Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

CHAPTER 14 Chemical Kinetics Learning Objectives: Factors Affecting Reaction Rate: Concentration State Surface Area Temperature Catalyst Collision Theory of Reactions and Effective Collisions Determining Reaction Order and Rate Law from Data Integrated Rate Laws Rate Law  Concentration vs Rate Integrated Rate Law  Concentration vs Time Units of Rate Constant and Overall Reaction Order Half Life vs Rate Constant (1st Order) Arrhenius Equation Mechanisms and Rate Laws Catalysts Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 14 Chemical Kinetics Lecture Road Map: Factors that affect reaction rates Measuring rates of reactions Rate Laws Collision Theory Transition State Theory & Activation Energies Mechanisms Catalysts Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 14 Chemical Kinetics Integrated Rate Laws Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Rate law tells us how speed of reaction varies with concentrations. Integrated Rate Laws Concentration & Time Rate law tells us how speed of reaction varies with concentrations. Sometimes want to know Concentrations of reactants and products at given time during reaction How long for the concentration of reactants to drop below some minimum optimal value  Need dependence of rate on time Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

First Order Integrated Rate Law Integrated Rate Laws First Order Integrated Rate Law Corresponding to reactions A  products Integrating we get Rearranging gives Equation of line y = mx + b Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Slope = –k Integrated Rate Laws First Order Integrated Rate Law Yields straight line Indicative of first order kinetics Slope = –k Intercept = ln [A]0 If we don't know already Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

2nd Order Integrated Rate Law Integrated Rate Laws 2nd Order Integrated Rate Law Corresponding to special second order reaction 2B  products Integrating we get Rearranging gives Equation of line y = mx + b Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Slope = +k Integrated Rate Laws 2nd Order Integrated Rate Law Yields straight line Indicative of 2nd order kinetics Slope = +k Intercept = 1/[B]0 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Graphically determining Order Integrated Rate Laws Graphically determining Order Make two plots: ln [A] vs. time 1/[A] vs. time If ln [A] is linear and 1/[A] is curved, then reaction is 1st order in [A] If 1/[A] plot is linear and ln [A] is curved, then reaction is 2nd order in [A] If both plots give horizontal lines, then 0th order in [A] Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Graphically determining Order Integrated Rate Laws Graphically determining Order Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Integrated Rate Laws Example: SO2Cl2  SO2 + Cl2 Time, min [SO2Cl2], M ln[SO2Cl2] 1/[SO2Cl2] (L/mol) 0.1000 -2.3026 10.000 100 0.0876 -2.4350 11.416 200 0.0768 -2.5666 13.021 300 0.0673 -2.6986 14.859 400 0.0590 -2.8302 16.949 500 0.0517 -2.9623 19.342 600 0.0453 -3.0944 22.075 700 0.0397 -3.2264 25.189 800 0.0348 -3.3581 28.736 900 0.0305 -3.4900 32.787 1000 0.0267 -3.6231 37.453 1100 0.0234 -3.7550 42.735 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Reaction is 1st order in SO2Cl2 Integrated Rate Laws Example: SO2Cl2  SO2 + Cl2 Reaction is 1st order in SO2Cl2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Time (s) [HI] (mol/L) ln[HI] 1/[HI] (L/mol) Integrated Rate Laws Example: HI(g)  H2(g) + I2(g) Time (s) [HI] (mol/L) ln[HI] 1/[HI] (L/mol) 0.1000 -2.3026 10.000 50 0.0716 -2.6367 13.9665 100 0.0558 -2.8860 17.9211 150 0.0457 -3.0857 21.8818 200 0.0387 -3.2519 25.840 250 0.0336 -3.3932 29.7619 300 0.0296 -3.5200 33.7838 350 0.0265 -3.6306 37.7358 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Reaction is second order in HI. Integrated Rate Laws Example: HI(g)  H2(g) + I2(g) Reaction is second order in HI. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ? A. Concentration B. ln of Concentration C. 1/Concentration D. 1/ ln Concentration Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Half Life (t1/2) for first order reactions Integrated Rate Laws Half Life (t1/2) for first order reactions Half-life = t½ We often use the half life to describe how fast a reaction takes place First Order Reactions Set Substituting into Gives Canceling gives ln 2 = kt½ Rearranging gives Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Half Life (t1/2) for First Order Reactions Integrated Rate Laws Half Life (t1/2) for First Order Reactions Observe: t½ is independent of [A]o For given reaction (and T) Takes same time for concentration to fall from 2 M to 1 M as from 5.0  10–3 M to 2.5  10–3 M k1 has units (time)–1, so t½ has units (time) t½ called half-life Time for ½ of sample to decay Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Does this mean that all of sample is gone in two half-lives (2 × t½)? Integrated Rate Laws Half Life (t1/2) Does this mean that all of sample is gone in two half-lives (2 × t½)? No! In 1st t½, it goes to ½[A]o In 2nd t½, it goes to ½(½[A]o) = ¼[A]o In 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o In nth t½, it goes to [A]o/2n Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Integrated Rate Laws Half Life (t1/2) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Half Life (t1/2): First Order Example Integrated Rate Laws Half Life (t1/2): First Order Example 131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s–1? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem The half-life of I-132 is 2.295 h. What percentage remains after 24 hours? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Half Life (t1/2): Carbon-14 Dating Integrated Rate Laws Half Life (t1/2): Carbon-14 Dating Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Half Life (t1/2): Second Order Reactions Integrated Rate Laws Half Life (t1/2): Second Order Reactions How long before [A] = ½[A]o? t½, depends on [A]o t½, not useful quantity for a second order reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem The rate constant for the second order reaction 2A → B is 5.3 × 10–5 M–1 s–1. What is the original amount present if, after 2 hours, there is 0.35 M available? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 14 Chemical Kinetics Collision Theory Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

