Rates of Reactions

1. Rates of Reactions

2. Reaction Rates • The amount of time required for a chemical rxn to come to completion can vary tremendously • When you strike a match it flames up instantly • Coal is made over millions of years from very slow chemical reactions • Chemists find it useful, although difficult, to study a reactions progress over a period of time, which is called Kinetics.

3. Reaction Rates • A reaction rate measures how quickly the reactant is consumed as it coverts into product • Rates measure the speed of any change that occurs within an interval of time • The interval of time may range from fractions of a second to centuries • Rates of chemical change usually are expressed as the change in conc of reactant over time.

5. Collision Theory: Orientation • Rates of chemrxns are related to the properties of atoms, ions, and molecules through a model called collision theory • According to collision theory, atoms, ions, and molecules can react to form products when they collide • provided that the particles have enough kinetic energy • And that the molecules are oriented properly

7. Collision Theory: Energy • The minimum amount of energy that the particles or reactants must have in order to react is called the rxn’sactivation energy. • In a sense the activation energy is a barrier that reactants must get over to be converted to products • The higher the barrier the larger the investment of energy in order to get the rxn to proceed • And typically the slower the rxn

8. Activated complex

9. Collision Theory: Progress • During a rxn, a particle that is neither reactant nor product forms momentarily, called an activated complex • if there is sufficient energy • and if the atoms are oriented properly • An activated complex is a kind of transition molecule which has similarities to reactants & products • An activated complex is the arrangement of atoms at the peak of the activation-energy barrier.

11. Collision Theory • Collision theory explains why some naturally occurring rxns are immeasurably slow at room temp. • Carbon and Oxygen react when charcoal burns, but this reaction has a high activation energy • At room temp, the collisions of oxygen and carbon molecules aren’t energetic enough to react • But the rxn can be helped along a number of ways

12. Manipulating the Rate • It is possible to vary the conditions of the rxn, the rate of almost any rxn can be modified • collision theory can help explain why the rates can be modified • Several strategies can be used: • Increase the temperature • Increase the concentration • Decrease the particle size • Employ a catalyst

13. Temperature • Increasing the temp speeds up the rxn, while lowering the temp slows down the rxn • Increasing the temp increases the frequency of the collisions • Collisions taking place more often more likely they are to stick • And the extra energy increases the power of the collisions • Also increasing the likelihood of a successful collision

14. Just sitting out, charcoal does not react at a measurable rate • However, when a starter flame touches the charcoal, atoms of reactants collide with higher energy and greater frequency • Some of the collisions are high enough in energy that the product CO2 is formed • The energy released by the rxn then supp-lies enough energy to get more C and O2 over the activation-energy barrier • Evidence of this would be if you remove the starter flame, the rxn will continue on its own.

15. Concentration • The more reacting particles you have in a given volume, the higher the rate of rxn. • Cramming more particles into a fixed volume increases the concentration of reactants, • Increasing the concentration, increases the frequency of the collisions, and therefore increases the reaction rate.

16. Particle Size • The smaller the particle size, the larger the surface area for a given mass of particles • The total surface area of a solid or liquid reactant has an important effect on the rate of reaction. • An increase in surface area increases the amount of the reactant exposed for collision to take place… • Which increases the collision frequency and the reaction rate.

17. Particle Size • One way to increase the surface area of solid reactants is to dissolve them • which separates the particles and makes them more accessible to other reactants. • Grinding solids into a fine powder also increases the surface area of reactants • Small dust-like particles can be very dangerous, can be highly explosive

18. Catalyst • An increase in temp is not always the best way to increase the rate of rxn • A catalystis often better. • A catalyst is a substance that increases the rate of a rxn without being changed during the rxn • They permit rxns to proceed at lower energy than is normally required • With a lower activation energy more reactants can form products in a given amount of time.

20. Catalyst

21. Catalyst • Since catalysts are not consumed during a rxn, they do not appear as reactants or products in the chemeqn • Often written above the rxn arrow(s) • Catalysts are crucial for many life processes. • Your body temp is only 37°C and cannot be raised significantly without danger • Without catalysts, few rxns in the body would proceed fast enough at that temp • Enzymes, biological catalysts, increase the rates of biological rxns

22. When you eat a meal containing protein, enzymes in your digestive tract break down the protein molecules in a few hrs.. • Without enzymes, the digestion of proteins at 37C takes yrs • An inhibitor is a substance that interferes with the action of a catalyst • An inhibitor could work by reacting with or “poisoning” the catalyst itself

24. Rate Laws • The rate of a rxn depends in part on the concentration of the reactants • Concentration is a measure of how much stuff is available to react • For a rxn in which reactant A reacts to form product B in 1 step, you can write a simple rxneqn: A  B • The speed that A forms B is dependent on how the conc of A changes over time • As the conc of A decreases the rate of the rxn generally decreases

25. Rate Laws • You can express the rate as the disappearance of A (DA) with respect to the change in time (Dt) • The rate of disappearance of A is there-fore, proportional to the concentration or molarity (# ofmoles/Liter) of reactant A • This proportionality can be expressed as a constant (k) multiplied by the concentration of reactant A • m is a category that the rxn fits in called an order k[A]m

