Collision Theory. Unit 12: Chapter 18. Reaction Rates. Activation Energy. Reversible Reactions. Reversible Reactions. Reaction Rates and Equilibrium

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1 Collision Theory For reactions to occur collisions between particles must have Unit 12: Chapter 18 Reaction Rates and Equilibrium the proper orientation enough kinetic energy See Both In Action 1 2 Activation Energy Minimum amount of energy required for particles to react This is an exothermic reaction. What would an endo look like? Activated Complex (transition state) Reaction Rates Reaction Rates: Speed at which reaction proceeds Factors Particle Size Temperature Concentration or Pressure for gases Catalyst Lycopodium: Size Antacids: Temp HCl and Mg: Conc. H2O2: Catalyst 3 4 Reversible Reactions Reversible Reactions Reversible Reactions: a reaction that occurs simultaneously in both directions 2SO 2 + O 2 2SO 3 SO 2 and O 2 (not at equilibrium) 2SO 2 + O 2 2SO 3 (at equilibrium SO 3 (not a equilibrium) 5 6

2 Reversible Reactions Chemical Equilibrium: when the forward and reverse reactions take place at the same rate DOES NOT MEAN that the AMOUNTS of reactants and products are the same! Reversible Reactions Equilibrium Position: indicates whether the reactants or products have the larger concentration at equilibrium. A B Product Favored A Catalysts do not affect the amount of reactants and products present, they decrease the time needed to establish equilibrium B Reactant Favored 7 8 Equilibrium Constants K can tell you the equilibrium position = how much of reactants or products do you have Equilibrium Constant ( ): the ratio of product concentration to reactant concentration at equilibrium K = [Products] [Reactants] aa + bb k eq = C A [ ] c [ D] d [ ] a [ B] b cc + dd [ ] = concentration of A,B,C,D a,b,c,d = coefficients > 1, products favored < 1, reactants favored small K intermediate K > 1, products favored < 1, reactants favored large K 9 10 Practice Problems N 2 O 4, a colorless gas, and NO 2, a dark brown gas, exists at equilibrium with each other. N 2 O 4(g) 2NO 2(g) A liter of the gas mixture at equilibrium contains mol N 2 O 4 and mol NO 2. Write the expression for the equilibrium constant and calculate. [ ] 2 [ ] = [.030]2 [.0045] = = NO 2 N 2 O 4 Heterogeneous Equilibria Example: Decomposition of calcium carbonate CaCO 3 (s) CaO (s) + CO 2 (g) The straight forward application of the law of equilibrium would be: However, the concentration of solids and liquids are not included in the equilibrium expression for the reaction Therefore: [ ][ CaO] [ ] = CO 2 CaCO 3 [ ] K = CO 2 12

3 Question: Write Keq expression for the following reactions: BaCl2(aq) + Na2SO4(aq) 2NaCl(aq) + BaSO4(s) 2KNO 3 (s) 2KNO 2 (s) + O 2 (g) = [ O 2 ] H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) [ Na = 2 SO 4 ] 1 [ H 2 SO 4 ] 1 NaOH [ ] 2 [ NaCl] 2 = [ BaCl 2 ] 1 Na 2 SO 4 [ ] 1 Practice Problems One mol of colorless hydrogen gas and one mole of violet iodine vapor are sealed in a 1.00 L flask and allowed to react at 450ºC. At equilibrium, 1.56 mol of colorless hydrogen iodide gas is present, together with 0.22 mol of each reactant gas. Calculate for the reaction. In a 1.00 Liter container BrCl gas is allowed to break down into Cl2 gas and Br2 gas at C. Once at equilibrium 4.00 mol of Cl2 are present. The equilibrium constant is known to be 6.25 at C. What is the concentration of the BrCl gas at equilibrium? : if stress is applied to a system in dynamic equilibrium, the system changes to relieve the stress Concentration Temperature Pressure Le Chatelier s Applet Website Concentration Example: Carbonic Acid H 2 CO 3 What will adding CO 2 do? What will removing CO 2 do? Add CO 2 Remove CO 2 CO 2 + H 2 O What would happen to the amount of CO2 if you added water? Temperature Adding heat will cause the reaction to shift to the side that ABSORBS heat Example: formation of SO 3 2SO 2 + O 2 Add heat Remove heat 2SO 3 + heat Exothermic Reaction: increase temperature = shift toward reactants (think of heat as a product) Endothermic Reaction: increase temperature = shift toward products (think of heat as a reactant) Pressure Only affects equilibrium that has an unequal number of moles of gaseous reactants and products Adding pressure: shift to side with less moles removing pressure: shift to side with more moles Example: Formation of ammonia N 2(g) + 3H 2(g) Decrease Pressure Increase pressure 2NH 3(g) 17 18

