Chapter 14
14.1 Factors that Affect Reaction Rates 14.2 Reaction Rates 14.3 Concentration and Rate Laws 14.4 The Change of Concentration with Time 14.5 Temperature and Rate 14.6 Reaction Mechanisms 14.7 Catalysis 2014 Pearson Education, Inc. Kerosene + O 2 (l)
In chemical kinetics we study the rate (or speed) at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism, a molecular-level view of the path from reactants to products.
Factors that Affect Reaction Rate 1. Physical state of the reactants 2. Reactant concentrations 3. Reaction temperature 4. Presence of a catalyst The greater the frequency of collisions, the higher the reaction rate.
Physical State of the Reactant The more readily the reactant molecules collides with one another, the more rapidly they react. Homogeneous reactions are often faster. Heterogeneous reactions that involve solids are faster if the surface area is increased; fine powders react faster than pellets or tablets.
Reactant Concentration Increasing reactant concentration generally increases reaction rate. Since there are more molecules, more collisions occur.
Temperature Reaction rate generally increases with increased temperature. Kinetic energy of molecules is related to temperature. At higher temperatures, molecules move more quickly, increasing numbers of collisions and the energy the molecules possess during the collisions.
Presence of a Catalyst Catalysts affect rate without being in the overall balanced equation. Catalysts affect the kinds of collisions, changing the mechanism (individual reactions that are part of the pathway from reactants to products). Catalysts are critical in many biological reactions.
Rate Rate is how much a quantity changes in a given period of time. The speed you drive your car is a rate the distance your car travels (miles) in a given period of time (1 hour). So, the rate of your car has units of mile/h. H 2 I 2 HI 2014 Pearson Education, Inc.
Reaction Rate Rate is a change in concentration over a time period: [ ]/ t. M/s means change in. [ ] means molar concentration. t represents time. Types of rate measured: average rate instantaneous rate initial rate
Sample Exercise 14.1 Calculating an Average Rate of Reaction From the data in Figure 14.3, calculate the average rate at which A disappears over the time interval from 20 s to 40 s. Solution Analyze We are given the concentration of A at 20 s (0.54 M) and at 40 s (0.30 M) and asked to calculate the average rate of reaction over this time interval. Plan The average rate is given by the change in concentration, Δ[A], divided by the change in time, Δt. Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity. Solve Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Change of Rate with Time The average rate is calculated by the (change in [C 4 H 9 Cl]) (change in time). It is typical for rates to decrease as a reaction proceeds because the concentration of reactants decreases.
Plotting Rate Data The slope of the curve at one point in time gives the instantaneous rate. The instantaneous rate at time zero is called the initial rate; this is often the rate of interest to chemists.
Sample Exercise 14.2 Calculating an Instantaneous Rate of Reaction Using Figure 14.4, calculate the instantaneous rate of disappearance of C 4 H 9 Cl at t = 0 s (the initial rate). Solve The tangent line falls from [C 4 H 9 Cl] = 0.100 M to 0.060 M in the time change from 0 s to 210 s. Thus, the initial rate is Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Reaction Rate and Stoichiometry As was said, rates are followed using a reactant or a product. Rate is dependent on stoichiometry. If we followed use of C 4 H 9 Cl and compared it to production of C 4 H 9 OH, the values would be the same. Note that the change would have opposite signs one goes down in value, the other goes up.
