From the Arrhenius equation we have: 301

From the Arrhenius equation we have: 302

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

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

305

306

307

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

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

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

Take the ratio 311

Take the ratio Now so that 312

Take the ratio Now so that 313

Now take the natural log of both sides. 314

Now take the natural log of both sides. 315

Now take the natural log of both sides. That is: 316

Now take the natural log of both sides. That is: (recall that ) 317

The preceding result simplifies to give: 318

The preceding result simplifies to give: The numerical value for E a is given as: = 52 kJ mol

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

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

This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well. 322

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

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

NO 2 Cl + Cl NO 2 + Cl 2 325

NO + NO 3 2NO 2 Red = oxygen blue = nitrogen 326

For reactions involving only atoms P = 1; for reactions involving simple small molecules, P varies between approximately 0.2 and For reactions involving complex polyatomic molecules, P can be as small as

Reaction Mechanisms 328

Reaction Mechanisms Reaction Mechanism: The sequence of elementary steps that leads to product formation. 329

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

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

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

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

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

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

The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different. 336

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

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

Question: What might case 3 be? 339

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

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

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

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

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

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

and are the rate constants for the two individual steps. 346

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

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

For the reaction: 2 N 2 O 2 N 2 + O 2 349

For the reaction: 2 N 2 O 2 N 2 + O 2 the experimental rate law is: overall rate 350

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

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

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

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

Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: 355

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

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

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

359 2 NO 2 + F 2 2 NO 2 F

Summary comments on mechanism For a reaction mechanism to be viable, two main conditions apply. 360