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Goes with Chapter 16: Silberberg Principles of General Chemistry AP chemistry Mrs. Laura Peck 1.

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Presentation on theme: "Goes with Chapter 16: Silberberg Principles of General Chemistry AP chemistry Mrs. Laura Peck 1."— Presentation transcript:

1 Goes with Chapter 16: Silberberg Principles of General Chemistry AP chemistry Mrs. Laura Peck 1

2 Identify factors which affect reaction rates Calculate the rate of production of a product or consumption of a reactant using mole ratios and the given rate Determine the rate law for a reaction from given data, overall order, and value of the rate constant, inclusive of units Determine the instantaneous rate of a reaction Use integrated rate laws to determine concentrations at a certain time, t, and create graphs to determine the order of a reaction. Also, determine the half-life of a reaction. Write the rate law from a given mechanism given the speeds of each elementary step Write the overall reaction for a mechanism and identify catalysts and intermediates present Determine the activation energy for the reaction using the Arrhenius equation Graphically determine the activation energy using the Arrhenius equation. 2

3 Increasing the concentrations of reactants increases the chances for more molecular collisions which make products. Generally, the increase in [reactants] will increase the rate of a reaction Increasing the surface are of the reactants increase the number of collisions. Example, granular zinc will react more quickly with HCl than a strip of zinc Increasing the temperature of a reaction will also speed it up. Increasing the T increases the rate of  and  rxns, increasing the number of collisions over a certain period of time. 3

4 The collision theory model accounts for the observed characteristics of reaction rates. For a reaction to occur, the molecules must collide in the correct orientation with sufficient energy. The theory assumes that all but the simplest reactions take place in a series of two-particle collisions. This sequence of two-particle collisions is called the reaction mechanism 4

5 The potential energy diagram for a reaction, the reaction progress, is shown in the graphs above. At the top of the ‘hill’, or barrier, is the activated complex, or transition state. Once the energy in the transition state, the activation energy, is overcome, the reactants can become products. Activation energy is the amount of energy required to form an activated complex. 5

6 Because products have a higher potential energy than reactants, the reaction represented in the above diagram is endothermic. ΔΗ rxn = ΔH products – ΔH reactants The difference between the potential energies of the products and the reactants gives the enthalpy of the reaction. If asked to draw the reaction progress for an exothermic reaction, the energy of the products would be lower than that of the reactants. The diagram shows the pathway for both catalyzed and uncatalyzed reactions. A catalyst speeds up a reaction by lowering the activation energy of the reaction without being consumed itself. 6

7 The reaction rate is the change in the concentration of a reactant or product per unit time Consider the reaction, A  B Rate = - Δ[A] Δt The concentration of A in moles per liter is represented by [A]. The change in time is represented by Δt. The quantity, -Δ[A] / Δt, is negative because the reactants are disappearing and forming products. 7

8 Consider the reaction 4PH 3(g)  P 4(g) + 6H 2(g) If mol of PH3 is consumed in a 2.0 L container during each second of the reaction, what are the rates of production for P4 and H2? The rate at which PH3 is being consumed is – Δ[PH3] Δt mol PH3/(2.0 L x s) = mol/L*s PH3 The rate at which P4 and H2 are being produced can be Determined by using mole ratios mol PH3 x 1 mol P4 = mol/L*s P4 L x s 4 mol PH3 Ph3 x 6 mol H2 = mol/L*s H2 Lx s 4 mol PH3 8

9 Rate laws show how the rate of a rxn depends on the [reactants]. Except for an elementary rxn (one in which the balanced equation represents the mechanism), the rate law cannot be determined from the balanced equation For the rxn: 2NO 2(g)  2NO (g) + O 2(g) The rate law can be written as: Rate = k[NO 2 ] n The proportionality constant, k, is called the rate constant and is determined by experiment. For a given rxn at a given temperature, this value is constant. Its units depend on the order of the reactants. 9

10 The order, n, of the reactant must also be determined by experiment. It is the power to which the [reactant] must be raised in the rate law. If the rxn A  B is first order, then the rate law is Rate = k[A]; doubling the [reactant] doubles the rate of the rxn. If the rxn is second order, the Rate = k[A] 2 ; doubling the [reactant] will result in the rate quadrupling. If the [reactant] is changed and the rate is not affected, the order of the reactant is zero. The rate law would be Rate= k[A] 0 10

