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Rate Laws The rate of a reaction can be expressed in a second way. For the hydrolysis of acetyl chloride, we can write That is: 181.

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Presentation on theme: "Rate Laws The rate of a reaction can be expressed in a second way. For the hydrolysis of acetyl chloride, we can write That is: 181."— Presentation transcript:

1 Rate Laws The rate of a reaction can be expressed in a second way. For the hydrolysis of acetyl chloride, we can write That is: 181

2 Rate Laws The rate of a reaction can be expressed in a second way. For the hydrolysis of acetyl chloride, we can write That is: The exponent n is determined by experiment. k is called the rate constant. Note that this constant does not depend on the concentration of acetyl chloride. 182

3 For a more general reaction such as: a A + b B c C + d D 183

4 For a more general reaction such as: a A + b B c C + d D The forward rate (left right) is given by: 184

5 For a more general reaction such as: a A + b B c C + d D The forward rate (left right) is given by: Note that x and y are NOT necessarily related to the stoichiometric coefficients a and b in the above equation. They are numbers which are constants for a given reaction. 185

6 Rate Law: An expression relating the rate of a reaction to the rate constant and the concentrations of the reactants raised to the appropriate powers. 186

7 Rate Law: An expression relating the rate of a reaction to the rate constant and the concentrations of the reactants raised to the appropriate powers. Reaction Order: The power to which the concentration of a reactant needs to be raised in the rate law expression. For a reaction with more than one species present, it is the sum of the reaction orders of the individual species in the rate law. 187

8 Suppose in the preceding reaction that x = 1 and y = 2. The reaction is said to be first-order in species A and second-order in species B. 188

9 Suppose in the preceding reaction that x = 1 and y = 2. The reaction is said to be first-order in species A and second-order in species B. The reaction would be described as third-order (sum of the reaction orders of the individual species reacting). 189

10 Important Point The exponents x and y that determine the order of the reaction are not found from the stoichiometric coefficients in the overall balanced equation. 190

11 Important Point The exponents x and y that determine the order of the reaction are not found from the stoichiometric coefficients in the overall balanced equation. These exponents must be determined experimentally. 191

12 Some examples: 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) 192

13 Some examples: 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) The rate law is given by: 193

14 Some examples: 2 N 2 O 5(g) 4 NO 2(g) + O 2(g) The rate law is given by: Note that the rate is NOT given by: 194

15 For the reaction H 2(g) + I 2(g) 2 HI (g) 195

16 For the reaction H 2(g) + I 2(g) 2 HI (g) The rate of reaction is: 196

17 For the reaction H 2(g) + I 2(g) 2 HI (g) The rate of reaction is: It is just a coincidence that the exponents on the concentration terms in this rate law match the stoichiometric coefficients in the overall balanced equation for this example. 197

18 For the reaction CHCl 3(g) + Cl 2(g) CCl 4(g) + HCl (g) 198

19 For the reaction CHCl 3(g) + Cl 2(g) CCl 4(g) + HCl (g) The rate of reaction is: 199

20 For the reaction CHCl 3(g) + Cl 2(g) CCl 4(g) + HCl (g) The rate of reaction is: The exponents are not required to be integers, though this is often the case. 200

21 Integrated rate laws: Concentration changes over time 201

22 Analysis of some simple classes of reaction by Order 202

23 Analysis of some simple classes of reaction by Order Zero-order reactions 203

24 Analysis of some simple classes of reaction by Order Zero-order reactions The simplest class of reactions to treat are zero- order. In practical applications, this case does not arise very frequently, but it is simple to treat. 204

25 For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. 205

26 For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. If the following reaction: A P follows zero-order kinetics, 206

27 For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. If the following reaction: A P follows zero-order kinetics, then 207

28 For a zero-order reaction, the rate of reaction is independent of the concentrations of the reactants. If the following reaction: A P follows zero-order kinetics, then So the rate is a fixed constant k and does not change with time. 208

29 Notation: The concentration of reactant A at time t is denoted as and the concentration of reactant A at the start of the reaction, taken as t = 0, is denoted by, which is the initial concentration of A. 209

30 Notation: The concentration of reactant A at time t is denoted as and the concentration of reactant A at the start of the reaction, taken as t = 0, is denoted by, which is the initial concentration of A. From the expression we can write 210

31 Recall that the equation of a straight line is 211

32 Recall that the equation of a straight line is So comparison with, indicates that a plot of versus t will yield a straight line with a slope = - k. 212

33 Recall that the equation of a straight line is So comparison with, indicates that a plot of versus t will yield a straight line with a slope = - k. 213

34 A plot of the rate versus time would look like: 214

35 First-order reactions From a practical standpoint, this is an important case, and occurs commonly. 215

36 First-order reactions From a practical standpoint, this is an important case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then 216

37 First-order reactions From a practical standpoint, this is an important case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then 217

38 First-order reactions From a practical standpoint, this is an important case, and occurs commonly. If the following reaction: A P follows first-order kinetics, then The equation can be solved (with a slight modification and using calculus). 218

39 The result is: 219

40 The result is: Because of the following property of logs: 220

41 The result is: Because of the following property of logs: then the above equation can be written as 221

42 The preceding equation can be put in the form (by taking antilogs of both sides of the equation): 222

43 The preceding equation can be put in the form (by taking antilogs of both sides of the equation): that is: 223

44 The preceding equation can be put in the form (by taking antilogs of both sides of the equation): that is: You can convert the preceding equation back into by taking ln of both sides of the equation. 224

45 Note that the equation is that of a straight line if we identify y with the term and x with the time t (the constant b would be zero). 225

46 Note that the equation is that of a straight line if we identify y with the term and x with the time t (the constant b would be zero). 226

47 An alternative plot can be made: Rewriting as so the following plot can be made: 227

48 An alternative plot can be made: Rewriting as so the following plot can be made: 228

49 A plot of versus t will show the following appearance: 229

50 A plot of versus t will show the following appearance: This is an exponential fall-off of the concentration with time. 230

51 Second-order reactions 231

52 Second-order reactions We will examine two cases: Both cases arise fairly commonly. 232

53 Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. 233

54 Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. If the following reaction: A P follows second-order kinetics, 234

55 Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. If the following reaction: A P follows second-order kinetics, then 235

56 Second-order reactions We will examine two cases: Both cases arise fairly commonly. First case – a single reactant species. If the following reaction: A P follows second-order kinetics, then This equation can be solved (using calculus) for to yield the following result. 236

57 237

58 This is of the form of a straight line if you identify y with and x with the time t. 238

59 This is of the form of a straight line if you identify y with and x with the time t. The constant b would be and the slope = k. 239

60 240


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