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14.1 Factors that affect reaction rates

Chemical kinetics: The area of chemistry that is concerned with the speeds or rates of reactions is called chemical kinetics. 14.1 Factors that affect reaction rates.

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14.1 Factors that affect reaction rates

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  1. Chemical kinetics: The area of chemistry that is concerned with the speeds or rates of reactions is called chemical kinetics 14.1 Factors that affect reaction rates • 1.The physical state of reactants: Reactant must come together to react. The more readily molecules collide with each other, the more rapidly they react. • 2.The concentrations of reactants: As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. • The temperature at which the reaction occurs: The rates of chemical reactions increase as temperature is increased. At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. • 4.The presence of catalyst: Catalysts are agents that increase reaction rates without being used up

  2. 14.2 Reaction rates Let’s consider a simple hypothetical reaction: A B Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

  3. The rate of the reaction can be expressed either as the rate of disappearance of reactant A or as the rate of appearance of product B. Change in concentration of B Average rate of appearance of B = Change in time [B] at t2 [B] at t1 [B] = = t2 t1 t Average rate of appearance of B after 20 sec. 0.46 M 0.0 M = 2.3  10-2M /s = 20 s  0 s [A] Average rate of disappearance of A = t Average rate of disappearance of A after 20 sec. 0.54 M 1.0 M = 2.3  10-2M /s = 20 s  0 s

  4. Example: For the reaction pictured, calculate the average rate of appearance of B over the time interval from 0 to 40 s. [B] Average rate of appearance of B = t 0.70 M  0.0 M = 40.0 s 0.0 s 0.70 M = 40.0 s = 1.8  10-2 M /s

  5. Change of rate with time C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • Note that the average rate decreases as the reaction proceeds. • This is because the concentration of reactants decreases and as the reaction goes forward, there are fewer collisions between reactant molecules.  [ C4H9Cl ] average rate =   t 0.0905  0.1000 =  = 1.9  10-4M /s 50.0  0.0 s

  6. Instantaneous rate C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • A plot of concentration vs. time for this reaction yields a curve like this. • The slope of a line tangent to the curve at any point is the instantaneous rate at that time.  [ C4H9Cl ] Instantaneous rate =   t 0.017  0.042 =  = 6.3  10-5M /s (800  400) s

  7. aA + bB cC + d D Reaction rates and stoichiometry • What happen if the stoichiometric relationship are not one to one ? 2 HI(g) H2(g) + I2(g) • 2 mol of HI disappear for each mole of H2 or I2 that forms  [HI] Rate =  = =  t 1 • To generalize, for the reaction  [H2]  [I2] 2  t  t Reactants (decrease) Products (increase)

  8. Example: (a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction (b) If the rate at which O2 appears, [O2] /t, is 6.0  10-5M /s at a particular instant, at what rate is O3 disappearing at this same time -[O3] /t ? 2O3(g)  3O2(g) (a) Rate =  [O3]  [O3] 2 1 2 ( 6.0  10-5M /s )   1 = = = 3 3 3  t  t  [O2]  [O2] 2  t  t (b) = 4.0  10-5M /s

  9. Example: The decomposition of N2O5 proceeds according to the following equation: If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2  10-7M /s, what is the rate of appearance of (a) NO2, (b) O2 ? 2N2O5(g)  4NO2(g) + O2(g) Rate =  [N2O5]  [N2O5]  [N2O5] 1    4 1 1 = = 4  t  t  t (a)  [NO2]  [O2]  [O2]  [NO2] = = 2 (4.2  10-7M /s) 2 2 2  t  t  t  t = 8.4  10-7 M /s (b) = (4.2  10-7 M /s)/2 = 2.1  10-7M /s =

  10. 14.3 The rate law: The effect of concentration on rate Rate Law - A mathematical equation that shows how the rate of a reaction depends on the concentration of reactants NH4+(aq) + NO2(aq) N2(g) + 2H2O(l) Rate = k[NH4+][NO2] • k is a constant, called “rate constant” that has a specific value for each reaction. • The value of k is determined experimentally. • k is independent of reactant concentrations but is dependent on temperature.

