Rate of Reaction and Chemical Equilibrium

2 Collision Theory Molecules must collide to react Effective collisions lead to products being formed Ineffective collisions do not form products Reactants will form products if they collide with adequate kinetic energy Reactants will bounce apart if they do not have adequate kinetic energy

3 A) Two BrNO molecules approach each other at high speed B) The collision occurs. C) The energy of the collision causes Br-N bonds to break and Br-Br bonds to form. D) The products: one Br 2 and two NO molecules.

4 Reaction progress of two BrNO molecules. Activation Energy Activation Energy: The minimum amount of energy particles must have in order to react Activated complex: Reactants that have attained the activation energy level. Unstable with short lifetime ~ s Sometimes called transition state

5 Factors Affecting Reaction Rate TemperatureConcentration Particle Size Catalyst

6 Comparison of the activation energies for an uncatalyzed reaction and for the same reaction with a catalyst present.

The relationship between the rate of a reaction and the concentration of one reactant is determined experimentally. A series of experiments reveals how the concentration of each reactant affects the reaction rate. An equation that relates reaction rate and concentrations of reactants is called the rate law for the reaction. Rate Laws for Reactions Chapter 17 Section 2 Reaction Rate It is applicable for a specific reaction at a given temperature.

The general form for the rate law is given by the following equation where A and B are reactants: R = k[A] n [B] m The rate law is applicable for a specific reaction at a given set of conditions and must be determined from experimental data. The power to which a reactant concentration is raised is called the order in that reactant. The value of n is said to be the order of the reaction with respect to [A], so the reaction is said to be “nth order in A.” Rate Laws for Reactions, continued Using the Rate Law Chapter 17 Section 2 Reaction Rate The value of k usually increases as the temperature increases, but the relationship between reaction rate and concentration almost always remains unchanged.

An order of one for a reactant means that the reaction rate is directly proportional to the concentration of that reactant. An order of two means that the reaction rate is directly proportional to the square of the reactant. An order of zero means that the rate does not depend on the concentration of the reactant, as long as some of the reactant is present. Rate Laws for Reactions, continued Using the Rate Law, continued Chapter 17 Section 2 Reaction Rate

The sum of all of the reactant orders is called the order of the reaction, or overall order. Rate Laws for Reactions, continued Using the Rate Law, continued Chapter 17 Section 2 Reaction Rate NO 2 (g) + CO(g)NO(g) + CO 2 (g) R = k[NO 2 ] 2 second order in NO 2 zero order in CO second order overall The orders in the rate law may or may not match the coefficients in the balanced equation.

The specific rate constant (k) is the proportionality constant relating the rate of the reaction to the concentrations of reactants. 1.Once the reaction orders (powers) are known, the value of k must be determined from experimental data. 2.The value of k is for a specific reaction; k has a different value for other reactions, even at the same conditions. Rate Laws for Reactions, continued Specific Rate Constant Chapter 17 Section 2 Reaction Rate

3.The units of k depend on the overall order of the reaction. 4.The value of k does not change for different concentrations of reactants or products. So, the value of k for a reaction remains the same throughout the reaction and does not change with time. Rate Laws for Reactions, continued Specific Rate Constant, continued Chapter 17 Section 2 Reaction Rate

5.The value of k is for the reaction at a specific temperature; if we increase the temperature of the reaction, the value of k increases. 6.The value of k changes (becomes larger) if a catalyst is present. Rate Laws for Reactions, continued Specific Rate Constant, continued Chapter 17 Section 2 Reaction Rate

Sample Problem B Three experiments that have identical conditions were performed to measure the initial rate of the reaction 2HI(g) H 2 (g) + I 2 (g) The results for the three experiments in which only the HI concentration was varied are as follows: Write the rate law for the reaction. Find the value and units of the specific rate constant. Experiment[HI] (M)Rate (M/s) × 10 − × 10 − × 10 −3 Chapter 17 Section 2 Reaction Rate

Sample Problem B Solution Rate Laws for Reactions, continued Chapter 17 Section 2 Reaction Rate The general rate law for this reaction is R = k[HI] n Concentration ratio: rate ratio: rate law: R = k[HI] 2

Sample Problem B Solution, continued Rate Laws for Reactions, continued Chapter 17 Section 2 Reaction Rate

Sample Problem C Three experiments were performed to measure the initial rate of the reaction A + B C Conditions were identical in the three experiments, except that the concentrations of reactants varied. The results are as follows: Write the rate law for the reaction. Find the value and units of the specific rate constant. Experiment[A] (M)[B] (M)Rate (M/s) × 10 – × 10 − × 10 −7 Chapter 17 Section 2 Reaction Rate

