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## Chapter 11

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**Chapter 11**Chemical Kinetics**Chemical Kinetics**• Chemical Kinetics – the study of the rates of chemical rxns, the factors that affect rxn rates, and the mechanisms (the series of steps) by which the rxns occur. • Rates of reaction – describes how fast reactants are used up and products are formed. • In thermodynamics we learned to assess whether a particular rxn will occur. • Kinetics determines if a substantial rxn will occur in a certain time period.**Chemical Kinetics**• If a rxn is not thermodynamically favored it will not occur under the given condition. • Or even if it is thermodynamically favored it might not occur at a measurable rate. • Examples: • C(diamond) + O2(g) CO2(g) ΔGorxn = -397 kj/mol • Does not occur at an observable rate at RT. • C(graphite) + O2(g) CO2(g) ΔGorxn = -394 kj/mol • Once started occurs rapidly at RT. Kinetics explains this!**Chemical Kinetics**• The rate of reaction • Rate of Rxn usually expressed in moles/L·time or M/time • With the balanced equation the rate of the rxn can be determined by following the change in concentration of any product or reactant that can be detected quantitatively. • Examples: • Measuring color change with spectroscopy • If a gas is produced, measure the change in pressure**Chemical Kinetics**• Calculating Rate of Rxn: Sample Rxn: aA + bB cC + dD Rate of rxn = -1 Δ[A] = -1 Δ[B] = 1 Δ[C] = 1 Δ[D] aΔt bΔt cΔt dΔt Coefficients from balanced equation Reactants – negative b/c they are being used up Products – positive because they are being created * We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.**Chemical Kinetics**• Calculating Rate of Rxn: Rxn: 2N2O5(g) 4 NO2(g) + O2(g) NO2 is forming at a rate of 0.0072 mol/L·s a. What is the rate change of [O2] at this time? b. What is the rate change of [N2O5] at this time? c. What is the rate of reaction, at this time?**Chemical Kinetics**Rxn: 2N2O5(g) 4 NO2(g) + O2(g) NO2 is forming at a rate of 0.0072 mol/L·s • What is the rate change of [O2] at this time? Mole ratio 1 mole O2: 4 moles NO2 Rate change of [O2] = Δ[O2] = 0.0072 mol NO2 x 1 mol 02 Δt L·s 4 mol NO2 = 0.0018 mols O2/L·s *The rate of change is positive b/c it is being produced**Chemical Kinetics**Rxn: 2N2O5(g) 4 NO2(g) + O2(g) NO2 is forming at a rate of 0.0072 mol/L·s b. What is the rate change of [N2O5] at this time? Mole ratio 2 mole N2O5: 4 moles NO2 Rate change of [N2O5] = -Δ[N2O5] = 0.0072 mol NO2 x -2 mol N205 Δt L·s 4 mol NO2 = - 0.0036 mols N2O5/L·s *The rate of change is negative b/c it is being used up.**Chemical Kinetics**Rxn: 2N2O5(g) 4 NO2(g) + O2(g) c. What is the rate of rxn, at this time? You can use any of the rate changes for the reactants or products to calculate the rate of rxn. Using Rate change of [N2O5] Rate of rxn = -1 Δ[N2O5] = -1 (-0.0036 mol/L·s) 2 Δt 2 = 0.0018 mols /L·s**Chemical Kinetics**Rxn: 2N2O5(g) 4 NO2(g) + O2(g) c. What is the rate of rxn, at this time? You can use any of the rate changes for the reactants or products to calculate the rate of rxn. Using Rate change of [NO2] Rate of rxn = 1 Δ[NO2] = 1 (0.0072 mol/L·s) 4 Δt 4 = 0.0018 mols /L·s**Chemical Kinetics**Rxn: 2N2O5(g) 4 NO2(g) + O2(g) c. What is the rate of rxn, at this time? You can use any of the rate changes for the reactants or products to calculate the rate of rxn. Using Rate change of [O2] Rate of rxn = 1 Δ[O2] = 1 (0.0018 mol/L·s) 1 Δt 1 = 0.0018 mols /L·s Notice that no matter which rate of change you use you get the same rxn rate.**Factors that Affect Reaction Rates**• Four Factors • Nature of the reactants • Concentration of reactants • Temperature • Presence of a catalyst • Nature of the Reactants A. Physical states of reacting substances are important in determining their reactivity’s.