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<p>Chapter 4Kinetics: Rates and Mechanisms of Chemical Reactions</p> <p>16-1</p> <p>Kinetics: Rates and Mechanisms of Chemical Reactions16.1 Factors That Influence Reaction Rate 16.2 Expressing the Reaction Rate</p> <p>16.3 The Rate Law and Its Components16.4 Integrated Rate Laws: Concentration Changes over Time 16.5 The Effect of Temperature on Reaction Rate</p> <p>16.6 Explaining the Effects of Concentration and Temperature 16.7 Reaction Mechanisms: Steps in the Overall Reaction16.8 Catalysis: Speeding Up a Chemical Reaction</p> <p>16-2</p> <p>Figure 16.1</p> <p>Reaction rate: the central focus of chemical kinetics.</p> <p>16-3</p> <p>Figure 16.2</p> <p>The wide range of reaction rates.</p> <p>16-4</p> <p>Factors That Influence Reaction Rate</p> <p>Under a specific set of conditions, every reaction has its own characteristic rate, which depends upon the chemical nature of the reactants.</p> <p>Four factors can be controlled during the reaction: 1. Concentration - molecules must collide to react; 2. 3. 4. Physical state - molecules must mix to collide; Temperature - molecules must collide with enough energy to react; The use of a catalyst.</p> <p>16-5</p> <p>Figure 16.3</p> <p>The effect of surface area on reaction rate.</p> <p>16-6</p> <p>Figure 16.4</p> <p>Collision energy and reaction rate.</p> <p>16-7</p> <p>Expressing the Reaction Ratereaction rate - changes in the concentrations of reactants or products per unit time reactant concentrations decrease while product concentrations increase for</p> <p>A</p> <p>B=-</p> <p>Rate of reaction = -</p> <p>change in concentration of A change in time</p> <p>conc A2 - conc A1t2 - t1</p> <p>-</p> <p> (conc A) t</p> <p>16-8</p> <p>Table 16.1 Concentration of O3 at Various Times in its Reaction with C2H4 at 303 KC2H4(g) + O3(g) C2H4 O(g) + O2(g)</p> <p>Time (s)</p> <p>Concentration of O3 (mol/L) 3.20x10-5 2.42x10-5 1.95x10-5 1.63x10-5 1.40x10-5 1.23x10-5 1.10x10-5</p> <p>-</p> <p> (conc A) t</p> <p>0.010.0 20.0 30.0 40.0 50.0 60.0</p> <p>16-9</p> <p>Figure 16.5</p> <p>The concentration of O3 vs. time during its reaction with C2H4.C2H4(g) + O3(g) C2H4 O(g) + O2(g)</p> <p>rate = [C2H4] t</p> <p>=</p> <p>-</p> <p> [O3] t</p> <p>16-10</p> <p>Figure 16.6</p> <p>Plots of [C2H4] and [O2] vs. time.</p> <p>Tools of the Laboratory</p> <p>16-11</p> <p>In general, for the reaction aA + bB cC + dD</p> <p>rate =</p> <p>-</p> <p>1 a</p> <p>[A]t</p> <p>= -</p> <p>1 b</p> <p>[B]t</p> <p>= +</p> <p>1 c</p> <p>[C]t</p> <p>= +</p> <p>1 d</p> <p>[D]t</p> <p>The numerical value of the rate depends upon the substance that serves as the reference. The rest is relative to the balanced chemical equation.</p> <p>16-12</p> <p>Sample Problem 16.1 PROBLEM:</p> <p>Expressing Rate in Terms of Changes in Concentration with Time</p> <p>Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and may be used by earthbound engines in the near future. 2H2(g) + O2(g) 2H2O(g)</p> <p>(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.