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Chapter 13 Chemical Kinetics

CHEMISTRY. Chapter 13 Chemical Kinetics. Factors that Affect Reaction Rates. Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of reactions: reactant concentration, temperature, action of catalysts, and surface area.

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Chapter 13 Chemical Kinetics

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  1. CHEMISTRY Chapter 13Chemical Kinetics

  2. Factors that Affect Reaction Rates • Kinetics is the study of how fast chemical reactions occur. • There are 4 important factors which affect rates of reactions: • reactant concentration, • temperature, • action of catalysts, and • surface area. • Goal: to understand chemical reactions at the molecular level.

  3. Reaction Rates • Change of Rate with Time • Consider: • C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

  4. Reaction Rates • For the reaction A  B there are two ways of measuring rate: • the speed at which the products appear (i.e. change in moles of B per unit time), or • the speed at which the reactants disappear (i.e. the change in moles of A per unit time).

  5. Reaction Rates • Reaction Rate and Stoichiometry • In general for: • aA + bB cC + dD

  6. Problem 1 In the Haber process for the production of ammonia, N2(g) + 3H2(g)  2NH3(g) What is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

  7. 2N2O5(g)  4NO2(g) + O2(g) At a particular instant rate of (N2O5) = 4.2 x 10−7 mol·L−1·s−1 What are the rates of appearance of NO2 and O2 ?

  8. Problem 3. Consider the reaction 4PH3(g)  P4(g) + 6H2(g) If, in a certain experiment, over a specific time period, 0.0048 mol PH3 is consumed in a 2.0-L container each second of reaction, what are the rates of production of P4 and H2 in this experiment?

  9. Take Note! • Since the direction of equilibrium changes as more product is produced, rates have to be determined as soon as the experiment has begun.

  10. This is why…….. • …… even if the rate of production of product is of interest, rate expression still uses the starting reagents.

  11. 2N2O5 4NO2 + O2

  12. Rate is (-) if reagent is consumed. • Rate is (+) if compound is produced. • Rate will ultimately be (+) because change in concentration will be negative. Two (-)’s become (+).

  13. Differential Rate Law - is a rate law that expresses how rate is dependent on concentration Example: Rate = k[A]n

  14. Differential First Order Rate Law • First Order Reaction • Rate dependent on concentration • If concentration of starting reagent was doubled, rate of production of compounds would also double

  15. Concentration and Rate • Using Initial Rates to Determines Rate Laws • A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. • A reaction is first order if doubling the concentration causes the rate to double. • A reacting is nth order if doubling the concentration causes an 2n increase in rate. • Note that the rate constant does not depend on concentration.

  16. Differential Rate Law • For single reactants: A  C Rate = k[A]n • For 2 or more reactants: A + B  C Rate = k[A]n[B]m Rate = k[A]n[B]m[C]p

  17. Problem • NH4+ + NO2- N2 + 2H2O • Give the general rate law equation for rxn. • Derive rate order. • Derive general rate order. • Solve for the rate constant k.

  18. To Determine the Orders of the Reaction (n, m, p, etc….) • 1. Write Rate law equation. • 2. Get ratio of 2 rate laws from successive experiments. • Ratio = rate Expt.2 = k2[NH4+]n[NO2-]m rate Expt.1 k1[NH4+]n[NO2-]m • 3. Derive reaction order. • 4. Derive overall reaction order.

  19. Experimental Data

  20. A + B  C • Determine the differential rate law • Calculate the rate constant • Calculate the rate when [A]=0.050 mol·L-1 and [B]=0.100 mol·L-1

  21. Use the data in table 12.5 to determine 1) The orders for all three reactants 2) The overall reaction order 3) The value of the rate constant

  22. 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) • Determine the differential rate law • Calculate the rate constant • Calculate the rate when [NO]=0.050 mol·L-1 and [H2]=0.150 mol·L-1

  23. Sample Problem:. Consider the general reaction aA + bB cC and the following average rate data over some time period Δt: Determine a set of possible coefficients to balance this general reaction.

  24. Problem • Reaction: A + B  C obeys the rate law: Rate = k[A]2[B]. • A. If [A] is doubled (keeping B constant), how will rate change? • B. Will rate constant k change? Explain. • C. What are the reaction orders for A & B? • D. What are the units of the rate constant?

  25. You now know that…. • The rate expression correlates consumption of reactant to production of product. For a reaction: 3A  2B - 1D[A] = 1D[B] 3 Dt 2 Dt • The differential rate law allows you to correlate rate with concentration based on the format: Rate = k [A]n

  26. You also know that… • 1. Rate of consumption of reactant decreases over time because the concentration of reactant decreases. Lower concentration equates to lower rate. • 2. If a graph of concentration vs. time were constructed, the graph is not a straight line

  27. How can we make the line straight?What is the relationship between concentration and time?

  28. By graphing concentration vs. 1/time?

  29. The Integrated Rate Law makes this possible!

  30. Integrated Rate Law • Expresses the dependence of concentration on time

  31. Integrated Rate Laws • Zero Order: [A]t = -kt + [A]o • First Order: ln[A]t = -kt + ln[A]o • Second Order: 1 = kt + 1 [A]t [A]o where [A]o is the initial concentration and [A]t is the final concentration.

  32. Integrated First-Order Rate Law • ln[A]t = -kt + ln[A]0 • Eqn. shows [concn] as a function of time • Gives straight-line plot since equation is of the form y = mx + b

  33. The Change of Concentration with Time • Zero Order Reactions • A plot of [A]tversus t is a straight line with slope -k and intercept [A]0.

  34. The Change of Concentration with Time • First Order Reactions • A plot of ln[A]t versus t is a straight line with slope -k and intercept ln[A]0.

  35. The Change of Concentration with Time • Second Order Reactions • A plot of 1/[A]tversus t is a straight line with slope kand intercept 1/[A]0.

  36. Integrated Rate Law 2N2O5 NO2 + O2 Rate = - D[N2O5] = k[N2O5] n D t If n = 1, upon integration: ln [N2O5]t = -kt + ln [N2O5]0 initial concentration at t=0

  37. 2N2O5 4NO2 + O2

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