Créer une présentation
Télécharger la présentation

Télécharger la présentation
## AP Notes Chapter 15

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**AP Notes Chapter 15**Principles of Reactivity: Chemical Kinetics**Chemical Kinetics**• The study of how fast chemical reactions occur. • How fast species are appearing and disappearing. • It has nothing to do with the extent to which a reaction will occur – that is thermodynamics.**There is a difference between the probability that molecules**will react when they meet each other and once they do react how stable is the product. • First, probability of reaction has to do with overcoming an energy barrier or activation energy. • Second, has to do with the differences between reactants and products.**Thermodynamics:**• Spontaneity deals with how likely a reaction will take place. • Spontaneous reactions are reactions that will happen - but we can’t tell how fast. • Diamond will spontaneously turn to graphite – eventually.**Factors Affecting Rates**1. Surface area Powders 2. Temperature Dye 3. Pressure 4. Catalyst /Inhibitor 5. Concentration H202 & I2**Reaction Rate**The change in concentration of a reactant or product per unit time For a reaction A B Average Rate = Change in Moles of B Change in time Rate = D [B] = - D [A] D t D t**Reaction Rate**• Rate = Conc. of A at t2 -Conc. of A at t1t2- t1 • Rate =D[A]Dt • Change in concentration per unit time • Changes in Pressure to changes in concentration**Concentration**N2 + 3H2 2NH3 • As the reaction progresses the concentration H2 goes down [H2] Time**Concentration**N2 + 3H2 2NH3 • As the reaction progresses the concentration N2 goes down 1/3 as fast [N2] [H2] Time**Concentration**N2 + 3H2 2NH3 • As the reaction progresses the concentration NH3 goes up. [N2] [H2] [NH3] Time**Calculating Rates**• Average rates are taken over long intervals • Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time • Derivative and Integrated Rates.**Concentration**• Average slope method D[H2] Dt Time**Concentration**• Instantaneous slope method. D[H2] D t Time**Defining Rate**• We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. • In our example N2 + 3H2 2NH3 • -D[N2] = - D[H2] = D[NH3] Dt 3 Dt 2 Dt**RATE LAW**Mathematical model that contains specific information about what reactants determine the rate and how the rate is controlled**Rate Laws**• Reactions are reversible. • As products accumulate they can begin to turn back into reactants. • Early on the rate will depend on only the amount of reactants present. • We want to measure the reactants as soon as they are mixed. • This is called the Initial rate method.**Rate Laws**• Two key points • The concentration of the products do not appear in the rate law because this is an initial rate. • The order must be determined experimentally, • can’t be obtained from the equation**2 NO2 2NO + O2**• You will find that the rate will only depend on the concentration of the reactants. • Rate = k[NO2]n • This is called a rate law expression. • k is called the rate constant. • n is the order of the reactant -usually a positive integer.**2 NO2 2NO + O2**• The rate of appearance of O2 can be said to be. • Rate' = D[O2] = D[NO2] = k'[NO2]Dt 2Dt • Because there are 2 NO2 for each O2 • Rate = 2 x Rate' • So k[NO2]n= 2 x k'[NO2]n • So k = 2 x k'**Types of Rate Laws**• Differential Rate law - describes how rate depends on concentration. • Integrated Rate Law - Describes how concentration depends on time. • For each type of differential rate law there is an integrated rate law and vice versa. • Rate laws can help us better understand reaction mechanisms.**Determining Rate Laws**• The first step is to determine the form of the rate law (especially its order). • Must be determined from experimental data. • For this reaction 2 N2O5 (aq) 4NO2 (aq) + O2(g)The reverse reaction won’t play a role**[N2O5] (mol/L) Time (s)**1.00 0 0.88 200 0.78 400 0.69 600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30 2000 Now graph the data**To find rate we have to find the slope at two points**• We will use the tangent method.