Chapter 12 Reaction Rates and Chemical Equilibrium Collision Theory Conditions That Affect Reaction Rates 12- Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Reaction Rate A measure of how fast a reaction occurs Conditions that affect reaction rate: Temperature Higher temperatures generally cause reactions to move faster Reactant concentration Increasing the concentration of a reactant generally increases the reaction rate 12-

Conditions that affect reaction rate cont’d: Surface area Increasing the surface area increases the reaction rate if the reactant is a solid Presence of a catalyst Increases the rate of the reaction 12-

Reaction Rate 12-

Collision Theory States that in order for a reaction to occur, reactant molecules must collide in the proper orientation and with sufficient energy Which of the pictures on the right has the molecules in the proper orientation to collide? 12-

Energy Diagrams Figure 12.6 12-

Activation Energy Reactants must overcome an energy barrier before they can change to products Energy is required to break bonds in reactants before the reactants can be converted into products The minimum amount of energy needed to overcome the energy barrier is called the activation energy, Ea Reactions with large activation energies tend to be slow because a relatively small fraction of reactants have sufficient energy for an effective collision Reactions with small activation energies tend to be fast because a large fraction of reactants have sufficient energy for an effective collision 12-

Activated Complex Activated complex Short-lived, unstable, high-energy chemical species that must be achieved before products can form Formed from reactant molecules that collide with the proper orientation and sufficient energy Actual structure is unknown Each reaction has its own reaction diagram, which shows the amount of energy required to form the activated complex as the reaction progresses 12-

Energy Diagram for a Reaction 12-

Practice – Energy Diagrams The following reaction is an endothermic reaction: 2 NO2(g)  2 NO(g) + O2(g) Draw an energy diagram that shows the relative energies of the reactants, products, and the activated complex. Label the diagram with molecular representations of reactants, products, and a possible structure for the activated complex. 12-

Practice Solutions – Energy Diagrams 12-

Conditions that Affect Reaction Rates Concentration An increase in concentration of one or more of the reactants increases the number of reactants per unit volume More molecules are thus closer together and the number of collision per unit time increases As total collisions increase, the number of molecules with the energy and orientation required for the reaction also increases The fraction of effective collisions remains the same, because the temperature and kinetic energy are constant 12-

Concentration and Reaction Rates Figure 12.8 12-

Conditions that Affect Reaction Rates Temperature The average kinetic energy of a substance increases as the temperature increases. An increase in kinetic energy causes the reaction rate to increase. 12-

Conditions that Affect Reaction Rates - Temperature An increase in kinetic energy causes the reaction rate to increase in two ways: Increases collision rate Molecules move faster at higher temperatures, and therefore collide more frequently Increases the fraction of effective collisions More of the molecules attain activation energy because the average kinetic energy of the molecules increases 12-

Temperature and Collision Theory 12-

How to Tame Allergic Reactions How can you slow down a histamine attack? Histamine attacks are greater when you are hot. Cooling down affected areas can reduce allergy symptoms.

Practice – Collision Rates Suppose the collision rate between molecules A and B at 25°C is 10,000 collisions per second. The number of effective collisions at this same temperature is 100 collisions per second. How will each of the following affect the total number of collisions and the fraction of effective collision between molecules A and B? The temperature decreases The concentration of reactant B decreases 12-

Practice Solutions – Collision Rates The temperature decreases When the temperature decreases, Average kinetic energy decreases Average velocity of the molecules decreases Should decrease the number of collisions and the fraction of effective collisions Therefore, we expect the total number of collisions to be less than 10,000 and the fraction of effective collisions to be less than 100. 12-

Practice Solutions – Collision Rates The concentration of reactant B decreases The number of collisions between A and B decreases Less B molecules can make contact with A molecules. The fraction of effective collisions should remain the same The average kinetic energy and temperature of the molecules remains the same. Therefore, we expect the total number of collisions to be less than 10,000 and the fraction of effective collisions to be 100 (same as before). 12-

What makes Switzerland unique? Activation Energy What makes Switzerland unique?

Activation Energy Chemical Reactions must go over an energy hill like a mountain (Swiss Alps).