11 One way to determine the rate of a reaction at a particular time, the instantaneous rate, is to plot the [reactant] vs time and take the slope of the tangent to the curve at time t. If the slopes of tangents to the curve at two different [] are calculated, the rate law of a rxn can be determined by comparing the Δ rate to the Δ [concentration]. In the graph above, when the [reactant] is halved, the rate is also halved. The rxn is first order. The rate law for the rxn is: Rate=k[reactants] 11

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13 The initial rate of a rxn is the instantaneous rate just after the rxn begins (just after t = 0 and before the initial [reactants] have changed) The rate law of a rxn can be determined by performing a few trials with different [reactants] and measuring the initial rate for each trial. To find the order of one reactant, chage its concentration while holding the concentration of the other reactants constant. 13

14 Determine the rate law and the value of the rate constant for the reaction at -10*C: 2NO(g) + Cl2(g)  2NOCl(g) 14 Trial[NO]0(mol/ L) [Cl2]0 (mol/L) Initial rate (mol/L*s) First, write the general form of the rate law: Rate = k[NO] x [Cl2] y -Your goal is to determine x and y -To find x, the order with respect to NO, pick two trials in which NO changes and Cl2 remains the same. (compare trial 3 to 2) -write a ratio of the rate law of trial 3 to the rate of the rate law For trial 2. Substitute the values for the rates and known [] and Solve for the order, x Rate (trial 3) = k[NO] x [Cl2] y  1.45 = (0.20) x  4.0 = 2 x  x = 2.0 Rate (trial 2) k[NO] x [Cl2] y 0.36 (0.10) x The order, x, with respect to NO is 2. This means that when the [NO] Is doubled, the rate quadruples. Not that the rate constant and the [Cl2] Cancel out because they are the same in trials 2 and 3. To find y, the order with respect to Cl2, pick two trials in which Cl2 Changes and NO remains the same. So compare trials 2 and 1 Rate(trial 2) = k[NO] x [Cl2] y  0.36 = (0.20) y  2.0 = 2 y  y = 1 Rate(trial 1) k[NO] x [Cl2] y 0.18 (0.10) y The order, y, with respect to Cl2, is 1. This means that when the [Cl2] is Doubled, the rate also doubles. Note that the rate constant and the [NO] Cancel out because they are the same in Trials 1 and 2. - thus, the rate law for this rxn is Rate= k[NO] 2 [Cl2] The overall order of a rxn is the sum of the reaction orders. For this example, the overall order is = 3 -To determine the value of the rate constant, including its units, Use the rate law and experimental data from any given trial. - k = Rate/([NO] 2 [Cl2]) Using the values from trial 1: = (0.18M/s)/(0.10M) 2 (0.10M) = 180L 2 /(mol 2 s)

15 An integrated rate law, derived from the differential rate law, expresses the [reactant] as a function of time. The table summarizes the integrated rate laws for the rxn: A  Products 15 Order: zeroOrder: firstOrder: second Rate lawRate=kRate = k[A]Rate=k[A]2 Integrated[A]=-kt+ln[A]0Ln[A]=-kt+ln[A]01/[A]=kt+(1/[A]0) Straight line plot[A] vs tln[A] vs t1/[A] vs t Relationship of Rate constant to slope Slope = - k Slope = k Half-lifeT1/2=[A]0/2kT1/2=0.693/kT1/2=1/k[A]0 You can determine the order of the reactant graphically if you know The [A] at various times, t, during the rxn. For example, if you plot ln[A] vs t and obtain a straight line, the rxn is first order in A. If the Graph is not linear, then the rxn is not first order. The table above Summarizes what is graphed to test for the order of the reactant.

16 Example #3: The rate of the reaction NO2(g) + CO(g)  NO(g) + CO2(g) Depends only on the concentration of nitrogen dioxide below 225*C. At a temperature below 225*C, the following data were collected. 16 Time (s)[NO2](M)Ln[NO2]1/[NO2]/M x10^ x10^ x10^ x10^ x10^ Assume that the data are first order and see if the plot of ln[NO2] vs time Is linear. If this isn’t linear, try the second-order plot of 1/[NO2] vs time.

