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Chapter 5 Rates of Chemical Reaction

Chapter 5 Rates of Chemical Reaction. 5-1 Rates and Mechanisms of Chemical Reactions 5-2 Theories of Reaction Rate 5-3 Reaction Rates and Concentrations 5-4 Effect of Temperature on Reaction Rates 5-5 Effect of Catalyst on Reaction Rates.

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Chapter 5 Rates of Chemical Reaction

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  1. Chapter 5Rates of Chemical Reaction

  2. 5-1 Rates and Mechanisms of Chemical Reactions 5-2 Theories of Reaction Rate 5-3 Reaction Rates and Concentrations 5-4 Effect of Temperature on Reaction Rates 5-5 Effect of Catalyst on Reaction Rates

  3. 5-1 Rates and Mechanisms of Chemical Reactions 5-1.1 The Rate of Chemical Reaction reaction rates: the change in the concentration of a reactant or a product with time. The rate is defined to be a positive number.

  4. The rate of a chemical reaction is measured by the decrease in concentration of a reactant or the increase in concentration of a product in a unit of time.

  5. (5-2) A+3B → 2D (5-1) or

  6. △[A] 1 △[B] 1 △[D] Rate = - = － = + △t 3 △t 2 △t The units of the rate are usually mol·L-1·s-1 ; mol·L-1·min-1 ; mol·L-1·h-1

  7. average reaction rateis obtained by dividing the change in concentration of a reactant or product by the time interval over which the change occurs. vrefers to average reaction rate, Δc refers to change in concentration and Δtrefers to change in time.

  8. Let us look at a specific example: N2 + 3H2 = 2NH3 c( mol/L, t=0) 1.0 3.0 0 c( mol/L, t=2s) 0.8 2.4 0.4 vNH3= (0.4-0)/2= 0.2 (mol·L-1·s-1) or vH2= -(2.4-3.0)/2= 0.3 (mol·L-1·s-1) vN2= -(0.8-1.0)/2= 0.1 (mol·L-1·s-1) vN2 ：vH2 ：vNH3= 1 ：3：2 or

  9. Instantaneous rate: the average rate over an arbitrary short period of time 2NO2 = 2NO + O2

  10. Determination ofinstantaneous rate

  11. 5-1.2 The Mechanisms of Chemical Reactions • Reaction Mechanismsis a description of the path that a reaction takes. • Elementary reaction:A reaction can complete directly by only one step or reactants canconvert into products. 2NO2 = 2NO + O2 2I + H2 = 2HI

  12. Overall reaction:A reaction was completed • through several elementary reactions. Rate controlling step For example H2(g) + I2(g) 2HI(g) First step I2(g) 2I(g) (fast) Second step H2(g) + 2I(g) 2HI(g) (slow) termolecular • Types of Elementary Reactions SO2 + Cl2 unimolecular SO2Cl2 NO + CO2 bimolecular NO2 + CO 2HI 2I + H2

  13. 5-2 Theories of Reaction Rate 5-2.1 Collision Theory and Activation Energy ●Contents of Collision Theory ⑴ reacting molecules must come so close that they collide. ⑵ not every collision between molecules creates products, only few collisions between reactant molecules will react. effective collision: a collision that leads to a reaction

  14. ⑶ enough energy; proper orientation ineffective collision effective collision 2NOCl ─→ 2NO + Cl2

  15. (a)ineffective collision (b)effective collision

  16. For a collision to result in reaction, the molecules must be properly oriented. For the reaction CO(g) + NO2(g) → CO2(g) + NO(g) the carbon atom of the CO molecule must strike an oxygen atom of the NO2 molecule, forming CO2 as one product, NO as the other.

  17. Collisions must occur with enough energy to break the bonds in the reactants so that new bonds can form in the products.

  18. Ee Ea E1 • Activation molecule is the molecule have enough energy and can produce effective collision

  19. ●activation energy (Ea) :The minimum energy of a collision that leads to a reaction. It has the symbol Eaand is expressed inkilojoules.

  20. Figure: As the activation energy of a reaction decreases, the number of molecules with at least this much energy increases, as shown by the yellow shaded areas. In general: ＜ 40 kJ/mol very fast ＞120 kJ/mol slow Ea : 40～400kJ/mol

  21. 5-2.2 The Transition StateTheory Transition state theory (TST) is also called activated complex theory. reactants pass through high-energy transition states before forming products, they are associated in an unstableentity called an activated complex, then change into products.

  22. Give off energy Absorb energy Example 1: HI + HI → I•••H •••H ••• I → H2 + I2activated complex Activated process Activation energy

  23. 5-3 Reaction Rates and Concentrations Chemical reactions are faster when the concentrations of the reactants are increased because more molecules will exist in a given volume. More collisions will occur and the rate of a reaction will increase.

  24. 5-3.1 The Rate Law The rate of a reaction is proportional to the product of the concentrations of the reactants raised to some power. Consider the reaction: a A+b B → c C+d D v∝[A]m[B]n v = k[A]m[B]n

  25. where k is the rate constant; [A], [B] are the concentration of A and B; m and n are themselves constants for a given reaction Notice: ① when [A]=[B]=1mol·L-1, v=k ② the greater the k , the faster the rate ③ m and n must be determined experimentally, in general, m andn are not equal to the stoichiometric coefficients aand b

  26. 5-3.2 Order of A Reaction The order of a reaction with respect to one of the reactantsis equal to the power to which the concentration of that reactant is raised in the rate equation. The sum of the powers to which all reactant concentrations appearing in the rate law are raised is called the overall reaction order.

