12-3-2011

1. 12-3-2011

2. Concentration-Time Equation Order Rate Law Half-Life 1 1 - = kt [A]t [A]0 = [A]0 t½ = t½ t½ = Ln 2 2k k 1 k[A]0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions [A]t - [A]0 = - kt rate = k 0 ln[A]t - ln[A]0 = - kt 1 rate = k [A] 2 rate = k [A]2 13.3

3. The Collision Theory of Chemical Kinetics Gas molecules frequently collide with one another • Expect that the rate of a reaction is equivalent to the number of collisions • Reaction rate is dependent on concentration • Rate will also be dependent on orientation The rate of reaction is proportional to the number of molecular collisions per second, or the frequency of molecular collisions. Rate α number of collisions/s

4. In many cases, collisions alone do not guarantee that a reaction will take place, and collisions can be effective when: 1- Colliding species have the correct orientation (how reacting molecules are oriented relative to each other. This is neglected in reactions between atoms.

5. Calculations based on the kinetic molecular theory show that at 1 atm and 298K, 1X1027 collisions per second occur between two molecules placed in 1 mL of volume. Even more collisions per second occur in liquids. If collision only was responsible for chemical reaction to occur, almost all reactions would proceed instantaneously, which is not the case.

8. 2- Colliding species should have enough kinetic energy, because, when molecules collide, part of their kinetic energy is converted to vibrational energy. If the initial kinetic energy is large, then the colliding molecules will vibrate strongly as to break some of the chemical bonds. The bond fracture is the first step toward product formation.

9. In other word, the colliding molecules must have a total kinetic energy equal to or greater than the activation energy, Ea,which is the minimum energy required to initiate the chemical reaction. Normally, only a small fraction of the colliding molecules (the fastest moving ones) have enough kinetic energy to exceed the activation energy. These molecules can therefore take part in the reaction.

10. The Collision Theory of Chemical Kinetics Activation Energy (Ea):the minimum amount of energy required to initiate a chemical reaction. Activated Complex (Transition State):a temporary species formed by the reactant molecules as a result of the collision before they form the product. The activated complex: is a highly unstable species with a high potential energy.

12. Effect of temperature on reaction rates The kinetic energy (speed) distribution of gas molecules at two different temperatures: The speed of the molecules obey the Maxwell distribution as shown below: Compare the speed distributions at two different temperatures. Because more high-energy molecules are present at higher temperature, the rate of product formation is also greater at the higher temperature.

14. The Collision Theory of Chemical Kinetics What does this have to do with temperature? If molecules are oriented correctly, high energy molecules and high temperatures should result in increased product formation

15. -Ea /RT K = Ae The Arrhenius Equation A relation between the rate constant of a reaction and temperature: Where A is the collision frequency Relation between activation energy and temperature. lnk = - (Ea/R) x (1/T) + lnA

20. Remember, we can not take the logarithm of a unit therefore, lnk is unitless

21. Temperature Dependence of the Rate Constant k = A •exp( -Ea/RT ) (Arrhenius equation) Eais the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor 13.4

22. Rate Constants and Temperature • ln (k1/k2) = (Ea/R){(1/T2) - (1/T1)}

23. Solution: ln (k1/k2) = (Ea/R){(1/T2) - (1/T1)}

25. The decomposition of C2H5Cl is a first order reaction having k = 0.032 s-1 at 550 oC and 0.093 s-1 at 575 oC. What is the activation energy, in kJ/mol, for this reaction?

26. ln (k1/k2) = (Ea/R){(1/T2) - (1/T1)} T1 = 550 + 273 = 823 K; T2 = 575 + 273 = 848 K ln (0.032/0.093) = (Ea/8.314 J mol-1 K-1){(1/848 K) - (1/823 K)} -1.1 = (Ea/8.314 J mol-1 K-1){(1/848 K) - (1/823 K)} Ea = 2.48x105 J mol-1 Ea = 248 kJ mol-1

27. At 300 oC the rate constant for the conversion of cyclopropane to propylene is 2.41x10-10 s-1. At 400 oC k = 1.16x10-6 s-1. Find the activation energy, in kJ/mol, and A for this reaction. Solution: ln (k1/k2) = (Ea/R){(1/T2) - (1/T1)} T1 = 300 + 273 = 573 K; T2 = 400 + 273 = 673 K ln (2.41x10-10 s-1/1.16x10-6 s-1) = (Ea/8.314 J mol-1 K-1){(1/673 K) - (1/573 K)}

28. -8.479 = (Ea/8.314 J mol-1 K-1){(1/673 K) - (1/573 K)} Ea = 2.7x105 J mol-1 Ea = 270 kJ mol-1 k = Ae-Ea/RT, or, ln k = ln A – Ea/RT Substitution using either rate constant with corresponding temperature gives: ln (2.41x10-10) = ln A – 2.7x105 J mol-1/(8.314 J mol-1K-1 * 573 K) A = 9.9x1014 s-1

29. Activation Energy, Reaction Rates and Temperature As stated earlier, for a reaction to take place, molecules must posses enough kinetic energy. Kinetic energy must be higher than Ea. Each reaction takes place at a specific temperature……but what happens if we adjust this temp.?

30. Activation Energy, Reaction Rates and Temperature • Increasing Temperature leads to: • Molecules reach high kinetic energy faster • Number of molecules with high enough kinetic energy increases • Reaction rate increases