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Chapter 13 Chemical Kinetics
Presented by: Sudhir Kumar Maingi PGT (Chemistry) K.V. No. 1 PATHANKOT, AIR FORCE STN, JAMMU REGION
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Kinetics kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. experimentally it is shown that there are 4 factors that influence the speed of a reaction: nature of the reactants, temperature, catalysts, concentration
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Defining 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 mi/hr
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Defining Reaction Rate
the rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time or product concentration increases for reactants, a negative sign is placed in front of the definition
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Reaction Rate Changes Over Time
as time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases. at some time the reaction stops, either because the reactants run out or because the system has reached equilibrium.
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at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1
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at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2
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at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8 at t = 48 [X] = 5 [Y] = 5 [Z] = 3
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Hypothetical Reaction Red Blue
Time (sec) Number Red Number Blue 100 5 84 16 10 71 29 15 59 41 20 50 25 42 58 30 35 65 70 40 75 45 21 79 18 82 in this reaction, one molecule of Red turns into one molecule of Blue the number of molecules will always total 100 the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time
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Hypothetical Reaction Red Blue
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Hypothetical Reaction Red Blue
ro, Chemistry: A Molecular Approach
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Reaction Rate and Stoichiometry
in most reactions, the coefficients of the balanced equation are not all the same H2 (g) + I2 (g) 2 HI(g) for these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made therefore the rate of change will be different in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient
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Average Rate the average rate is the change in measured concentrations in any particular time period linear approximation of a curve the larger the time interval, the more the average rate deviates from the instantaneous rate
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Hypothetical Reaction Red Blue
Avg. Rate Time (sec) Number Red Number Blue (5 sec intervals) (10 sec intervals) (25 sec intervals) 100 5 84 16 3.2 10 71 29 2.6 2.9 15 59 41 2.4 20 50 1.8 2.1 25 42 58 1.6 2.3 30 35 65 1.4 1.5 70 1 40 75 45 21 79 0.8 18 82 0.6 0.7
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H2 I2 HI Avg. Rate, M/s Time (s) [H2], M [HI], M -D[H2]/Dt
1/2 D[HI]/Dt 0.000 1.000 10.000 0.819 0.362 0.0181 20.000 0.670 0.660 0.0149 30.000 0.549 0.902 0.0121 40.000 0.449 1.102 0.0100 50.000 0.368 1.264 0.0081 60.000 0.301 1.398 0.0067 70.000 0.247 1.506 0.0054 80.000 0.202 1.596 0.0045 90.000 0.165 1.670 0.0037 0.135 1.730 0.0030 Avg. Rate, M/s Time (s) [H2], M [HI], M -D[H2]/Dt 1/2 D[HI]/Dt 0.000 1.000 10.000 0.819 20.000 0.670 30.000 0.549 40.000 0.449 50.000 0.368 60.000 0.301 70.000 0.247 80.000 0.202 90.000 0.165 0.135 Avg. Rate, M/s Time (s) [H2], M [HI], M -D[H2]/Dt 0.000 1.000 10.000 0.819 0.362 0.0181 20.000 0.670 0.660 0.0149 30.000 0.549 0.902 0.0121 40.000 0.449 1.102 0.0100 50.000 0.368 1.264 0.0081 60.000 0.301 1.398 0.0067 70.000 0.247 1.506 0.0054 80.000 0.202 1.596 0.0045 90.000 0.165 1.670 0.0037 0.135 1.730 0.0030 Avg. Rate, M/s Time (s) [H2], M [HI], M -D[H2]/Dt 1/2 D[HI]/Dt 0.000 1.000 10.000 0.819 0.362 20.000 0.670 0.660 30.000 0.549 0.902 40.000 0.449 1.102 50.000 0.368 1.264 60.000 0.301 1.398 70.000 0.247 1.506 80.000 0.202 1.596 90.000 0.165 1.670 0.135 1.730 Stoichiometry tells us that for every 1 mole/L of H2 used, 2 moles/L of HI are made. The average rate is the change in the concentration in a given time period. Assuming a 1 L container, at 10 s, we used moles of H2. Therefore the amount of HI made is 2(0.181 moles) = moles In the first 10 s, the D[H2] is M, so the rate is At 60 s, we used moles of H2. Therefore the amount of HI made is 2(0.699 moles) = moles
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average rate in a given time period = slope of the line connecting the [H2] points; and ½ +slope of the line for [HI] the average rate for the first 80 s is M/s the average rate for the first 10 s is M/s the average rate for the first 40 s is M/s