**Factors that Affect Reaction Rates**• Examples: • Gasoline liquid can burn smoothly vs. gasoline vapor which can burn explosively. • Immiscible liquids react slowly at their interface, but if mixed the reaction speeds up. • K2SO4(s) + Ba(NO3)2(s) no rxn for years K2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2 KNO3(aq) occurs rapidly**Factors that Affect Reaction Rates**• Examples: • White phosphorus and red phosphorus are allotropes of elemental phosphorus. White phosphorus ignites when exposed to oxygen while red phosphorus can be kept exposed to oxygen for long periods of time. • Allotropes – different forms of the same element in the same physical state • Examples: • O2 (oxygen) & O3 (ozone) • C (diamond) & C (graphite)**Factors that Affect Reaction Rates**• Nature of the Reactants B. Chemical Identities of Elements & Compounds Affect Rxn Rates • Examples: • Na(s) metal + H2O(l) at RT reacts rapidly low ionization energy • Ca(s) metal + H2O(l) at RT reacts slowly higher ionization energy**Factors that Affect Reaction Rates**• Examples: • Acids & Bases react rapidly • Reactions involving electrostatic interactions b/t ions in aqueous solution react more rapidly than rxns involving the breaking of covalent bonds C. Reaction rates depend on surface area & degree of reactant subdivisions • Example: • Large chunks of metal do not burn but powdered metals with large surface areas (more atoms exposed to O2) burn easily.**Factors that Affect Reaction Rates**• The ultimate degree of subdivisions would make all reactant molecules (or atoms or ions) accessible at the same time. This situation can be achieved when the reactants are in the gaseous state or in solution. • Examples: • Medicines – rapid relief vs. extended release**The Rate Law Expression**• Concentration of Reactants – The Rate Law Expression • As the concentration of reactants change at constant temperature the rate of rxn changes. • Rate Law Expression (AKA - Rate Law) • For Rxn A + B products • Rate = k[A]x[B]y…**The Rate Law Expression**• Rate Law Expression • For Rxn A + B products • Rate = k[A]x[B]y… • k = specific rate constant at a particular temperature • x & y = the order of the rxn with respect to A & B; can be integers, zero, fractions or negative; must be determined experimentally has nothing to do with coefficients from the balanced equations. • x + y = the overall order of the rxn**The Rate Law Expression**• Important Points Regarding Specific Rate Constant (k) • Once the rxn orders are known, experimental data must be used to determine the value of k for the rxn at appropriate conditions. • The value is for a specific reaction, represented by a balanced equation • Units of k depends on the overall order of the rxn • k does not change w/ concentration of either reactants or products • k does not change with time • k refers to the rxn at a particular temperature & changes if we change the temperature • k value depends on whether a catalyst is present**The Rate Law Expression**• Example: • Rxn: A + B + C products • Rate law = k[A][B]2 • What happens to the rxn with the following concentration changes: • Double concentration A; B & C stay the same • Double concentration B; A & C stay the same • Double concentration C; A & B stay the same • Double concentration A, B & C simultaneously**The Rate Law Expression**• Example: • Rxn: A + B + C products • Rate law = k[A][B]2 • What happens to the rxn with the following concentration changes: • Double concentration A; B & C stay the same. - rate is directly proportional to [A]; so rate would increase by 2. • Double concentration B; A & C stay the same - rate is directly proportional to [B] to the power of 2; so the rate would increase by 22 = 4. • Double concentration C; A & B stay the same - rate is independent of [C]; so changing C causes no change in rxn • Double concentration A, B & C simultaneously rate - rate increases by a factor of 2 due to change in [A]; and by a factor of 4 due to the change in [B]; and is unaffected by change in C. The result: the rxn rate increase by 2 x 4 = 8.**The Rate Law Expression**• Determine the Rate Law from Initial Rates • Example: • Rxn: A + 2B C Rate = k[A]x[B]y**The Rate Law Expression**• Determine the Rate Law from Initial Rates • Example: • Rxn: A + 2B C • Rate = k[A]x[B]y Look at experiment 1 & 2 [A] stays the same, so the change in Initial Rate of Formation of C must be due to the change in [B]. a. How has [B] changed? 