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KINETICS. What is Kinetics?  Kinetics is the study of the speed of reactions and is largely an experimental science. Some general qualitative ideas about.

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Presentation on theme: "KINETICS. What is Kinetics?  Kinetics is the study of the speed of reactions and is largely an experimental science. Some general qualitative ideas about."— Presentation transcript:

1 KINETICS

2 What is Kinetics?  Kinetics is the study of the speed of reactions and is largely an experimental science. Some general qualitative ideas about reaction speed may be developed, but accurate quantitative relationships require that experimental data be collected.  For a chemical reaction to occur, there must be a collision between reactant particles. That collision is necessary to transfer kinetic energy needed to break reactant chemical bonds and reform product bonds. If the collision does not provide enough energy, no reaction will occur. The collision must also take place with the proper orientation at the correct place on the molecule, called the reactive site.

3 Five factors that affect rates of chemical reactions:  1. Nature of the reactants: Large, complex molecules tend to react more slowly than smaller ones because statistically there is a greater chance of collisions occurring somewhere else on the molecule, rather than at the reactive site.  2. An increase in temperature increases the kinetic energy, providing more energy transfer and speeding up the number of collisions.  3. The higher the concentration of reactants, the greater the chance of collision and the greater the reaction rate. In gases, the pressure is directly related to the concentration; the greater the pressure, the greater the rate of collision.  4. Reactants in the same physical state react more readily with one another. Gases and liquids tend to react faster than solids because of increased surface area.  5. Catalysts are substances that speed up the reaction rate without affecting the reaction. Catalysts work by reducing the activation energy needed to start a reaction.

4 Orientation Click

5 Types of Rate Laws  Differential rate law shows how the rate is dependent on the concentration.  Integrated rate law shows how the concentrations of species in the reaction depend on time.  Our rate laws will involve only concentrations of reactants.  Experimental convenience usually dictates which type of rate law is determined experimentally.  Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps

6 Rates of ReactionRates of Reaction (3:28)  The rate (speed) of a reaction is related to the change in concentration of either a reactant or product over time.  For the reaction: 2A + B  C + 3D  As the reaction proceeds, the concentrations of A and B will decrease and the concentrations of C and D will increase. Thus, the rate can be expressed as: 1 Δ[A] Δ[B] Δ[C] 1 Δ[D] Rate = - 2 Δt = - Δt = Δt = 3 Δt The negative signs for the reactants indicate their concentrations are decreasing with time. The brackets indicate concentration (molarity). Since the rate of reaction decreases during the course of the reaction, the above calculation will indicate the average rate of reaction over a given time frame, or more commonly, as the initial reaction rate—the rate of reaction at the instant the reactants are mixed.

7 Instantaneous rate of reaction: N 2 O 5  2NO 2 + ½ O 2  The instantaneous rate of reaction is the rate at any point in time and is represented by the instantaneous slope of the curve at that point.  You can determine the instantaneous rate by calculating the slope of the tangent to the curve at the point of interest.  I.R.= - Δ[N 2 O 5 ] =.0019 M = 1.9 x ΔT 100 s  I.R.= + 1 Δ[NO 2 ] =.0037 M = 1.9 x ΔT 100 s

8 Rate equation (law)  Rate of reaction may depend on reactant concentration, product concentration (rare) and temperature.  For the general reaction, aA + bB  cC + dD where the lower case letters are the coefficient and the upper-case letters are the reactants, the rate law is written: Rate = k [A] m [B] n …  k is the rate constant—a constant for each chemical reaction at a given temperature.  The exponents m and n are called the orders of reaction. They indicate what effect a change in concentration of that reactant species will have on the reaction rate.  Example: if m=1 and n=2, that means that if the concentration of reactant A is doubled, then the rate will also double ( [2] 1 = 2), and if the concentration of reactant B is doubled, then the rate will increase fourfold ( [2] 2 = 4). We say that it is first order with respect to A and second order with respect to B. If the concentration of a reactant is doubled and that has no effect on the rate of reaction, then the reaction is zero order with respect to that reactant ([2] 0 = 1).  Many times the overall order of reaction is calculated; it is simply the sum of the individual coefficients, third order in this example.  Rate = k[A][B] 2 where the exponent 1 is understood

9 Reaction Rates  The reaction rate of a chemical reaction is defined as the change in concentration of a reactant or product per unit of time: Concentration of A at time t 2 – concentration of A at time t 1  Rate = t 2 – t 1 Δ[A]  Therefore, Rate = Δt

10  It is important to realize that the rate law (the rate, rate constant, and orders of reaction) is determined experimentally. Do not use the balanced chemical equation to determine the rate law.  Ways to determine rate of reaction can be measured in many ways including taking the slope of the concentration versus time plot for the reaction.  Once the orders of reaction have been determined, it is easy to calculate the rate constant (k).