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Instantaneous Rate the instantaneous rate is the change in concentration at any one particular time slope at one point of a curve determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function for you calculus fans
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H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is: Using [HI], the instantaneous rate at 50 s is:
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Ex 13. 1 - For the reaction given, the [I] changes from 1. 000 M to 0
Ex For the reaction given, the [I] changes from M to M in the first 10 s. Calculate the average rate in the first 10 s and the D[H+]. H2O2 (aq) + 3 I(aq) + 2 H+(aq) I3(aq) + 2 H2O(l) Solve the equation for the Rate (in terms of the change in concentration of the Given quantity) Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value
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Measuring Reaction Rate
in order to measure the reaction rate you need to be able to measure the concentration of at least one component in the mixture at many points in time there are two ways of approaching this problem (1) for reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration, or (2) for reactions that happen over a very long time, sampling of the mixture at various times can be used when sampling is used, often the reaction in the sample is stopped by a quenching technique
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Continuous Monitoring
polarimetry – measuring the change in the degree of rotation of plane-polarized light caused by one of the components over time spectrophotometry – measuring the amount of light of a particular wavelength absorbed by one component over time the component absorbs its complimentary color total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction
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Sampling gas chromatography can measure the concentrations of various components in a mixture for samples that have volatile components separates mixture by adherence to a surface drawing off periodic aliquots from the mixture and doing quantitative analysis titration for one of the components gravimetric analysis
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Factors Affecting Reaction Rate Nature of the Reactants
nature of the reactants means what kind of reactant molecules and what physical condition they are in. small molecules tend to react faster than large molecules; gases tend to react faster than liquids which react faster than solids; powdered solids are more reactive than “blocks” more surface area for contact with other reactants certain types of chemicals are more reactive than others e.g., the activity series of metals ions react faster than molecules no bonds need to be broken
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Factors Affecting Reaction Rate Temperature
increasing temperature increases reaction rate chemist’s rule of thumb - for each 10°C rise in temperature, the speed of the reaction doubles for many reactions there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later
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Factors Affecting Reaction Rate Catalysts
catalysts are substances which affect the speed of a reaction without being consumed. most catalysts are used to speed up a reaction, these are called positive catalysts catalysts used to slow a reaction are called negative catalysts. homogeneous = present in same phase heterogeneous = present in different phase how catalysts work will be examined later
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Factors Affecting Reaction Rate Reactant Concentration
generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the gas higher pressure = higher concentration concentration of solutions depends on the solute to solution ratio (molarity)
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The Rate Law the Rate Law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well the rate of a reaction is directly proportional to the concentration of each reactant raised to a power for the reaction aA + bB products the rate law would have the form given below n and m are called the orders for each reactant k is called the rate constant
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Reaction Order the exponent on each reactant in the rate law is called the order with respect to that reactant the sum of the exponents on the reactants is called the order of the reaction The rate law for the reaction: 2 NO(g) + O2(g) ® 2 NO2(g) is Rate = k[NO]2[O2] The reaction is second order with respect to [NO], first order with respect to [O2], and third order overall
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The reaction is autocatalytic, because a product affects the rate.
Sample Rate Laws The reaction is autocatalytic, because a product affects the rate. Hg2+ is a negative catalyst, increasing its concentration slows the reaction.
