Lecture 12

# Lecture 12 - Rates of reac ons can be determined...

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Unformatted text preview: Rates of reac+ons can be determined by monitoring the change in concentra+on of either reactants or products as a func+on of +me.  ­ Δ[A] or Δ[B] vs Δt (A B) Each reac+on has its own equa+on that gives its rate as a func+on of reactant concentra+ons. ⇒Rate Law To determine the rate law we measure the rate at diﬀerent star+ng concentra+ons. Concentra+on and Rate when [NH4+] doubles (keeping NO2 ­ ﬁxed), the ini+al rate doubles. Concentra+on and Rate As [NO2 ­ doubles, the ini+al rate also doubles. Rate Law & Rate Constant This equa+on is called the rate law, and k is the rate constant. Rate Laws •  A rate law shows the rela+onship between the reac+on rate and the concentra+ons of reactants. –  For gas ­phase reactants use PA instead of [A]. •  k is a constant that has a speciﬁc value for each reac+on. •  The value of k is determined experimentally. k is unique for each rxn k changes with temperature Rate Laws •  Exponents tell the order of the reac+on with respect to each reactant. •  This reac+on is First ­order in [NH4+] First ­order in [NO2−] •  The overall reac+on order can be found by adding the exponents on the reactants in the rate law. •  Reac+on order is not related to coeﬃcients in the general balanced equa+on. •  This reac+on is second ­order overall. Sample Problem 16.2 PROBLEM: Determining Reac+on Order from Rate Laws For each of the following reac+ons, use the given rate law to determine the reac+on order with respect to each reactant and the overall order: (a) 2NO(g) + O2(g) 2NO2(g); rate = k[NO]2[O2] (b) CH3CHO(g) CH4(g) + CO(g); rate = k[CH3CHO]3/2 (c) H2O2(aq) + 3I ­(aq) + 2H+(aq) I3 ­(aq) + 2H2O(l); rate = k[H2O2][I ­ PLAN: Look at the rate law and not the coeﬃcients of the chemical reac+on. SOLUTION: (a) The reac+on is 2nd order in NO, 1st order in O2, and 3rd order overall. 3 3 (b) The reac+on is 2 order in CH3CHO and order overall. 2 (c) The reac+on is 1st order in H2O2, 1st order in I ­, zero order in H+, and 2nd order overall. Determining Reac+on Orders Using ini+al rates  ­ Run a series of experiments, each of which starts with a diﬀerent set of reactant concentra+ons, and from each obtain an ini+al rate. See Table 16.2 for data on the reac+on O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n Compare 2 experiments in which the concentra+on of one reactant varies and the concentra+on of the other reactant(s) remains constant. rate2 k [O2]2m[NO]2n = rate1 k [O2]1m[NO]1n 6.40 x 10 ­3 mol/L•s 3.21 x 10 ­3 mol/L•s = = [O2]2m [O2]1m 2.20 x 10 ­2 mol/L 1.10 x 10 ­2 mol/L = [O2]2 m [O2]1 m ; Do similar calcula+ons for the other reactant(s). 2 = 2m m = 1 Sample Problem 16.3 PROBLEM: Determining Reac+on Orders from Ini+al Rate Data Many gaseous reac+ons occur in a car engine and exhaust systems. One of these is rate = k[NO2]m[CO]n Use the following data to determine the individual and overall reac+on orders: Experiment 1 2 3 PLAN: SOLUTION: Ini+al Rate (mol/L•s) 0.0050 0.080 0.0050 Ini+al [NO2] (mol/L) 0.10 0.40 0.10 Ini+al [CO] (mol/L) 0.10 0.10 0.20 Solve for the order with respect to each reactant using the general rate law using the method described previously. rate = k [NO2]m[CO]n First, choose two experiments in which [CO] remains constant and the [NO2] varies. Sample Problem 16.3 Determining Reac+on