Presentation on theme: "The Progress of Chemical Reactions"— Presentation transcript:
1 The Progress of Chemical Reactions Chemistry 122
2 Rate LawsOne of the variables influencing the rate, or speed of a reaction, is concentrationDue to effective collisions, the number of reactant particles eventually decrease and the reaction slows downWhen a reactant’s concentration decreases, the products concentration increasesThis makes the rate of a reaction always positive
3 A → BWhen there is only one reactant (decomposition), its concentration can be written into the formula Rate = k[A]Ex. H2O2 → H2 + O2Rate = k[A]m= k[H2O2]1But what does it mean?
4 The significance of ‘k’ When ‘k’ is large, the rate of reaction is fast and products form quicklyUnits include L/(mol∙s), L2/(mol2∙s), and s-1k is unique for every reaction and must be supplied if we are to determine the rate of the reaction
5 Reaction orderRate = k[A]; it is understood that the notation [A] means the same as [A]1This exponent ‘1’ is referred to as its reaction order, not the coefficientIt is used to define how the rate is affected by the concentration of the reactantWhen it is in the first order, or exponent = 1, the rate changes proportionally to the concentration
6 As the concentration changes, so does the reaction rate First order reactionsRate = k[A]mAs the concentration changes, so does the reaction rateIf the concentration decreases by ¼, the reaction rate slows down by ¼.If the concentration doubles, so does the rate of reactionPage 577
7 Second order reactions In double replacement reactions, there are two reactantsTherefore, both concentrations have to be considered when determining the rate of reactionThe rate is calculated as follows:Rate = k[A]m[B]nThe rate has to be determined through experimentation since it is not always the case that the exponents in the rate law correspond to the coefficients in a chemical equation
8 Finding the order of a reaction 2NO(g) + 2H2(g) → N2(g) + 2H2O(g)This reaction occurs in more than one stepRate = k[NO]2[H2]1This has been determined experimentally, it is not a reflection of the balanced chemical equationThe reaction is described as second order in NO, first order in H2, and third order overall because the sum of the orders for the individual reactants is = 3.
9 In the real world of chemistry Most reactions are more complex than one-step reactionsAs a result, reactions cannot be predicted solely on the basis of a balanced equation; it must be determined experimentallyThis is because the rate depends primarily on the slowest step in the reaction mechanism
10 Using a graph to determine the rate of reaction When the rate of reaction produces a curve, you must draw a tangent to the curve in order to determine kTake the slope of the line at two points along the curve and determine the rate (concentration/time)
11 Determining reaction order The method of initial rates determines reaction order by comparing the initial rates of reaction carried out with varying reactant concentrationsUse the expression aA + bB → products as well as rate = k[A]m[B]n
12 Example Trial Initial [A] (M) Initial [B] (M) Initial Rate 1 0.100 Compare the concentration of Trial 1 and 2 to their respective rates.Compare Trial 2 and 3 to their respective rates.What do you notice?Which reactant is first order? Which is second?What is the order of the overall reaction?Write the rate in the form of an equation.TrialInitial [A] (M)Initial [B] (M)Initial Rate10.1002.00 x 10-320.2004.00 x 10-3316.0 x 10-3Example
13 Questions1. Write the rate law for the reaction aA →bB if the reaction is third order in A. [B] is not part of the rate law. 2. Given the following experimental data, use the method of initial rates to determine the rate law for the reaction aA + bB → products. Hint: any number to the zero power equals one. For example, (0.22)0 = 1 and (55.6)0 = 1.TrialInitial [A] (M)Initial [B] (M)Initial Rate10.1002.00 x 10-320.20034.00 x 10-3
14 Questions3. Given the following experimental data, use the method of initial rates to determine the rate law for the reaction CH3CHO(g) → CH4(g) + CO(g)TrialInitial [CH3CHO] (M)Initial Rate (mol/(L∙s)12.00 x 10-32.70 x 10-1124.00 x 10-310.8 x 10-1138.00 x 10-343.2 x 10-11
15 Reaction mechanismsIf we were to know all the changes that occur as reactants turn into products, we would be able to graph the transformationThis is known as a reaction progress curveThe simplest curve is for an elementary reactionSuch reactions convert reactants to products in a single stepThere is only one activated complex
16 Reaction mechanismMost chemical reactions occur as a series of steps which produces many peaks and valleysThe series of reactions that take place constitute a reaction mechanismThe peaks are the individual activated complexes and the valleys are the energies of the intermediatesAn intermediate is a temporary product; becoming a reactant in the next step of the overall reactionFigure 18.28, p. 578
17 For the remainder of class Guided reading for section 18.5Questions 36, 37, p. 577Important points to consider:Rate in most cases is mol/L/s or mol∙s/L,Anything is square brackets [A] is typically in molarity (mol/L).Q. 38 – 42, p. 579.
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