1 / 102

Chemistry 1011 Y8Y,U Paul G. Mezey

Chapter 13: Chemical Kinetics. Chemistry 1011 Y8Y,U Paul G. Mezey. Reaction Rates. Reaction rate is concentration change divided by time change Reaction rate = D [X] / D t. Changes! Final minus initial. D [X] = [X] final – [X] initial D t = t final – t initial.

Télécharger la présentation

Chemistry 1011 Y8Y,U Paul G. Mezey

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 13: Chemical Kinetics Chemistry 1011 Y8Y,U Paul G. Mezey

  2. Reaction Rates • Reaction rate is concentration change divided by time change • Reaction rate= D[X] /Dt Changes! Final minus initial D[X] = [X]final – [X]initial Dt = tfinal – tinitial We most often use molL-1as units of concentration This means that rate often has units molL-1s-1

  3. Reaction Rates • Reaction rate is concentration change divided by time change • Reaction rate= D[X] /Dt Changes! Final minus initial D[X] = [X]final – [X]initial Dt = tfinal – tinitial We most often use molL-1as units of concentration This means that rate often has units molL-1s-1

  4. Reaction Rate • Thereaction rateis defined either as theincrease in the concentration of a productover time, or thedecrease in the concentration of a reactantover time. A + B  C + D rate = -D[A] / Dt = -D[B] / Dt = +D[C] / Dt = +D[D] / Dt Rate is always positive, so we must put negative signs in front of reactant concentration changes!

  5. 2 N2O5 (g)  4 NO2 (g) + O2 (g)

  6. Be Careful! 2 N2O5 (g)  4 NO2 (g) + O2 (g) Between 300 and 400 seconds: Rate of decomposition of N2O5 = -D[N2O5]/ Dt = -(0.0101 molL-1 – 0.0120 molL-1) / (400 s – 300 s) = 1.9 x 10-5 mol(L·s)-1

  7. Be Careful! 2 N2O5 (g)  4 NO2 (g) + O2 (g) Between 300 and 400 seconds: Rate of formation of NO2 = +D[NO2]/ Dt = +(0.0197 molL-1 – 0.0160 molL-1) / (400 s – 300 s) = 3.7 x 10-5 mol(L·s)-1

  8. Be Careful! 2 N2O5 (g)  4 NO2 (g) + O2 (g) Between 300 and 400 seconds: Rate of formation of O2 = +D[O2]/ Dt = +(0.0049 molL-1 – 0.0040 molL-1) / (400 s – 300 s) = 9 x 10-6 mol(L·s)-1

  9. . • Average reaction rate • Slopes • Time = 0

  10. 2 N2O5 (g)  4 NO2 (g) + O2 (g) • The three values for rate that we calculated are not the same! • Why? • We have different molar amounts. • But therelative rates ARE THE SAME!

  11. 2 N2O5 (g)  4 NO2 (g) + O2 (g) • The relative rate of formationofO2 is • (1/1)9 x 10-6 mol(L·s)-1=9 x 10-6 mol(L·s)-1 • The relative rate of formationofNO2 is • (1/4)3.7 x 10-5 mol(L·s)-1=9.3 x 10-6 mol(L·s)-1 • The relative rate of decompositionofN2O5is • (1/2) 1.9 x 10-5 mol(L·s)-1=9.5 x 10-6 mol(L·s)-1

  12. Instantaneous Reaction Rates • What’s happening at “this instant in time”? We can use instantaneous reaction rates. The initial rate is theinstantaneous reaction ratefor a reactionat time zero.

  13. Problem • Consider the following reaction • 3 I- (aq) + H3AsO4 (aq) + 2 H+ (aq) • → I3– (aq) + H3AsO3 (aq) + H2O (l) • a) If –D[I-]/Dt = 4.8 x 10-4 mol(L·s)-1, what is the value of D[I3-]/Dt during the same time interval? • b) What is the average rate of consumption of H+ during the same time interval?

  14. Be Careful! A reactantmight not affect the rate, regardless of its concentration. Rate Laws and Reaction Order • The rate of a chemical reaction depends on the concentration of some or all of the reactants.

  15. Rate laws • The rate lawfor a reaction • is the equation showing the dependence of thereaction rateon the • concentrations of the reactants.

  16. aA + bB  products • Rate =k [A]m[B]n • k is a constant for the reaction at a given temperature, and is called the rate constant. Be Careful! m does not have to equal a n does not have to equal b

  17. “Sensitivity” to concentration change

  18. Reaction order • Reaction order • with respect to a given reactant • is the value of the exponent of the rate law equation for the specific reactant only. • The overall reaction order is the • sum of the reaction orders for • all reactants.

  19. Reaction order example • rate = k [A]2[B] • The reaction order with respect to A is 2 • or the reaction is second order in A • The reaction order with respect to B is 1 • or the reaction is first order in B • The overall reaction order is 3 (2 + 1 = 3) • or the reaction is third order overall

  20. Problem • Consider three reactions with their given rate laws below. What is the order of each reaction in the various reactants, and what is the overall reaction order for each reaction?