  27. For equation v = k[A]m[B]n m is the order of the reaction with respect to A,n is the order of the reaction with respect to B. The overall order of the reaction is the sum of m and n. the exponents m and n are not necessarily related to the stoichiometric coefficients in the balanced equation, that is, in general it is not true that fora A + b B → c C + d D, a = m and b = n

  28. For the thermal decomposition of N2O5 2N2O5(g) →4NO2(g) + O2(g) the rate law is v = k[N2O5] and not v = k[N2O5]2, as we might have inferred from the balanced equation

  29. For the follow reaction • C2H6(g) → 2CH3(g) • The rate expression has the form • v = k [C2H6]2 so that n = 2 even though the coefficient of C2H6 in the chemical equation is 1. Thus, the decomposition of N2O5 is first order, whereas that of C2H6 is second order.

  30. Example 5-1:The reaction of nitric oxide with hydrogen at 1280℃ is The follow example illustrates the procedure for determining the rate law of a reaction. N2(g) + 2H2O(g) 2NO(g) + 2H2(g) From the following data collected at this temperature, determine the rate law and calculate the rate constant.

  31. Experiment [NO] [H2] Initial Rate (mol/L s) 1 5.0×10-3 2.0×10-3 1.3×10-5 2 10.0×10-3 2.0×10-3 5.0×10-5 3 10.0×10-3 4.0×10-3 10.0×10-5 Reasoning and Solution: We assume that the rate law takes the form v = k[NO]m[H2]n

  32. Experiments 1 and 2 shows that when we double the concentration of NO at constant concentration of H2, the rate quadruples. Thus the reaction is second order in NO. Experiments 2 and 3 indicate that doubling [H2] at constant [NO] doubles the rate; the reaction is first order in H2. The rate law is given by v = k[NO]2[H2] which shows that it is a (1+2) or third-order reaction overall. The rate constant k can be calculated using the data from any one of the experiments. Since

  33. v k=-------------- [NO]2[H2] data from experiment 2 gives us 5×10-5 k=---------------------------- (10 ×10-3)2(2 ×10-3) =2.5 ×102/(mol/L)2·s Comment: Note that the reaction is first order in H2, whereas the stoichiometric coefficient for H2 in the balanced equation is 2

  34. Example 5-2: Given the following data, what is the rate expression for the reaction between hydroxide ion and chlorine dioxide? 2ClO2(a q) + 2OH-(a q) →ClO3-(a q) + ClO2-(a q) + H2O [ClO2] (mol l-1) [OH-] (mol l-1) Rate (mol L-1 s-1) 0.010 0.030 6.00×10-4 0.010 0.075 1.50×10-3 0.055 0.030 1.82×10-2

  35. Solution: v3/v1= ([ClO2]3 / [ClO2]1)m 1.82×10-2/6.00×10-4 = (0.055/0.010)m 30.3 = (5.5)m By inspection, m = 2. The reaction is 2nd order in ClO2 v2/v1 = ([OH-]2/[OH-]1)n 1.50×10-3/6.00×10-4 = (0.075/0.030)n 2.5 = (2.5)n By inspection, n = 1 The overall rate expression is therefore v = k[ClO2]2[OH-]

  36. First-order reactions A → product the rate is A first-order reactionis a reaction whose rate depends on the reactant concentration raised to the first power.

  37. Also, from the rate law we know that Thus Integrate the left side from c = c0 to c and the right from t = 0 to t.

  38. The characteristics of first-order reactions： 1.A plot of logcversus t (time) gives a straight line with a slope of -k/2.303.

  39. 2.The rate constant, k, has units of [time]-1. 3. half-life (t1/2) : is the time it takes for the concentration of a reactant A to fall to one half of its original value. • By definition, when t = t1/2, c = c0/2, so

  40. Example 5-3(a) What is the rate constant k for the first-order decomposition of N2O5(g) at 25℃ if the half-life of N2O5(g) at that temperature is 4.03×104 seconds?(b) Under these conditions, what percent of the N2O5 molecules will not have reacted after one day? Solution: (a)

  41. (b) Putting in the value for k and substituting t = 8.64×104 seconds (one day has 86,400 seconds) gives

  42. Hence Therefore, 22.6% of the N2O5 molecules will not have reacted after one day at 25℃.

  43. Example 5-4SO2Cl2 decomposes to sulfur dioxide and chlorine gas. The reaction is first order: If it takes 13.7 hours for a 0.250mol/L solution of SO2Cl2 to decompose into a 0.117mol/L solution, what is the rate constant for the reaction and what is the half-life of SO2Cl2 decomposition? Solution:

  44. k=0.0554 h-1

  45. Second - order reactions A second-order reactionis a reaction whose rate depends on reactant concentration raised to the second power or on the concentrations of two different reactants, each raised to the first power. v=k[A][B] v=k[A]2

  46. The characteristics of second- order reactions： 1.A graph of 1/c against timeis a straight line , the slope of which gives the rate constant for the reaction；

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