Reaction Rate and Stoichiometry What if the equation is not 1:1? What will the relative rates be for: 2 O 3 (g) 3 O 2 (g)
Sample Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear (a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2 O 3 (g) 3O 2 (g)? (b) If the rate at which O 2 appears, Δ[O 2 ]/Δt, is 6.0 10 5 M/s at a particular instant, at what rate is O 3 disappearing at this same time, Δ[O 3 ]/Δt? Solution Analyze We are given a balanced chemical equation and asked to relate the rate of appearance of the product to the rate of disappearance of the reactant. Plan We can use the coefficients in the chemical equation as shown in Equation 14.4 to express the relative rates of reactions. Solve (a) Using the coefficients in the balanced equation and the relationship given by Equation 14.4, we have: (b) Solving the equation from part (a) for the rate at which O 3 disappears, Δ[O 3 ]/Δt, we have: Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Determining Concentration Effect on Rate We keep every concentration constant except for one reactant and see what happens to the rate. Then, we change a different reactant. We do this until we have seen how each reactant has affected the rate. x 2 x 1 x 2 x 1 x 2 x 2 Rate = k [NH 4+ ] [NO 2 ]
Rate Constant k is the rate constant. It is a temperature-dependent quantity. Rate = k [NH 4+ ] [NO 2 ]
Reaction Order The exponents tell the order of the reaction with respect to each reactant. In our example from the last slide: Rate = k [NH 4+ ] [NO 2 ] The order with respect to each reactant is 1. (It is first order in NH 4+ and NO 2 ) The overall reaction order is second order (1 + 1 = 2; add up all of the reactants orders to get the reaction s order). The rate law must be determined experimentally.
Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate Consider a reaction A + B C for which rate = k[a][b] 2. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. Solution Analyze We are given three boxes containing different numbers of spheres representing mixtures containing different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to rank the mixtures in order of increasing reaction rates. Plan Because all three boxes have the same volume, we can put the number of spheres of each kind into the rate law and calculate the rate for each box. Solve Box 1 contains 5 red spheres and 5 purple spheres, giving the following rate: Box 1: Rate = k(5)(5) 2 = 125k Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate Continued Box 2 contains 7 red spheres and 3 purple spheres: Box 2: Rate = k(7)(3) 2 = 63k Box 3 contains 3 red spheres and 7 purple spheres: Box 3: Rate = k(3)(7) 2 = 147k The slowest rate is 63k (Box 2), and the highest is 147k (Box 3). Thus, the rates vary in the order 2 < 1 < 3. Check Each box contains 10 spheres. The rate law indicates that in this case [B] has a greater influence on rate than [A] because B has a larger reaction order. Hence, the mixture with the highest concentration of B (most purple spheres) should react fastest. This analysis confirms the order 2 < 1 < 3. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data The initial rate of a reaction A + B C was measured for several different starting concentrations of A and B, and the results are as follows: Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050M and [B] = 0.100 M. Solution Analyze We are given a table of data that relates concentrations of reactants with initial rates of reaction and asked to determine (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentrations not listed in the table. Plan (a) We assume that the rate law has the following form: Rate = k[a] m [B] n. We will use the given data to deduce the reaction orders m and n by determining how changes in the concentration change the rate. (b) Once we know m and n, we can use the rate law and one of the sets of data to determine the rate constant k. (c) Upon determining both the rate constant and the reaction orders, we can use the rate law with the given concentrations to calculate rate. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.6 Determining a Rate Law from Initial Rate Data Continued Solve (a) If we compare experiments 1 and 2, we see that [A] is held constant and [B] is doubled. Thus, this pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0). In experiments 1 and 3, [B] is held constant, so these data show how [A] affects rate. Holding [B] constant while doubling [A] increases the rate fourfold. This result indicates that rate is proportional to [A] 2 (that is, the reaction is second order in A). Hence, the rate law is (b) Using the rate law and the data from experiment 1, we have (c) Using the rate law from part (a) and the rate constant from part (b), we have Because [B] is not part of the rate law, it is irrelevant to the rate if there is at least some B present to react with A. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
2014 Pearson Education, Inc. Reactant Concentration & Time
2014 Pearson Education, Inc. Rate Law & Integrated Rate Law
