Chemical Kinetics Chapter 13. Chemical Kinetics  Chemical reactions occur when the reactant species strike each other and interact to form products 

Chemical Kinetics Chapter 13

Chemical Kinetics  Chemical reactions occur when the reactant species strike each other and interact to form products  Chemical kinetics = the study of rates of chemical reactions and their reaction mechanisms

13.1 The Rate of Reaction Objectives  To define reaction rate and differentiate between instantaneous and average rates  To calculate the rate of reaction from experimental data  To express the rate of reaction in terms of changes in concentration of reactants and products

13.1 The Rate of Reaction  The rate of a chemical reaction is the change in concentration per unit time rate = Δc/Δt  The units used to expression reaction rates are typically mol/Ls (or atm/s for gases).  Reaction rates can be determined by measuring the decrease in concentration of reactants or the increase in concentration of products over time.  By convention, the rate of reaction is always expressed as a positive number

The change in concentration of a reactant or product per unit of time For the reaction: A → B

13.1 The Rate of Reaction  Instantaneous & Average Rates  The average rate of reaction is equal to the change in concentration divided by the time interval.  As the interval between measurements becomes smaller, the average rate approaches the instantaneous rate.  The instantaneous rate = the slope of the tangent to the curve at a particular time point.

Instantaneous Rate

13.1 The Rate of Reaction  Instantaneous & Average Rates  A car may travel at an average rate of 50 mph but at any given time could have had an instantaneous speed of 75 mph or 35 mph.  Note that rates of reaction typically decrease as a reaction progresses.

Instantaneous Rate

13.1 The Rate of Reaction  The rate of reaction does not depend on which species is measured.  The rate of reaction is the absolute value of the rate of change of a substance divided by its coefficient in the chemical equation

2NO 2 (g)  2NO(g) + O 2 (g) Reaction Rates: 2. Can measure appearance of products 1. Can measure disappearance of reactants 3. Are proportional stoichiometrically

2NO 2 (g)  2NO(g) + O 2 (g) Reaction Rates: 4. Are equal to the slope tangent to that point  [NO 2 ] tt 5. Change as the reaction proceeds, if the rate is dependent upon concentration

13.1 The Rate of Reaction  Consider the formation of ammonia from the elements: N 2 (g) + 3H 2 (g) → 2NH 3 (g)  The rate of disappearance of N 2 was 0.14 mol/Ls of N 2. What is the rate of disappearance of hydrogen? What is the rate of reaction? A: Rate of disappearance of H 2 = 0.42 mol/Ls A: The rate of reaction is 0.14 mol/Ls

13.2 Relationships Between Rate and Concentration Objectives  To define rate law, rate constant, and reaction order  To use initial concentrations and initial rates of reactions to determine the rate law and rate constant

13.2 Relationships Between Rate and Concentration  The rate of a reaction is strongly influenced by the concentrations of the reacting species.  Experimental Rate Laws For the equation: aA + bB → products The rate of the reaction is proportional to the product of the concentrations of the reactants each raised to some power Rate = k[A] x [B] y is the rate law where k = rate constant and the exponents x and y represent the orders of the reaction

13.2 Relationships Between Rate and Concentration  Experimental Rate Laws Orders of the reaction are usually a small positive integer but can sometimes be zero, negative, or fractional The overall order of a reaction is determined by adding x + y Note that the orders x and y are not necessarily the coefficients of the balanced chemical equation. Orders of reactions are determined experimentally!

13.2 Relationships Between Rate and Concentration  The initial rate method is often used to determine the order of a reaction.  Method Repeat an experiment several times with different known ratios of reactants and evaluate how the reaction rate changes with concentration Important to measure the initial rate of the reaction

Writing a Rate Law 2 NO(g) + Cl 2 (g)  2 NOCl(g) Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 11.4 x 10 -6

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 1.14 x 10 -5 In experiment 1 and 2, [Cl 2 ] is constant while [NO] doubles. R = k[NO] x [Cl 2 ] y The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO] 2 [Cl 2 ] y

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 1.14 x 10 -5 R = k[NO] 2 [Cl 2 ] y In experiment 2 and 4, [NO] is constant while [Cl 2 ] doubles. The rate doubles, so the reaction is first order with respect to [Cl 2 ]  R = k[NO] 2 [Cl 2 ]

Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 R = k[NO] 2 [Cl 2 ]

Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO] 2 [Cl 2 ] Overall order is the sum of the exponents, or orders, of the reactants 2+1 = 3  The reaction is 3 rd order

13.2 Relationships Between Rate and Concentration  It is very important to note that the rate law cannot be predicted from the reaction stoichiometry!!!  Calculate the rate law and rate constant for (CH 3 ) 3 CBr + OH - → (CH 3 ) 3 COH + Br - based on the following data: Experiment [(CH 3 ) 3 CBr](mol/L) [OH - ](mol/L)RateMol/L·s 10.0500.010 4.1 x 10 -8 20.0800.020 6.6 x 10 -8 30.0800.010

13.2 Relationships Between Rate and Concentration  A: Should note that a change in [OH] - did not affect reaction rate so rate = k[(CH 3 ) 3 CBr] k = 8.2 x 10 -7 s -1

13.3 Dependence of Concentrations on Time Objectives  To compare the concentration-time behaviors of reactions with different rate laws  To determine the rate law and rate constant from experimental concentration vs. time data  To describe the relationships between reaction rate, rate constant, and half-life  To calculate the concentration-time behavior from the rate law and the rate constant or half-life

13.3 Dependence of Concentrations on Time  In addition to using initial rate studies, rate laws can also be determined by examining how the concentration of a reactant changes with time during the course of a single experiment  Reactions of different orders behave quite differently  Units of the rate constant depend on the order of the reaction

13.3 Dependence of Concentrations on Time  Zero-Order Reactions Time Concentration

13.3 Dependence of Concentrations on Time  Zero-Order Rate Laws Some reactions show rates that are independent of the concentration of reactants Reaction rate = k The units of a zero-order rate constant are mol/Ls If the graph of reactant concentration vs. time is a straight line, the reaction obeys zero-order kinetics Ex: metabolism of ethyl alcohol in humans and numerous other biochemical reactions involving enzymes

13.3 Dependence of Concentrations on Time  First-Order Reactions Time Concentration Time ln[concentration]

13.3 Dependence of Concentrations on Time  First-Order Rate Laws Some reactions show rates that are proportional to the concentration of the reactant Reaction rate = -Δ[A]/ Δ t = k[A] The units of a first-order rate constant are s -1 If the graph of the natural log (ln) of reactant concentration vs. time is a straight line, the reaction obeys first-order kinetics

13.3 Dependence of Concentrations on Time  First-Order Rate Laws Reaction rate = -Δ[A]/ Δ t = k[A] is referred to as the differential form of the rate law Can also write the rate law in an integrated form [A] =[A] 0 e -kt where A = concentration at time t A 0 = concentration of A at t = o e = base of the natural logarithm k = rate constant and t = time (usually in sec)

13.3 Dependence of Concentrations on Time  Note that the equation [A] =[A] 0 e -kt describes exponential decay  Other forms of the same equation that may be easier to use include: ln[A] = ln[A] 0 - kt or ln[A/A 0 ] = -kt

13.3 Dependence of Concentrations on Time  First-Order Reactions, cont’d  Note that the slope of the line = -k  Note that the y-intercept = [A] 0 Time ln[concentration]

13.3 Dependence of Concentrations on Time  Practice Problems: Time (s) Concentration (M) ln[concentration] 0 0.22-1.51 0.001 0.15-1.90 0.002 0.10-2.30 0.003 0.07-2.66 0.004 0.045-3.10 0.005 0.030-3.51 What is the rate equation for A → product? What is the value of k?

13.3 Dependence of Concentrations on Time  Practice Problems: The pesticide fenvalerate degrades in the environment with first-order kinetics and a rate constant of 3.9 x 10 -7 s -1. An accidental discharge of 100 kg of fenvalerate into a holding pond results in a fenvalerate concentration of 1.3 x 10 -5 M. Calculate the concentration left after 1 month (2.6 x 10 6 sec) Equation ln[A] = ln[A] 0 – kt Answer: A = 4.7 x 10 -6 M

13.3 Dependence of Concentrations on Time  Practice Problems: N 2 O 5 decomposes to nitrogen dioxide and oxygen. When the reaction takes place in carbon tetrachloride, both nitrogen oxides are soluble, but the oxygen escapes as a gas. The rate of reaction can be measured by monitoring the volume of gas that is produced by the reaction. Quantitative chemical analysis of the gas generated in such an experiment shows that the rate law is first order in N 2 O 5 with a rate constant of 8.1 x 10 -5 s -1. If the initial N 2 O 5 concentration is 0.032 M, how long does it take for its concentration to decrease to 0.015 M? Equation ln[A] = ln[A] 0 – kt Answer: A = 9.4 x 10 3 s