  21. Experimental Determination of a Rate Law • Reaction rate lawscan only be determined experimentally! • We most commonly carry out a series of experiments in which the • initial rate of the reaction is measured • as a function of • different initial concentrations of reactants

  22. Method of initial rates If you see a table like this with chemical concentrations or pressures and rate data, chances are good the question is a method of initial rates problem.

  23. Method of initial rates • Focus on the chemicals in the TABLE. • You always require at least one more experimental reaction than your number of chemicals given in your table! • Sometimes we are given a table with an extra experiment which we can use to check if we’ve done everything correctly.

  24. 2 NO (g) + O2 (g)  NO2 (g) • Since rate laws are always expressed in terms of reactants (and sometimescatalysts – we’ll see these later), lets create a general form of the rate law for this reaction based on what chemicals the TABLE tells us are involved in the rate of the reaction.

  25. 2 NO (g) + O2 (g)  NO2 (g) • rate =k [NO]m[O2]n

  26. Method of initial rates • For our initial reactant order determination we need to choose a pair of reactions where only one reactant concentration changes. Experiments #1 and #2 fulfill this condition.

  27. Method of initial rates rate =k [NO]m[O2]n • Since k is a constantthen • k for experiment 1 • IS EQUAL TO • k for experiment 2! • k =rate / [NO]m[O2]n

  28. Reaction order w.r.t. NO

  29. Reaction order w.r.t. O2

  30. Our rate law rate =k [NO]2[O2]1

  31. Rate constant using experiment 1 k =rate / [NO]2[O2]1

  32. Rate constant using experiment 2 k =rate / [NO]2[O2]1 The rate constant is the same, as it should be!

  33. Check using extra experiment rate =(1.42 x 104 M-2s-1) [NO]2[O2]1 The rate is the same as the experimentally observed rate (within rounding errors). We MUST have done everything right!

  34. Units of rate constants • Rate always has the units • mol(L·s)-1 • To ensure we get the right units for rate means the rate constant must have different units depending on the overall reaction order. Be Careful!

  35. Problem • H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) •  I3- (aq) + 2 H2O (l) • D[I3-]/Dt can be determined by measuring the rate of appearance of the colour.

  36. Problem a) What is the rate law for the formation of I3-? b) What is the value for the rate constant? • c) What is the initial rate of formation of triiodide when the concentrations are [H2O2] = 0.300 molL-1 and [I-] = 0.400 molL-1?

  37. Reaction Rates and Temperature • Increasing the temperature increases a chemical reactions rate. • In general, reaction rates approximately double if you increase the temperature by 10 °C.

  38. Bumper cars A gasp of surprise could be a “reaction” when riding in a bumper car. “Reactions” occur ONLY when the bumps are “very hard”and occur “from behind”.

  39. Collision theory • A + BC  AB + C • If this reaction occurs in a single step, then at some point in time, the B-C bond starts to break, while theA-B bond starts to form. • At this point,all three nuclei are weakly linked together.

  40. Collision theory • Molecules tend to repel each other when they get close. • We must insert energy to force the molecules close together. This is like forcing together the north poles of two magnets. • This inserted energy is the kinetic energy of the molecules. It becomes potential energy as the molecules get closer.

  41. Collision theory • A---B---C‡ • has a higher potential energy than either • A + B-C orA-B + C • A---B---C‡ • is thetransition state • or theactivated complex

  42. Figure on Reaction Barrier

  43. Figure There are two useful energy differences in the Figure. The difference in energy betweenproductsand reactants isDH The difference in energy between thetransition stateand thereactantsis Ea – the activation energy

  44. Activation energy • The activation energy (Ea)of a reaction is thewill always be positive! The energy of collision between two molecules must be AT LEASTas big asEaotherwise we cannot make it to the transition state.

  45. Collisions between molecules at higher temperatures are more likelyto have collision energy GREATER THAN theactivation energy. • Higher temperaturesmean higher rates of reaction!

  46. Collisions • An individual molecule collides with other molecules about once every billionth of a second (one billion collisions per second). • If every collision was successful in creating products, then every reaction would be almost instantaneous. This is not the case. • Not every collision breaks theactivation energy barrier!

  47. Collisions • The fraction of collisions that have enough energy to break the activation barrier is given by • f=e-Ea/RT e is approximately 2.7183, Ea is the activation energy, T is the temperature in Kelvin, R is the gas law constant (8.314 JK-1mol-1) Be Careful!

  48. .

  49. Bumper cars and energy • bumper car - a more energetic collision is more likely to make us gasp (our “reaction”) • molecular collisions -higher energy collisions are more likely to lead to reaction • (by overcoming the activation energy)

  50. Bumper cars and orientation • You are also more likely to gasp if you are hit from behind by another bumper car. The orientation of how the collision occurs is also important to get a “reaction.” • The same is true for molecules where the fraction of collisions that have the right orientation isp. We call this fraction p the steric factor.

More Related