Catalysts Lower the activation energy for the reaction by forming new activated complexes with lower activation energies Remain unchanged after the reaction 12-

Enzymes Molecules that catalyze specific reactions within living organisms Most enzymes are large protein molecules with molar masses between 12,000 and 40,000 g/mol Enzymes contain depressions, or holes, called active sites The shape of an active site is unique to only one specific kind of reactant molecule, called a substrate Figure 12.12 12-

Enzyme Catalysis 12-

Practice – Catalysis The decomposition of HI is an exothermic reaction: 2 HI(g) H2(g) + I2(g) Draw an energy diagram for the uncatalyzed reaction. Label reactants and products. Sketch a possible activated complex. Use a dotted line to show the energy changes when platinum metal, a catalyst that increases the rate of the reaction, is added to the system. 12-

Practice Solutions – Catalysis 12-

Reaction Intermediates A molecule or compound that forms temporarily during a reaction In any reaction that occurs in more than one step, intermediates and catalysts are not part of the net (or overall) equation. PLEASE SPECIFY WHAT FIGURE TO PUT HERE! 12-

Practice – Identifying Intermediates and Catalysts Ethene (H2C=CH2) can be converted to ethanol (CH3CH2OH) by a three-step process. Identify any intermediates or catalysts. H2C=CH2 + H3O+ H3C-CH2+ + H2O H3C-CH2+ + H2O CH3CH2OH2+ CH3CH2OH2+ + H2O CH3CH2OH + H3O+ 12-

Practice Solutions – Identifying Intermediates and Catalysts Identify any intermediates or catalysts. H2C=CH2 + H3O+ H3C-CH2+ + H2O H3C-CH2+ + H2O CH3CH2OH2+ CH3CH2OH2+ + H2O CH3CH2OH + H3O+ Intermediates in this process include H3C-CH2+ and CH3CH2OH2+. H3C-CH2+ is formed in step 1, then used in step 2. CH3CH2OH2+ is formed in step 2, then used in step 3. H3O+ is used in step 1, then regenerated in step 3, so it is therefore a catalyst. 12-

Chapter 12 Reaction Rates and Chemical Equilibrium The Equilibrium Constant Le Chatelier’s Principle 12- Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chemical Equilibrium A state reached by a chemical reaction where there is no change in the concentrations of reactants and products Established when a single reaction occurs in which reactants are converted to products, and those products are converted back to reactants in a reverse process at an equal rate. 12-

Chemical Equilibrium N2O4(g) 2 NO2(g) The rates of the forward and reverse reactions are equal; there is no net change in the concentrations of reactants and products. True equilibria are obtained in closed containers, where reactants and products cannot escape. Reactions that can reach equilibrium must be reversible reactions. Equilibrium is represented with an equilibrium arrow ( ) as in the example below: N2O4(g) 2 NO2(g) 12-

Chemical Equilibrium 12-

Position of Equilibrium When a reaction reaches equilibrium, amounts of reactants and products may be about equal (they usually are not though). When we describe the equilibrium in terms of which side it favors, products or reactants, we are describing the position of equilibrium. 12-

Position of Equilibrium Two cases can occur when products and reactants are not equal: Large amount of reactants and a small amount of products We say equilibrium favors reactants in this case. Small amount of reactants and a large amount of products We say equilibrium favors products in this case. 12-

Equilibrium Constant aA + bB cC + dD Keq = [C]c[D]d [A]a[B]b A constant (Keq) for a specific reaction whose value is always the same at a specified temperature. For a reaction with the general form: aA + bB cC + dD The equilibrium constant expression is: Keq = [C]c[D]d [A]a[B]b where [A], [B], [C], and [D] are the molar concentrations of the reactants and products at equilibrium, and the exponents a, b, c, and d are the values of the coefficients in the balanced chemical equation. 12-

Equilibrium Constant The value of the equilibrium constant tells us about the position of equilibrium. When the value is much greater than 1, there are more products than reactants at equilibrium. When the value is less than 1, there are more reactants than products at equilibrium. 12-

Table 12.2 The Meaning of the Value of Keq Value of the Equilibrium Constant, Keq Position of Equilibrium Keq >> 1 Lies to right. Reaction is product favored. Keq << 1 Lies to left. Reaction is reactant favored. Keq = 1 Lies in middle. Similar amounts of reactants and products. 12-

Practice – Equilibrium Constants At 25°C a pure sample of N2O4 is placed into a reaction container and allowed to reach equilibrium: 2 NO2(g) N2O4(g) The equilibrium concentrations are determined to be: [NO2] = 0.0750 M [N2O4] = 1.25 M Write the equilibrium constant expression for this reaction. Calculate the value of the equilibrium constant at 25°C. Describe the position of equilibrium. 12-