13.3 Dependence of Concentrations on Time  Half-Life A large value for k implies a fast reaction Half-life, t ½, also describes the speed of a reaction  = the time needed for the concentration of a reactant to decrease to ½ its original value  A short half-life indicates a rapid reaction For a first-order reaction the relationship between half-life and k is given by: t ½ = 0.693/k For a first-order reaction note that half-life is independent of the concentration of the reactant

Visualization of Half-Life

13.3 Dependence of Concentrations on Time  Radioactive decays are often first-order reactions  Carbon-14 is a radioactive isotope with a half-life of 5730 years. If a particular sample has decayed so that it has only 25% of its original amount of 14 C, how old is it?  Equations needed: t ½ = 0.693/k & ln[A] = ln[A] 0 – kt  Answer: t = 1.1 x 10 4 years

13.3 Dependence of Concentrations on Time  Second-Order Reactions Time Concentration or ln[concentration] Time 1/[concentration] Some form of curved line

13.3 Dependence of Concentrations on Time  Second-Order Rate Laws Some reactions show rates that are proportional to the concentration of the reactant raised to the second power Reaction rate = -Δ[A]/ Δ t = k[A] 2 The units of a second-order rate constant are L/mols If the graph of 1/reactant concentration vs. time is a straight line, the reaction obeys second-order kinetics

13.3 Dependence of Concentrations on Time  Second-Order Rate Laws Can also write the rate law in an integrated form 1/[A] = 1/[A] 0 + kt where A = concentration at time t A 0 = concentration of A at t = o k = rate constant (slope of line) t = time t ½ = 1/k[A]

13.3 Dependence of Concentrations on Time  To determine a reaction’s rate law…. First, plot concentration vs time. If a straight line results, the system is zero-order. If the line curves, construct a plot of natural logarithm of concentration as a function of time If this line is straight the reaction is first order. If the line curves, construct a plot of 1/concentration as a function of time If this line is straight the reaction is second order.

Rate Laws Summary or p. 544 2 nd ed Zero Order First Order Second Order Rate Law Rate = kRate = k[A]Rate = k[A] 2 Integrated Rate Law [A] = -kt + [A] 0 ln[A] = -kt + ln[A] 0 Plot that produces a straight line [A] versus tln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life

Solving an Integrated Rate Law Time (s) [H 2 O 2 ] (mol/L) 01.00 1200.91 3000.78 6000.59 12000.37 18000.22 24000.13 30000.082 36000.050 Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!!

Time vs. [H 2 O 2 ] Time (s) [H 2 O 2 ] 01.00 1200.91 3000.78 6000.59 12000.37 18000.22 24000.13 30000.082 36000.050 y = ax + b a = -2.64 x 10 -4 b = 0.841 r 2 = 0.8891 r = -0.9429 Regression results:

Time vs. ln[H 2 O 2 ] Time (s) ln[H 2 O 2 ] 00 120-0.0943 300-0.2485 600-0.5276 1200-0.9943 1800-1.514 2400-2.04 3000-2.501 3600-2.996 Regression results: y = ax + b a = -8.35 x 10 -4 b = -.005 r 2 = 0.99978 r = -0.9999

Time vs. 1/[H 2 O 2 ] Time (s) 1/[H 2 O 2 ] 01.00 1201.0989 3001.2821 6001.6949 12002.7027 18004.5455 24007.6923 300012.195 360020.000 y = ax + b a = 0.00460 b = -0.847 r 2 = 0.8723 r = 0.9340 Regression results:

And the winner is… Time vs. ln[H 2 O 2 ] 1. As a result, the reaction is 1 st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

Finding the Rate Constant, k Method #1: Calculate the slope from the Time vs. ln[H 2 O 2 ] table. Time (s) ln[H 2 O 2 ] 00 120-0.0943 300-0.2485 600-0.5276 1200-0.9943 1800-1.514 2400-2.04 3000-2.501 3600-2.996 Now remember:  k = -slope k = 8.32 x 10 -4 s -1

Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis. Now remember:  k = -slope k = 8.35 x 10 -4 s -1 Regression results: y = ax + b a = -8.35 x 10 -4 b = -.005 r 2 = 0.99978 r = -0.9999

13.4 Models of Reaction Rates Objectives  To determine the influence of temperature on the rate of reaction  To define collision theory and use it to explain observations of chemical kinetics  To determine how activation energy and collision frequency influence the rate of reaction  To draw the energy-level diagram for a reaction  To evaluate experimental data to measure the activation energy

Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

13.4 Models of Reaction Rates  Influence of Temperature on Rate Constant Collisions between molecules are necessary for reactions to occur The rates of most reactions increase dramatically with temperature Is the rate increase due to increased kinetic energy and movement of particles, thereby increasing the chance of a successful collision?