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Reactant Concentration vs. Time A Products
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Half-Life the half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value the half-life of the reaction depends on the order of the reaction
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Zero Order Reactions [A] Rate = k[A]0 = k constant rate reactions
[A] = -kt + [A]0 graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0 t ½ = [A0]/2k when Rate = M/sec, k = M/sec [A]0 [A] time slope = - k
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First Order Reactions Rate = k[A] ln[A] = -kt + ln[A]0
graph ln[A] vs. time gives straight line with slope = -k and y-intercept = ln[A]0 used to determine the rate constant t½ = 0.693/k the half-life of a first order reaction is constant the when Rate = M/sec, k = sec-1
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ln[A]0 ln[A] time slope = −k
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Half-Life of a First-Order Reaction Is Constant
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Rate Data for C4H9Cl + H2O ® C4H9OH + HCl
Time (sec) [C4H9Cl], M 0.0 0.1000 50.0 0.0905 100.0 0.0820 150.0 0.0741 200.0 0.0671 300.0 0.0549 400.0 0.0448 500.0 0.0368 800.0 0.0200 0.0000
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C4H9Cl + H2O C4H9OH + 2 HCl
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C4H9Cl + H2O C4H9OH + 2 HCl
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C4H9Cl + H2O C4H9OH + 2 HCl slope = -2.01 x 10-3 k = 2.01 x 10-3 s-1
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Second Order Reactions
Rate = k[A]2 1/[A] = kt + 1/[A]0 graph 1/[A] vs. time gives straight line with slope = k and y-intercept = 1/[A]0 used to determine the rate constant t½ = 1/(k[A0]) when Rate = M/sec, k = M-1∙sec-1
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slope = k 1/[A] l/[A]0 time
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Partial Pressure NO2, mmHg
Rate Data For 2 NO2 ® 2 NO + O2 Time (hrs.) Partial Pressure NO2, mmHg ln(PNO2) 1/(PNO2) 100.0 4.605 30 62.5 4.135 60 45.5 3.817 90 35.7 3.576 120 29.4 3.381 150 25.0 3.219 180 21.7 3.079 210 19.2 2.957 240 17.2 2.847
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Rate Data Graphs For 2 NO2 ® 2 NO + O2
Tro, Chemistry: A Molecular Approach
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Rate Data Graphs For 2 NO2 ® 2 NO + O2
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Rate Data Graphs For 2 NO2 ® 2 NO + O2
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Determining the Rate Law
can only be determined experimentally graphically rate = slope of curve [A] vs. time if graph [A] vs time is straight line, then exponent on A in rate law is 0, rate constant = -slope if graph ln[A] vs time is straight line, then exponent on A in rate law is 1, rate constant = -slope if graph 1/[A] vs time is straight line, exponent on A in rate law is 2, rate constant = slope initial rates by comparing effect on the rate of changing the initial concentration of reactants one at a time
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Tro, Chemistry: A Molecular Approach
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Practice - Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Prod
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Practice - Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Prod
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Practice - Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Prod
-D[A] Dt = 0.1 [A]2 the reaction is second order,
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Ex – The reaction SO2Cl2(g) SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10-4 s-1 at a given set of conditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]0 = M Given: Find: [SO2Cl2]0 = M, t = 865, k = 2.90 x 10-4 s-1 [SO2Cl2] Concept Plan: Relationships: [SO2Cl2] [SO2Cl2]0, t, k Solution: Check: the new concentration is less than the original, as expected
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Initial Rate Method another method for determining the order of a reactant is to see the effect on the initial rate of the reaction when the initial concentration of that reactant is changed for multiple reactants, keep initial concentration of all reactants constant except one zero order = changing the concentration has no effect on the rate first order = the rate changes by the same factor as the concentration doubling the initial concentration will double the rate second order = the rate changes by the square of the factor the concentration changes doubling the initial concentration will quadruple the rate
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Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not
Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Write a general rate law including all reactants Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Determine by what factor the concentrations and rates change in these two experiments.
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Determine to what power the concentration factor must be raised to equal the rate factor.