  11. For a general reaction, the rate law has the general form reaction rate = k [A]m [B ]n • exponents m and n are determined experimentally, and are typically small whole numbers (usually 0, 1, or 2) • [The rate law exponents are usually not the same as the coefficients in the balanced chemical equation. ] • The rate law for a reaction cannot be predicted from the chemical equation. It must be experimentally determined. a A + b B cC + dD

  12. Concentration and Rate This equation is called the rate law, and k is the rate constant.

  13. Reaction orders: The exponents in the rate law The rate laws for most reactions have the general form Rate = k[reactant 1]m[reactant 2]n ……… The exponents m and n in the rate law are called reaction orders with respect to each reactant • This reaction is • First-order in [NH4+] • First-order in [NO2−] • The overall reaction order can be found by adding (1+1=2) the exponents on the reactants in the rate law. • This reaction is second-order overall.

  14. Example: Consider a reaction for which rate =k[A][B]2. Each of the following boxes represents a reaction mixtures in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction. A + B  C (1) (2) (3) rate =k[A][B]2 Box 1: rate = k(5)(5)2 =125k Box 2: rate = k(7)(3)2 =63k Box 3: rate = k(3)(7)2 =147k Rate in order: box 2 < box 1 < box 3

  15. Units of rate constants The units of rate constant depend on the overall reaction order of the rate law. Reaction that is second order overall Units of rate = (units of rate constant)(units of concentration)2 units of rate units of rate constant = (units of concentration)2 M /s = = M 1s1 M 2 Reaction that is first order Units of rate = (units of rate constant)(units of concentration) units of rate M /s = units of rate constant = = s1 units of concentration M

  16. Example: (a) What is the reaction order of the reactant H2 in the following reaction? (b) What are the units of the rate constant ? rate =k[H2][I2] H2(g) + I2(g) 2 HI(g) (a) rate =k[H2][I2] reaction order of the reactant H2 is first order units of rate (b) units of rate constant = (units of concentration)2 M /s = = M 1s1 M 2

  17. Using initial rates to determine rate laws The rate laws for most reactions have the general form Rate = k[reactant 1]m[reactant 2]n ……… The exponents m and n in the rate law are called reaction orders with respect to each reactant In most reactions the reaction orders are 0, 1 or 2 • If a reaction is zero order in a particular reactant, changing its concentration will have no effect on rate • When a reaction is first order in a reactant, changes in the concentration of that reactant will produce proportional changes in the rate • When a reaction is second order in a reactant, doubling or tripling its concentration increases the rate by a factor 22=4 or 32=9

  18. Concentration and rate NH4+(aq) + NO2(aq) N2(g) + 2H2O(l) Compare experiments 1 and 2: when [NH4+] doubles, the initial rate doubles. First order in [NH4+]

  19. Concentration and rate NH4+(aq) + NO2(aq) N2(g) + 2H2O(l) Likewisecompare experiments 5 and 6: when [NO2] doubles, the initial rate doubles. First order in [NO2]

  20. A + B  C Example: The initial rate of a reaction was measured for several different starting concentrations of A and B, and the results are as follows: Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] =0.100 M • Experiment Initial rate • Number [A] (M ) [B] (M ) (M /s) • 0.100 0.100 4.0  10-5 • 0.100 0.200 4.0  10-5 • 0.200 0.100 16.0  10-5 comparing exp.1 and 2, n = 0 comparing exp.1 and 3 m = 2 (a) Rate = k[A]m[B]n = k[A]2[B]0 K[0.100]m[0.200]n Rate 2 4.0  10-5 M/s Rate 2 = = 2n = = 1 K[0.100]m[0.100]n Rate 1 4.0  10-5 M/s Rate 1 n =0 2n =1 K[0.200]m[0.100]n Rate 3 16.0  10-5 M/s Rate 3 = = 2m = = 4 K[0.100]m[0.100]n 4.0  10-5 M/s Rate 1 Rate 1 m =2 2m =4 Rate = k[A]2[B]0