Practice Solutions – Equilibrium Constants Write the equilibrium constant expression for this reaction. Keq = [N2O4] [NO2]2 Calculate the value of the equilibrium constant at 25°C. Keq = [N2O4] = 1.25 = 16.7 [NO2]2 0.0750 Describe the position of equilibrium. Because Keq is much larger than 1, the reaction is product favored and the position of equilibrium lies to the right. 12-

Predicting the Direction of a Reaction Suppose we start a reaction with a mixture of reactants and products. If the relative amounts of reactants and products are at equilibrium concentrations, the system will remain at equilibrium and no net reaction will occur. If the relative amounts are not at equilibrium concentrations, a forward or reverse reaction will occur until concentrations are equilibrium concentrations, as described by the equilibrium constant. 12-

Predicting the Direction of a Reaction Start with more products than there should be at equilibrium The reaction will proceed in the reverse direction until the system reaches equilibrium. Start with more reactants than there should be at equilibrium The reaction will proceed in the forward direction until the system reaches equilibrium. 12-

Practice - Predicting the Direction of a Reaction Consider this reaction and its equilibrium constant: S2Cl2(g) + Cl2(g) 2 SCl2(g) Keq = 4 Suppose we start with the following initial conditions of reactants and products: [S2Cl2]initial = 0.10 M [Cl2]initial = 0.10 M [SCl2]initial = 0.30 M Determine whether the system is at equilibrium. If it is not, predict the direction in which the reaction will proceed to reach equilibrium. What will happen to the concentration of Cl2? 12-

Practice Solutions - Predicting the Direction of a Reaction Determine whether the system is at equilibrium. If it is not, predict the direction in which the reaction will proceed to reach equilibrium. Keq = [SCl2]2 = (0.30)2 = 9 [S2Cl2][Cl2] (0.10)(0.10) The Keq stated in the problem was 4. The calculated value of Keq is not equal to the value given in the problem and the system is not at equilibrium. 9 is also greater than 4; therefore, there is too much product and not enough reactant. The reaction will proceed in the reverse direction until the system reaches equilibrium. 12-

Practice Solutions - Predicting the Direction of a Reaction What will happen to the concentration of Cl2? As the reaction proceeds in the reverse direction, the concentration of Cl2 will increase. 12-

Heterogeneous Equilibrium Homogeneous equilibrium An equilibrium in which reactants and products are in the same physical state Heterogeneous equilibrium An equilibrium in which reactants and products are not in the same physical states When finding the equilibrium constant expression for any equilibrium, omit pure liquids and solids. Use only gases (g) and dissolved substances (aq) in the equilibrium constant expression. 12-

Heterogeneous Equilibrium Consider the reaction: Br2(l) Br2(g) 12-

Practice – Equilibrium Constant Expressions Write the equilibrium constant expression for the following reactions. Mg(s) + CO2(g) MgO(s) + CO(g) PbCl2(s) Pb2+(aq) + 2 Cl-(aq) 12-

Practice Solutions – Equilibrium Constant Expressions Write the equilibrium constant expression for the following reactions. Mg(s) + CO2(g) MgO(s) + CO(g) Solids are not included in the equilibrium constant expression, so: Keq = [CO] [CO2] PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Keq = [Pb2+][Cl-]2 12-

Le Chatelier’s Principle States that if a system at equilibrium is disrupted, it shifts to establish a new equilibrium. Changes that can disrupt a system include: Changes in the concentration of a reactant or product Changes in the volume of a gas-reaction container Temperature changes 12-

Changes in Concentration For a system at equilibrium, when the concentration of a reactant or product is increased, the equilibrium will shift to consume the added substance. When the concentration of a reactant or product is reduced, the equilibrium will shift to produce more of the removed substance. Figure 12.20 12-

Le Chatelier’s Principle Equilibrium shifts to counter a disturbance. Hills and Valleys! O3 (g) + Cl (g) O2 (g) + OCl(g)

Table 12.3 Equilibrium Shift Due to Concentration Changes General Reaction Add reactant Add product Remove reactant Remove product A(g) + B(g) C(g) + D(g) Shift right Shift left 12-