Sample Problem C Solution The general rate law for this reaction is R = k[A] n [B] m.The general rate law for this reaction is R = k[A] n [B] m. To find m, compare Experiments 1 and 2, which have the same [A].To find m, compare Experiments 1 and 2, which have the same [A]. Rate Laws for Reactions, continued Chapter 17 Section 2 Reaction Rate Concentration ratio: rate ratio: m is 1, and the reaction is first order in B

Sample Problem C Solution, continued To find n, compare Experiments 1 and 3, which have the same [B].To find n, compare Experiments 1 and 3, which have the same [B]. Rate Laws for Reactions, continued Chapter 17 Section 2 Reaction Rate Concentration ratio: rate ratio: n is 2, and the reaction is second order in A The rate law is R k[A] 2 [B].

Sample Problem C Solution, continued Rate Laws for Reactions, continued Chapter 17 Section 2 Reaction Rate

21 Catalysts & Inhibitors Catalysts increase the rate of a chemical reaction without being used up in the reaction –Permit the reaction to occur with lower activation energy than normal Pt Pt 2H 2 (g) + O 2 (g)  2H 2 O(l) –Your body makes extensive use of catalysts or enzymes to allow bodily functions to progress rapidly Inhibitors are substances that interfere with the action of a catalyst –Poisons the catalyst

22 The amount of the substance in the vapor state becomes constant and reaches equilibrium How would we reverse this process?

23 Reversible Reactions A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur simultaneously. When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium. At chemical equilibrium, no net change occurs in the actual amounts of the components of the system. The relative concentrations of the reactants and products at equilibrium constitute the equilibrium position of a reaction.

24 Reversible Reactions At equilibrium, all three types of molecules are present. SO 2 and O 2 react to give SO 3 SO 3 decomposes to SO 2 and O 2

25 Changes in Concentration of Reactants and Products Initially SO 2 Initially SO 3 Present O 2 Present

26 Le Chatelier’s Principle 1884 When a system at chemical equilibrium is disturbed, the system will adjust itself to minimize the disturbance –Excellent guide for estimating how a chemical reaction will shift in response to changes Henri Le Chatelier

27 Factors Affecting Equilibrium: Le Châtelier’s Principle Concentration Rapid breathing during and after vigorous exercise helps reestablish the body’s correct CO 2 :H 2 CO 3 equilibrium, keeping the acid concentration in the blood within a safe range.

28 Factors Affecting Equilibrium: Le Châtelier’s Principle Temperature Dinitrogen tetroxide is a colorless gas; nitrogen dioxide is a brown gas. The flask on the left is in a dish of hot water; the flask on the right is in ice. Warm Cool

29 Factors Affecting Equilibrium: Le Châtelier’s Principle Pressure Pressure affects a mixture of nitrogen, hydrogen, and ammonia at equilibrium

30 Equilibrium Concentration aA + bB  cC + dD K eq = [C] c [D] d K eq = [C] c [D] d [A] a [B] b [A] a [B] b K eq > 1 products favored at equilibrium K eq > 1 products favored at equilibrium K eq < 1 reactants favored at equilibrium K eq < 1 reactants favored at equilibrium Square brackets indicate concentration or “molarity”

31 The Solubility Product Constant K sp, solubility product constant equals the product of the concentrations of the ions, each raised to a power equal to the coefficient of the ion in the dissociation equation. The smaller the numerical value of the solubility product constant, the lower the solubility of the compound. AgCl (s) ↔ Ag+ (aq) + Cl- (aq) K sp = [Ag+] 1 [Cl-] 1

32 The Solubility Product Constant

33 The Solubility Product Constant 18.3

34 System initially at equilibrium. CaCO 3  CO 2 + CaO

35 a) A mixture of NH 3 (g), N 2 (g), and H 2 (g) at equilibrium. b) The volume is suddenly decreased. c) The new equilibrium position. N 2 + 3H 2 2NH 3

36 Le Chatelier’s Principle and Temperature Exothermic Reaction H 2 + I 2  2HI + heat Heating drives reaction to the left Cooling drives reaction to the right Endothermic Reaction heat + NH 4 Cl  NH 3 + HCl Heating drives reaction to the right Cooling drives reaction to the left

37 Rules for assigning oxidation states and catalyzed reactions.

38 Some Reaction are Reversible

39 (a)Initial equilibrium mixture (b) Addition of N 2. (c) New equilibrium position.

40 (a) Equal numbers of moles of H 2 O and CO are mixed in a closed container. (b) The reaction begins to occur. (c) The reaction continues, and more reactants are changed to products. (d) No further changes are seen as time continues to pass.