ch. 13: chemical kinetics dr. namphol sinkaset chem 201: general chemistry ii
TRANSCRIPT
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Ch. 13: Chemical KineticsCh. 13: Chemical Kinetics
Dr. Namphol Sinkaset
Chem 201: General Chemistry II
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I. Chapter OutlineI. Chapter Outline
I. Introduction
II. The Rate of a Chemical Reaction
III. Reaction Rate Laws
IV. Integrated Rate Laws
V. Temperature and Rate
VI. Reaction Mechanisms
VII. Catalysis
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I. IntroductionI. Introduction
• Some reactions are quick (explosions) while others are slow (rusting of iron).
• Knowing the rate of a reaction and what factors influence it allow chemists to plan accordingly.
• If we understand what contributes to the rate, we can control the reaction.
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I. IntroductionI. Introduction
• Balanced equations only give net change.• Equations tell us nothing about how the
reaction happens.• One possibility for the above: simultaneous
collision of 6 molecules.• Unlikely, so reaction must occur in a series of
small steps that leads to final products.• We study mechanisms later in the chapter.
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
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II. Reaction RatesII. Reaction Rates
• Rates are generally change of something divided by change in time.
• Reaction rates are no different.• The rate of a reaction can be written
with respect to any compound in that reaction.
• However, there can only be one numerical value for a rate of reaction.
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II. Average Rates of ReactionII. Average Rates of Reaction
H2(g) + I2(g) 2HI(g)
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II. General Reaction RatesII. General Reaction Rates
aA + bB cC + dD
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II. Some Rate DataII. Some Rate Data
• If we plot average rate data as a function of time, we see that the reaction rate constantly changes.
• Thus, rate depends on concentration of reactants!
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III. Rate LawsIII. Rate Laws
• If the rate depends on concentration of reactants, then we should be able to write an equation.
• A rate law describes the mathematical relationship between the concentration of reactants and how fast the reaction occurs.
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III. A Simple Rate LawIII. A Simple Rate Law
• Consider a decomposition reaction where A products
• If the reverse reaction is negligible, then the rate law is: Rate = k[A]n. k is called the rate constant n is called the reaction order
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III. Reaction OrdersIII. Reaction Orders
• The reaction order, n, determines how the rate depends on the concentration of the reactant. For the previous reaction, if… n = 0, zero order, rate is independent of [A] n = 1, first order, rate is directly
proportional to [A] n = 2, second order, rate is proportional to
the square of the [A]
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III. Reaction Orders and RateIII. Reaction Orders and Rate
• The rate law for the decomposition can then be either: Rate = k[A]0 = k Rate = k[A]1
Rate = k[A]2
• Each will have a different type of curve when graphed.
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III. Determining OrdersIII. Determining Orders
• Reaction orders can only be determined by experiment!!
• Reaction orders are not related to the stoichiometry of a reaction!
• If reaction orders match a reaction’s stoichiometry, it is just a coincidence.
• Therefore, orders cannot be determined without experimental data!
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III. Sure-fire MethodIII. Sure-fire Method
[A] (M) Initial Rate (M/s)
0.10 0.015
0.20 0.060
0.40 0.240
For the reaction, A Products, we have the following data:
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III. More Complex ReactionsIII. More Complex Reactions
• What if we have a more complicated reaction like: aA + bB cC + dD?
• Writing the general rate law is easy. Simply include all reactants, each with its own order. Rate = k[A]m[B]n
• If there are more reactants, there are more terms in the rate law.
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III. Example ReactionIII. Example Reaction
• 2H2(g) + 2NO(g) N2(g) + 2H2O(g)
• After looking at experimental data, the rate law was found to be Rate = k[H2][NO]2.
• We say the reaction is 1st order in H2, 2nd order in NO, and 3rd order overall.
• Note that Rate always has units of M/s, so the units on k will depend on the rate law.
• What are the units of k for the rate law above?
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III. Steps for Finding Rate LawIII. Steps for Finding Rate Law
1) Pick two solutions where one reactant stays same, but another changes.
2) Write rate law for both w/ as much information as you have.
3) Ratio the two and solve for an order.
4) Repeat for another pair of solutions.
5) Use any reaction to get value of k.
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III. Sample ProblemIII. Sample Problem
[CHCl3] (M) [Cl2] (M) Initial Rate (M/s)
0.010 0.010 0.0035
0.020 0.010 0.0069
0.020 0.020 0.0098
0.040 0.040 0.027
Determine the complete rate law for the reaction CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) using the data below.
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III. Sample ProblemIII. Sample Problem
[NO] (M) [H2] (M) Initial Rate (M/s)
0.10 0.10 0.00123
0.10 0.20 0.00246
0.20 0.10 0.00492
Sometimes, rate laws can be found by inspection. Determine the rate law for the reaction 2NO(g) + 2H2(g) N2(g) + H2O(g) using the data below.
