1. Reaction Rates & Equilibrium Unit 13 - Chapter 18

2. Reaction Rate Reaction rate – how fast reactants disappear and how fast product appears

3. A B 13.1 rate = -  [A] tt rate = [B][B] tt time

4. Reaction Rate Reaction Rate = ∆ [A] ∆ t Example: CO (g) + NO 2(g)  CO 2(g) + NO (g) - at t = 4.0 min, [CO 2 ] =.12 mol/L - at t = 8.0 min, [CO 2 ] =.24 mol/L - reaction rate =.24 mol/L -.12 mol/L 8.0 min – 4.0 min = 0.030 mol/L. min Unit for reaction rate = conc. with some time unit Products have a (+) rate Reactants have a (-) rate

5. Collision Theory of Kinetics Kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. In order for a reaction to take place, the reacting molecules must collide into each other. Once molecules collide they may react together or they may not, depending on two factors - ¬Whether the collision has enough energy to "break the bonds holding reactant molecules together"; ­Whether the reacting molecules collide in the proper orientation for new bonds to form. orientation

7. Effective Collisions Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions. The higher the frequency of effective collisions the faster the reaction rate. When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex It is a transition state between reactant and product It has a very short lifetime (10 -13 s) Has to form for product to be formed

8. Activated Complex The difference in potential energy between the reactant molecules and the activated complex is called the activation energy, E a This is the minimum amount of energy that particles must have in order to react. The larger the activation energy, the slower the reaction The energy to overcome the activation energy comes from the kinetic energy of the collision being converted into potential energy, or from energy available in the environment, i.e. heat. Different reactions have different activated complexes and therefore different activation energies

9. Energy Diagram Energy of products is lower than energy of reactants energy lost, exothermic, -∆H Energy of products is higher than energy of reactants energy gained, endothermic, +∆H What is this called?

10. Factors Affecting Reaction Rate 1.Nature of the Reactants Cl 2(g) + CH 4(g)  CH 3 Cl (g) + HCl (g) Cl 2  Cl + Cl(fast) Cl + CH 4  CH 3 Cl + H(slow) H + Cl  HCl(very fast) individual steps = elementary steps all steps together = reaction mechanism the slowest step determines the rate of the reaction called the rate determining step eliminating the intermediates allows us to write the balanced equation of the mechanism Intermediates – product in one step, reactant in another

11. Factors Affecting Reaction Rate For the following mechanism: Write the balanced equation ; draw and label the energy diagram (Be sure to include reactants, products, change in enthalpy, E a, the activated complex and assume the reaction is exothermic). Br 2 → 2BrFast Br + H 2 → HBr + HSlow H + Br → HBrVery Fast

12. Factors Affecting Reaction Rate 2.Concentration The larger the concentration of reactant molecules, the faster the reaction will go. Increases the frequency of reactant molecule collisions Different reactants have different effects on reaction rates based on the order of the reactant rate expression – represents the rate of a reaction in terms of the concentration of reactants Example H 2 + O 2  H 2 O Rate = k[H 2 ] 2 [O 2 ] k = rate constant, specific for a particular reaction and at certain temperature, large k = fast reaction

13. The Method of Initial Rates The table below gives initial reaction rates for the reaction 2 NO(g) + O 2 (g) → 2 NO 2 (g) NOO2O2 NO 2 10.01250.02530.0281 20.02500.02530.1124 30.01250.05060.0561 Experiment Initial concentrations (mol/L) Initial Rate (mol/L. s) 1. Determine the order for each reactant, the overall order and write the rate expression. 2. Calculate the value for the rate constant. 3. Determine the rate if the concentrations of NO and O 2 are 0.50 and 0.75 mol/L, respectively.

14. The Method of Initial Rates The table below gives initial reaction rates for the reaction 2 ClO 2 (aq) + 2 OH - (aq) → ClO 3 - (aq) + ClO 2 - (aq) + H 2 O(l) ClO 2 OH - ClO 3 - 10.0200.0300.00276 20.0600.0300.02484 30.0200.0900.00828 Experiment Initial concentrations (mol/L) Initial Rate (mol/L. s) Perform the same operations as in 1 – 3 on the previous slide, except use 0.050 M for ClO 2 and 0.075 M for OH - to calculate the rate. Answers: 1. rate = k[ClO 2 ] 2 [OH - ]; 2. 75 L 2 /mol 2. s 3. 0.014 mol/L. s

15. Factors Affecting Reaction Rate 3. Particle Size (Surface Area) more particles on the surface = more particles available for collisions more collisions = more act. complex = more product smaller particles give you more surface area 4. Agitation this puts more liquid/gas particles in contact with the solid = ↑ collisions = ↑ act. complex = ↑ product 5. Pressure ↑ pressure by ↓ volume – puts particles closer together = ↑ collisions = ↑ act. complex = ↑ product All of these factors are similar, in terms of explanation, to concentration!!!

