bài giảng hóa bằng tiếng anh Chemical Kinetics hay

Chemical Kinetics Kinetics • kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds • experimentally it is shown that there are factors that influence the speed of a reaction: nature of the reactants, temperature, catalysts, concentration Defining Rate • rate is how much a quantity changes in a given • period of time the speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour)  so the rate of your car has units of mi/hr  distance Speed   time Defining Reaction Rate • the rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time  or product concentration increases • for reactants, a negative sign is placed in front of the definition  concentrat ion Rate   time  [product]  [reactant] Rate    time  time Reaction Rate Changes Over Time • as time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases • at some time the reaction stops, either because the reactants run out or because the system has reached equilibrium at t = [A] = [B] = [C] = at t = [X] = [Y] = [Z] = at t = 16 [A] = [B] = [C] = at t = 16 [X] = [Y] = [Z] = C C2  C1  Rate   t  t1  t Z Z2  Z1  Rate   t  t1  t Rate  Rate  4  0  0.25 16  0 1  0  0.0625 16  0 at t = 16 [A] = [B] = [C] = at t = 16 [X] = [Y] = [Z] = at t = 32 [A] = [B] = [C] = at t = 32 [X] = [Y] = [Z] = C C2  C1  Rate   t  t1  t   X 2  X 1  X  Rate    t  t1  t Rate  Rate   6  4  0.125 16  0 6    0.0625 16  0 at t = 32 [A] = [B] = [C] = at t = 32 [X] = [Y] = [Z] = at t = 48 [A] = [B] = [C] = at t = 48 [X] = [Y] = [Z] =   A 2  A 1  A  Rate    t  t1  t   X 2  X 1  X  Rate    t  t1  t Rate   Rate   0  2  0.125 16  0 5  6  0.0625 16  0 Hypothetical Reaction Red  Blue Time (sec) Number Number Red Blue 100 84 16 10 71 29 15 59 41 20 50 50 25 42 58 30 35 65 35 30 70 40 25 75 45 21 79 50 18 82 in this reaction, one molecule of Red turns into one molecule of Blue the number of molecules will always total 100 the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time Hypothetical Reaction Red  Blue Concentration vs Time for Red -> Blue 100 100 84 90 Number of Molecules 80 71 70 75 59 60 79 82 70 65 50 Number Red Number Blue 58 42 50 50 40 35 30 41 30 25 21 29 20 10 18 16 0 10 15 20 25 Time (sec) 30 35 40 45 50 10 Hypothetical Reaction Red  Blue Rate of Reaction Red -> Blue 4.5 Rate,  [Blue]/ t 3.5 5, 3.2 10, 2.6 2.5 15, 2.4 20, 1.8 25, 1.6 1.5 30, 1.4 35, 40, 45, 0.8 50, 0.6 0.5 0 10 20 30 time, (sec) 40 50 11 Reaction Rate and Stoichiometry • in most reactions, the coefficients of the balanced • equation are not all the same H2 (g) + I2 (g)  HI(g) for these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another  for the above reaction, for every mole of H2 used, mole of I2 will also be used and moles of HI made  therefore the rate of change will be different • in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient [H ] [I ]   [HI] Rate       t t   t 12 Average Rate • the average rate is the change in measured concentrations in any particular time period linear approximation of a curve • the larger the time interval, the more the average rate deviates from the instantaneous rate 13 Hypothetical Reaction Red  Blue Avg Rate Avg Rate Avg Rate (5 sec Number Number intervals) Red Blue Time (sec) 100 84 16 3.2 10 71 29 2.6 15 59 41 2.4 20 50 50 1.8 25 42 58 1.6 30 35 65 1.4 35 30 70 40 25 75 45 21 79 0.8 50 18 82 0.6 (10 sec intervals) (25 sec intervals) 2.9 2.1 2.3 1.5 0.7 14 H2 I2 HI Time (s) 0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 Avg Rate, M/s Avg Rate, M/s [H2], M [HI], M -[H2]/t 1/2 [HI]/t 1.000 0.000 0.819 0.362 0.0181 0.670 0.660 0.0149 0.549 0.902 0.0121 0.449 0.368 0.301 1.102 1.264 1.398 0.0100 0.0081 0.0067 0.247 0.202 1.506 1.596 0.0054 0.0045 0.165 1.670 0.0037 15 0.135 1.730 0.0030 2.000 1.800 concentration, (M) 1.600 1.400 Concentration vs Time for H2 + I2 > 2HI average rate in a given time period =  slope of the line connecting the [H2] points; and ½ +slope of the line for [HI] 1.200 the average rate for the first 10 s is 0.0181 M/s 1.000 [H2], M [HI], M 0.800 0.600 0.400 0.200 0.000 0.000 10.000 20.000 30.000 40.000 50.000 time, (s) 60.000 70.000 80.000 90.000 100.000 16 Instantaneous Rate • the instantaneous rate is the change in concentration at any one particular time slope at one point of a curve • determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function for you calculus fans 17 H2 (g) + I2 (g)  HI (g) Using [H2], the instantaneous rate at 50 s is: Rate    0.28 M 40 s Rate  0.0070 M s Using [HI], the instantaneous rate at 50 s is:   0.56 M Rate      40 s Rate  0.0070 M s 18 Ex 13.1 - For the reaction given, the [I] changes from 1.000 M to 0.868 M in the first 10 s Calculate the average rate in the first 10 s and the [H+] H2O2 (aq) + I(aq) + H+(aq)  I3(aq) + H2O(l) Solve the equation for the Rate (in terms of the change in concentration of the Given quantity) Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value    [I ]   0.868 M  1.000 M  Rate        10 s   t  3 Rate  4.40 10-3 M s    [H ] Rate       t [H  ]  2Rate t   [H  ] M M  2 4.40 10-3   8.80 10-3 s s t