</p> <p>(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H 2O] increasing? PLAN: Choose [O2] as a point of reference since its coefficient is 1. For every molecule of O2 which disappears, 2 molecules of H2 disappear and 2 molecules of H2O appear, so [O2] is disappearing at half the rate of change of H2 and H2O.SOLUTION:</p> <p>(a) (b) [O2] t</p> <p>rate = -</p> <p>1 2</p> <p>[H2] t</p> <p>=-</p> <p>[O2] t t</p> <p>1 =+ 2</p> <p>[H2O] t = 0.46 mol/L*s</p> <p>= - 0.23 mol/L*s = + 1 2</p> <p>[H2O]</p> <p>[H2O]t</p> <p>16-13</p> <p>Sample Problem 16.2 Determining Reaction Order from Rate Laws PROBLEM: For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. 2NO2(g); rate = k[NO]2[O2] CH4(g) + CO(g); rate = k[CH3CHO]3/2</p> <p>(a) 2NO(g) + O2(g) (b) CH3CHO(g)</p> <p>(c) H2O2(aq) + 3I-(aq) + 2H+(aq)</p> <p>I3-(aq) + 2H2O(l); rate = k[H2O2][I-]</p> <p>PLAN: Look at the rate law and not the coefficients of the chemical reaction. SOLUTION:</p> <p>(a) The reaction is 2nd order in NO, 1st order in O 2, and 3rd order overall.(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall. (c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+, while being 2nd order overall.</p> <p>16-14</p> <p>Table 16.2 Initial Rates for a Series of Experiments in the Reaction Between O2 and NO2NO(g) + O2(g) 2NO2(g)</p> <p>Initial Reactant Concentrations (mol/L)</p> <p>Experiment</p> <p>O2</p> <p>NO</p> <p>Initial Rate (mol/L*s) 3.21x10-36.40x10-3 12.8x10-3</p> <p>12 3</p> <p>1.10x10-22.20x10-2 1.10x10-2</p> <p>1.30x10-21.30x10-2 2.60x10-2</p> <p>45</p> <p>3.30x10-21.10x10-2</p> <p>1.30x10-23.90x10-2</p> <p>9.60x10-328.8x10-3</p> <p>16-15</p> <p>Determining Reaction OrdersUsing initial rates Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. See Table 16.2 for data on the reaction O2(g) + 2NO(g) 2NO2(g) rate = k [O2]m[NO]n</p> <p>Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. rate 2 k [O2]2m[NO]2n [O2]2m [O2]1mm</p> <p>=rate 1 k [O2]1m[NO]1n = 3.21x10-3 mol/L*s</p> <p>=</p> <p>=</p> <p>[O2]2</p> <p>m</p> <p>[O2]1</p> <p>6.40x10-3 mol/L*s</p> <p>2.20x10-2 mol/L1.10x10-2 mol/L</p> <p>;</p> <p>2 = 2m</p> <p>m=1</p> <p>Do a similar calculation for the other reactant(s).</p> <p>16-16</p> <p>Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data PROBLEM: Many gaseous reactions occur in a car engine and exhaust system. One of these isNO2(g) + CO(g) NO(g) + CO2(g)</p> <p>rate = k[NO2]m[CO]n</p> <p>Use the following data to determine the individual and overall reaction orders.Experiment Initial Rate (mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L) 0.0050 0.080 0.0050 0.10 0.40 0.10 0.10</p> <p>1 23</p> <p>0.100.20</p> <p>PLAN:</p> <p>Solve for each reactant using the general rate law using the method described previously.rate = k [NO2]m[CO]n</p> <p>SOLUTION:</p> <p>First, choose two experiments in which [CO] remains constant and the [NO2] varies.</p> <p>16-17</p> <p>Sample Problem 16.3 Determining Reaction Orders from Initial Rate Data</p> <p>rate 2</p> <p>rate 10.080 0.0050</p> <p>=</p> <p>k [NO2]m2[CO]n2</p> <p>k [NO20.40</p> <p>]mm</p> <p>1</p> <p>[CO]n</p> <p>=</p> <p>[NO2] 2</p> <p>m</p> <p>1</p> <p>[NO2] 1</p> <p>The reaction is 2nd order in NO2.</p> <p>=</p> <p>;</p> <p>16 = 4m and m = 2 [CO] 3[CO] 1 1 = 2n and n = 0n</p> <p>0.10</p> <p>rate 3rate 1 0.0050 =</p> <p>k [NO2]m3[CO]n3k [NO2]m1 [CO]n1 0.20n</p> <p>=</p> <p>The reaction is zero order in CO.