Chapter 15 Chemical Kinetics: The Rate of Chemical Reactions Kinetics In kinetics we study the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). 2009, Prentice-Hall, Inc.
Factors That Affect Reaction Rates Physical State of the Reactants In order to react, molecules must come in contact with each other. The more homogeneous the mixture of reactants, the faster the molecules can react. 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Concentration of Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.
2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Temperature At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Presence of a Catalyst Catalysts speed up reactions by changing the mechanism of the reaction. Catalysts are not consumed during the course of the reaction. 2009, Prentice-Hall, Inc.
Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. 2009, Prentice-Hall, Inc. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.
2009, Prentice-Hall, Inc. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate =
[CC4H9Cl] t 2009, Prentice-Hall, Inc. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, there are fewer collisions between
reactant molecules. 2009, Prentice-Hall, Inc. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) A plot of [CC4H9Cl] vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time.
2009, Prentice-Hall, Inc. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. 2009, Prentice-Hall, Inc.
Reaction Rates and Stoichiometry Reaction rates can be monitored by following either the loss (-) of reactants or the production (+) of products as generalized below: aA + bB cC + dD 1 [CA] 1 [CB] 1 [CC] Rate = a t = b t = c t 2009, Prentice-Hall, Inc.
1 [CD] = d t Reaction Rates and Stoichiometry Describe the stoichiometric relationship for the disappearance of reactants and the formation of product in the following balanced equation: 2HI (g) H2(g) + I2 (g) Concentration and Rate NH4+(aq) + NO2(aq)
N2(g) + 2 H2O(l) We can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration. If we compare Experiments 1 and 2, we see that when [CNH4+] doubles, the initial rate doubles. 2009, Prentice-Hall, Inc. Concentration and Rate NH4+(aq) + NO2(aq) N2(g) + 2 H2O(l) Likewise, when we compare Experiments 5 and 6,
we see that when [CNO2] doubles, the initial rate doubles. 2009, Prentice-Hall, Inc. Concentration and Rate This means Rate [CNH4+] Therefore, Rate [CNO2 ] Rate [CNH4+] [CNO2] which, when written as an equation, becomes
Rate = k [CNH4+] [CNO2] This equation is called the rate law, and k is the rate constant. 2009, Prentice-Hall, Inc. Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k [CNH ] [CNO2 ]
the reaction is First-order in [CNH4+] and + 4 First-order in [CNO2]. 2009, Prentice-Hall, Inc. Rate Laws Rate = k [CNH4+] [CNO2] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.
2009, Prentice-Hall, Inc. Rate Law and Order or Reaction 0 order- the rate is independent of [Creactant] 1st order- rate = [Creactant] 2nd order- rate [Creactant]2 3rd order- rate [Creactant]3 Rate Law and Oder of Reaction
Since rate is defined as change in concentration (molarity)/time (s). The unit for k will vary based on the order of the reaction. 0 order: k = mole/L s 1st order: k = 1/s 2nd order: k = L/mole s 3rd order: k = L2/mole2 s Rate Law and reactant concentration reaction: N2O5 4NO2 + O2 Rate Law Equation: Rate = k [CN2O5]1 At 64C, the rate constant = 4.82 x 10-3s-1
1. Determine the reaction rate when [CN2O5] = 0.0240M. 2. What is the rate when the concentration is doubled ? Determination of the rate law for a reaction: (role of concentration of reaction rate) reaction: 2A + 3B 2C Rate Law Equation: Rate = k [CA]m[CB]n Data: Trial [CA] [CB] 1
2 3 0.0010 0.0020 0.0010 0.0030 0.0030 0.0090 Solve for: m, n and k Initial Rate (mole/L.s)
0.0020 0.0040 0.018 Integrated Rate Law Relates concentration to reaction time and expresses the concentration of reactant as a function of time. Fits to a linear form ( y= mx +b) Based on natural long (ln) base number e = 2.7183 Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us Where
[CA]t ln [CA]0 = kt [CA]0 is the initial concentration of A, and [CA]t is the concentration of A at some time, t, during the course of the reaction. 2009, Prentice-Hall, Inc. Integrated Rate Laws Manipulating this equation produces [CA]t
ln [CA]0 = kt ln [CA]t ln [CA]0 = kt ln [CA]t = kt + ln [CA]0 which is in the form y = mx + b 2009, Prentice-Hall, Inc. Integrated Rate Laws Manipulating this equation produces [CA]t
ln [CA]0 = kt ln [CA]t ln [CA]0 = kt ln [CA]t = kt + ln [CA]0 which is in the form y = mx + b 2009, Prentice-Hall, Inc. First-Order Processes ln [CA]t = -kt + ln [CA]0
Therefore, if a reaction is first-order, a plot of ln [CA] vs. t will yield a straight line, and the slope of the line will be -k. 2009, Prentice-Hall, Inc. First-Order Processes CH3NC CH3CN This data was collected for this reaction at 198.9 C.