**At .90 M the rate is**(.98 - .76) = 0.22 =- 5.5x 10 -4 (0-400) -400**At .40 M the rate is**(.52 - .31) = 0.22 =- 2.7 x 10 -4 (1000-1800) -800**Since the rate at twice the concentration is twice as fast**the rate law must be.. • Rate = -D[N2O5] = k[N2O5]1 = k[N2O5] Dt • We say this reaction is first order in N2O5 • The only way to determine order is to run the experiment.**The method of Initial Rates**• This method requires that a reaction be run several times. • The initial concentrations of the reactants are varied. • The reaction rate is measured bust after the reactants are mixed. • Eliminates the effect of the reverse reaction.**Integrated Rate Law**• Expresses the reaction concentration as a function of time. • Form of the equation depends on the order of the rate law (differential). • Rate = D[A]nDt • We will only work with n=0, 1, and 2 although negative and fractional exist**General Rate Law**aA + bB cC Rate -[A] or -[B] r = k[A]m[B]n • Where: • m = order with respect to A • n = order wrt B • and n + m = order of reaction Notice we are primarily discussing the reactants**For example: aA + bB cC + dD**• Zero Order: r = k • First Order in A: r = k [A] • First Order in B: r = k [B] • Second Order in A: r = k [A]2 • Second Order in B: r = k [B]2 • First Order in A, first Order in B, for a second order overall: r = k [A][B]**Initial Rate Method to establish the exact rate law**R0 = k[A0]m[B0]n but experimental data must provide orders Use ratio method to solve for m and n R0 = k[A0]m[B0]n**R0 = k[A0]m[B0]n**Exp Rate [A0] [B0] 1 0.05 0.50 0.20 2 0.05 0.75 0.20 3 0.10 0.50 0.40 Thus, rate = k[A]0[B]1 or rate = k[B]**Where k is the rate constant , with appropriate units.**rate = k[B] Using Exp 1:**INTEGRATED RATE LAWS**Each “order” has a separate “law” based on the integration of the rate expression.**You NEED TO KNOW the rate expressions**& half-life formulas for 0, 1st, and 2nd order reactions. Third order and fractional order reactions exist, but are rare. Nuclear reactions are typically 1st order, therefore, use the first-order rate law and half-life to solve problems of this type**Zero Order Rate Law**• Most often when reaction happens on a surface because the surface area stays constant. • Also applies to enzyme chemistry.**Concentration**Time**Concentration**k = D[A]/Dt D[A] Dt Time**Zero Order Rate Law**• Rate = k[A]0 = k • Rate does not change with concentration. • Integrated [A] = -kt + [A]0 • When [A] = [A]0 /2 t = t1/2 • t1/2 = [A]0 /2k**For a zero-order reaction, the rate is independent of the**concentration. Expressed in terms of calculus, a zero - order reaction would be expressed as:**m = -k**[B] t**Time for the concentration to become one-half of its initial**value Half-life 0-Order Reaction**First Order**• For the reaction 2N2O5 4NO2 + O2 • We found the Rate = k[N2O5]1 • If concentration doubles rate doubles. • If we integrate this equation with respect to time we get the Integrated Rate Law • ln[N2O5] = - kt + ln[N2O5]0 • ln is the natural log • [N2O5]0 is the initial concentration.**First Order**• General form Rate = D[A] / Dt = k[A] • ln[A] = - kt + ln[A]0 • In the form y = mx + b • y = ln[A] m = -k • x = t b = ln[A]0 • A graph of ln[A] vs time is a straight line.**First Order**• By getting the straight line you can prove it is first order • Often expressed in a ratio**First Order**• By getting the straight line you can prove it is first order • Often expressed in a ratio**First OrderNuclear Decay and Half life**• The time required to reach half the original concentration. • If the reaction is first order • [A] = [A]0/2 when t = t1/2**The time required to reach half the original concentration.**• If the reaction is first order • [A] = [A]0/2 when t = t1/2 • ln(2) = kt1/2