</p> <p>0.0050</p> <p>=</p> <p>;</p> <p>0.10rate = k [NO2]2[CO]0 = k [NO2]2</p> <p>16-18</p> <p>Table 16.3 Units of the Rate Constant k for Several Overall Reaction Orders</p> <p>Overall Reaction Order 0</p> <p>Units of k (t in seconds) mol/L*s (or mol L-1 s-1)</p> <p>1</p> <p>1/s (or s-1)</p> <p>2</p> <p>L/mol*s (or L mol -1 s-1)</p> <p>3</p> <p>L2 / mol2 *s (or L2 mol-2 s-1)</p> <p>16-19</p> <p>Integrated Rate Laws[A]</p> <p>rate = -</p> <p>t</p> <p>= k [A]</p> <p>first order rate equation[A]0 [A]t</p> <p>ln[A]rate = t = k [A]2</p> <p>= kt</p> <p>ln [A]0 = kt + ln [A]t</p> <p>second order rate equation</p> <p>1 [A]t[A] rate = t = k [A]0</p> <p>-</p> <p>1 [A]0</p> <p>= kt</p> <p>1 [A]t</p> <p>= kt</p> <p>+</p> <p>1 [A]0</p> <p>zero order rate equation</p> <p>[A]t - [A]0 = - kt</p> <p>16-20</p> <p>Sample Problem 16.5 Determining the Reactant Concentration at a Given Time PROBLEM: At 1000oC, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 87 s -1, to two molecules of ethylene (C2H4). (a) If the initial C4H8 concentration is 2.00 M, what is the concentration after 0.010 s? (b) What fraction of C4H8 has decomposed in this time? PLAN: Find the [C4H8] at time, t, using the integrated rate law for a 1st order reaction. Once that value is found, divide the amount decomposed by the initial concentration.</p> <p>SOLUTION:</p> <p>ln(a)</p> <p>[C4H8]0[C4H8]t</p> <p>= kt ;</p> <p>ln</p> <p>2.00[C4H8]</p> <p>= (87 s-1)(0.010 s)</p> <p>[C4H8] = 0.83 mol/L(b) [C4H8]0 - [C4H8]t [C4H8]0 2.00 M - 0.87 M = 2.00 M</p> <p>= 0.58</p> <p>16-21</p> <p>Figure 16.7</p> <p>Integrated rate laws and reaction orders.</p> <p>1/[A]t = kt + 1/[A]0</p> <p>ln[A]t = -kt + ln[A]0</p> <p>[A]t = -kt + [A]0</p> <p>16-22</p> <p>Figure 16.8</p> <p>Graphical determination of the reaction order for the decomposition of N2O5.</p> <p>16-23</p> <p>Figure 16.9</p> <p>A plot of [N2O5] vs. time for three half-lives. for a first-order process t1/2 =</p> <p>ln 2k</p> <p>=</p> <p>0.693 k</p> <p>16-24</p> <p>Sample Problem 16.6 Using Molecular Scenes to Determine Half-Life PROBLEM: The 1st order process of compound A (red) converting to compound B (black) is depicted at 0.0 s and 30.0 s:</p> <p>(c)</p> <p>(a) Find the half-life, t1/2, of the reaction. (b) Calculate the rate constant, k. (c) Draw a scene that represents the reaction mixture at 2.00 min. PLAN: Number of spheres represents concentration. Half-life is constant for a 1st order reaction and the elapsed time when half the red spheres turn black. SOLUTION: (a) At t = 0, 8 A and 0 B. At t = 30.0 s, 6 A and 2 B. For 4 A and 4 B, t = 60.0 s = t1/2. (b) t1/2 = 0.693/k; k = 0.693/t1/2 = 0.693/60.0 s = 1.16 x 10-2 s-1 (c) The 2.00 min is 120. s or two half-lives represented by the above scene.</p> <p>16-25</p> <p>Sample Problem 16.7 Determining the Half-Life of a First-Order Reaction PROBLEM: Cyclopropane is the smallest cyclic hydrocarbon. Because its 60o bond angles allow poor orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000oC via the following first-order reaction: CH2 H3C CH CH2 (g) H2C CH2 (g)</p> <p>The rate constant is 9.2 s-1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? 