2009, Prentice-Hall, Inc. First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore, The process is first-order. k is the negative of the slope: 5.1 10-5 s1. 2009, Prentice-Hall, Inc. Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get
1 1 = kt + [CA]t [CA]0 also in the form y = mx + b 2009, Prentice-Hall, Inc. Second-Order Processes 1 1 = kt +
[CA]t [CA]0 So if a process is second-order in A, a plot 1 of [A] vs. t will yield a straight line, and the slope of that line is k. 2009, Prentice-Hall, Inc. Second-Order Processes The decomposition of NO2 at 300C is described by the equation 1 NO2 (g) NO (g) + 2 O2 (g)
and yields data comparable to this: Time (s) [NO2], M 0.0 0.01000 50.0 0.00787 100.0
0.00649 200.0 0.00481 300.0 0.00380 2009, Prentice-Hall, Inc. Second-Order Processes Plotting ln [NO2] vs. t yields the graph at the right.
The plot is not a straight line, so the process is not first-order in [A]. Time (s) [NO2], M ln [NO2] 0.0 0.01000 4.610
5.337 300.0 0.00380 5.573 2009, Prentice-Hall, Inc. Second-Order Processes Graphing ln [NO1 ] vs. t, however, gives this plot. 2
Time (s) [NO2], M 1/[NO2] 0.0 0.01000 100 50.0
0.00787 127 100.0 0.00649 154 200.0 0.00481 208
300.0 0.00380 263 Because this is a straight line, the process is secondorder in [A]. 2009, Prentice-Hall, Inc. Summary: Integrated rate law Order
Rate Equation Integrated Rate Equation Straight Line Plot Slope k Units 0 Rate = k[CR]0
[CR]o - [CR]t = kt [CR]t vs. t -k mol/Ls 1 Rate = k[CR]1 ln([CR]o/[CR]t) = kt ln[CR]t vs. t
-k s-1 2 Rate = k[CR]2 (1/[CR]t) - (1/[CR]o) = kt 1/[CR]t vs. t k
L/mols memorize this! Half-Life and First-Order Reactions: (radioactivity is a first-order reaction) ln becomes ln(2) = kt ln(2) = 0.693 so k = 0.693/ t and t = 0.693/k Sample problem: At 150C the decomposition of acetaldehyde CH3CHO to methane is a first order reaction. If the rate constant for the reaction at 150C is 0.029 min-1, how long does it take a concentration of 0.050 mol L-1 of
acetaldehyde to reduce to a concentration of 0.040 mol L-1? Half-Life Half-life is defined as the time required for one-half of a reactant to react. Because [CA] at t1/2 is one-half of the original [CA], [CA]t = 0.5 [CA]0. 2009, Prentice-Hall, Inc. Reaction Half-Life: 1st Order Kinetics
[A]0 [A]t at time = t 2 For a first order process the half life (t ) is found mathematically from: (1) ln A t kt ln A 0 Start with the integrated rate law expression for a 1st order process (2) ln A t ln A 0 kt
Bring the concentration terms to one side. A t (3) ln kt [ A]0 Express the concentration terms as a fraction using the rules of ln. Reaction Half-Life: 1st Order Kinetics A 0
(4) ln kt [A] A 0 (5) ln [A]0 2 exchange [A] with [A]0 to reverse the sign of the ln term and cancel the negative sign in front of k
kt Substitute the value of [A] at the 1 2 half-life ln2 0.693 t1 k k 2
Half-Life For a first-order process, this becomes ln 0.5 [CA]0 [CA]0 = kt1/2 ln 0.5 = kt1/2 0.693 = kt1/2 NOTE: For a first-order process, then, the half-life does not depend on [CA]0. 0.693
= t1/2 k 2009, Prentice-Hall, Inc. Half-Life For a second-order process, 1 0.5 [CA]0 1 = kt1/2 + [CA]0 2 [CA]0
Practice life problem: The gold-198 isotope has a half-life of 2.7 days. If you start with 10 mg at the beginning of the week, how much remains at the end of the week, seven days later? Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent. 2009, Prentice-Hall, Inc.