16. Factors Affecting Reaction Rate 6. Temperature most effective at speeding up a reaction ↑ temp. = ↑ KE (particles moving faster) particles move faster leading to more collisions the collisions are also harder these harder collisions contain the needed energy to overcome the E a therefore the reaction rate will increase

17. Factors Affecting Reaction Rate 7. Catalyst substance that speeds up a reaction, but isn’t used up in the reaction provides a “different pathway” that requires lower E a lower E a = more collisions having the proper amount of energy = ↑ act. complex = ↑ product

18. Reaction Dynamics If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up. However, if the products are allowed to accumulate; they will start reacting together to form the original reactants - called the reverse reaction. We show this reverse reaction by using a double arrow (H 2(g) + I 2(g)  2HI (g) )

19. Reaction Dynamics The forward reaction slows down as the amounts of reactants decreases because the reactant concentrations are decreasing At the same time the reverse reaction speeds up as the concentration of the products increases. Eventually the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium. rate forward = rate reverse Note: This equilibrium is dynamic

20. Chemical Equilibrium Dynamic Equilibrium can only be reached in a closed system!! When a system reaches equilibrium, the amounts of reactants and products in the system stays constant the forward and reverse reactions still continue, but because they go at the same rate the amounts of materials don't change. There is a mathematical relationship between the amounts of reactants and products at equilibrium

21. Equilibrium Expression Capital letters (A,B,C,D) – reactants or products Lowercase letter (a,b,c,d) – coefficients from the equation NOTE – products on top, reactants on bottom In this expression, K eq is a number called the equilibrium constant. ratio of product concentration to reactant concentration at equilibrium Do not include solids or liquids, only solutions and gases The value of K eq depends on temp. of the reaction – if temp. changes then the value of K eq changes. aA + bB  cC + dD = [C] c [D] d [A] a [B] b K eq

22. Example – Determine the value of the Equilibrium Constant for the Reaction 2 SO 2(g) + O 2(g)  2 SO 3(g) ¬Determine the Equilibrium Expression ­Plug the equilibrium concentrations into to Equilibrium Expression ®Solve the Equation 3.503.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical So what does this K eq value tell us???

23. Position of Equilibrium The size of the equilibrium constant shows whether products or reactants are favored at equilibrium. K eq > 1, products are favored at equilibrium K eq < 1, reactants are favored at equilibrium

24. ¬Determine the Equilibrium Expression ­Plug the equilibrium concentrations and Equilibrium Constant into the Equilibrium Expression ®Solve the Equation ?3.00SO 3 1.251.50O2O2 2.00SO 2 [Equilibrium][Initial]Chemical Example – If the value of the Equilibrium Constant for the Reaction 2 SO 2 + O 2  2 SO 3 is 4.36, Determine the Equilibrium Concentration of SO 3

25. More Equilibrium Practice 1.Write the equilibrium constant expression for the following reaction. 3H 2(g) + N 2(g) ↔ 2NH 3(g) 2.An analysis of an equilibrium mixture for this reaction in a 1.0 L flask at 300 o C gave the following results: 0.15 mol H 2, 0.25 mol N 2 and 0.10 mol NH 3. Calculate the K eq for this reaction. 3.2BrCl (g) ↔ Cl 2(g) + Br 2(g) The equilibrium constant for this reaction is 11.1. The equilibrium mixture contains 4.00 mol Cl 2 and 4.00 moles of Br 2. How many moles of BrCl are present?

26. Le Ch âtelier’s Principle Le Châtelier's Principle guides us in predicting the effect various changes have on the position of equilibrium it says that if stress is applied to a system in dynamic equilibrium, the system will change to relieve the stress. The position of equilibrium moves to counteract the change. Three common stressors: Concentration Temperature Pressure

27. Concentration Changes and Le Châtelier’s Principle A+ B ↔ C + D Adding a reactant – equilibrium shifts right Removing a reactant – equilibrium shifts left Adding a product – equilibrium shifts left Removing a product – equilibrium shifts right

28. Only affects a reaction involving gases with an unequal number of mole of reactants & products. Increasing the pressure on the system causes the position of equilibrium to shift toward the side of the reaction with the fewer gas molecules Decreasing pressure causes a shift toward the side with more gas molecules Example 3H 2(g) + N 2(g)  2NH 3(g) ↑ Pressure – shifts to the right ↓ Pressure – shift to the left Changing Pressure and Le Châtelier’s Principle

29. Increasing the temperature causes the reaction to shift away from the heat. For exothermic reactions - Think of heat as a product of the reaction Therefore shift the position of equilibrium toward the reactant side For endothermic reactions - Think of heat as a reactant The position of equilibrium will shift toward the products Cooling an exothermic or endothermic reaction will have the opposite effects. Changing Temperature and Le Châtelier’s Principle

30. Examples – Le Chatelier’s Principle What effect do the following changes have on the equilibrium position for the following reaction? 1.PCl 5(g) + heat ↔ PCl 3(g) + Cl 2(g) a. addition of Cl 2 b. increase in pressure c. removal of heat d. removal of PCl 3 as formed 2. C (s) + H 2 O (g) + heat ↔ CO (g) + H 2(g) a. Lowering the temperature b. Increasing the pressure c. Removal of H 2 as formed

31. Examples – Le Chatelier’s Principle 3. At 425 K – Fe 3 O 4(s) + 4H 2(g) ↔ 3Fe (s) + 4H 2 O (g) How would the equilibrium concentration of H 2 O be affected by the following: a. Adding more H 2 b. Adding more Fe (s) c. Decreasing the pressure d. Adding a catalyst