CHEM 106 1nd Edition Lecture 12Outline of Last Lecture I. Chemical KineticsII. Reactions RatesIII. Effect of ConcentrationOutline of Current Lecture II. Reaction order and Rate constantIII. Rate LawIV. Integrated Rate Lawsa. Calculus Current LectureChapter 15: Chemical Kinetics (continued)For the general reaction: aA + bB → cC + dDRate is given by: 1c ∆[C]∆ t = 1b ∆[B ]∆ t = 1c ∆[C]∆ t = 1d ∆[ D ]∆ t2O3(g) → 3O2(g)Rate = 12 ∆[O3]∆t = 13 ∆[O2]∆tLast time – 15.3 Concentration vs RateWe expect reactions rate to be dependent on concentrations, but we can’t necessary know in advance about how concentration affects rate!Reaction Order and Rate ConstantReaction Order: experimentally determined # that defines the dependence of the reaction rate on the concentration of the reactant.2NO + O2 → 2NO2Table 15.3 page 712Experiment # [NO]0 (M) [O2]0 (M) Rate (M/s)1 0.0100 0.0100 1 * 10-62 0.0100 0.0050 0.5 * 10-63 0.0050 0.0100 2.5 * 10-7The equation that describes mathematically how the rate depends on concentrations of reactants is called the Rate LawThe rate law has the general form:Rate ∝ [A]x Where x = integer; 0, 1, 2...X is the order of the reaction with respect to [A] X is determined experimentallySo, look at the date in table 15.3 and compare experiments 1 & 2[NO] constant; [O2] is 2x in experiment 1what happened to rate?When [O2] doubles, rate doubles OR when [O2] halved, rate is one-halfCan see a direct 1:1 dependence with respect to [O2]So we call this 1 st order! Rate ∝ [O2]1, e.g. x =1If [O2] is tripled, rate ↑ by a factor of 3!Now, let’s compare experiments 1 & 3:[O2] constant, when [NO] halved, rate ↓ by factor of 4!When [NO] doubled, rate increases by fourRate ∝ [NO]2, e.g. x=2 This is 2 nd order! It is treated like “squared exponent” When [NO] doubles, rate ↑ by 22 = 4When [NO] triples, rate ↑ by 32 = 9You can also have zero orderThis tells us that there is no dependence on rate with respect to that reactant So, what we know from the data from table 15.3Rate ∝ [NO]2[O2]1To get actual rate law, we need to include the rate constant, k, proportionally constant atgiven temperature.The general form:Rate = k [A]m [B]nk = rate constantm = order of reaction with regards to [A]n = order of reaction with regards to [B]k is temperature dependent! Any value of k is only valid at a specific temperature.For a given reaction at a specific temperature, value of k will not vary with the concentration of reactants if you know the order of reaction correctly for each reactant.For example: Rate = k [NO]m [O2]nWe can show this mathematically:If we compare experiments 1 & 2 again: Experiment 1: 1 * 10-6 M/s = k [0.0100)m (0.0100)nExperiment 2: 0.5 * 10-6 M/s = k (0.0100)m (0.0050)nCan take ratio: 0.0100¿¿0.5∗10−6=k ¿1∗10−6=k (0.0100)m(0.0100)n¿ 2.0 = (0.0100)n(0.0050)n2.0 = 2.0nlog 2.0 = n log 2.0n = log2.0log2.0 = 1 reaction order with respect to [O2]Can do same for experiments 1 & 3 to see reaction order with respect to [NO]1. Write out observed rate law for experiment 1 & 32. Take ratio of the two expressions 4.0 = (0.0100)m(0.0050)m4.0 = 2.0mlog 4.0 = m log 4.0m = 2 reaction order with respect to [NO]Therefore, for this reaction, rate = k [NO]2 [O2] We would also say that the overall order of the reaction is sum exponentsm + n = 2 + 1 = 3We can now determine the value of k:Rate = k [NO]2 [O2]For experiment 1:K = O[NO ]2[¿¿2]Rate¿ = 1∗10−6M /s(0.010m)2(0.010 m ) = 1.0 m-2 s-1 For experiment 2:K = O[NO ]2[¿¿2]Rate¿ = 0.5∗10−6M / s(0.010m)2(0.0 05 m ) = 1 m-2 s-1Remember:Reaction rate depends on the concentration of the reactantsRate constant (k) only depends on the temperatureIntegrated Rate Laws-allows us to determine the Rate Law and the rate constant using all experimental data, not just initial reaction rate data.General approach: test the data against expected trends for 0 order, 1st order, and 2nd orderChoose the one that best represents observations/experimental data.Easiest to show this with examples:Start with 1st order:O3(g) → O2(g) + O(g) Rate law: rate = k [O3]Also, rate = -∆[O3]∆t = k [O3]Using calculus, we can transform this expression into:OO3¿¿ln ¿ln[¿¿3]¿¿ = -k t integrated rate lawMore generally, for any 1st order processRate = k [A]ln [ A][ A ]0 = -k tln [A] = ln [A]0 – k ty = b - mxk = m = slope of lineb = ln [A]0 = y interceptx = t, independent variabley = ln [A], dependent variableRemember ln = natural log[A]0 = initial [A][A] = [A] at any time, tPlot ln [A] vs. t-if linear, 1st order!See Fig. 15.8 page 717Can calculate slope to get value of k for the reaction: Slope = - ∆[O3]∆t = −9.650−(−9.32)400−100 sk = 1.1 * 103 s-1Remember units, also k is never <