11 2NO (g) + O 2(g)  2NO 2(g) ExperimentInitial NO Initial O 2 Rate of NO 2 Formation M/s  If the numbers are simple, you can reason out the orders of reaction. From exp. 1 to exp. 2, [NO] is doubled and [O 2 ] stays the same. The rate increased fourfold. This means the reaction is second order with respect to NO. In exp. 1 and 3, the [O 2 ] was doubled, [NO] remained the same and the rate doubled. Therefore, the reaction is first order with respect to O 2 and the rate equation can be written as: Rate = k[NO] 2 [O 2 ]  The rate constant can be determined by substituting the values of the concentrations of NO and O 2 from any of the experiments into the rate equation above and solve for k.

12 Types of Reactions  Zero-order Reaction: rate = k[A] 0  The concentration of the reactant decreases linearly with time and the slope of the line is constant indicating constant rate.  Sublimation is an example of a zero rate reaction. As one layer of particles sublimes, the next layer is exposed and the number of particles available to sublime remains constant.

13 First order reaction  The rate of the reaction is directly proportional to the concentration of the reactant.  Rate = k[A] 1  The rate slows down as the reaction proceeds because the concentration of the reactants decreases.

14 Second-order Reaction  The rate of the reaction is proportional to the square of the concentration of the reactant.  Rate = k[A] 2  For a second-order reaction, the rate is even more sensitive to the reactant concentration. That’s why the curve (rate) flattens out more quickly than it does for a first-order reaction.

15 Determining the Order of a Reaction  The order can ONLY be determined by experimentation so you will need data!  Rate = k[A] m  Therefore: k = rate [A] m [A] (M)Initial rate (M/s)

16 [A] (M)Initial rate (M/s) [A] (M)Initial rate (M/s)

17 What if the numbers aren’t obvious?  If you are unsure about how the initial rate is changing with the initial reactant concentration, or if the numbers are not as obvious as the ones in the preceding examples, you can substitute any two initial concentrations and the corresponding initial rates into a ratio of the rate laws to determine order.  rate 2 = k[A] n  rate 1 k[A] n

18  The reaction A  B has been experimentally determined to be firs order. The initial rate is M/s at an initial concentration of A of 0.100M. What is the initial rate of [A] = 0.500M?  rate 2 = 0.500M  M/s 0.100M  =.0500M/s

19 Reaction order for multiple reactants  aA + bB  cC + dD  rate = k[A] m [B] n  The overall order is the sum of m + n.  For example, the reaction between hydrogen and iodine has been experimentally determined to be first order with respect to hydrogen and first order for iodine.  H 2 + I 2  2HI rate = k[H 2 ] 1 [I 2 ] 1  Thus it is second order overall.

20 NO 2(g) + CO (g)  NO (g) + CO 2(g)  From the data determine:  The rate law for the reaction  The rate constant (k) for the reaction. Experiment [NO 2 ] (M)[CO] (M)Initial rate (M/s)

21  Between the first two experiments the NO 2 concentration doubles and the initial rate quadruples suggesting NO 2 is second-order.  Between 2 nd and 3 rd experiments, the NO 2 stays constant, the CO doubles yet the initial rate remains constant which means CO is zero- order.  Overall rate = = 2  Rate = k[NO 2 ] 2 [CO] 0 = k[NO 2 ] 2  k = rate [NO 2 ] 2 k = M/s (0.10 M) 2 k = 0.21 M· s

22 CHCl 3(g) + Cl 2(g)  CCl 4(g) + HCl (g)  From the data, determine:  The rate law for the reaction  The rate constant (k) for the reaction Experiment[CHCl 3 ] (M)[Cl 2 ] (M)Initial rate (M/s)