13.4 Models of Reaction Rates  Collision Theory Basic assumption = molecules must collide in order to react Collision frequency = the number of molecular collisions per second  Z = Z 0 [A][B]  where Z 0 = proportionality constant that depends on the speeds and sizes of the molecules A & B However, experimental evidence shows that not every collision results in a chemical reaction

13.4 Models of Reaction Rates  Arrhenius began to look at other factors besides frequency of collisions to account for reaction rates  Activation Energy The minimum collision energy required for a reaction to occur Only collisions with enough energy to rearrange bonds can result in the formation of products If the total energy of colliding species is too small, the molecules simply bounce off each other

13.4 Models of Reaction Rates  The Activated Complex (denoted [X]*) Intermediate species in chemical reactions are higher in energy than either the reactants or products of a chemical reaction The high-energy activated complex is very unstable and is also referred to as the transition state The activation energy is the energy needed to form the activated complex from the reactants

Exothermic Reactions

Endothermic Reactions

13.4 Models of Reaction Rates  The Influence of Temperature on Kinetic Energy The number of collisions with energies that exceed E a grows exponentially with temperature The fraction of collisions with energy in excess of E a can be derived from f r = e -Ea/RT As T increases, f r increases and approaches 100%

13.4 Models of Reaction Rates  The Influence of Temperature on Kinetic Energy Can now predict that the rate of a reaction should equal the collision frequency times the fraction of collisions that have energies in excess of the activation energy Predicted Rate = Z x f r or Z 0 [A][B]e -Ea/RT If we set the predicted rate law from above with the experimental rate law, rate = k[A][B] then we can solve for k and k = Z 0 e -Ea/RT

13.4 Models of Reaction Rates  The Influence of Temperature on Kinetic Energy It turns out that the equation k = Z 0 e -Ea/RT correctly describes an increase in the rate constant with increasing temperature However, the rate constants predicted by the equation are much faster than those observed in the laboratory Therefore, another factor is involved in reaction rate….

13.4 Models of Reaction Rates  The Steric Factor Not all collisions with energies that exceed E a are productive The geometry of collisions must also be considered - not every collision occurs with the reactants in the correct orientation to produce products

13.4 Models of Reaction Rates  The Steric Factor Rate = p x Z 0 [A][B] x e -Ea/RT the steric x collision x fraction factor frequency exceeding Ea These equations can be rearranged to generate the Arrhenius Equation k = Ae -Ea/RT where A (referred to as the pre-exponential term) is determined experimentally

The Arrhenius Equation k = rate constant at temperature T  k = rate constant at temperature T  A = pre-exponential term  E a = activation energy  R = Gas constant, 8.31451 J/K·mol

13.4 Models of Reaction Rates  The Arrhenius Equation Often helpful to rewrite the Arrhenius equation as: ln k = ln A – (E a /R ) (1/T) A plot of ln k vs 1/T is called the Arrhenius plot and yields a straight line with a slope = -E a /R To calculate the E a from rate measurements at two temperatures use the following equation: ln (k 1 /k 2 ) = -(E a /R) (1/T 1 – 1/T 2 )

Fig. 13-11, p. 540 The Arrhenius Plot Slope = -Ea/R

13.4 Models of Reaction Rates  Consider the decomposition of nitrogen dioxide. At 650K, the rate constant is 1.66 units; at 700K it is 7.39 units. Use these rate constants to determine the activation energy  A: E a = 113 kJ/mol  Calculate the activation energy in kJ/mol for a reaction that has k = 1.0 x 10 8 s -1 at 250K and k = 3.8 x 10 8 s -1 at 350K.  A: E a = 9.7 kJ/mol

Collision Model Overview Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). Colliding particles must be correctly oriented to one another in order to produce a reaction. 1. 2.