  21. (b) Rate = k[A]2[B]0 rate K = [A]2 4.0  10-5 M/s = [0.100 M]2 = 4.0  10-3 M-1s-1 (c) Rate = k[A]2[B]0 = (4.0  10-3 M-1s-1)(0.050 M)2 = 1.0  10-5 M/s

  22. Example: The following data were measured for the reaction of nitric oxide with hydrogen: 2NO(g) + 2H2 (g) N2(g) +2H2O(g) (a) Determine the rate law for the reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H2] =0.150 M • Experiment Initial rate • Number [NO] (M ) [H2] (M ) (M /s) • 0.10 0.10 1.23  10-3 • 0.10 0.20 2.46  10-3 • 0.20 0.10 4.92  10-3 comparing exp.1 and 2, n = 1 comparing exp.1 and 3 m = 2 (a) Rate = k[NO]m[H2]n K[0.10]m[0.20]n Rate 2 2.46  10-3 M/s Rate 2 = = 2n = = 2 K[0.10]m[0.10]n Rate 1 1.23  10-3 M/s Rate 1 n =1 2n =2 K[0.20]m[0.10]n Rate 3 4.92  10-3 M/s Rate 3 = = 2m = = 4 K[0.10]m[0.10]n 1.23  10-3 M/s Rate 1 Rate 1 m =2 2m =4 Rate = k[NO]2[H2]

  23. (b) Rate = k[NO]2[H2] rate K = [NO]2[H2] 1.23  10-3 M/s = [0.10 M]2[0.10 M] = 1.2 M-2s-1 (c) Rate = k[NO]2[H2] = (1.2 M-2s-1)(0.050 M)2(0.150 M) = 4.5  10-4 M/s

  24. 14.4 The change of concentration with time First-order reactions: A first-order reaction is one whose rate depends on the concentration of a single reactant raised to the first power A products Rate law of first-order: = k[A]  [A] [A]t Rate =  ln This form of a rate law, which expresses how rate depends on concentration, is called the differential rate law  t [A]0 If the concentration of A at the start of the reaction and at any time t, are [A]0 and [A]t =  kt ln[A]tln[A]0 = ktor ln[A]t = kt + ln[A]0 This form of a rate law is called the integrated rate law

  25. ln[A]t = kt + ln[A]0 This equation has the form of the general equation for a straight line,y = mx +b ln[A]t = k.t + ln[A]0 We can use pressure as a unit of concentration for a gas because from the ideal-gas law the pressure is directly proportion to the number of moles per unit volume y = m.x + b For a first-order reaction, therefore, a graph of ln[A]t versus time gives a straight line with a slope of –k and y-intercept of ln[A]0. A reaction that is not first order will not yield a straight line (b) (a) (a) Variation in the partial pressure of methyl isonitrile (CH3NC) with time during the reaction CH3NC  CH3CN (b) A plat of natural logarithm of the CH3NC pressure as a function of time

  26. Example: The decomposition of a certain insecticide in water follows first-order kinetics with a rate constant of 1.45 yr-1 at 12C. A quantity of this insecticide is washed into a lake on June 1, leading to a of 5.010-7 g/cm3. Assume that the average temperature of the lake is 12C. (a) What is the concentration of the insecticide on June 1 of the following year? (b) How long will it take for the concentration of the insecticide to decrease to 3.010-7 g/cm3 (a) ln[A]tln[A]0 = kt ln[insecticide]t=1 yearln[insecticide]0 = kt ln[insecticide]t=1 yearln[5.010-7] = (1.45 yr -1)(1 yr) ln[insecticide]t=1 year ( 14.51) = 1.45 ln[insecticide]t=1 year = 15.96 [insecticide]t=1 year = e 15.96 = 1.2 10-7 g/cm3 ln[insecticide]tln[insecticide]0 = kt (b) ln[3.010-7] ln[5.010-7] = (1.45 yr -1)(t) (15.02)  (14.51)= (1.45 yr -1)(t) t = 0.51)/( 1.45 yr -1) = 35 yr