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Repeat for the other reactants
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Substitute the exponents into the general rate law to get the rate law for the reaction n = 2, m = 0
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Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g) given the data below. Expt. Number Initial [NO2], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.0021 2. 0.20 0.0082 3. 0.0083 4. 0.40 0.033 Substitute the concentrations and rate for any experiment into the rate law and solve for k
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Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1 ® N2 + 2 H2O given the data below. Expt. No. Initial [NH4+], M Initial [NO2-], M Initial Rate, (x 10-7), M/s 1 0.0200 0.200 10.8 2 0.0600 32.3 3 0.0202 4 0.0404 21.6
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Practice - Determine the rate law and rate constant for the reaction NH4+1 + NO2-1 ® N2 + 2 H2O given the data below. Rate = k[NH4+]n[NO2]m Expt. No. Initial [NH4+], M Initial [NO2-], M Initial Rate, (x 10-7), M/s 1 0.0200 0.200 10.8 2 0.0600 32.3 3 0.0202 4 0.0404 21.6
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The Effect of Temperature on Rate
changing the temperature changes the rate constant of the rate law Svante Arrhenius investigated this relationship and showed that: where T is the temperature in kelvins R is the gas constant in energy units, J/(mol∙K) A is a factor called the frequency factor Ea is the activation energy, the extra energy needed to start the molecules reacting
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Activation Energy and the Activated Complex
energy barrier to the reaction amount of energy needed to convert reactants into the activated complex aka transition state the activated complex is a chemical species with partially broken and partially formed bonds always very high in energy because partial bonds
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Isomerization of Methyl Isonitrile
methyl isonitrile rearranges to acetonitrile in order for the reaction to occur, the H3C-N bond must break; and a new H3C-C bond form
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Energy Profile for the Isomerization of Methyl Isonitrile
As the reaction begins, the C-N bond weakens enough for the CN group to start to rotate the activated complex is a chemical species with partial bonds the collision frequency is the number of molecules that approach the peak in a given period of time the activation energy is the difference in energy between the reactants and the activated complex
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The Arrhenius Equation: The Exponential Factor
the exponential factor in the Arrhenius equation is a number between 0 and 1 it represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier the higher the energy barrier (larger activation energy), the fewer molecules that have sufficient energy to overcome it that extra energy comes from converting the kinetic energy of motion to potential energy in the molecule when the molecules collide increasing the temperature increases the average kinetic energy of the molecules therefore, increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier therefore increasing the temperature will increase the reaction rate
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Arrhenius Plots the Arrhenius Equation can be algebraically solved to give the following form: this equation is in the form y = mx + b where y = ln(k) and x = (1/T) a graph of ln(k) vs. (1/T) is a straight line ( J/mol∙K)(slope of the line) = Ea, (in Joules) ey-intercept = A, (unit is the same as k)
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Ex Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data: Temp, K k, M-1∙s-1 600 3.37 x 103 1300 7.83 x 107 700 4.83 x 104 1400 1.45 x 108 800 3.58 x 105 1500 2.46 x 108 900 1.70 x 106 1600 3.93 x 108 1000 5.90 x 106 1700 5.93 x 108 1100 1.63 x 107 1800 8.55 x 108 1200 3.81 x 107 1900 1.19 x 109
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Ex Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data: use a spreadsheet to graph ln(k) vs. (1/T)
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Ex Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data: Ea = m∙(-R) solve for Ea A = ey-intercept solve for A
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Arrhenius Equation: Two-Point Form
if you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:
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Ex – The reaction NO2(g) + CO(g) CO2(g) + NO(g) has a rate constant of 2.57 M-1∙s-1 at 701 K and 567 M-1∙s-1 at 895 K. Find the activation energy in kJ/mol Given: Find: T1 = 701 K, k1 = 2.57 M-1∙s-1, T2 = 895 K, k2 = 567 M-1∙s-1 Ea, kJ/mol Concept Plan: Relationships: Ea T1, k1, T2, k2 Solution: Check: most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable
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Collision Theory of Kinetics
for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other. once molecules collide they may react together or they may not, depending on two factors - whether the collision has enough energy to "break the bonds holding reactant molecules together"; whether the reacting molecules collide in the proper orientation for new bonds to form.