2.0 x 10-2 = 2.0 = [B] ratio 1.0 x 10-2 b. How has the rate changed? 3.0 x 10-6 = 2.0 = rate ratio 1.5 x 10-6**The Rate Law Expression**• Determine the Rate Law from Initial Rates • Example: c. How to get y. rate ratio = ([B]ratio)y 2.0 = (2.0)y so y = 1 That means the reaction is first order in [B]. So far: rate = k[A]x[B]1 • To figure out x; we do the same for [A] using experiments 1 & 3. Since there is no change in [B] in experiment 1 & 3 then the rate difference observed must be solely to do with [A].**The Rate Law Expression**• Determine the Rate Law from Initial Rates • Example: • Rxn: A + 2B C • Rate = k[A]x[B]y a. How has [A] changed? 2.0 x 10-2 = 2.0 = [A] ratio 1.0 x 10-2 b. How has the rate changed? 6.0 x 10-6 = 4.0 = rate ratio 1.5 x 10-6**The Rate Law Expression**• Determine the Rate Law from Initial Rates • Example: c. How to get x. rate ratio = ([A]ratio)x 4.0 = (2.0)x so x = 2 That means the reaction is second order in [A]. So far: rate = k[A]2[B]1 • You can now solve for k by plugging in the values from any of the 3 experiments and solving for k. rate = k[A]2[B]1 k = rate [A]2[B]1**The Rate Law Expression**• Determine the Rate Law from Initial Rates • Example: k = rate [A]2[B]1 Let’s use the values from experiment 2 k = 3.0 x 10-6 M/s [1.0 x 10-2M]2 [2.0 x 10-2M]1 k = 1.5/M2·s or k = 1.5 L2/mol2·s**The Rate Law Expression**• Determine the Rate Law from Initial Rates • Example: rate = 1.5 [A]2[B]1 M2·s We can say the reaction is 2nd order in [A] and the reaction is 1st order in [B] and the overall reaction is 3rd order.**Integrated Rate Equation**• Concentration vs. Time – Integrated Rate Equation • The integrated rate equation (IRE) relates concentration and time and can be used to calculate the half-life (t1/2) of a reactant. • Half-Life (t1/2) – the time it takes for the reactant to be converted to product. • The IRE & t1/2 are different for rxns of different order.**Integrated Rate Equation**• First Order Rxn: • Rxn: aA products Conditions [A] is 1st order Rxn is 1st order overall Represents coefficients from balanced equation Rate constant IRE is ln [A]o = akt (first order) [A] time coefficient**Integrated Rate Equation**• First Order Rxn: • Rxn: aA products Conditions [A] is 1st order Rxn is 1st order overall IRE is ln [A]o = akt (first order) [A] [A]o is initial concentration of reactant A [A] is concentration at some time, t, after rxn begins**Integrated Rate Equation**• First Order Rxn: • Solving for t: • t = 1 ln [A]o [A] For t1/2: Definition: [A] = ½ [A]o at t = t1/2 t1/2 = 1 ln [A]o = 1 ln 2 = ln 2 = 0.693 ak ½[A]o ak ak ak ak First order**Integrated Rate Equation**• First Order Rxn: • Example: 1. Half Life: 1st Order Reaction Rxn: A B + C k = 0.045/s What is the half life at 25oC? Use t1/2 equation from earlier. a = 1; k is given • t1/2 = ln 2 = 0.693 = 15.4 s ak 1(0.045/s) Conditions [A] is 1st order Rxn is 1st order overall**Integrated Rate Equation**• First Order Rxn: • Example: 1. 1st Order Reaction using IRE Rxn: 2 N2O5(g) 2 N2O4(g) + 02(g) Rate = k[N2O5] k = 0.00840/s • If 2.50 mol N2O5 were placed in a 5.00L container at that temperature, how many moles of N2O5 would remain after 1.00 min? • How long would it take for 90% of the original N2O5 to react? Conditions [N2O5] is 1st order Rxn is 1st order overall**Integrated Rate Equation**Example: Rxn: 2 N2O5(g) 2 N2O4(g) + 02(g) Rate = k[N2O5] k = 0.00840/s • If 2.50 mol N2O5 were placed in a 5.00L container at that temperature, how many moles of N2O5 would remain after 1.00 min? Use IRE 1st Order Equation: ln [N2O5]o = akt [N2O5] Conditions [N2O5] is 1st order Rxn is 1st order overall**Integrated Rate Equation**Use IRE 1st Order Equation: ln [N2O5]o = akt [N2O5] Step 1: Determine [N2O5]o (initial molar concentration) [N2O5]o = 2.50 mol = 0.500 M 5.00 L Step 2: Solve for [N2O5]after 1.00 min. a = 2k = 0.00840/s t = 1.00 min = 60.0 s [N2O5] = ? Note: Be sure t & k are the same units**Integrated Rate Equation**Solve for unknown [N2O5]: Definition: ln x = ln x – ln y ln [N2O5]o = ln [N2O5]o – ln [N2O5] = akt [N2O5] ln [N2O5] = ln [N2O5]o – akt = ln(0.500) – (2)(0.00840/s)(60.0s) = - 0.693 – 1.008 ln [N2O5] = -1.701 (use the ex key times -1.701) [N2O5] = 1.82 x 10-1 M = .182 M after 1.00 min y**Integrated Rate Equation**Step 3: Calculate the number of moles of N2O5 left in the container. 