1. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 1 Collision Theory of Reactions A chemical reaction occurs when  collisions between molecules have sufficient energy to break the bonds in the reactants  molecules collide with the proper orientation  bonds between atoms of the reactants (N 2 and O 2 ) are broken and new bonds (NO) form

2. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 2 Collision Theory of Reactions (continued) A chemical reaction does not take place if the  collisions between molecules do not have sufficient energy to break the bonds in the reactants  molecules are not properly aligned

3. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 3 Activation Energy The activation energy  is the minimum energy needed for a reaction to take place upon proper collision of reactants

4. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 4 Reaction Rate and Temperature Reaction rate  is the speed at which reactant is used up  is the speed at which product forms  increases when temperature rises because reacting molecules move faster, thereby providing more colliding molecules with energy of activation

5. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 5 Reaction Rate and Concentration Increasing the concentration of reactants  increases the number of collisions  increases the reaction rate

6. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 6 Reaction Rate and Catalysts A catalyst  speeds up the rate of a reaction  lowers the energy of activation  is not used up during the reaction

7. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 7 Reversible Reactions In a reversible reaction, there are both forward and reverse reactions.  Suppose SO 2 and O 2 are present initially. As they collide, the forward reaction begins. 2SO 2 (g) + O 2 (g) 2SO 3 (g)  As SO 3 molecules form, they also collide in the reverse reaction that forms reactants. This reversible reaction is written with a double arrow. forward 2SO 2 (g) + O 2 (g) 2SO 3 (g) reverse

8. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 8 Chemical Equilibrium At equilibrium,  the rate of the forward reaction becomes equal to the rate of the reverse reaction  the forward and reverse reactions continue at equal rates in both directions

9. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 9 Chemical Equilibrium (continued) When equilibrium is reached,  there is no further change in the amounts of reactant and product

10. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 10 Equilibrium At equilibrium,  the forward reaction of N 2 and O 2 forms NO  the reverse reaction of 2NO forms N 2 and O 2  the amounts of N 2, O 2, and NO remain constant N 2 (g) + O 2 (g) 2NO(g)

11. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 11 Equilibrium Constants For the reaction aA bB  The equilibrium constant expression, K c, gives the concentrations of the reactants and products at equilibrium: K c = [B] b = [Products] [A] a [Reactants]  The square brackets indicate the moles/liter of each substance.  The coefficients b and a are written as superscripts that raise the moles/liter to a specific power.

12. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 12 Guide to Writing the K c Expression

13. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 13 Writing a K c Expression Write the K c expression for the following: STEP 1 Write the balanced equilibrium equation: 2CO(g) + O 2 (g) 2CO 2 (g) STEP 2 Write the product concentrations in the numerator and the reactant concentration in the denominator: K c = [CO 2 ] [products] [CO][O 2 ] [reactants] STEP 3 Write the coefficients as superscripts: K c = [CO 2 ] 2 [CO] 2 [O 2 ]

14. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 14 Heterogeneous Equilibrium In heterogeneous equilibrium,  Gases and solid and/or liquid states are part of the reaction. 2NaHCO 3 (s) Na 2 CO 3 (s) + CO 2 (g) + H 2 O(g)  The concentration of solids and liquids is constant.  The K c expression is written with only the compounds that are gases. K c = [CO 2 ][H 2 O]

15. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. Guide to Calculating the K c Value 15

16. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 16 Example of Calculating Equilibrium Constants What is the K c for the following reaction? H 2 (g) + I 2 (g) 2HI(g) Equilibrium concentrations: [H 2 ] = 1.2 moles/L [I 2 ] = 1.2 moles/L [HI] = 0.35 mole/L STEP 1 Write the K c expression: K c = [HI] 2 [H 2 ][I 2 ] STEP 2 Substitute the equilibrium concentrations in K c : K c = [0.35] 2 = 8.5 x 10 -2 [1.2][1.2]

17. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 17 Reaching Chemical Equilibrium A container initially filled with SO 2 (g) and O 2 (g) or only SO 3 (g)  contains mostly SO 2 (g) and small amounts of O 2 (g) and SO 3 (g) at equilibrium  reaches equilibrium in both situations

18. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 18 Equilibrium Can Favor Product If equilibrium is reached after most of the forward reaction has occurred,  the system favors the products

19. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 19 Equilibrium with a Large K c At equilibrium,  a reaction with a large K c produces a large amount of product; very little of the reactants remain K c = [NCl 3 ] 2 = 3.2 x 10 11 [N 2 ][Cl 2 ] 3  a large K c favors the products N 2 (g) + 3Cl 2 (g) 2NCl 3 (g) When this reaction reaches equilibrium, it will essentially consist of the product NCl 3.

20. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 20 Equilibrium Can Favor Reactant If equilibrium is reached when very little of the forward reaction has occurred,  the reaction favors the reactants

21. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 21 Equilibrium with a Small K c At equilibrium,  a reaction that produces only a small amount of product has a small K c K c = [NO] 2 =2.3 x 10 -9 [N 2 ][O 2 ]  a small K c favors the reactants N 2 (g) + O 2 (g) 2NO(g) When this reaction reaches equilibrium, it will essentially consist of the reactants N 2 and O 2.

22. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 22 Summary of K c Values A reaction  that favors products has a large K c  with about equal concentrations of products and reactants has a K c close to 1  that favors reactants has a small K c

23. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. Guide to Using the K c Value 23

24. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 24 Using K c to Solve for an Equilibrium Concentration At equilibrium, the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) has a K c of 4.2 x 10 –2 and contains [PCl 3 ] = [Cl 2 ] = 0.10 M. What is the equilibrium concentration of PCl 5 ?

25. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 25 Using K c to Solve for an Equilibrium Concentration (continued) STEP 1 Write the K c expression: K c = [PCl 3 ][Cl 2 ] [PCl 5 ] STEP 2 Solve for the unknown concentration: [PCl 5 ] = [PCl 3 ][Cl 2 ] K c STEP 3 Substitute the known values and solve: [PCl 5 ] = [0.10][0.10] = 0.24M 4.2 x 10 –2 STEP 4 Check answer by placing concentrations in K c : K c = [0.10][0.10] = 4.2 x 10 –2 [0.24]

26. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 26 Le Châtelier’s Principle Le Châtelier’s principle states that  any change in equilibrium conditions upsets the equilibrium of the system  a system at equilibrium under stress will shift to relieve the stress  there will be a change in the rate of the forward or reverse reaction to return the system to equilibrium

27. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. What happens when a system is at equilibrium and you upset the balance? 27

28. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. Le Chatelier’s Principle If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. 28

29. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. Le Chatelier’s Principle Used to predict how a equilibrium system will react to changes in concentration, pressure (volume) and temperature. 29

30. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 30 Effects of Changes on Equilibrium

31. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 31 Heat and Endothermic Reactions For an endothermic reaction at equilibrium,  a decrease in temperature (T) removes heat, and the equilibrium shifts toward the reactants  an increase in temperature adds heat, and the equilibrium shifts toward the products. CaCO 3 (s) + 133 kcal CaO(s) + CO 2 (g) Decrease T Increase T

32. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 32 Heat and Exothermic Reactions For an exothermic reaction at equilibrium,  a decrease in temperature removes heat, and the equilibrium shifts toward the products  an increase in temperature adds heat, and the equilibrium shifts toward the reactants. N 2 (g) + 3H 2 (g) 2NH 3 (g) + 22 kcal Decrease T Increase T

33. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 33 Saturated Solution A saturated solution  contains the maximum amount of dissolved solute  contains solid solute  is an equilibrium system: solid ions in solution

34. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 34 Solubility Product Constant The solubility product constant for a saturated solution  gives the ion concentrations at constant temperature  is expressed as K sp  does not include the solid, which is constant Fe(OH) 2 (s) Fe 2+ (aq) + 2OH − (aq) K sp = [Fe 2+ ] [OH − ] 2

35. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 35 Guide to Calculating K sp

36. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 36 Example of Calculating Solubility Product Constant Calculate the K sp of PbSO 4 (solubility 1.4 x 10 –4 M). STEP 1 Write the equilibrium equation for dissociation: PbSO 4 (s) Pb 2+ (aq) + SO 4 2− (aq) STEP 2 Write the K sp expression: K sp = [Pb 2+ ][SO 4 2− ] STEP 3 Substitue molarity values and calculate: K sp = (1.4 x 10 –4 ) x (1.4 x 10 –4 ) = 2.0 x 10 –8

37. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 37 Molar Solubility (S) The molar solubility (S) is  the number of moles of solute that dissolve in 1 L of solution  determined from the formula of the salt  calculated from the K sp

38. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 38 Calculating Molar Solubility (S)

39. General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 39 Solubility Calculation Determine the solubility (S) 2 of SrCO 3 (K sp = 5.4 x 10 –10 ). STEP 1 Write the equilibrium equation for dissociation: SrCO 3 (s) Sr 2+ (aq) + CO 3 2− (aq) STEP 2 Write the K sp expression: K sp = [Sr 2+ ][CO 3 2− ] STEP 3 Substitute S for the molarity of each ion into K sp : K sp = [Sr 2+ ][CO 3 2− ] = [S][S] = S 2 = 5.4 x 10 −10 STEP 4 Calculate the solubility, S: S = = 2.3 x 10 −5 M