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IV. Concentration and TimeIV. Concentration and Time
• Study and elucidation of rate laws allow the prediction of when a reaction will end.
• An integrated rate law for a chemical reaction is a relationship between the concentrations of reactants and time.
• Integrated rate laws depend on the order of the reaction; thus, we examine each separately.
• We will only consider reactions with one reactant.
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IV. 1IV. 1stst Order Integrated Rate Law Order Integrated Rate Law
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IV. 1IV. 1stst Order Integrated Rate Law Order Integrated Rate Law
• Notice this equation is in y = mx + b form.
• A plot of ln[A] vs. t for a 1st order reaction yields a straight line with m = -k and b = ln[A]0.
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IV. 2IV. 2ndnd Order Integrated Rate Law Order Integrated Rate Law
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IV. 2IV. 2ndnd Order Integrated Rate Law Order Integrated Rate Law
• Again, this equation is in y = mx + b form.
• A plot of 1/[A] vs. t yields a straight line with slope equal to k and y-intercept equal to 1/[A]0.
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IV. Zero Order Integrated Rate LawIV. Zero Order Integrated Rate Law
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IV. Zero Order Integrated Rate LawIV. Zero Order Integrated Rate Law
• Yet again in y = mx + b form!
• Plot of [A] vs. t results in a straight line with slope equal to -k and b = [A]0.
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IV. Reaction Half LivesIV. Reaction Half Lives
• The half-life, t1/2, of a reaction is the time required for the concentration of a reactant to decrease to half its initial value.
• Half life equations depend on the order of the reaction.
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IV. 1IV. 1stst Order Reaction Half Life Order Reaction Half Life
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IV. 1IV. 1stst Order Reaction Half Life Order Reaction Half Life
• Notice that the half life doesn’t depend on reactant concentration!
• Unique for 1st order.• The half life for a 1st order reaction is
constant.
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IV. 1IV. 1stst Order Half Lives Order Half Lives
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IV. 2IV. 2ndnd Order Reaction Half Life Order Reaction Half Life
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IV. 2IV. 2ndnd Order Reaction Half Life Order Reaction Half Life
• For 2nd order, the half life depends on initial concentration.
• As concentration decreases, half life gets longer and longer.
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IV. Zero Order Reaction Half LifeIV. Zero Order Reaction Half Life
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IV. Zero Order Reaction Half LifeIV. Zero Order Reaction Half Life
• We see that for zero order reactions, the half life depends on concentration as well.
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V. Temperature and RateV. Temperature and Rate
• In general, rates of reaction are highly sensitive to temperature.
• If Rate = k[A]n, where does the temperature factor in?
• It’s in the constant k!
• Generally, increasing temperature increases k.
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V. The Arrhenius EquationV. The Arrhenius Equation
• Note that R is the gas constant, and T is temperature in kelvin.
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V. Parameters of Arrhenius Eqn.V. Parameters of Arrhenius Eqn.
• We can describe the physical meanings of the aspects of the Arrhenius equation by considering a specific reaction.
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V. Activation EnergyV. Activation Energy
• To get to product state, reactant must go through high-energy activated complex, or transition state.
• Even though reaction is exo overall, it must go through an endo step.
• Higher Ea means slower reaction.
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V. Frequency FactorV. Frequency Factor• The frequency factor represents the number of
approaches to the activation barrier per unit time.
• For this reaction, it represents how often the NC part of the molecule vibrates.
• Note that not all approaches result in reaction due to not having enough energy.
• A frequency factor of 109/s means that there are 109 vibrations per second of the NC group.
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V. Exponential FactorV. Exponential Factor
• The exponential factor is a number between 0 and 1 that represents the fraction of molecules that successfully react upon approach.
• An exponential factor of 10-7 means that 1 out of every 107 molecules has enough energy to cross the energy barrier.
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V. Exponential Factor & TempV. Exponential Factor & Temp
• Since exponential factor = e-Ea/RT, temperature has a huge influence.
• As T 0, the factor goes to 0, and as T ∞, the factor goes to 1.
• Thus, higher temperatures mean more successful approaches because the molecules have more energy to overcome the activation barrier.
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V. Finding A and EV. Finding A and Eaa
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V. Arrhenius PlotsV. Arrhenius Plots
• If we have kinetic data at various temperatures, we can plot ln k vs. 1/T.
• We should get a straight line with m = -Ea/R and b = ln A.
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V. Two-Point FormV. Two-Point Form
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V. Sample ProblemV. Sample Problem
• The decomposition of HI has rate constants of k = 0.079 1/M·s at 508 °C and k = 0.24 1/M·s at 540 °C. What is the activation energy of this reaction in kJ/mole?