0.693 PLAN: Use the half-life equation, t1/2 = , to find the half-life. k One-quarter of the initial value means two half-lives have passed.</p> <p>SOLUTION:(a) t1/2 = 0.693/9.2 s-1 = 0.075 s (b) 2 t1/2 = 2(0.075 s) = 0.150 s</p> <p>16-26</p> <p>Table 16.4 An Overview of Zero-Order, First-Order, and Simple Second-Order ReactionsZero Order Rate lawUnits for k Integrated rate law in straight-line form Plot for straight line rate = k</p> <p>First Order rate = k[A] 1/s</p> <p>Second Order rate = k[A]2L/mol*s</p> <p>mol/L*s [A]t = -kt + [A]0 [A]t vs. t k, [A]0[A]0/2k</p> <p>ln [A]t = -kt + ln [A]0 1/[A]t = kt + 1/[A]0 ln [A]t vs. t -k, ln [A]0ln 2/k</p> <p>1/[A]t = tk, 1/[A]0 1/k[A]0</p> <p>Slope, y interceptHalf-life</p> <p>16-27</p> <p>Figure 16.10</p> <p>Dependence of the rate constant on temperature.</p> <p>16-28</p> <p>The Effect of Temperature on Reaction RateThe Arrhenius Equation E a / RT</p> <p>k Ae</p> <p>where k is the kinetic rate constant at T Ea is the activation energy</p> <p>R is the energy gas constant</p> <p>lnk = lnA - Ea/RT</p> <p>T is the Kelvin temperature A is the collision frequency factor</p> <p>ln</p> <p>k2 k1</p> <p>= -</p> <p>Ea R</p> <p>1 T2</p> <p>-</p> <p>1 T1</p> <p>16-29</p> <p>Figure 16.11 Graphical determination of the activation energy.</p> <p>ln k = (-Ea/R )(1/T) + lnA</p> <p>16-30</p> <p>Sample Problem 16.8 Determining the Energy of Activation PROBLEM: The decomposition of hydrogen iodide, 2HI(g) H2(g) + I2(g)</p> <p>has rate constants of 9.51x10-9 L/mol*s at 500. K and 1.10x10-5 L/mol*s at 600. K. Find Ea. PLAN: Use the modification of the Arrhenius equation to find Ea. -1</p> <p>SOLUTION:ln k2</p> <p>k1</p> <p>= -</p> <p>Ea</p> <p>1</p> <p>-</p> <p>1</p> <p>R</p> <p>T2</p> <p>T1</p> <p>Ea = - R ln 1 600 K</p> <p>k2</p> <p>1</p> <p>-</p> <p>1</p> <p>k11</p> <p>T2</p> <p>T1</p> <p>Ea = - (8.314 J/mol*K) ln</p> <p>1.10x10-5 L/mol*s 9..51x10-9 L/mol*s</p> <p>-</p> <p>500 K</p> <p>Ea = 1.76x105 J/mol = 176 kJ/mol</p> <p>16-31</p> <p>Figure 16.12</p> <p>Information sequence to determine the kinetic parameters of a reaction.Series of plots of concentration vs. time</p> <p>Initial rates</p> <p>Determine slope of tangent at t0 for each plot</p> <p>Reaction Rate constant orders (k) and actual Compare initial rate law rates when [A] Substitute initial rates, changes and [B] is orders, and concentrations Find k at held constant and into general rate law: varied T m[B]n vice versa rate = k [A]</p> <p>Plots of concentration vs. time</p> <p>Integrated rate law (half-life, t1/2)</p> <p>Rate constant and reaction orderRearrange to linear form and graph</p> <p>Activation energy, Ea Find k at varied T</p> <p>16-32</p> <p>Use direct, ln or inverse plot to find order</p> <p>Figure 16.13</p> <p>The dependence of number of possible collisions on the product of reactant concentrations.</p> <p>A</p> <p>B 4 collisions</p> <p>A</p> <p>B A</p> <p>Why concentrations Are Multiplied in the Rate Law ?