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Effective Collisions collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions the higher the frequency of effective collisions, the faster the reaction rate when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state
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Effective Collisions Kinetic Energy Factor
for a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide it can form the activated complex
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Effective Collisions Orientation Effect
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Collision Theory and the Arrhenius Equation
A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier there are two factors that make up the frequency factor – the orientation factor (p) and the collision frequency factor (z)
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Orientation Factor the proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form the more complex the reactant molecules, the less frequently they will collide with the proper orientation reactions between atoms generally have p = 1 reactions where symmetry results in multiple orientations leading to reaction have p slightly less than 1 for most reactions, the orientation factor is less than 1 for many, p << 1 there are some reactions that have p > 1 in which an electron is transferred without direct collision
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Reaction Mechanisms we generally describe chemical reactions with an equation listing all the reactant molecules and product molecules but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules describing the series of steps that occur to produce the overall observed reaction is called a reaction mechanism knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism
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An Example of a Reaction Mechanism
Overall reaction: H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) Mechanism: H2(g) + ICl(g) HCl(g) + HI(g) HI(g) + ICl(g) HCl(g) + I2(g) the steps in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps
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Elements of a Mechanism Intermediates
H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) H2(g) + ICl(g) HCl(g) + HI(g) HI(g) + ICl(g) HCl(g) + I2(g) notice that the HI is a product in Step 1, but then a reactant in Step 2 since HI is made but then consumed, HI does not show up in the overall reaction materials that are products in an early step, but then a reactant in a later step are called intermediates
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Molecularity the number of reactant particles in an elementary step is called its molecularity a unimolecular step involves 1 reactant particle a bimolecular step involves 2 reactant particles though they may be the same kind of particle a termolecular step involves 3 reactant particles though these are exceedingly rare in elementary steps
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Rate Laws for Elementary Steps
each step in the mechanism is like its own little reaction – with its own activation energy and own rate law the rate law for an overall reaction must be determined experimentally but the rate law of an elementary step can be deduced from the equation of the step H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl] HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]
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Rate Laws of Elementary Steps
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Rate Determining Step in most mechanisms, one step occurs slower than the other steps the result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction we call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy the rate law of the rate determining step determines the rate law of the overall reaction
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Another Reaction Mechanism
NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2 NO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 slow NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] fast The first step is slower than the second step because its activation energy is larger. The first step in this mechanism is the rate determining step. The rate law of the first step is the same as the rate law of the overall reaction.
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Validating a Mechanism
in order to validate (not prove) a mechanism, two conditions must be met: the elementary steps must sum to the overall reaction the rate law predicted by the mechanism must be consistent with the experimentally observed rate law
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Mechanisms 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
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An Example 2 NO(g) N2O2(g) Fast
H2(g) + N2O2(g) H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2] H2(g) + N2O(g) H2O(g) + N2(g) Fast k1 k-1 2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2 for Step 1 Rateforward = Ratereverse
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Ex Show that the proposed mechanism for the reaction 2 O3(g) 3 O2(g) matches the observed rate law Rate = k[O3]2[O2]-1 O3(g) O2(g) + O(g) Fast O3(g) + O(g) 2 O2(g) Slow Rate = k2[O3][O] k1 k-1 for Step 1 Rateforward = Ratereverse
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Catalysts catalysts are substances that affect the rate of a reaction without being consumed catalysts work by providing an alternative mechanism for the reaction with a lower activation energy catalysts are consumed in an early mechanism step, then made in a later step mechanism without catalyst O3(g) + O(g) 2 O2(g) V. Slow mechanism with catalyst Cl(g) + O3(g) O2(g) + ClO(g) Fast ClO(g) + O(g) O2(g) + Cl(g) Slow
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Ozone Depletion over the Antarctic
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Energy Profile of Catalyzed Reaction
polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals
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Catalysts homogeneous catalysts are in the same phase as the reactant particles Cl(g) in the destruction of O3(g) heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system
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Types of Catalysts
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Catalytic Hydrogenation H2C=CH2 + H2 → CH3CH3
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Ethene to Ethane animation
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Enzymes because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate protein molecules that catalyze biological reactions are called enzymes enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction
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Enzyme-Substrate Binding Lock and Key Mechanism
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Enzymatic Hydrolysis of Sucrose
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Thank You Presented by: Sudhir Kumar Maingi PGT CHEMISTRY
K.V. No. 1 PATHANKOT, JAMMU REGION Phone: Acknowledgements : Prentice Hall Publications Steven S. Zumdahl
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