4 Determining if a reaction will happen naturally Free Energy: Energy available to do work. Any reaction that releases free energy will happen spontaneously. Spontaneous reactions: Will happen naturally and are product favored. (Release free energy) Nonspontaneous reactions: Do not favor product production and do not occur naturally. (Absorb free energy) Reactions at equilibrium For every reaction at equilibrium one direction is favored. The favored direction is spontaneous. heat + H2O(g) + CO2(g) C6H18(g) + O2(g) Which direction is spontaneous? Reactions can be made spontaneous by changing temperature, pressure or other variables (such as sunlight in photosynthesis) So is exothermic always spontaneous and endothermic nonspontaneous? Is enthalpy the only sheriff in town? Entropy (S) The disorder of a system The natural tendency for molecules is to move towards increasing entropy. Think of your bedroom Entropy contributes to Free Energy Entropy For chemical or physical processes certain rules apply to determine what is higher in entropy. gas > liquid > solid more particles = more entropy (dissolving or crushing) entropy increases in reactions when more molecules (mols) of product form. 2H2O(l) 2H2(g) + O2(g) (2:3) Higher temperature = more entropy General Rule: As system becomes less organized entropy increases Calculating Changes in Entropy Standard Entropy (S 0 ) : Quantitative measure of disorder in a system. Molar value based off energy and absolute temperature Label: J/K*mol ΔS 0 = S 0 (products) - S 0 (reactants) What is the standard entropy change when liquid water is broken down into hydrogen gas and oxygen gas? You will need to look up Standard Entropy values for homework. I will provide for tests: See page 558. Enthalpy Changes in enthalpy (ΔH) +ΔH: endothermic -ΔH: exothermic Regarding Enthalpy: reactions that release energy as heat (-ΔH) tend to be spontaneous. Entropy Changes in entropy (ΔS) +ΔS: more disorder -ΔS: less disorder Regarding Entropy: reactions that become more disordered (+ΔS) tend to be spontaneous

5 When Combined enthalpy (ΔH) entropy (ΔS) Spontaneous? + + Temperature dependent (Higher temp more likely to be spontaneous) + - Always nonspontaneous - - Temperature dependent (Lower temp more likely to be spontaneous) - + Always spontaneous I bet you are just dying for an equation to calculate this! Gibbs Free Energy Change Gibbs Free Energy: Energy that becomes available to do work from a spontaneous reaction. (Total Energy Released) based off the enthalpy and entropy changes of a reaction Label: KJ ΔG = ΔH - TΔS T: Temperature must be in KELVINS! Using Tables 17.4 and 19.2 in book, determine the ΔG for the production of iron (III) oxide from the oxidation of metallic iron at 25 0 C. (Make sure you convert units to match!!!!) enthalpy (ΔH) Adding ΔG to table entropy (ΔS) Gibbs Free Energy (ΔG) or - Spontaneous? As Temp increases entropy has bigger effect. It can overcome the + enthalpy to give a -ΔG and make reaction spontaneous Always nonspontaneous or - ΔG must be negative (-) to have a Spontaneous Reaction! As Temp decreases entropy effect diminishes Enthalpy can overcome the negative entropy to give a -ΔG and make reaction spontaneous Always spontaneous Wouldn t it be great if we had standard values for G? Standard Gibbs Free Energy Calculations Similar to Standard entropy, and standard heat of formation we have done previously. Label for standard values: KJ/mol ΔG 0 = G 0 f (products) - G 0 f (reactants) What is the Standard Gibbs Free Energy change for the combustion of methane? Is this reaction spontaneous at the temperature given on table? You will need to look up Standard Gibbs values for homework. I will provide for tests: See page Reaction Mechanisms Looking at Reaction Mechanisms Graphically Many reactions proceed in multiple steps. This occurs when reactants form intermediate products and these are used to generate another product until a final reaction produces your end product. Each step has its own reaction rate & activation energy. The slowest step is what dictates overall reaction rate (Rate Determining Step) The highest activation energy is what dictates how much external energy is required to initiate reaction. Intermediate Products Initial Reactants Activated Complexes Final Products How many elementary reactions are part of this overall mechanism? Assuming Y is free energy, is this mechanism spontaneous at this temperature? 29 30

6 Organic Mechanism Rate Laws Calculating how fast a reaction will proceed. k = specific rate constant for the reaction. For single reactant, the rate is directly proportional to concentration of the reactant if the reaction is a single step. Note: exponent (m) is NOT the coefficient If m is 1 then we have a 1st order reaction 2: 2nd order 3: 3rd order A B Rate = ΔA Δt = k [ A ] m Graphical Analysis of first order reactions 33