Zero Order Reaction 1. Rate = k[a] 0 = k 2. [A] t = kt + [A] initial 3. Graph of [A] versus time straight line with slope = k and y-intercept = [A] initial 4. t ½ = [A initial ]/2k 5. When Rate = M/sec, k = M/sec [A] initial [A] t Time
First Order Reaction 1. Rate = k[a] 1 2. ln[a] t = kt + ln[a] initial 3. Graph ln[a] versus time straight line with slope = k and y-intercept = ln[a] initial 4. t ½ = 0.693/k 5. The half-life of a first order reaction is constant. 6. When Rate = M/sec, k = s 1
Finding the Rate Constant, k Besides using the rate law, we can find the rate constant from the plot of ln[a] vs. t. The plot will give a line. Its slope will equal k. ln[a] initial ln[a] t Time
1 st Order: Methyl Isonitrile to Acetonitrile The equation for the reaction: CH 3 NC CH 3 CN It is first order. Rate = k [CH 3 NC]
Half-life Definition: The amount of time it takes for one-half of a reactant to be used up in a chemical reaction. ln[a] = kt + ln[a] 0
Sample Exercise 14.7 Using the Integrated First-Order Rate Law The decomposition of a certain insecticide in water at 12 follows first-order kinetics with a rate constant of 1.45 yr 1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 10 7 g/cm 3. Assume that the temperature of the lake is constant (so that there are no effects of temperature variation on the rate). (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the insecticide concentration to decrease to 3.0 10 7 g/cm 3? Solution Analyze We are given the rate constant for a reaction that obeys firstorder kinetics, as well as information about concentrations and times, and asked to calculate how much reactant (insecticide) remains after 1 yr. We must also determine the time interval needed to reach a particular insecticide concentration. Because the exercise gives time in (a) and asks for time in (b), we will find it most useful to use the integrated rate law, Equation 14.13. Plan (a) We are given k = 1.45 yr 1, t = 1.00 yr, and [insecticide] 0 = 5.0 10 7 g/cm 3, and so Equation 14.13 can be solved for [insecticide] t. (b) We have k = 1.45 yr 1, [insecticide] 0 = 5.0 10 7 g/cm 3, and [insecticide] t = 3.0 10 7 g/cm 3, and so we can solve Equation 14.13 for time, t. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.7 Using the Integrated First-Order Rate Law Continued Solve (a) Substituting the known quantities into Equation 14.13, we have We use the ln function on a calculator to evaluate the second term on the right [that is, ln(5.0 10 7 )], giving To obtain [insecticide] t = 1 yr, we use the inverse natural logarithm, or e x, function on the calculator: Note that the concentration units for [A] t and [A] 0 must be the same. (b) Again substituting into Equation 14.13, with [insecticide] t = 3.0 10 7 g/cm 3, gives Solving for t gives ln[a] = kt + ln[a] 0 ln[insecticide] t = 1 yr = (1.45 yr 1 )(1.00 yr) + ln(5.0 10 7 ) ln[insecticide] t = 1 yr = 1.45 + ( 14.51) = 15.96 [insecticide] t = 1 yr = e 15.96 = 1.2 10 7 g/cm 3 ln(3.0 10 7 ) = (1.45 yr 1 )(t) + ln(5.0 10 7 ) t = [ln(3.0 10 7 ) ln(5.0 10 7 )]/1.45 yr 1 = ( 15.02 + 14.51)/1.45 yr 1 = 0.35 yr Check In part (a) the concentration remaining after 1.00 yr (that is, 1.2 10 7 g/cm 3 ) is less than the original concentration (5.0 10 7 g/cm 3 ), as it should be. In (b) the given concentration (3.0 10 7 g/cm) 2 is greater than that remaining after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is a reasonable answer. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction The reaction of C 4 H 9 Cl with water is a first-order reaction. (a) Use Figure 14.4 to estimate the half-life for this reaction. (b) Use the halflife from (a) to calculate the rate constant. Solution Analyze We are asked to estimate the half-life of a reaction from a graph of concentration versus time and then to use the half-life to calculate the rate constant for the reaction. Plan (a) To estimate a half-life, we can select a concentration and then determine the time required for the concentration to decrease to half of that value. (b) Equation 14.15 is used to calculate the rate constant from the half-life. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.9 Determining the Half-Life of a First-Order Reaction Continued Solve (a) From the graph, we see that the initial value of [C 4 H 9 Cl] is 0.100 M. The half-life for this first-order reaction is the time required for [C 4 H 9 Cl] to decrease to 0.050 M, which we can read off the graph. This point occurs at approximately 340 s. (b) Solving Equation 14.15 for k, we have Check At the end of the second half-life, which should occur at 680 s, the concentration should have decreased by yet another factor of 2, to 0.025 M. Inspection of the graph shows that this is indeed the case. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
1. Rate = k[a] 2 Second Order Reaction 2. 1/[A] t = kt + 1/[A] initial 3. Graph 1/[A] versus time straight line with slope = k and y-intercept = 1/[A] initial 4. t ½ = 1/(k[A 0 ]) 5. When Rate = M/sec, k = M 1 s 1 1/[A] t 1/[A] initial Time
2 nd Order: Decomposition of NO 2 Equation: NO 2 NO + ½ O 2 A plot following NO 2 decomposition shows that it must be second order because it is linear for 1/[NO 2 ], not linear for ln[no 2 ].
