Half-life Equations

Half-lives Half-lives The time it takes for a reactant to reach half of its initial amount Are there any differences in half-lives over time between zero-, first-, and second order reactions? 16M 8M 4M 2M 7 1M 7 6 6 1st Order 5 25 % 12.5 % t1/2 t1/2 t1/2 1 half-life 2 half-lives 3 half-lives 6.25 % [A] (mol/L) 50 % 100 % 4 1st half-life 3 2nd half-life 2 t1/2 4thhalf-life 0.5 1 1.5 Where “x” = number of half-lives 3 2nd half-life 2 3rd half-life 1 4thhalf-life 0 1 1.5 0 0.5 1 4thhalf-life 1.5 2 2.5 3 Time (s) Half-lives increase over time 2 2.5 3 General Half-life Equation With first-order half-life problems, you can sometimes use short cuts (such as the general rate law equation) because successive half-lives are the same time period. 1 [A]t  [A]0   2 x Where “x” = number of half-lives Integrated Half-life Equations Time (s) Half-Life Examples A certain reaction has a rate constant of 1.25 hr-1. What is the half life of this reaction in hours? Minutes? Half-lives are related to the rate constant, k, by the following equations  [A]0    2  l n(2) 1st Order ln    kt t 1/2 [A]0 k t 1/2 l n(2) k t 1/2 1 k[A]0 [A]0 t 1/2 2k 2nd Order 1 1  kt  [A]0  [A]0    2   0th Order  [A]0     [A]0  kt  2  t 1/2 t 1/2 [A]0 2k Graphing Integrated Rate Laws slope Equation of a straight line 0.555hr 1st-order rate law ln[A]0 y-intercept y = m x+b l n[A]t   k t  l n[A]0 ln[A]t vs. t time How many half lives will it take for the same reaction above to go from 6.0M to 0.020M? 1/[A]0 2nd-order rate law 1  [A]t k t  1 [A]0 1/[A]t 8.2 half-lives 1 k[A]0 ln[A]t Half lives decrease With second- and zeroorder problems that involve time, you must use the integrated half-life equations because successive half-lives have different time periods. 1/[A] t vs. t time If a 2nd order reaction has a half life of 3.1 days, how many hours will it take for 10.0% of the reactant to disappear if the concentration starts at 1.0M? [A]0 [A]t [A] (mol/L) 1st half-life 0.5 3 Half-life Equations 0th Order 0 2.5 Half lives are equal Half-lives 4 2 This equation can be used for any order as long as no time component is involved. 7 5 3rd half-life Time (s) General Equation for Half-life 6 2nd half-life 0 0 x 1st half-life 3 1 0 1 [A]t1 /2  [A]0   2 4 2 3rd half-life 1 4 half-lives 2nd Order 5 [A] (mol/L) Original concentration 0th-order rate law [A]t   k t  [A]0 [A] t vs. t time 1 Multistep Reactions – Reaction Mechanisms Determining order by graphing To determine the “order” of the reactant, you must graph the data three different ways: Time vs. Conc.; Time vs. ln(Conc.); and Time vs. 1/Conc. Time (s) 0 1 2 3 4 5 6 7 8 Time (s) 0 1 2 3 4 5 6 7 8 [A] (M) 1 0.5 0.333333 0.25 0.2 0.166667 0.142857 0.125 0.111111 Time vs. Conc. ln[A] 0 -0.69315 -1.09861 -1.38629 -1.60944 -1.79176 -1.94591 -2.07944 -2.19722 Time vs. ln(Conc.) Time (s) 1/[A] 1/M 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 NO2 + CO → NO + CO2 NO2 + CO NO + CO2 ? Time vs. 1/(Conc.) This reaction could happen in just one step…or in multiple steps. How can we figure it out? First we need to learn about these steps. Multistep Reactions – Reaction Mechanisms Molecularity and Rate Laws of Elementary steps Reactions that have multiple steps are made of individual steps called elementary steps. Elementary steps are single (concerted) steps. Step 1: 2NO2 → NO + NO3 Step 2: NO3 + CO → NO2 + CO2 Rate laws for elementary steps can be determined directly from the balanced equation The molecularity of an elementary step is based on the rate law for that step Intermediates are chemical species that are created during the reaction, but not a part of the overall reaction equation. Intermediates are created in an elementary step and consumed in another NO2 + CO → NO + CO2 Molecularity Equation Rate Law Unimolecular A → Product(s) Rate = k[A] Bimolecular A + A → Product(s) A + B → Product(s) Rate = k[A]2 Rate = k[A][B] Termolecular A + A +A → Product(s) A + B + B → Product(s) A + B + C → Product(s) Rate = k[A]3 Rate = k[A][B]2 Rate = k[A][B][C] Elementary steps must add up to the overall reaction equation. Since elementary steps are single steps, rate laws for each step can be determined from the balanced chemical equation Step 1: N2O4 →NO3 + NO Step 2: NO3+ CO → NO2 + CO2 These are not plausible steps for the reaction because the steps add up to a different overall reaction Step 1: 2NO2 → NO + NO3 Elementary Step Rate = k1 [NO2 N2O4 + CO → NO+ NO2 + CO2 Step 2: NO3 + CO → NO2 + CO2 ]2 Elementary Step Rate = k2[NO3][CO] Both of these elementary steps are bimolecular Overall reaction rates of multi-step reactions Multiple-step Reactions – Elementary Steps – Rate-limiting Step 1st Slow-step 1st Two-step reaction Overall Chemical Formula 2NO2 → NO + NO3 Elementary Steps 2NO2 → NO + NO3 NO3 + CO → NO2 + CO2 NO3 + CO → NO2 + CO2 NO2 + CO → NO + CO2 6,000 kisses per hour Slow Step Rate 1 < Rate 2 20,000 kisses per hour Fast Step How many kisses can be processed per hour overall??? The overall reaction can go no faster than the slow step. This step is called the rate-determining or rate-limiting step. Rate laws for elementary steps can be determined directly from the balanced equation 2NO2 → NO + NO3 NO3 + CO → NO2 + CO2 The elementary-step formulas must add up to the overall chemical formula If the slow, rate-determining, step is first, the overall rate law is the same as the slow step Step 1 Rate Law Rate = k1 [NO2]2 Just remember that reaction rates are not static; reaction rates are variable based on concentration. k1 [NO2]2 = Overall Rate Law Rate = k1 [NO2]2 k2[NO3][CO] Notice that the concentrations of “NO3” and “CO” do not affect the rate of reaction 2 Overall reaction rates of multi-step reactions Multiple-step Reactions – Elementary Steps – Rate-limiting Step 1st Elementary Steps Two-step reaction Slow-step 2nd Overall Chemical Formula 2NO2 → NO + NO3 2NO2 ⇌ NO + NO3 NO3 + CO → NO2 + CO2 NO3 + CO → NO2 + CO2 NO2 + CO → NO + CO2 Rate 1 500,000 kisses per hour Fast RXN Although it seems like the overall rate should = the slow step, in reality, reaction rates are variable based on concentrations of reactants. 500,000 kisses per hour 2NO2 → NO + NO3 NO3 + CO → NO2 + CO2 k1 [NO2]2 k2[NO3][CO] But what about that first step? Since there is an equilibrium taking place in the first step…. Can we use the slow-step rate law as the overall rate law in this case?? What’s wrong with it. Slow-step 2nd…3rd… 2NO2 ⇌ NO + NO3 We can actually write two rate laws for the first step Slow Step (Step 2) Rate Law NO3 + CO → NO2 + CO2 k2[NO3][CO] Rate 2 Products in Step 1 revert back into reactants before continuing to step 2 Remember, If the slow step is first, we can just use the first step rate law. What if the slow step is second? Rate = k-1[NO][NO3] Problem!! Neither rate law by itself can be correct because they both contain intermediates!!! Remember NO3…? We can’t just look the slow step in this case…. Step 1 Rate Law Step 2 Rate Law k1 [NO2]2 = k-1[NO][NO3] k2[NO3][CO] Catalysts #2 – Mechanisms Catalyzed Ea (new rxn route) Ea Step 2 Rate Law Combined Rate Law  k [NO2 ]2  k [CO] Rate   1  k -1[NO]  2   H2O2 + IO- → I- + H2O + O2 Substitute this whole expression for the intermediate Notice that the catalyst, I-, is regenerated in the end. There is no net change in catalyst in the reaction. Energy [NO2 ]2 [CO] Rate  k [NO] Area = number of particles that react at lower temp 2 1 Rate  k[NO2 ] [CO][NO] 2H2O2 → 2H2O + O2 H2O2 + I- → IO- +H2O # of particles k2[NO3][CO] Solve for intermediate k [NO2 ]2 [NO3 ]  1 k -1[NO] Step 1 Rate Law (reverse direction) Rate = k1 [NO2]2 k1 [NO2]2 = k-1[NO][NO3] We need to get rid of the intermediate. We can do this through mathematical substitution. k1 [NO2]2 = k-1[NO][NO3] Step 1 Rate Law (forward direction) Intermediates cannot be in the overall rate law. The overall rate law cannot be based solely on the Step 2 rate law. Multiple-step Reactions – Elementary Steps – Ratelimiting Step 2nd Step 1 Rate Law If the slow step is 2nd, determining the rate law will be more complicated……. Multiple-step Reactions – Elementary Steps – Ratelimiting Step 2nd If the slow step is second, the bottleneck creates an equilibrium in the first step. > The elementary-step formulas must add up to the overall chemical formula Rate laws for elementary steps can be determined directly from the balanced equation Multiple-step Reactions – Elementary Steps – Rate-limiting Step 2nd Rate 1 Rate 2 Slow RXN How many kisses can be processed per hour overall??? 2NO2 ⇌ NO + NO3 > 10,000 kisses per hour More reactant particles reach Ea = faster reaction The intermediate, IO-, is completely used in the reaction (as it should be). Area = number of particles that react at higher temp 3 Arrhenius Equation Svante Arrhenius determined that activation energy (Ea) of a reaction is related to the reaction constant. Arrhenius Equation We can also do calculations without graphing, but combining two equations into one. - Ea k  Ae R T That is…at any temperature the lower the activation energy, the faster a reaction will proceed. lnk1  lnA - Ea RT1 lnk2  lnA - Ea RT2 If we know both k1 and k2, along with T1 and T2 we calculate the energy of activation E lnk  lnA - a RT ln k1 E a  1 1      k 2 R  T2 T1  R  8.314 molJK In a certain equation, the rate constant at 701K was measured as 2.57M -1s-1 and that at 895 K was measured as 556M-1s-1. Find the activation energy of the reaction. lnk  y = Ea 1  lnA R T ln 1 556 Ms Ea 1   1     1 2.57 Ms 8.314 molJ K  701K 895K  5.38   Ea 3.09x10 -4 8.314 molJ K 1 K  1.45x105 J/mol m x + b 4