</p> <p>Add another molecule of A</p> <p>B</p> <p>AB A</p> <p>6 collisions</p> <p>A Add another molecule of B A A</p> <p>B B B</p> <p>16-33</p> <p>Figure 16.14</p> <p>The effect of temperature on the distribution of collision energies.The temperature rise enlarges the fraction of collisions with enough energy to exceed the Ea Energy required to activate the molecules into a state from which reactant bonds can change into product bonds</p> <p>16-34</p> <p>Table 16.5 The Effect of Ea and T on the Fraction (f) of Collisions with Sufficient Energy to Allow ReactionEa (kJ/mol) 5075 100 TIncreasing Ea by 25 kJ/mol</p> <p>f (at T = 298 K) 1.70x10-97.03x10-14 2.90x10-18 f (at Ea = 50 kJ/mol)Decreased</p> <p>25oC (298 K) 35oC (308 K) 45oC (318 K)</p> <p>1.70x10-9 3.29x10-9 6.12x10-9</p> <p>16-35</p> <p>Figure 16.15</p> <p>Energy-level diagram for a reaction.</p> <p>ACTIVATED STATE</p> <p>Collision Energy</p> <p>Ea (reverse)</p> <p>REACTANTS</p> <p>PRODUCTS The forward reaction is exothermic because the reactants have more energy than the products.</p> <p>16-36</p> <p>Collision Energy</p> <p>Ea (forward)</p> <p>Figure 16.16</p> <p>An energy-level diagram of the fraction of collisions exceeding Ea.</p> <p>Larger</p> <p>Smaller</p> <p>16-37</p> <p> In both reaction direction, a larger fraction of collision exceeds the activation energy at the higher temperature, T2; higher T increase reaction rate</p> <p>16-38</p> <p>Figure 16.17</p> <p>The importance of molecular orientation to an effective collision.</p> <p>NO + NO3</p> <p>2 NO2</p> <p>A is the frequency factor A = pZ where Z is the collision frequency p is the orientation probability factor</p> <p>16-39</p> <p>Figure 16.18</p> <p>Nature of the transition state in the reaction between CH3Br and OH-.CH3Br + OHCH3OH + Br -</p> <p>transition state or activated complex</p> <p>16-40</p> <p>Figure 16.19 Reaction energy diagram for the reaction of CH3Br and OH-.</p> <p>Forms only if the molecules collide In an effective orientation and with energy than Ea</p> <p>Transition state/activated complex Extremely unstable species (has very high potential energy)</p> <p>Thus, the Ea is the quantity needed To stretch and deform bonds in order to reach the transition state</p> <p>16-41</p> <p>Sample Problem 16.9 Drawing Reaction Energy Diagrams and Transition States PROBLEM: A key reaction in the upper atmosphere is O3(g) + O(g) 2O2(g)</p> <p>The Ea(fwd) is 19 kJ, and the Hrxn for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate Ea(rev). PLAN: Consider the relationships among the reactants, products and transition state. The reactants are at a higher energy level than the products and the transition state is slightly higher than the reactants.Potential Energy</p> <p>SOLUTION:</p> <p>Ea= 19 kJ</p> <p>O3+O</p> <p>transition state Ea(rev)= (392 + 19) kJ = 411kJ2O2breaking bond</p> <p>Hrxn = -392 kJ</p> <p>O</p> <p>O</p> <p>OO</p> <p>16-42</p> <p>Reaction progress</p> <p>forming bond</p> <p>REACTION MECHANISMS Table 16.6 Rate Laws for General Elementary StepsElementary Step A product Molecularity Unimolecular Rate Law Rate = [A]</p> <p>2A</p> <p>product</p> <p>Bimolecular</p> <p>Rate = k[A]2</p> <p>A+B</p> <p>product</p> <p>Bimolecular</p> <p>Rate = k[A][B]</p> <p>2A + B</p> <p>product</p> <p>Termolecular</p> <p>Rate = k[A]2[B]</p> <p>16-43</p> <p>S...</p>