Half-Life and Second Order Reactions Using the integrated rate law, we can see how half-life is derived: 1/[A] = 1/[A] o + k t 1/([A] o /2) = 1/[A] o + k t ½ 2/[A] o 1/[A] o = k t ½ t ½ = 1 / (k [A] o ) So, half-life is a concentration dependent quantity for second order reactions!
Sample Exercise 14.8 Determining Reaction Order from the Integrated Rate Law The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300, Is the reaction first or second order in NO 2? Solution Analyze We are given the concentrations of a reactant at various times during a reaction and asked to determine whether the reaction is first or second order. Plan We can plot ln[no 2 ] and 1/[NO 2 ] against time. If one plot or the other is linear, we will know the reaction is either first or second order. Solve To graph ln[no 2 ] and 1/[NO 2 ] against time, we first make the following calculations from the data given: Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Factors That Affect Reaction Rate 1. Temperature 2. Frequency of collisions 3. Orientation of molecules 4. Energy needed for the reaction to take place (activation energy)
Temperature and Rate Generally, as temperature increases, rate increases. The rate constant is temperature dependent: it increases as temperature increases. Rate constant doubles (approximately) with every 10 ºC rise.
Frequency of Collisions The collision model is based on the kinetic molecular theory. Molecules must collide to react. If there are more collisions, more reactions can occur. On average, about 10 9 collisions per second. Also, if the temperature is higher, molecules move faster, causing more collisions and a higher rate of reaction.
Orientation of Molecules Molecules can often collide without forming products. Aligning molecules properly can lead to chemical reactions. Bonds must be broken and made, and atoms need to be in proper positions.
Activation Energy (E a ) The minimum energy needed for a reaction to take place is called activation energy. An energy barrier must be overcome for a reaction to take place, much like the ball must be hit to overcome the barrier in the figure below. The organization of the atoms at this highest energy state is called the transition state (or activated complex).
Plots are made to show the energy possessed by the particles as the reaction proceeds. Reaction Progress At the highest energy state, the transition state is formed. The rate depends on the magnitude of E a ; generally, the lower the value of E a is, the faster the reaction.
Sample Exercise 14.10 Activation Energies and Speeds of Reaction Consider a series of reactions having these energy profiles: Rank the reactions from slowest to fastest assuming that they have nearly the same value for the frequency factor A. Solution The lower the activation energy, the faster the reaction. The value of ΔE does not affect the rate. Hence, the order from slowest reaction to fastest is 2 < 3 < 1. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Distribution of the Energy of Molecules Gases have an average temperature, but each individual molecule has its own energy. At higher energies, more molecules possess the energy needed for the reaction to occur.