  27. Example: The decomposition of dimethyl ether, (CH3)2O, at 510C is a first-order process with a rate constant of 6.8 10-4 s-1. If the initial pressure of (CH3)2O(g) is 135 torr, what is its pressure after 1420 s? (CH3)2O(g) CH4(g) + H2(g) + CO(g) ln[A]t = kt + ln[A]0 ln[(CH3)2O]t=1420s =  (6.8 10-4 s-1)(1420 s) + ln[135] ln[(CH3)2O]t=1420s = 0.9656 + 4.91 = 3.94 [(CH3)2O]t=1420s = e3.94 = 51.4 torr

  28. Second-order reactions: A second-order reaction is one whose rate depends on the reactant concentration raised to the second power or on the concentration of two different reactants, each raised to the first power. A + B products A products  [A] 1 1 1 1 Rate =  Rate law of second-order: = k[A]2 [A]t [A]t [A]0  t [A]0 This form of a rate law, which expresses how rate depends on concentration, is called the differential rate law If the concentration of A at the start of the reaction and at any time t, are [A]0 and [A]t , by integrating  = kt or = kt + This form of a rate law is called the integrated rate law

  29. = k.t + y = m.x + b For a second-order reaction, a graph of 1/[A]t versus time gives a straight line with a slope equal to k and y-intercept of 1/[A]0. 1 1 [A]t [A]0 • One way to distinguish between first- and second-order rate laws is to graph both ln[A]t and 1/[A]t against t. If the ln[A]t ploet is linear, the reactions is first order, if the 1/[A]t plot is linear, the reaction is second order.

  30. Example: The following data were obtained for the gas-phase decomposition of nitrogen dioxide at 300C, NO2(g) NO(g) + 1/2O2(g) Is the reaction is first order or second order in NO2? (a) (b) (a) Not linear, (b) linear, the reaction is second order in NO2

  31. Half-life The half-life of a reaction, t1/2, is the time required for the concentration of a reactant to reach one-half of its initial value, [A]0 1 1 1 1 [A]t = [A]0 For a first-order process, set [A]t = [A]0 in integrated rate equation: 2 2 2 2 1/2 [A]t ln [A]0 ln =  kt1/2 [A]0 0.693 t1/2 = k ln =  kt1/2 =  kt NOTE: For a first-order process, the half-life does not depend on [A]0.  0.693 =  kt1/2

  32. In a first-order reaction, the concentration of the reactant decreases by ½ in each of a series of regular spaced time intervals, namely, t1/2 2 H2O2 → 2 H2O + O2 decomposition of H2O2. 1st order reaction

  33. Example: Radioactive decay is a first order process. Tritium  electron + helium 3H 0-1e3He If you have 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years [A]t = ? [A]0 = 1.50 mg t = 49.2 years [A]t [A]t [A]t Need k, so we calc k from: k = 0.693 / t1/2 k = 0.0564 y-1 ln ln [A]0 [A]0 [A]0 =  kt= - (0.0564 y-1) • (49.2 y) = - 2.77 =  kt = e- 2.77 = 0.0627 Take antilog: [A]t = 0.0627  1.50 mg = 0.094 mg

  34. 14.5 Temperature and Rate • Generally, as temperature increases, so does the reaction rate. • This is because k is temperature dependent. Temperature affects the rate of the chemiluminescence reaction Dependence of rate constant on temperature. The data show the variation in the first-order rate constant for the rearrangement of methyl isonitrile as a function of temperature

  35. The collision model • In a chemical reaction, bonds are broken and new bonds are formed. • Molecules can only react if they collide with each other. • Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. Effective collision Ineffective collision