13.5 Catalysts Objectives  To define catalysts and identify heterogeneous, homogeneous, and enzymatic catalysts  To determine how catalysts influence chemical reactions  To draw energy-level diagrams for catalyzed and uncatalyzed reactions

13.5 Catalysts As we just learned, rates of reactions can be increased by increasing the frequency of productive collisions: Remember Rate = p x Z 0 [A][B] x e -Ea/RT the steric x collision x fraction factor frequency exceeding Ea The collision frequency and average kinetic energies of collisions can be increased by increasing temperature In addition, we can make a larger fraction of the collisions productive by adding a catalyst

Catalysts Increase the Number of Effective Collisions

13.5 Catalysts  Catalysts are substances that speed up the rate of reaction without getting consumed in the reaction  Catalyzed reactions often proceed by a different set of steps than uncatalyzed reactions. Catalysts are often described as providing a shortcut, a path with a lower energy barrier to the same product  Catalysts generally lower activation energy and increase reaction rate at any given temperature Some catalysts may increase the steric factor

Lowering of Activation Energy by a Catalyst

13.5 Catalysts  Types of catalysts include: Homogeneous Catalysts  Present in the same phase as the reacting molecules Heterogeneous Catalysts  Present in a different phase than the reacting molecules Enzymes  Large molecules (usually proteins) that catalyze specific biological reactions

13.5 Catalysts  Homogeneous Catalysts (same phase) Example: 2H 2 O 2 (aq) → 2H 2 O (l) + O 2 (g) The bromide catalyzed decomposition of hydrogen peroxide is believed to occur in two steps. Step 1 H 2 O 2 (aq) + 2Br - (aq) + 2H + (aq) → 2H 2 O (l) + Br 2 (aq) Step 2 2H 2 O 2 (aq) + Br 2 (aq) → 2H 2 O (l) + O 2 (g)

13.5 Catalysts  Heterogeneous Catalysts (different phases) Numerous gaseous reactions are catalyzed via adsorption to the surface of a solid metal catalyst, typically platinum or palladium Examples include  the hydrogenation of ethylene to produce ethane C 2 H 4 + H 2 → C 2 H 6  Catalytic converters in cars that help convert CO to CO 2 and nitrogen oxide to N 2

An Example of Heterogeneous Catalysis Step #1: Adsorption and activation of the reactants. Carbon monoxide and nitrogen monoxide adsorbed on a platinum surface

An Example of Heterogeneous Catalysis Step #2: Migration of the adsorbed reactants on the surface. Carbon monoxide and nitrogen monoxide arranged prior to reacting

An Example of Heterogeneous Catalysis Step #3: Reaction of the adsorbed substances. Carbon dioxide and nitrogen form from previous molecules

An Example of Heterogeneous Catalysis Step #4: Escape, or desorption, of the products. Carbon dioxide and nitrogen gases escape (desorb) from the platinum surface

13.5 Catalysts  Enzymes: Enzymes interact with reactant molecules in a way that places them in the correct geometry to form the products Can increase reaction rates up to 10 14 Often end with an “ase” suffix Often named for the reactions they catalyze  Ex: sucrase, alcohol dehydrogenase

13.6 Reaction Mechanisms Objectives  To express a chemical reaction as a sequence of elementary processes  To identify the rate limiting step of a reaction and determine its molecularity  To predict the experimental rate law, given the chemical equation for the rate-limiting step  To differentiate among possible reaction mechanisms by examining experimental rate data

13.6 Reaction Mechanisms  Why do chemists care about the mechanism of a reaction? Reactions can often be manipulated to decrease side reactions and improve yields if we know the appropriate reaction mechanism

13.6 Reaction Mechanisms  Are cars manufactured so that all parts are assembled in one single step?  Chemical equations are often written in a way that suggests that a reaction occurs in a single step  This is typically not the case  Just think of the equation we all learn for cellular respiration…..Does cellular respiration occur in one- single step? C 6 H 12 O 6 + 6 O 2 → 6CO 2 + 6H 2 O

13.6 Reaction Mechanisms  A reaction mechanism is the sequence of elementary steps that lead from reactants to products Some reactions require a single step Other reactions require multiple steps with intermediates that are produced, then consumed, during the course of the reaction Sometimes more than one reaction mechanism is possible and further experimentation can help determine which mechanism is most likely to occur Catalysts may allow for a different set of elementary steps to proceed as compared to those observed for an uncatalyzed reaction

13.6 Reaction Mechanisms  For a reaction mechanism to be valid: The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law

