Second Order Kinetics A (+ other reactants) → products - ∆A rate = = k [A]2 ∆t 1 1 1 Second Order Integrated = k2t + [A]t [A]o Rate Equation [A]t = concentration of A at some time = t [A]o = concentration of A at time t = 0 (initial concentration) k2 = second order rate constant (units = M-1s-1, M-1min-1, M-1h-1, etc) 1 Some mathematics: Half Life and Second Order Reactions 1 1 [A]t = half life = ½[A]o = k2t + [A]t [A]o 2 Example 13.7 - Iodine atoms combine to form molecular iodine in the gas phase - I(g) + I(g) → I2(g) This reaction follows second-order kinetics and has the high rate constant 7.0 x 109/M·s at 23 oC. (a) If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min. (b) Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M. Example 13.7 - solution 3 Zero Order Kinetics A (+ other reactants) → products - ∆A rate = = k [A]0 = k ∆t 0 0 4 [A] is the concentration of A at any time t [A] = [A]0 - kt [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 [A] t = 0 ½ 2k Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Concentration-Time Order Rate Law Equation Half-Life [A] [A] = [A] - kt t = 0 0 rate = k 0 ½ 2k 0.693 1 rate = k [A] ln[A] = ln[A] - kt t½ = 0 k 1 1 2 1 2 rate = k [A] = + kt t½ = [A] [A]0 k[A]0 5 Activation Energy and Temperature Dependence of Rate Constants Reaction rates increase with temperature. The reason is explained by Collision Theory. The Collision Theory of Chemical Reactions Molecules in the gas and liquid states are in constant random movement (Kinetic Molecular Theory) 1. The reacting molecules must collide with one another (at 25 oC and 1 atm in the gas phase there are about 1030 collisions/L•s). Since most reaction rates are much smaller than this, there must be other factors involved...... 2. The reacting molecules must collide with sufficient energy to break bonds. 3. The molecules must collide in an orientation that can lead to rearrangement of the atoms. 6 Effect of Concentration (Collisions) on Rate number of collisions rate α sec AB (a) 4 collisions between A and B (b) double A or B and now there are 8 collisions, so rate doubles For this example, rate = k[A][B] Activation Energy The minimum amount of energy required to initiate a chemical reaction. When molecules collide they form an "activated complex" or "transition state", a temporary species formed during the collision where bonds are being broken and formed simultaneously. activated activated complex) complex) 7 Temperature, Reaction Rate, and Activation Energy For example, if the activation energy is 266 kJ/mol, then raising the temperature increases the number of molecules with this energy, and the reaction rate increases. o NO2 + CO → NO + CO2 ∆H = -226 kJ/mol 8 Forward Reaction Ea = 132 kJ/mol ∆Ho = -226 kJ/mol (exothermic) Reverse Reaction ' Ea = 132 kJ/mol + 226 kJ/mol = 358 kJ/mol ∆Ho = +226 kJ/mol (endothermic) The Transition State or “Activated Complex” H2(g) + I2(g) → 2 HI(g) ∆H = -13 kJ II Transition state or activated complex HH H2 + I2 2 HI 9 Transition state - bonds between reactants are breaking at the same time that bonds are forming between the products Bonds forming II I - I I I →→ H - H H H HH Bonds breaking Effect of Molecular Orientation 2 AB → A2 + B2 (2 HCl → H2 + Cl2) 10 Another example: - - F + CH3 -Cl → F - CH3 + Cl The Arrhenius Equation Relationship between reaction rate, temperature and activation energy. k = A·e-Ea/RT k = rate constant (M-1s-1) A = frequency of collisions with correct geometry Ea = activation energy (kJ/mol) R = ideal gas law constant = 8.314 X 10-3 kJ/mol•K T = Kelvin temperature 11 Increasing T increases the fraction of molecules with E > Ea Ea = 20 kJ/mol Ea = 30 kJ/mol Ea = 40 kJ/mol Linearization of the Arrhenius Equation Reaction rates are measured at a variety of temperatures; the activation energy and collisional frequency factor can be determined graphically. -Ea/RT k = A·e 12 Example 13.8 - The rate constants for the decomposition of acetaldehyde CH3CHO(g) → CH4(g) + CO(g) were measured at five different temperatures. The data are shown in the table. Plot ln(k) versus 1/T and determine the activation energy (in kJ/mol) for the reaction. Note that the 1/2 reaction is "3/2" order in CH3CO, so k has units of 1/M ·s. k (1/M1/2·s) T (K) ln(k) 1/T (K-1) 0.011 700 0.035 730 0.105 760 0.343 790 0.789 810 Example 13.8 -solution 13 We can also measure rate constants at two different temperatures and derive a formula relating them to the activation energy- Example 13.9 - The rate constant of a first order reaction is 3.46 x 10-2 s-1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? 