  36. Activation energy • The minimum energy required to initiate a chemical reaction is called the activation energy, Ea. • The value of Ea varies from reaction to reaction • Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

  37. The particular arrangement of atoms at the top of the barrier is called the activated complex, or transition state. • The high point on the diagram is the transition state. • The species present at the transition state is called the activated complex. • The energy gap between the reactants and the activated complex is the activation energy barrier. Energy profile for methyl isonitrile isomerization

  38. Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. • Thus at higher temperatures, a larger population of molecules has higher energy. • This fraction of molecules that has an energy equal to or greater than Ea is given by the expression: • where R is the gas constant and T is the temperature in Kelvin . The effect of temperature on the distribution of kinetic energies.

  39. Arrhenius Equation Arrhenius developed a mathematical relationship between k and Ea: where K is rate constant, Ea is activation energy, R is gas constant T is the absolute temperature A is the frequency factor, a number that is related to the frequency of collisions and the probability that the collisions are favorably oriented for reaction The reaction rates decrease as Ea increases

  40. Example: Consider a series of reactions having the following energy profiles: • Rank the reactions from slowest to fastest assuming that they have nearly the same frequency factors. • Imagine that these reactions are reversed. Rank these reverse reactions from slowest to fastest. • The lower the activation energy, the faster the reaction. The value of E does not affect the rate • Hence the order is 2 < 3 < 1 (1) Ea =25 kJ/mol ; (2) Ea= 40 kJ; (3) Ea = 15 kJ Hence the order is 2 < 1 < 3

  41. Determining the activation energy Arrhenius equation: • Taking the natural logarithm of both sides, the equation becomes y = mx + b When k is determined experimentally at several temperatures, a plot of ln k vs. 1/T will be a line with a slope equal to Ea /R and a y-intercept equal to lnA

  42. Ea can be evaluated in a nongraphical way if the rate constant of a reaction is known at two temperatures Suppose that at two different teperatures, T1 and T2, a reaction has rate constants k1 and k2, then Ea Ea + LnA Lnk2 = and + LnA Lnk1 = RT2 RT1 Subtracting lnk2 from lnk1 gives ( ( Ea Ea ( ( + LnA Lnk1 Lnk2 = + LnA RT2 RT1 ( k1 Ea 1 1 ( Ln = k2 T1 T2 R

  43. Example: The following tables shows the rate constants for the rearrangement of methyl isonitrile at various temperatures: • From these data, calculate the activation energy for the reaction. • What is the value of the rate constant at 430.0 K ? (a) T (K) 1/T (K-1) lnk 462.9 2.160  10-3 - 10.589 472.1 2.118  10-3 - 9.855 503.5 1.986  10-3 - 7.370 524.4 1.907  10-3 - 5.757 Ea can be determined from the slope of a graph of lnk versus 1/T Temperature (C) k (s-1) 189.7 2.52  10-5 198.7 2.25  10-5 230.3 6.30  10-4 251.2 3.16  10-3

  44. (0.00195, - 6.6) (0.00215, - 10.4) - 6.6 - (-10.4) y = - 1.9  104 = Slope = x 0.00195 - 0.00215 Ea Slope = - Ea = - (slope)(R ) R = - (- 1.9  104)(8.314 J/mol.K) = 15.8  104 J/mol.K = 1.6  102 kJ/mol.K = 160 kJ/mol.K)

  45. Ea = 16 104 J/mol k1= ? k2 = 2.52  10-5 (s-1) T1 =430.0 K T2 = 462.9 K (b) ( ( ( ( ( ( = - 3.18 = e- 3.18 = 4.5 10-2 k1= (4.5 10-2)(2.52 10-5 s-1) ( k1 k1 Ea 1 k1 1 ( 1 1 16 104 J/mol Ln Ln = = k2 T1 T2 R 8.314 J/mol.K 430.0 K 2.52  10-5 s-1 2.52  10-5 s-1 462.9 K = 1 10-6 s-1

  46. Outline: Kinetics

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