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From the Arrhenius equation we have: 301
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From the Arrhenius equation we have: 302
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From the Arrhenius equation we have: Recall the equation of a straight line: where y = ln k, b= ln A, m = -E a /R, and x = T -1. 303
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So if we have temperature dependent data for the rate constant, we can make the following plot to obtain E a : 304 T -1 ln k slope = -E a /R
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305
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Sample problem: Some reactions double their reaction rates with every 10 o C rise in temperature. Assume a reaction to take place at 295 K and at 305 K. What must the activation energy be for the rate constant to exactly double? 308
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Sample problem: Some reactions double their reaction rates with every 10 o C rise in temperature. Assume a reaction to take place at 295 K and at 305 K. What must the activation energy be for the rate constant to exactly double? Approach: Start with the Arrhenius equation applied to the two separate conditions, where T 2 is the higher temperature. 309
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Sample problem: Some reactions double their reaction rates with every 10 o C rise in temperature. Assume a reaction to take place at 295 K and at 305 K. What must the activation energy be for the rate constant to exactly double? Approach: Start with the Arrhenius equation applied to the two separate conditions, where T 2 is the higher temperature. 310
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Take the ratio 311
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Take the ratio Now so that 312
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Take the ratio Now so that 313
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Now take the natural log of both sides. 314
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Now take the natural log of both sides. 315
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Now take the natural log of both sides. That is: 316
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Now take the natural log of both sides. That is: (recall that ) 317
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The preceding result simplifies to give: 318
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The preceding result simplifies to give: The numerical value for E a is given as: = 52 kJ mol -1 319
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The Arrhenius equation is quite useful when studying reactions involving simple species (atoms or diatomic molecules). For more complex systems, the Arrhenius equation is modified to the form: 320
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The Arrhenius equation is quite useful when studying reactions involving simple species (atoms or diatomic molecules). For more complex systems, the Arrhenius equation is modified to the form: where P, the probability factor, accounts for the fact that in order to react, molecules must be properly oriented with respect to each other during a collision. 321
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This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well. 322
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This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well. Example: NO 2 Cl + Cl NO 2 + Cl 2 323
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This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well. Example: NO 2 Cl + Cl NO 2 + Cl 2 The two Cl atoms must come into “contact” for reaction to occur. 324
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NO 2 Cl + Cl NO 2 + Cl 2 325
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NO + NO 3 2NO 2 Red = oxygen blue = nitrogen 326
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For reactions involving only atoms P = 1; for reactions involving simple small molecules, P varies between approximately 0.2 and 0.001. For reactions involving complex polyatomic molecules, P can be as small as 10 -6. 327
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Reaction Mechanisms 328
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Reaction Mechanisms Reaction Mechanism: The sequence of elementary steps that leads to product formation. 329
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Reaction Mechanisms Reaction Mechanism: The sequence of elementary steps that leads to product formation. Elementary Step: (Single step reaction) A reaction that occurs on the molecular level exactly as written. 330
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Reaction Mechanisms Reaction Mechanism: The sequence of elementary steps that leads to product formation. Elementary Step: (Single step reaction) A reaction that occurs on the molecular level exactly as written. An overall reaction may involve one or several elementary steps. 331
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Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH 332
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Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH For this reaction there are two obvious ways to arrive at CH 3 OH. 333
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Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH For this reaction there are two obvious ways to arrive at CH 3 OH. O CH 3 C O CH 3 case 1 334
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Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH For this reaction there are two obvious ways to arrive at CH 3 OH. O O CH 3 C O CH 3 or CH 3 C O CH 3 case 1 case 2 335
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The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different. 336
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The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different. case 1 O O CH 3 C O CH 3 + H 2 18 O CH 3 C 18 OH + CH 3 OH 337
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The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different. case 1 O O CH 3 C O CH 3 + H 2 18 O CH 3 C 18 OH + CH 3 OH case 2 O O CH 3 C O CH 3 + H 2 18 O CH 3 C OH + CH 3 18 OH 338
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Question: What might case 3 be? 339
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No CH 3 18 OH is found in the experiment, that means case 1 is the correct bond breaking step. O O CH 3 C O CH 3 + H 2 18 O CH 3 C 18 OH + CH 3 OH 340
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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: 341
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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate 342
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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate (the O atom is a reaction intermediate) 343
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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate (the O atom is a reaction intermediate) N 2 O + O N 2 + O 2 rate 344
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Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate (the O atom is a reaction intermediate) N 2 O + O N 2 + O 2 rate Overall reaction: 2 N 2 O 2 N 2 + O 2 345
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and are the rate constants for the two individual steps. 346
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and are the rate constants for the two individual steps. Key Point : The exponents in the rate law for an elementary process are equal to the coefficients obtained from the chemical equation for that elementary process. 347
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and are the rate constants for the two individual steps. Key Point : The exponents in the rate law for an elementary process are equal to the coefficients obtained from the chemical equation for that elementary process. Important reminder : You cannot get the rate law exponents for the overall reaction by looking at the balanced equation. 348
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For the reaction: 2 N 2 O 2 N 2 + O 2 349
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For the reaction: 2 N 2 O 2 N 2 + O 2 the experimental rate law is: overall rate 350
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For the reaction: 2 N 2 O 2 N 2 + O 2 the experimental rate law is: overall rate Notice that this is exactly the same as the rate law for the first elementary step. 351
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For the reaction: 2 N 2 O 2 N 2 + O 2 the experimental rate law is: overall rate Notice that this is exactly the same as the rate law for the first elementary step. The observed rate can be explained by assuming that the second step is faster than the first step, i.e. 352
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Thus, the overall rate of decomposition is then completely controlled by the rate of the first step, which is called the rate-determining step. 353
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Thus, the overall rate of decomposition is then completely controlled by the rate of the first step, which is called the rate-determining step. Rate-determining step : The slowest step in the sequence of steps leading to the formation of products. 354
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Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: 355
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Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: NO 2 + F 2 NO 2 F + F slow step 356
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Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: NO 2 + F 2 NO 2 F + F slow step NO 2 + F NO 2 F fast step 357
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Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: NO 2 + F 2 NO 2 F + F slow step NO 2 + F NO 2 F fast step Note that the two elementary steps add to the overall chemical equation. 358
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359 2 NO 2 + F 2 2 NO 2 F
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Summary comments on mechanism For a reaction mechanism to be viable, two main conditions apply. 360
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