14 Reaction Mechanisms The study of reaction KINETICS gives insight into reaction MECHANISMS - the sequence of bond-making and bond-breaking steps in a chemical reaction. 15 Most reactions involve a sequence of steps - Step 1 Br2(g) + NO(g) → Br2NO(g) Step 2 Br2NO(g) + NO(g) → 2 BrNO(g) Overall Br2(g) + 2 NO(g) → 2 BrNO(g) Reaction “intermediate” Each step in the mechanism is called an “elementary step” and has it’s own Ea and rate constant, k. Ea1 Ea2 Br2(g) + NO(g) Br2NO(g) [+ NO(g)] 2 BrNO(g) Reaction coordinate → 16 Molecularity of Elementary Steps 1 molecule = unimolecular 2 molecules = bimolecular 3 molecules = termolecular 1. Molecularities greater than termolecular are improbable because it's difficult to have 4 or more molecules colliding simultaneously with the correct energy and orientation. 2. The exponents in the rate expressions for elementary steps ARE the same as the stoichiometric coefficients in each elementary step. The reason why the exponents in the overall reaction are not necessarily the same as in the rate law is because of the presence sometimes of a SLOW rate-determining step. 17 Reaction mechanisms have a RATE DETERMINING STEP - the slowest step in the mechanism. - - 2 NH3 + OCl → N2H4 + Cl + H2O - - Step 1 NH3 + OCl → NH2Cl + OH fast + - Step 2 NH2Cl + NH3 → N2H5 + Cl slow + - Step 3 N2H5 + OH → N2H4 + H2Ofast Overall 2 NH + OCl- → N H + Cl- + H O Reaction 3 2 4 2 The overall rate expression must be consistent with the SUM of the elementary steps leading up to and including the slow step. - - Step 1 NH3 + OCl → NH2Cl + OH fast + - Step 2 NH2Cl + NH3 → N2H5 + Cl slow Overall up to - and including 2 NH3 + OCl → products the slow step 2 - so the rate law HAS to be rate = k[NH3] [OCl ] so fast it has no + - Step 3 N2H5 + OH → N2H4 + H2O effect on overall reaction rate 18 Writing plausible reaction mechanisms: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The sum of the elementary steps up to and including the rate-determining step should predict the same rate law that is determined experimentally. • the detection of an intermediate supports a given mechanism •The rate-determining step is the slowest step in the sequence of steps leading to product formation. Example 13.10 - The gas-phase decomposition of nitrous oxide (N2O) is believed to occur via two elementary steps: Step 1: N2O → N2 + O Step 2: N2O + O → N2 + O2 Experimentally the rate law is found to be rate = k[N2O]. (a) Write the equation for the overall reaction. (b) Identify the intermediates. (c) What can you say about the relative rates of steps 1 and 2? 19 Catalysts There are two ways to increase the reaction rate - 1. Increase the temperature - reaction rate increases because the number of molecules with energies > Ea increases 2. Lower the Ea by adding a CATALYST Catalyst = substance that increases the reaction rate by providing a lower energy reaction “pathway”. The catalyst is not consumed as part of the reaction, although there can be chemical transformations during the reaction. At the end of the reaction the catalyst is regenerated. 20 Homogeneous Catalyst The catalyst and reactants are in the same phase. MnO2(s) KClO3(s) 2 KCl(s) + 3 O2(g) O H+ (aq) H -C -O -H (aq) H2O(l) + CO(g) Heterogeneous Catalysts The catalyst and reactants are in different phases. Platinum-rhodium gauze catalyst used in the production of HNO3 21 Production of HNO3 Using a Heterogeneous Catalyst Solid Pt/Rh catalyst 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) 2 NO(g) + O2(g) 2 NO2(g) 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) Automobile Catalytic Converter Catalyst = Pt-NiO, finely divided powder. Higher surface area provides more sites for adsorption and reaction 22 The heterogeneous catalyst provides a surface where the reactants can be ADSORBED on an ACTIVE SITE - Industrial production of ammonia (Haber Process)- Fe 3 H2(g) + N2(g) → 2 NH3(g) 23 Enzyme Catalysis 24 enzyme uncatalyzed catalyzed 25.