Activation Energy & Temperature Arrhenius noted relationship between activation energy and temperature, Arrhenius equation: k = Ae Ea/RT Activation energy can be determined graphically by reorganizing the equation: ln k = E a /RT + ln A Frequency factor Activation energy Exponential factor
Activation Energy & Temperature Arrhenius noted relationship between activation energy and temperature, Arrhenius equation: k = Ae Ea/RT Activation energy can be determined graphically by reorganizing the equation: ln k = E a /RT + ln A
Sample Exercise 14.11 Determining the Activation Energy The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures (these are the data points in Figure 14.14): (a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at 430.0 K? Solution Analyze We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, E a, and the rate constant, k, at a particular temperature. Plan We can obtain E a from the slope of a graph of ln k versus 1/T. Once we know E a, we can use Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K. Solve (a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k. This gives us the table shown at the right: Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.11 Determining the Activation Energy Continued A graph of ln k versus 1/T is a straight line ( Figure 14.19). The slope of the line is obtained by choosing any two well-separated points and using the coordinates of each: Because logarithms have no units, the numerator in this equation is dimensionless. The denominator has the units of 1/T, namely, K 1. Thus, the overall units for the slope are K. The slope equals E a /R. We use the value for the gas constant R in units of J/mol-K (Table 10.2). We thus obtain We report the activation energy to only two significant figures because we are limited by the precision with which we can read the graph in Figure 14.19. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.11 Determining the Activation Energy Continued (b) To determine the rate constant, k 1, at T 1 = 430.0 K, we can use Equation 14.21 with E a = 160 kj/mol and one of the rate constants and temperatures from the given data, such as k 2 = 2.52 10 5 s 1 and T 2 = 462.9 K: Thus, Note that the units of k 1 are the same as those of k 2. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, 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.
Molecularity The molecularity of an elementary reaction tells how many molecules are involved in that step of the mechanism.
Termolecular? Termolecular steps require three molecules to simultaneously collide with the proper orientation and the proper energy. These are rare, if they indeed do occur. These must be slower than unimolecular or bimolecular steps. Nearly all mechanisms use only unimolecular or bimolecular reactions.
Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates It has been proposed that the conversion of ozone into O 2 proceeds by a two-step mechanism: O 3 (g) O 2 (g) + O(g) O 3 (g) + O(g) 2 O 2 (g) (a) Describe the molecularity of each elementary reaction in this mechanism. (b) Write the equation for the overall reaction. (c) Identify the intermediate(s). Solution Analyze We are given a two-step mechanism and asked for (a) the molecularities of each of the two elementary reactions, (b) the equation for the overall process, and (c) the intermediate. Plan The molecularity of each elementary reaction depends on the number of reactant molecules in the equation for that reaction. The overall equation is the sum of the equations for the elementary reactions. The intermediate is a substance formed in one step of the mechanism and used in another and therefore not part of the equation for the overall reaction. Solve (a) The first elementary reaction involves a single reactant and is consequently unimolecular. The second reaction, which involves two reactant molecules, is bimolecular. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates Continued (b) Adding the two elementary reactions gives 2 O 3 (g) + O(g) 3 O 2 (g) + O(g) Because O(g) appears in equal amounts on both sides of the equation, it can be eliminated to give the net equation for the chemical process: 2 O 3 (g) 3 O 2 (g) (c) The intermediate is O(g). It is neither an original reactant nor a final product but is formed in the first step of the mechanism and consumed in the second. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
What Limits the Rate? The overall reaction cannot occur faster than the slowest reaction in the mechanism. We call that the rate-determining step.
A Mechanism with a Slow Initial Step Overall equation: NO 2 + CO NO + CO 2 Rate law: Rate = k [NO 2 ] 2 If the first step is the rate-determining step, the coefficients on the reactants side are the same as the order in the rate law! So, the first step of the mechanism begins: NO 2 + NO 2
A Mechanism with a Slow Initial Step The easiest way to complete the first step is to make a product: NO 2 + NO 2 NO + NO 3 We do not see NO 3 in the stoichiometry, so it is an intermediate, which needs to be used in a faster next step. NO 3 + CO NO 2 + CO 2
A Mechanism with a Slow Initial Step Since the first step is the slowest step, it gives the rate law. If you add up all of the individual steps (2 of them), you get the stoichiometry. This is a plausible mechanism.
Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N 2 O, is believed to occur by a two-step mechanism: Solution N 2 O(g) N 2 (g) + O(g) (slow) N 2 O(g) + O(g) N 2 (g) + O 2 (g) (fast) (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. Analyze Given a multistep mechanism with the relative speeds of the steps, we are asked to write the overall reaction and the rate law for that overall reaction. Plan (a) Find the overall reaction by adding the elementary steps and eliminating the intermediates. (b) The rate law for the overall reaction will be that of the slow, rate-determining step. Solve (a) Adding the two elementary reactions gives 2 N 2 O(g) + O(g) 2 N 2 (g) + 2 O 2 (g) + O(g) Omitting the intermediate, O(g), which occurs on both sides of the equation, gives the overall reaction: 2 N 2 O(g) 2 N 2 (g) + O 2 (g) (b) The rate law for the overall reaction is just the rate law for the slow, rate-determining elementary reaction. Because that slow step is a unimolecular elementary reaction, the rate law is first order: Rate = k[n 2 O] Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
A Mechanism with a Fast Initial Step Equation for the reaction: 2NO + Br 2 2NOBr The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] The rate law indicates that a quickly established equilibrium is followed by a slow step. Step 1: NO + Br 2 NOBr 2 (fast) Step 2: NOBr 2 + NO 2 NOBr (slow)
What is the Rate Law? The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?
A Mechanism with a Fast Initial Step When a mechanism contains a fast initial step, the rate limiting step may contain intermediates. When a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal, so the concentrations of reactants and products of the step are related and the product is an intermediate. Substituting into the rate law of the RDS will produce a rate law in terms of just reactants.
[NOBr 2 ] as an Intermediate
Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the experimentally observed one: Solution Analyze We are given a mechanism with a fast initial step and asked to write the rate law for the overall reaction. Plan The rate law of the slow elementary step in a mechanism determines the rate law for the overall reaction. Thus, we first write the rate law based on the molecularity of the slow step. In this case, the slow step involves the intermediate N 2 O 2 as a reactant. Experimental rate laws, however, do not contain the concentrations of intermediates; instead they are expressed in terms of the concentrations of starting substances. Thus, we must relate the concentration of N 2 O 2 to the concentration of NO by assuming that an equilibrium is established in the first step. Solve The second step is rate determining, so the overall rate is Rate = k 2 [N 2 O 2 ][Br 2 ] We solve for the concentration of the intermediate N 2 O 2 by assuming that an equilibrium is established in step 1; thus, the rates of the forward and reverse reactions in step 1 are equal: K 1 [NO] 2 = k 1 [N 2 O 2 ] Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step Continued Solving for the concentration of the intermediate, N 2 O 2, gives Substituting this expression into the rate expression gives Thus, this mechanism also yields a rate law consistent with the experimental one. Remember: There may be more than one mechanism that leads to an observed experimental rate law! Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.
Catalysts Catalysts work by providing an alternative mechanism for the reaction with a lower activation energy. Catalysts change the mechanism by which the process occurs. Homogeneous catalysts Heterogeneous catalysts Enzymes
Catalytic Reaction Lowers the overall E a!!!
The catalyst is in a different phase than the reactants. Heterogeneous Catalyst Often, gases are passed over a solid catalyst. The adsorption of the reactants is often the rate-determining step.
Enzymes Enzymes are biological catalysts. They have a region where the reactants attach. That region is called the active site. The reactants are referred to as substrates.
Lock-and-Key Model In the enzyme substrate model, the substrate fits into the active site of an enzyme, much like a key fits into a lock. They are specific.
Basic Integration Rule where k is a constant for x > 0 This definition means that e is the unique number with the property that the area of the region bounded by the hyperbola y = 1/x, the x- axis, and the vertical line x = 1 and x = e is 1. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.
When e y = x Then base e logarithm of x is ln(x) = log e (x) = log e (e y ) = y When 10 y = x Then base 10 logarithm of x is log 10 (x) = log 10 (10 y ) = y The e constant or Euler's number is: e 2.71828183 ln log log log The natural logarithm function ln(x) is the Inverse function of the exponential function e x. For x>0, f -1 (f (x)) = ln(e x ) = x Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.
Integrated Rate Laws For the reaction A products, the rate law depends on the concentration of A. Applying calculus to integrate the rate law gives another equation showing the relationship between the concentration of A and the time of the reaction; this is called the integrated rate law. 2014 Pearson Education, Inc.