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Chapter 13 Chemical Kinetics

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, andsurface area.Goal: to understand chemical reactions at the molecular level.. Factors that Affect Reaction Rates.

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Chapter 13 Chemical Kinetics

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    1. Chapter 13 Chemical Kinetics

    12. Rate Comparison 2NO2 ? 2NO + O2 *If main interest is on the consumption of starting reagent, then: Rate = - D[NO2] = D[NO] = 2 D[O2] D t D t D t Since 2 NO2 molecules are consumed for every O2

    13. 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.

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

    15. 2N2O5 4NO2 + O2

    16. 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 (+).

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

    18. 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

    20. 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

    24. 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.

    25. 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.

    26. Experimental Data

    31. 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?

    33. 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

    34. 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

    35. Experiment 23 You varied concentration of KIO3 and held the concentration of NaHSO3 constant. From Expt. A ? E, you increased the amount of KIO3. Observed result: The time it took for the reaction to occur DECREASED.

    36. Conclusion Higher concentration of KIO3 lead to faster rate of reaction.

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

    40. By graphing concentration vs. 1/time?

    41. The Integrated Rate Law makes this possible!

    43. Integrated Rate Law Expresses the dependence of concentration on time

    45. 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.

    46. 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

    50. 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

    51. 2N2O5 4NO2 + O2

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