Rate = k[A]m[B]n[C]p • If m = 1 (1st order) Rate = k [A]1 If [A] doubles, then the rate doubles (goes up by a factor of 2) • If m = 2 (2nd order) Rate = k [A]2 If [A] doubles, then rate quadruples (increases rate by a factor of 4) • If m = 0 (zero order) Rate = k [A]0 If [A] doubles, rate does not change!

  11. Rate Constant, k Relates rate and concentration at a given temperature. General formula for units of k: M(1- overall order) time-1

  12. Rate Law Problem: The initial rate of decomposition of acetaldehyde, CH3CHO, was measured at a series of different concentrations and at a constant temperature. Using the data below, determine the order of the reaction – that is, determine the value of m in the equation CH3CHO(g)  CH4(g) + CO(g) Rate = k[CH3CHO]m

  13. Strategy Use the equation: Pick any two points from the given data!

  14. Deriving Rate Laws Rate of rxn = k[CH3CHO]2 Here the rate goes up by FOURwhen the initial concentration doubles. Therefore, we say this reaction is SECONDorder overall.

  15. Example: Using the same set of data from the previous example, and knowing the order of the reaction, determine: b) the value of the rate constant, k (w/ units!) c) the rate of the reaction when [CH3CHO] = 0.452 mol/L Strategy: • Use any set of data to find k. • Solve for rate using k, rate order equation, and given concentration.

  16. The data below is for the reaction of nitrogen (II) oxide with hydrogen at 800oC. 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) Determine the order of the reaction with respect to both reactants, calculate the value of the rate constant, and determine the rate of formation of product when [NO]=0.0024 M and [H2]=0.0042 M. Strategy: Choose two experiments where concentration of one reactant is constant and other is changed; solve for m and n separately!

  17. Example: The initial rate of a reaction A + B  C was measured with the results below. State the rate law, the value of the rate constant, and the rate of reaction when [A] = 0.050 M and [B] = 0.100 M.

  18. Potential Energy Diagrams Molecules need a minimum amount of energy for a reaction to take place. • Activation energy (Ea) – the minimum amount of energy that the reacting species must possess to undergo a specific reaction • Activated complex - a short-lived molecule formed when reactants collide; it can return to reactants or form products. • Formation depends on the activation energy & the correct geometry (orientation)

  19. Potential Energy Diagram

  20. Potential Energy Diagrams

  21. Potential Energy Diagrams

  22. Catalyzed Pathway Catalysts lower activation energy!!!

  23. Reaction Mechanisms Mechanism – how reactants are converted to products at the molecular level Most reactions DO NOT occur in a single step! They occur as a series of elementary steps (a single step in a reaction).

  24. Rate Determining Step Rate determining step – the slowest step in a reaction COCl2 (g)  COCl (g) + Cl (g) fast Cl (g) + COCl2 (g)  COCl (g) + Cl2 (g) slow 2 COCl (g)  2 CO (g) + 2 Cl (g) fast 2 Cl (g)  Cl2 (g) fast

  25. Getting the Overall Reaction COCl2 (g)  COCl (g) + Cl (g) fast Cl (g) + COCl2 (g)  COCl (g) + Cl2 (g) slow 2 COCl (g)  2 CO (g) + 2 Cl (g) fast 2 Cl (g)  Cl2 (g) fast 2 COCl2 (g)  2 Cl2 (g) + 2 CO (g) Adding elementary steps gives the net (or overall) reaction!

  26. Intermediates • Intermediates are produced in one elementary step but reacted in another NO (g) + O3 (g)  NO2 (g) + O2 (g) NO2 (g) + O (g)  NO (g) + O2 (g) O3 (g) + O (g)  2 O2 (g)

  27. Catalysts • Catalyst – a reactant in an elementary step but unchanged at the end of the reaction • A substance that speeds up the reaction but is not permanently changed by the reaction • Both an original reactant and a final product NO (g) + O3 (g)  NO2 (g) + O2 (g) NO2 (g) + O (g)  NO (g) + O2 (g) O3 (g) + O (g)  2 O2 (g)

  28. Example Cl2 (g)  2 Cl (g)Fast Cl (g) + CHCl3 (g)  CCl3 (g) + HCl (g)Slow CCl3 (g) + Cl (g)  CCl4 (g) Fast • Identify: • The rate determining step • The overall (net) reaction • The identity of any intermediates • The identity of any catalysts

  29. Example H2O2(aq) + I1-(aq)  H2O(l) + IO1-(aq)Slow H2O2(aq) + IO1-(aq)  H2O(l) + O2(g) + I1- (aq)Fast • Identify: • The rate determining step • The overall (net) reaction • The identity of any intermediates • The identity of any catalysts

  30. Example O3 (g) + Cl (g)  O2 (g) + ClO (g)Slow ClO (g) + O (g)  Cl (g) + O2 (g)Fast • Identify: • The rate determining step • The overall (net) reaction • The identity of any intermediates • The identity of any catalysts

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