Practice – Changes in Concentration Consider the following system at equilibrium: AgI(s) Ag+(aq) + I-(aq) Predict the effect of the following changes when the system is initially in a state of equilibrium. Assume the volume of solution remains constant. Ag+ is removed from the system by the addition of NaOH. Solid AgNO3 is added to the system. (AgNO3 is water soluble) Solid NaI is added to the system. (NaI is water soluble) 12-

Practice Solutions – Changes in Concentration AgI(s) Ag+(aq) + I-(aq) Ag+ is removed from the system by the addition of NaOH. Ag+ is a product, and if it is removed from the system, the system will shift the equilibrium to produce more of the removed substance. Thus, the system will shift to the right (towards products). 12-

Practice Solutions – Changes in Concentration AgI(s) Ag+(aq) + I-(aq) b) Solid AgNO3 is added to the system. (AgNO3 is water soluble.) AgNO3 will produce Ag+ ions in solution, thereby adding more product. The system will shift to consume the added product, thus shifting to the left (towards reactants). c) Solid NaI is added to the system. (NaI is water soluble.) NaI will produce I- ions in solution, thereby adding more product. The system will shift to consume the added product, thus shifting to the left (towards reactants). 12-

Changes in Volume Because gases expand to fill a container, changes in volume affect the concentrations of any gases in the reaction container. 12-

Changes in Volume Figure 12.22 12-

Table 12.4 Equilibrium Shifts Due to Changes in Volume Relative Number of Gaseous Molecules Increase Volume Decrease Volume Reactants < Products Shift right Shift left Reactants > Products Reactants = Products No shift 12-

Practice – Changes in Volume How will a decrease in container volume affect the following reactions at equilibrium? Explain. 2 NOBr(g) 2 NO(g) + Br2(g) CuO(s) + H2(g) Cu(s) + H2O(g) 12-

Practice Solutions – Changes in Volume How will a decrease in container volume affect the following reactions at equilibrium? Explain. 2 NOBr(g) 2 NO(g) + Br2(g) A decrease in the volume of the container increases both the concentrations of reactants and products. Because there are relatively more gaseous product molecules than reactant molecules, the system will shift to consume the extra product concentration. Thus, the decrease in volume will shift the reaction to the left (towards reactants). 12-

Practice Solutions – Changes in Volume CuO(s) + H2(g) Cu(s) + H2O(g) A decrease in the volume of the container increases both the concentrations of reactants and products. Because there are relatively the same number of gaseous reactant and product molecules, the system will not shift, but simply maintain the same equilibrium. 12-

Changes in Temperature When a reaction is endothermic (heat requiring), you can think of the heat required as an additional reactant. Therefore, put heat as a reactant on the left side of the arrow. heat + N2O4(g) 2 NO2(g) When a reaction is exothermic (heat producing), you can think of the heat produced as an additional product. Put heat as a product on the right side of the arrow. N2(g) + 3 H2(g) 2 NH3(g) + heat 12-

Industrial Application-The Manufacture of Ammonia N2(g) + 3H2(g) 2NH3(g) ΔH = -92.4 kJ mol-1 To increase production how would you manipulate the equilibrium? Lower Volume 2. Lower Temperature 3. Remove Product

Changes in Temperature 12-

Table 12.5 Equilibrium Shifts Due to Temperature Changes Type of reaction Equation Increase temperature Decrease temperature endothermic Heat + A + B C + D Shift right Keq increases Shift left Keq decreases exothermic A + B C + D + heat 12-

Practice – Temperature Changes Consider the following equilibrium: Fe3+(aq) + NCS-(aq) FeNCS+2(aq) When the temperature is increased, the solution turns darker, indicating a higher concentration of the FeNCS2+ product. Is this reaction endothermic or exothermic? 12-

Practice Solutions – Temperature Changes Consider the following equilibrium: Fe3+(aq) + NCS-(aq) FeNCS+2(aq) When the temperature is increased, the solution turns darker, indicating a higher concentration of the FeNCS2+ product. Is this reaction endothermic or exothermic? If the temperature increase results in a higher concentration of product, then the heat added forces the equilibrium to the right (towards products). This means that the heat existed as a reactant on the left; thus, the reaction is endothermic. 12-

Catalysts and Increasing Product Yield Do not change the position of equilibrium or affect a system that is in a state of equilibrium Catalysts are neither reactants nor products in the net reaction Increasing product yield Le Chatelier’s Principle can be used to impose conditions in the reaction environment that shift the equilibrium toward the product side of the reaction. 12-