The Collision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other. 2009, Prentice-Hall, Inc. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. 2009, Prentice-Hall, Inc.
Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, Ea. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. 2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a
reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. 2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams The diagram shows the energy of the reactants and products (and, therefore, E). The high point on the diagram is the transition
state. The species present at the transition state is called the activated complex. The energy gap between the reactants and the activated complex is the activation energy barrier. 2009, Prentice-Hall, Inc. Activation Energy & Coordinate Diagram Label the following diagram: MaxwellBoltzmann Distributions Temperature is defined as a measure of the average kinetic
energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies. 2009, Prentice-Hall, Inc. MaxwellBoltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of
molecules has higher energy. 2009, Prentice-Hall, Inc. MaxwellBoltzmann Distributions This fraction of molecules can be found through the expression -Ea f=e RT where R is the gas constant and T is the Kelvin temperature.
2009, Prentice-Hall, Inc. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and Ea: -Ea k=Ae RT where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. 2009, Prentice-Hall, Inc.
Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes Ea 1 ln k = ( T ) + ln A R y = m x + b
Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the 1 T slope of a plot of ln k vs. . 2009, Prentice-Hall, Inc. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. 2009, Prentice-Hall, Inc.
Reaction Mechanisms Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process. 2009, Prentice-Hall, Inc. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process. 2009, Prentice-Hall, Inc.
Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step. 2009, Prentice-Hall, Inc. Reaction Mechanism guidelines: a) Molecularity describes the number of reactant molecules (or ions , atoms) that react in an elementary step. It is rare to have an elementary step that involves three reactants (termolecular) b) The sum of the elementary steps must describe the chemical reaction. c) The proposed reaction mechanism must support the experimentally determined reaction rate.
d) The rate of a reaction is determined by the slowest elementary step in the process. e) For elementary steps, the coefficients describe the rate law exponents. Slow Initial Step CO (g) + NO2 (g) NO (g) + CO2 (g) The rate law for this reaction is found experimentally to be Rate = k [CNO2]2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. 2009, Prentice-Hall, Inc.
Possible reaction mechanisms: Mechanism 1: single elementary step CO(g) + NO2(g) CO2(g) + NO(g) rate equation: Rate = k [CCO][CNO2] Mechanism 2: two step (slow) NO2 + NO2 NO3 +NO (fast) NO3 + CO NO2 + CO2 rate equation: Rate = k[CNO2]2 Mechanism 3: two step (slow) NO2 NO + O (fast) CO + O CO2 rate equation: Rate = k [CNO2] Slow Initial Step A proposed mechanism for this reaction is
Step 1: NO2 + NO2 NO3 + NO (slow) Step 2: NO3 + CO NO2 + CO2 (fast) The NO3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. 2009, Prentice-Hall, Inc. What happens when there is an initial Fast Step? 2 NO (g) + Br2 (g) 2 NOBr (g) The rate law for this reaction is found to be Rate = k [CNO]2 [CBr2] Because termolecular processes are rare,
this rate law suggests a two-step mechanism. 2009, Prentice-Hall, Inc. Fast Initial Step A proposed mechanism is Step 1: NO + Br2 NOBr2 Step 2: NOBr2 + NO 2 NOBr (fast) (slow)
Step 1 includes the forward and reverse reactions. 2009, Prentice-Hall, Inc. Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k2 [CNOBr2] [CNO] But how can we find [CNOBr2]? 2009, Prentice-Hall, Inc. Fast Initial Step NOBr2 can react two ways:
With NO to form NOBr By decomposition to reform NO and Br2 The reactants and products of the first step are in equilibrium with each other. Therefore, Ratef = Rater 2009, Prentice-Hall, Inc. Fast Initial Step Because Ratef = Rater , k1 [CNO] [CBr2] = k1 [CNOBr2] Solving for [CNOBr2] gives us k1 k1
[CNO] [CBr2] = [CNOBr2] 2009, Prentice-Hall, Inc. Fast Initial Step Substituting this expression for [CNOBr2] in the rate law for the rate-determining step gives Rate = k2k1 k1 [CNO] [CBr2] [CNO]
= k [CNO]2 [CBr2] 2009, Prentice-Hall, Inc. Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. Catalysts change the mechanism by which the process occurs. 2009, Prentice-Hall, Inc. Enzymes Enzymes are catalysts in biological systems.
The substrate fits into the active site of the enzyme much like a key fits into a lock. 2009, Prentice-Hall, Inc.
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