5.00 L x 0.182 mol = 0.910 mol N2O5 L • IRE involves ratios of concentration, we don’t need the numeric value s for the concentration of N2O5 when 90% has reacted 10.0% remains. Step 1: [N2O5] = (0.100) [N2O5]o 10.0% original concentration**IRE - First Order Reactions**Step 2: Substitute into IRE: ln [N2O5]o = akt [N2O5] ln [N2O5]o = (2)(0.00840/s)t (0.100) [N2O5]o ln(10.0) = (0.0168/s)t 2.302 = (0.0168/s)t t = 2.302 = 137 seconds 0.0168/s Remember a=2 k = 0.00840/s t = ?**IRE - Second Order Reaction**Conditions [A] is 2nd order Rxn is 2nd overall • Rxn: aA product 2nd Order IRE is: 1 - 1 = akt [A] [A]o Initial concentration of reactants**IRE - Second Order Reaction**Conditions [A] is 2nd order Rxn is 2nd overall • Rxn: aA product Half-life 2nd Order Equation: t1/2 = 1 ak[A]o Examples: Rxn: A + B C + D rate = k[A]2 k = 0.622 L/mol·min at 30oC Conditions [A] is 2nd order Rxn is 2nd overall**IRE - Second Order Reaction**• What is the half life of A when 4.10 x 10-2 M A is mixed with excess B • Look at the rate equation: rate = k[A]2 • The rate is independent of B, and as long as there is some of B present to react with A the reaction proceeds at the given rate. • So use the 2nd Order Half Life Equation: • t1/2 = 1 = 1 ak[A]o (1)(0.622/M·min)(4.10x10-2M) = 39.2 min Initial concentrationof A (given) Coefficient from balanced equation k (given)**IRE - Second Order Reaction**• Concentration vs. Time: 2nd Order Rxn – IRE • Rxn: 2NOBr(g) 2NO(g) + Br2(g) • rate = k[NOBr]2 • k = 0.810/m·s at 10.0oC • [NOBr]o = 4.00x10-3 M at 10.0oC • How many seconds does it take to use up 1.50x10-3M of NOBr • Step 1: Determine [NOBr] after 1.50x10-3M is used up. • ? M NOBr remaining = (0.00400 – 0.00150)M = 0.00250 M NOBr**IRE - Second Order Reaction**• Step 2: Use 2nd Order IRE: • 1 - 1 = akt (solve for t) [NOBr] [NOBr]o t = 1 1 - 1 ak [NOBr] [NOBr]o = 1 1 - 1 (2)(0.810/M·s) 0.00250M 0.00400M = 1 (400/M – 250/M) 1.62/M·s = 92.6 seconds**IRE - Second Order Reaction**• Example 2 • Same Rxn: 2NOBr(g) 2NO(g) + Br2(g) • rate = k[NOBr]2 • k = 0.810/m·s at 10.0oC • [NOBr]o = 2.40x10-3 M • What is the concentration of NOBr after 5.00 min. • Step 1: Use 2nd Order IRE t = 5.00 min • 1 - 1 = akt [NOBr] [NOBr]o**IRE - Second Order Reaction**What is the concentration of NOBr after 5.00 min. • Step 1: Use 2nd Order IRE t = 5.00 min • 1 - 1 = akt [NOBr] [NOBr]o 1 - 1 = (2)(0.810/M·s)(5.00min)(60s/min) [NOBr] 2.40x10-3M 1 - 4.17x102 = 486 [NOBr] M M 1 = 486 + 417 = 903/M [NOBr] M M [NOBr] = 1 = 1.11x10-3M 903/M**IRE - Zero Order Reaction**• Rxn: aA products Zero order reaction • Therefore rxn rate is independent of concentration • IRE for Zero Order Rxn: • rate = k • [A] = [A]o – akt (Zero Order) • Half Life Equation for Zero Order Rxn: • t1/2 = [A]o (Zero Order) 2ak**Problem Solving Tips: Which Equation Do You Use?**• Step 1: You must decide if you use the Rate Law Expression (RLE) or the Integrated Rate Equation (IRE) • The RLE relates rate and concentration • The IRE relates time and concentration • So if the problem asks you to make calculations involving rxn rate and concentration use RLE • If the problem asks you to make calculations involving time you have to use the IRE.**Problem Solving Tips: Which Equation Do You Use?**• Step 2: You must decide which form of the RLE or IRE (Zero, First or Second Order) that is right for the given equation. • Hints: • The problem may state the order of the rxn • The RLE may be given; so you can tell the order of the rxn from the exponents in the expression • The units of the specific rate constant (k) may be given; you can interpret the stated units to figure out the order