2What 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.
3Five 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.
5Types of Rate LawsDifferential 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
6Rates 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 + 3DAs 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] Δ[D]Rate = Δt = Δt = Δt = ΔtThe 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.
7Instantaneous rate of reaction: N2O5 2NO2 + ½ O2 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[N2O5] = M = 1.9 x 10-5ΔT sI.R.= + 1 Δ[NO2 ] = M = 1.9 x 10-52 ΔT s
8Rate 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 ( 1 = 2), and if the concentration of reactant B is doubled, then the rate will increase fourfold ( 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 (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
9Reaction RatesThe 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 t2 – concentration of A at time t1Rate = t2 – t1Δ[A]Therefore, Rate = Δt
10It 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).
112NO(g) + O2(g) 2NO2(g) Experiment Initial NO O2 Rate of NO2 If the numbers are simple, you can reason out the orders of reaction. From exp. 1 to exp. 2, [NO] is doubled and [O2] stays the same. The rate increased fourfold. This means the reaction is second order with respect to NO. In exp. 1 and 3, the [O2] was doubled, [NO] remained the same and the rate doubled. Therefore, the reaction is first order with respect to O2 and the rate equation can be written as: Rate = k[NO]2[O2]The rate constant can be determined by substituting the values of the concentrations of NO and O2 from any of the experiments into the rate equation above and solve for k.ExperimentInitialNOO2Rate ofNO2Formation M/s10.010.0520.020.2030.10
12Types 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.
13First order reactionThe rate of the reaction is directly proportional to the concentration of the reactant.Rate = k[A]1The rate slows down as the reaction proceeds because the concentration of the reactants decreases.
14Second-order Reaction The rate of the reaction is proportional to the square of the concentration of the reactant.Rate = k[A]2For 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.
15Determining the Order of a Reaction The order can ONLY be determined by experimentation so you will need data!Rate = k[A]mTherefore: k = rate[A]m[A] (M)Initial rate (M/s)0.100.0150.200.0300.400.060
17What 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]nrate k[A]n
18The reaction AB has been experimentally determined to be firs order The reaction AB 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 = 0.500M0.0100M/s M= .0500M/s
19Reaction order for multiple reactants aA + bB cC + dDrate = k[A]m [B]nThe 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.H2 + I2 2HI rate = k[H2]1[I2]1Thus it is second order overall.
20NO2(g)+ CO(g) NO(g) + CO2(g) From the data determine:The rate law for the reactionThe rate constant (k) for the reaction.Experiment[NO2] (M)[CO] (M)Initial rate (M/s)10.100.002120.200.008230.008340.400.033
21Between the first two experiments the NO2 concentration doubles and the initial rate quadruples suggesting NO2 is second-order.Between 2nd and 3rd experiments, the NO2 stays constant, the CO doubles yet the initial rate remains constant which means CO is zero- order.Overall rate = = 2Rate = k[NO2]2 [CO]0 = k[NO2]2k = rate[NO2]2k = M/s(0.10 M)2k = 0.21M· s
22CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) From the data, determine:The rate law for the reactionThe rate constant (k) for the reactionExperiment[CHCl3] (M)[Cl2] (M)Initial rate (M/s)10.0100.003520.0200.006930.013840.0400.027
23Algebraically rate2 = k[A]m[B]n rate1 k[A]m[B]n 1.38 x = k[0.020][0.020]n6.9 x k[0.020][0.010]n2= 2n ** log xn = n log x**log 2= n · log 2n = log 2log 2n= 1
24Solve for k:Rate = k[A]m[B]nk = rate[A]m[B]nk = M/s[.010M][.010M]k = 35 M -1 s-1
25The 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]0Where [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
26For 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.
27SO2Cl2(g) SO2(g) + Cl2(g) The concentration of SO2Cl2 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)[SO2Cl2] (M).100800.0793100.0971900.0770200.09441000.0748300.09171100.0727400.08901200.0706500.08651300.0686600.08401400.0666700.08161500.0647
28If 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 10-4s-1, the rate constant is x 10-4s-1.Now use your graph to predict the concentration at 1900 seconds.
29Practice SO2Cl2(g) SO2(g) + Cl2(g) Using the rate constant from the last problem, (+2.90 x 10-4 s-1) if the first- order reaction is carried out at the same temperature, and the initial concentration of SO2Cl2 is M, what will the SO2Cl2 concentration be after 865 s?Given: k = x 10-4 s-1[SO2Cl2]0 = MFind: [SO2Cl2] at t = 865 sUse first –order integrated rate law.Equation: ln[A] = -kt + ln[A]0
30Solution:ln [SO2Cl2]t = -kt + ln[SO2Cl2]0ln[SO2Cl2]t = -(2.90 x 10-4 s-1)865s + ln(0.0225)ln[SO2Cl2]t =[SO2Cl2]t = e-4.04= M
31Second-Order Integrated Rate Law Rate = k[A]2Since rate = -Δ[A] = k[A]2Δ tThe above can be integrated to obtain:1/[A]t = kt + 1/[A]0The second-order integrated rate law is also in the form of an equation for a straight line: y = mx + b
32NO2(g) NO(g) + O(g)The concentration of NO2 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)[NO2] (M)50050550100600150650200700250750300800350850400900450950
33The plot should NOT be linear so it is not a first order. Next prepare a graph of 1/[NO2] versus time.
340.05 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 104/M2 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.
35Keywords and Equations ln[A]t – ln[A]0 = -kt (first order) [A]t [A]0 = kt (second order) ln 2 = t1/2 = k k -E a ln k = RT + ln AT = time (seconds)Ea = activation energyK = rate constantA = frequency factorGas constant, R = J/ mol K