17 The plot of 1/[NO2] vs time should be the linear plot. So the slope of this line gives the value of k Slope = k = Δy = (5.75 – 2.00)M -1 = 2.08x10 -4 L/mol*s Δx (1.8x10 4 )s To determine [NO2] at 2.70x10 4 s, use the integrated rate law where 1/[NO2]0 = 1/0.500M = 2.00 M -1 1/[NO2] = kt + 1/[NO2]0 1/[NO2] = 2.08x10 -4 L x 2.70x10 4 s M-1 mol s  1/[NO2] = 7.62  [NO2] = M 17

18 A mechanism is a series of steps by which a rxn occurs. The elementary steps, each step in the mechanism, must add up to give the overall balanced equation for the mechanism. The slowest step in the mechanism must agree with the experimentally determined rate law. Once the rate law has been determined experimentally, then chemists propose several mechanisms consistent with that rate law and design experiments to determine which matches the data. You will have to know how to identify catalysts and intermediates in a mechanism and write the overall equation for a mechanism. You will also need to be able to write a rate law from a given mechanism. 18

19 Consider the balanced equation for the hypothetical reaction: 2A + 2B  2C + D2 The experimentally determined rate law is Rate= k[A]2[B]. The two following mechanisms were proposed for this reaction: I. A + B  E + C (slow) E + A  E2 (fast) E2 + B  D2 + C (fast) II. A + B  E (fast, equilibrium) E + A  E2 + C (slow) E2 + B  D2 + C (fast) Identify the presence of catalysts or intermediates in each of the mechanisms. In the two mechanisms above, there are no catalysts. 19

20 An intermediate is produced in one elementary step and then consumed in the next. It also does not appear in the overall reaction for the mechanism because it cancels out. In both mechanisms, E is an intermediate. It is produced in the first step and consumed in the second. Which of the two mechanisms is consistent with the rate law? For the proposed mechanism to be correct, the overall reaction and the rate law for the mechanism must agree with the given reaction and the experimentally determined rate law. Write the overall react for each mechanism. Add up the elementary steps. Cancel out the catalysts and intermediates that appear. 20

21 I. A + B  E + C (slow) E + A  E2 (fast) E2 + B  D2 + C (fast) 2A + 2B  2C + D2 is the overall rxn II. A + B  E (fast, equilibrium) E + A  E2 + C (slow) E2 + B  D2 + C (fast) 2A + 2B  2C + D2 is the overall rxn The overall reaction for each mechanism matches the given reaction. 21

22 Determine the rate law from the Mechanism Begin by writing the rate law for the slow step of each mechanism The rate law for the slow step of mechanism I is Rate=k[A][B] which is not consistent with the rate law determined by experiment. Mechanism I is not possible for this reaction. The rate law for the slow step of mechanism II is Rate = K[E][A]. [E] is an intermediate and may not be included in the rate law. Because the first step in the mechanism is reversible and fast, the rate of the forward reaction equals the rate of the reverse reaction. Rate (forward) = k f [A][B] Rate (reverse) = k r [E] Rate (forward) = Rate (reverse), therefore, K f [A][B] = k r [E]  [E] = (k f /k r )[A][B]  k’[A][B] Substitute for [E] in the rate law for the slow step in mechanism II to get Rate=kk’[A] 2 [B] or k”[A] 2 [B] where k’ is the rate constant of the experimentally determined rate law Mechanism II is possible because its rate law agrees with the experimentally determined rate law. 22

23 A catalyst is present in the reactants and then appears in the products. Recall that the catalyst is not consumed in the reaction. The catalyst does not appear in the overall reaction. An example of a mechanism involving a catalyst involves the decomposition of ozone by atoms of chlorine. Cl (g) + O 3(g)  ClO (g) + O 2(g) O (g) + ClO (g)  Cl (g) + O 2(g) O (g) + O 3(g)  2O 2(g) So Cl atoms are the catalysts. 23

24 Activation energy can be determined using the Arrhenius equation: ln(k) = - Ea x 1 + ln[A] R T The rate constant is represented by k Ea is the activation energy R is the gas constant, J/(molK) T is the Kelvin temperature One way to determine the activation energy, Ea, is to measure the rate constant, k, at several different temperatures, and then graph ln(k) vs 1/T which gives a straight line with the slope equal to –Ea/R 24

25 Activation energy can also be calculated from the values of k at only two temperatures using the equation: ln(K2/K1) = Ea/R(1/T1 – 1/T2) Example #5: A first order reaction has constants of 4.6x s -1 and 8.1x10 -2 s -1 at 0*C and 20*C, respectively. Calculate the value of the activation energy. ln[(8.1x10 -2 s -1 )/(4.6x10 -2 s -1 ) = Ea/(8.3145J/molK) * [1/273K – 1/293K] 0.57 = Ea/8.3145(2.5x10-4) Ea = 1.9x104 J/mol = 19 kJ/mol 25

26 ‘This was child’s play…’ 26

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