23 Algebraically  rate 2 = k[A] m [B] n  rate 1 k[A] m [B] n  1.38 x = k[0.020][0.020] n  6.9 x k[0.020][0.010] n  2= 2 n ** log x n = n log x**  log 2= n · log 2  n = log 2 log 2 n= 1

24  Solve for k:  Rate = k[A] m [B] n  k = rate [A] m [B] n k =.0035 M/s [.010M][.010M] k = 35 M -1 s -1

25 The Integrated Rate Law: The Dependence of Concentration on Time  The integrated rate law for a chemical reaction is a relationship between the concentrations of the reactants and time.  First order Integrated Rate Law is where the rate is directly proportional to the concentration of A: rate = k[A]  -Δ[A] ΔT = k[A] ln[A] t = -kt + ln[A] 0 or ln [A] t = - kt [A] 0 Where [A] t is the concentration of A at any time t, k is the rate constant, and [A] 0 is the initial concentration of A. Notice that the integrated rate law has the form of an equation for a straight line: y = mx + b

26  For a first-order reaction, a plot of the natural log of the reactant concentration as a function of time yields a straight line with a slope of –k and a y-intercept of ln[A] 0. The slope may be negative but the rate constant (k) is always positive.

27 SO 2 Cl 2(g)  SO 2(g) + Cl 2(g) The concentration of SO 2 Cl 2 was monitored at a fixed temperature as a function of time during the decomposition reaction, and the following data was tabulated. Show that the reaction is first order, and determine the rate constant for the reaction. Time (s)[SO 2 Cl 2 ] (M)Time (s)[SO 2 Cl 2 ] (M)

28  If the plot is linear, then it is a first-order reaction.  To obtain the rate constant, fit the data to a line. The slope of the line will be equal to –k.  Since the slope of the best fitting line is x s -1, the rate constant is x s -1.  Now use your graph to predict the concentration at 1900 seconds.

29 Practice SO 2 Cl 2(g)  SO 2(g) + Cl 2(g)  Using the rate constant from the last problem, (+2.90 x s -1 ) if the first- order reaction is carried out at the same temperature, and the initial concentration of SO 2 Cl 2 is M, what will the SO 2 Cl 2 concentration be after 865 s?  Given: k = x s -1  [SO 2 Cl 2 ] 0 = M  Find: [SO 2 Cl 2 ] at t = 865 s  Use first –order integrated rate law.  Equation: ln[A] = -kt + ln[A] 0

30  Solution:  ln [SO 2 Cl 2 ] t = -kt + ln[SO 2 Cl 2 ] 0  ln[SO 2 Cl 2 ] t = -(2.90 x s -1 )865s + ln(0.0225)  ln[SO 2 Cl 2 ] t =  [SO 2 Cl 2 ] t = e  = M

31 Second-Order Integrated Rate Law  Rate = k[A] 2  Since rate = -Δ[A] = k[A] 2 Δ t The above can be integrated to obtain: 1/[A] t = kt + 1/[A] 0 The second-order integrated rate law is also in the form of an equation for a straight line: y = mx + b

32 NO 2(g)  NO (g) + O (g)  The concentration of NO 2 is at a fixed temperature as a function of time. Show by graphical analysis that the reaction is not first order and that it is second order. Determine the rate constant for the reaction. Time (s)[NO 2 ] (M)Time (s)[NO 2 ] (M)

33  The plot should NOT be linear so it is not a first order.  Next prepare a graph of 1/[NO 2 ] versus time.

34  0.05 M/s = k(0.01 M) 2 (0.01 M)  k = 0.05 M/s ÷ (0.01 M) 2 (0.01 M)  k = 5 x 10 4 /M 2 s  **In choosing experiments to compare, choose two in which the concentration of only one reactant has changed while the other has remained constant.  Go to page 8 in your packet.

35 Keywords and Equations ln[A] t – ln[A] 0 = -kt (first order) [A] t [A] 0 = kt (second order) ln 2 = t 1/2 = k k -E a ln k = RT + ln A  T = time (seconds)  E a = activation energy  K = rate constant  A = frequency factor  Gas constant, R = J/ mol K


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