13.6 Reaction Mechanisms  For example… Overall equation: NO 2 + CO → NO + CO 2 Experimentally derived rate law: rate = k[NO 2 ] 2 Proposed mechanism:  NO 2 + NO 2 → NO 3 + NO slow  NO 3 + CO → NO 2 + CO 2 fast

13.6 Reaction Mechanisms Proposed mechanism:  NO 2 + NO 2 → NO 3 + NO slow  NO 3 + CO → NO 2 + CO 2 fast  Overall: NO 2 + CO → NO + CO 2  Note that the first criteria for a reaction mechanism is met, the sum of the elementary steps does give the overall balanced equation for the reaction.  Note that NO 3 is a reaction intermediate that is made in step one and consumed in step 2  Is the second criteria met? Does the predicted rate law agree with the experimental rate law? Let’s explore further….

13.6 Reaction Mechanisms  Writing rate laws for elementary steps….  As defined previously, an elementary step is an equation that describes an actual molecular level event  Unlike rate laws generated for overall rate equations, the rate laws for elementary reactions depend on the number of species involved in the single elementary step and do make use of the coefficients in a balanced equation

13.6 Reaction Mechanisms For a unimolecular reaction, such as the decomposition of A → B, the elementary step involves a single molecule and a first order rate law describes the kinetics: rate = k[A] For a bimolecular reaction, A + B → C, the elementary step involves two molecules and a second order rate law describes the kinetics: rate = k[A][B] or if 2A → B the rate law is written: rate = k[A] 2 For a termolecular reaction, the elementary step involves three molecules and a third order rate law describes the kinetics: For 2A + B → C, the rate = k[A] 2 [B] or for A + B + C → D, the rate = k[A][B][C]

13.6 Reaction Mechanisms  Back to our original problem: Experimentally derived rate law: rate = k[NO 2 ] 2 Proposed mechanism:  NO 2 + NO 2 → NO 3 + NO slow rate = k[NO 2 ] 2  NO 3 + CO → NO 2 + CO 2 fast rate = k[NO 3 ][CO]  Note how the experimentally derived rate law is in agreement with the rate law of the rate-limiting slow step in the proposed reaction mechanism  Note: Reactions cannot proceed any faster than the speed of their rate-limiting step…..

The Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

13.6 Reaction Mechanisms  So let’s review… Overall equation: NO 2 + CO → NO + CO 2 Experimentally derived rate law: rate = k[NO 2 ] 2 Proposed mechanism:  NO 2 + NO 2 → NO 3 + NO slow rate = k[NO 2 ] 2  NO 3 + CO → NO 2 + CO 2 fast So the sum of the elementary steps gives the overall balanced equation for the reaction and the mechanism agrees with the experimentally determined rate law

Another example…. For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g)  N 2 O(g) is an intermediate

Identifying the Rate-Determining Step For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) The experimental rate law is: R = k[NO] 2 [H 2 ] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) Step #1 agrees with the experimental rate law

13.6 Reaction Mechanisms  A discussion of complex reaction mechanisms (TB pp. 553-555) will be discussed in class but will not be tested….  Sample problems to test basic understanding of reaction mechanisms include: Ch 13 # 77-80 (81& 82 are more challenging dealing with the reversible reactions)

Reaction Mechanisms Summary To determine whether a proposed mechanism is plausible: Is the rate-limiting step consistent with the observed rate law? Does the sum of all the steps provide the correct stoichiometry? If one or more mechanisms pass these tests, other factors including reasonable intermediates and analogy with known reactions will be used to propose a likely reaction mechanism

To Summarize Factors Affecting Rate Increasing temperature always increases the rate of a reaction.  Particles collide more frequently  Particles collide more energetically Increasing surface area increases the rate of a reaction Increasing Concentration USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

Chemical Kinetics Chapter 13. Chemical Kinetics  Chemical reactions occur when the reactant species strike each other and interact to form products 

Similar presentations

Presentation on theme: "Chemical Kinetics Chapter 13. Chemical Kinetics  Chemical reactions occur when the reactant species strike each other and interact to form products "— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chemical Kinetics Chapter 13. Chemical Kinetics  Chemical reactions occur when the reactant species strike each other and interact to form products 

Similar presentations

Presentation on theme: "Chemical Kinetics Chapter 13. Chemical Kinetics  Chemical reactions occur when the reactant species strike each other and interact to form products "— Presentation transcript:

Similar presentations

About project

Feedback