As the concentration of reactants increase Collision Theory Reaction Rates Collision Theory As the concentration of reactants increase The number of collisions increases Reaction rate increases As temperature increases Molecular speed increases Higher proportion of collisions with enough force (energy) There are more collisions per second Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

One that gives rise to product e.g. At room temperature and pressure Collision Theory Reaction Rates Rate of reaction proportional to number of effective collisions/sec among reactant molecules Effective collision One that gives rise to product e.g. At room temperature and pressure H2 and I2 molecules undergoing 1010 collisions/sec Yet reaction takes a long time Not all collisions lead to reaction Only very small percentage of all collisions lead to net change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Molecular Orientation Collision Theory Molecular Orientation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

So more molecules undergo reaction Collision Theory Temperature As T increases More molecules have Ea So more molecules undergo reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Activation Energy (Ea) Collision Theory Activation Energy (Ea) Molecules must possess certain amount of kinetic energy (KE) in order to react Activation Energy, Ea = Minimum KE needed for reaction to occur Get energy from collision with other molecules If molecules move too slowly, too little KE, they just bounce off each other Without this minimum amount, reaction will not occur even when correctly oriented Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 14 Chemical Kinetics Transition State Theory Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Molecular Basis of Transition State Theory KE KE KE decreasing as PE increases Is the combined KE of both molecules enough to overcome Activation Energy PE KE KE Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Hreaction = Hproducts – Hreactants Transition State Molecular Basis of Transition State Theory Activation energy (Ea) = hill or barrier between reactants and products Potential Energy Heat of reaction (H) = difference in PE between products and reactants Hreaction = Hproducts – Hreactants Products Reaction Coordinate (progress of reaction) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Products lower PE than reactants Transition State Exothermic Reactions Exothermic reaction Products lower PE than reactants Potential Energy Exothermic Reaction H = – Products Reaction Coordinate (progress of reaction) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Hreaction < 0 (negative) Decrease in PE of system Transition State Exothermic Reactions Hreaction < 0 (negative) Decrease in PE of system Appears as increase in KE So the temperature of the system increases Reaction gives off heat Can’t say anything about Ea from size of H Ea could be high and reaction slow even if Hrxn large and negative Ea could be low and reaction rapid Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Hreaction = Hproducts – Hreactants Transition State Endothermic Reactions Endothermic Reaction H = + Hreaction = Hproducts – Hreactants Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Endothermic Reactions Transition State Endothermic Reactions Hreaction > 0 (positive) Increase in PE Appears as decrease in KE So temperature of the system decreases Have to add E to get reaction to go Ea  Hrxn as Ea includes Hrxn If Hrxn large and positive Ea must be high Reaction very slow Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Arrangement of atoms at top of activation barrier Transition State Activated Complex Arrangement of atoms at top of activation barrier Brief moment during successful collision when bond to be broken is partially broken and bond to be formed is partially formed Example Transition State Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CH3CH2O- + H3O+  CH3CH2OH + H2O Group Problem Draw the transition state complex, or the activated complex for the following reaction: CH3CH2O- + H3O+  CH3CH2OH + H2O Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

CHAPTER 14 Chemical Kinetics Activation Energies Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Equation expressing temperature-dependence of k Ea Arrhenius Equation The rate constant is dependent on Temperature, which allows us to calculate Activation Energy, Ea Arrhenius Equation: Equation expressing temperature-dependence of k A = Frequency factor has same units as k R = gas constant in energy units = 8.314 J mol–1 K–1 Ea = Activation Energy—has units of J/mol T = Temperature in K Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Calculating Activation Energy Ea Calculating Activation Energy Method 1. Graphically Take natural logarithm of both sides Rearranging Equation for a line y = b + mx Arrhenius Plot Plot ln k (y axis) vs. 1/T (x axis)  yield a straight line Slope = -Ea/R Intercept = A Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Ea Arrhenius Equation: Graphing Example k (M/s) T, ˚C T, K 0.000886 25 298 0.000894 50 348 0.000908 100 398 0.000918 150 448 ? 75 Given the following data, predict k at 75 ˚C using the graphical approach ln (k) = –36.025/T – 6.908 ln (k) = –36.025/(348) – 6.908 = – 7.011 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Arrhenius Equation: Graphing Example Ea Arrhenius Equation: Graphing Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Sometimes a graph is not needed Only have two k s at two Ts Ea Arrhenius Equation Sometimes a graph is not needed Only have two k s at two Ts Here use van't Hoff Equation derived from Arrhenius equation: Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Arrhenius Equation: Ex Vant Hoff Equation Ea Arrhenius Equation: Ex Vant Hoff Equation CH4 + 2 S2  CS2 + 2 H2S k (L/mol s) T (˚C) T (K) 1.1 = k1 550 823 = T1 6.4 = k2 625 898 = T2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Group Problem Given that k at 25 ˚C is 4.61 × 10–1 M/s and that at 50 ˚C it is 4.64 × 10–1 M/s, what is the activation energy for the reaction? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

A. Rate increases approximately 1.5 times Group Problem A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C? A. Rate increases approximately 1.5 times B. Rate increases approximately 5000 times C. Rate does not increase D. Rate increases approximately 3 times Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E