Order from Ini+al Rate Data con+nued rate 2 rate 1 0.080 0.0050 rate 3 rate 1 0.0050 0.0050 = = = = k [NO2]m2[CO]n2 k [NO2]m1 [CO]n1 0.40 = m k [NO2]21 [CO]n1 0.20 0.10 n m The reac+on is 2nd order in NO2. [NO2] 1 16 = 4.0m and m = 2.0 0.10 k [NO2]23[CO]n3 [NO2] 2 = [CO] 3 n [CO] 1 1 = 2.0n and n = 0 rate = k [NO2]2[CO]0 = k [NO2]2 The reac+on is second order overall. The reac+on is zero order in CO. Integrated Rate Laws rate =  ­ Δ[A] Δt ln rate =  ­ Δ[A] Δt [A]t rate =  ­ Δt [A]0 [A]t = k [A]0 =  ­ kt ln [A]0  ­ ln [A]t = kt Second-order rate equation = k [A]2 1 Δ[A] First-order rate equation = k [A]  ­ 1 [A]0 = kt Zero-order rate equation [A]t  ­ [A]0 =  ­ kt 1 [A]t = kt + 1 [A]0 Sample Problem 16.4 PROBLEM: Determining Reac+on Orders from a Series of Molecular Scenes At a par+cular temperature, two gases, A (red) and B (blue), react to form products. The following molecular scenes represent four experiments run at the same volume, labeled 1 through 4 with their ini+al rates (in mol/L•s): (a)  What is the reac+on order with respect to A? With respect to B? The overall order? (b) Write the rate law for the reac+on. (c) Predict the ini+al rate of experiment 4. Sample Problem 16.4 con+nued PLAN: Determining Reac+on Orders from a Series of Molecular Scenes (a)  Determine the change in each reactant by coun+ng the number of spheres and see how the change aﬀected the rate. (b) Use the orders from (a) to write the rate law. (c) Use the results from Expts 1 and 3 and the rate law from part (b) to determine ini+al rate of Expt 4. Determining Reac+on Orders from a Series of Molecular Scenes Sample Problem 16.4 con+nued SOLUTION: (a)  From Expts 1 and 2, A doubles while B is constant and the rate also doubles. The reac+on is ﬁrst order with respect to A. From Expts 1 and 3 B doubles, A is constant and the rate quadruples. The reac+on is second order with respect to B. The overall order is (1 + 2), 3rd order. (b)  Rate = k[A][B]2 (c)  Comparing Expts 3 and 4 we see that A doubles. Comparing Expts 2 and 4 A is constant and B doubles, so the rate should quadruple. Rate 4 = 4.0 x 10 ­4 mol/L•s Sample Problem 16.5 PROBLEM: Determining the Reactant Concentra+on at a Given Time At 1000oC, cyclobutane (C4H8) decomposes in a ﬁrst ­order reac+on, with the very high rate constant of 87 s ­1, to two molecules of ethylene (C2H4). (a) If the ini+al C4H8 concentra+on is 2.00 M, what is the concentra+on aker 0.010 s? (b) What frac+on of C4H8 has decomposed in this +me? PLAN: Find the [C4H8] at +me, t, using the integrated rate law for a 1st order reac+on. Once that value is found, divide the amount decomposed by the ini+al concentra+on. SOLUTION: (a) ln [C4H8]0 [C4H8]t = kt ; ln 2.00 [C4H8] = (87 s ­1)(0.010 s) [C4H8] = 0.83 mol/L (b) [C4H8]0  ­ [C4H8]t [C4H8]0 = 2.00 M  ­ 0.83 M 2.00 M = 0.58 Rate Laws •  Exponents tell the order of the reac+on with respect to each reactant. •  This reac+on is First ­order in [NH4+] First ­order in [NO2−] •  The overall reac+on order can be found by adding the exponents on the reactants in the rate law. •  This reac+on is second ­order overall. Figure 16.6 Graphical determina+on of the reac+on order for the decomposi+on of N2O5. ...
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## This note was uploaded on 02/18/2012 for the course CHEM 6C taught by Professor Hoeger during the Spring '08 term at UCSD.

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