26. Rate Laws • This mathematical expression is called a differentialrate law or just, rate law • A rate law is an expression which relates the rate of a rxn to the conc of reactants • The magnitude of the rate constant (k) depends on the conditions at which the rxn is conducted • If reactant A reacts to form product B quickly, the value of k will be large • If reactant A reacts to form B slowly, the value of k will be small

27. Rate Laws • Rxns are classified as either zero-order, first-order, second-order, or mixed order (higher order) rxns. • The rate of chemical rxns and the size of the rate constant (k) is dependent on the “order” of the rxn • Zero-Order Rxns • (m = 0) have a constant rate. This rate is independent of the conc of the reactants. The rate law is: k, with k having the units of M/sec. • The half-life of a zero-order reaction steadily decreases

29. Rate Laws • First-Order Reactions • (m = 1) has a rate proportional to the conc of one of the reactants. A common example of a first-order rxn is the phenomenon of radioactive decay. The rate law is: k[A]1 (or B instead of A), with k having the units of sec-1. • The half life is a constant.

31. Rate Laws • Second-Order Reactions • (order = 2) has a rate proportional to the conc of the square of a single reactant or the product of the conc of two reactants that are first-order each. • Rate law =k[A]2 (or substitute B for A or k multiplied by the concentration of A, [A], times the concentration of B, [B]), with the units of the rate constant M-1sec-1 • The half life steadily increases for second-order reactions

33. Determining Rate Laws • Rate laws can only be determined experimentally. • One way to experimentally determine the order and the rate constant is to determine a class of rate laws called Integrated Rate Laws. • Integrated Rate Laws are determined through graphical analysis. • Integrated rate laws can help us write the rate law for a reaction

34. Determining the Rate Law • We first gather data on how the [A] changes over time • We then graph the data 3 different ways • And we look for a pattern with the data • The Zero order integrated rate law shows that its rate is independent of the [A] • Where graphing [A] vs. time is a straight line with a slope of -k Integrated Rate Law: Zero Order

36. Integrated Rate Law: First Order • The first order integrated rate law can be used to determine [A] at any time. • Where graphing the ln[A] vs. time produces a straight line • If the graph is a straight line, then the rxn has first-order kinetics and it’s rate constant (k) = slope x (-1)

38. Integrated Rate Law: Second Order • The second-order integrated rate law is be used to determine the [A] at any time. • If we achieve a straight line when the inverse of [A] is graphed vs. time • If 1/[A] vs time produces a linear function, the kinetics is 2nd order and the rate constant is the slope of the line

41. Example: Data was collected for the rxn SO2Cl2 SO2 + Cl2: Determine the rate law for the reaction. Determine the time required for the SO2Cl2conc to drop from 0.100 mol/L to 0.050 mol/L. How long does it take for the conc to drop from 0.050 mol/L to 0.025 mol/L?

42. Graph the data 3 different ways Choose the one that makes a straight line

43. Example: The ln[SO2Cl2] vs time produced a straight line when graphed. So the rxn is a first order rxn which fits the integrated law: ln[SO2Cl2] = -kt + ln[SO2Cl2]0 This indicates a 1st order rxn, so our rate law is so far k[SO2Cl2]1 k can be determined by negative of the slope, according to the regression line our rate law ends up as: 0.002 L/mol•sec [SO2Cl2]1 The ½ life equation that fits the integrated law for the first order rxn is… t1/2=0.693/k

44. Example: So the time it takes to go from 0.1 mol/L to 0.05 mol/L is calculated by… t1/2 = 0.693/0.002 = 345.6 sec To go from 0.05 mol/L to 0.025 mol/L it takes another 345.6 sec, because the half-life of a 1st order reaction is constant

45. Rate Laws • In some kinds of rxns, such as double replacement, 2 substances react to give products • The coefficients in the eqn for such a rxn can be represented by lower-case letters: aA + bB  cC + dD • For a 1 step rxn of A+B, the rate of rxn is dependent on the concentrations of reactants A & B • It’s rate law would follow the eqn: Rate = k[A]a[B]b

46. Rate Laws • When each of the exponents a & b in the rate law equals 1, the rxn is said to be 1st order in A, 1st order in B, & 2nd order overall • The overall order of a rxn is the sum of the exponents for the individual reactants • The rate law gives a chemist valuable information about the mechanism of a reaction • The mechanism of the rxn is the path a rxn might take to produce products

47. Reaction Mechanisms • A graph can be produced to track energy changes verses time, which is called a rxn progress curve • The simplest curve and therefore, simplest rxn would be a one-step, elementary rxn • Reactants form products in a single step • 1 activated complex • 1 energy peak

48. Reaction Mechanisms • For a more complex rxn, or a higher order rxn, the rxn progress curve resembles a series of hills & valleys • The peaks correspond to the energies of the activated complexes • Each valley represents an intermediate productwhich becomes a react of the next stage of the rxn • Intermediates have a significant lifetime compared with an activated complex • They have real ionic or molecular structures and some stability

49. Reaction Mechanism • Intermediate products do not appear in the overall eqn for a rxn • For example in the following overall rxn: H2(g) + 2ICl(g) <==> I2(g) + 2HCl(g) • This rxn is believed to be a 2 part rxn • There is an intermediate reaction in between the reactants and products. 1) H2(g) + 2ICl(g) ICl(g) + HCl(g) + HI(g) 2) ICl(g) + HCl(g) + HI(g)  I2(g) +2HCl(g)

50. Reaction Mechanism