AP Chem

 Area of chemistry that deals with rates or speeds at which a reaction occurs  The rate of these reactions are affected by several factors

Such as: Concentration Color Bubbles Temp pH Whatever is appropriate: Hours Minutes Seconds

 Candle Wax Example

 Collisions cause reactions! ◦ Breaking of bonds directly linked to rate ◦ Must overcome repulsion of electron clouds  Also correct orientation sometimes needed ◦ Example: Chalk dropping

 Concentration of reactants ◦ With gases, pressure used instead  Temperature at which the reaction occurs  The presence of a catalyst  Surface area of solid or liquid reactants

 Demo: ◦ Chalk + 1 M HCl / Chalk + 1 M Acetic Acid  Prediction?  Factor? ◦ Chalk + 1 M HCl / Chalk + 6 M HCl  Prediction?  Factor? ◦ Chalk + 1 M HCl (Room Temp) / Chalk + 1 M HCl (Heated)  Prediction?  Factor?

 Speed of reaction or reaction rate is the time over which a change occurs  Consider the reaction A  B  Reaction rate is a measure of how quickly A is consumed or B is produced

 Average rate of reaction can be written:  This is a measure of the average rate of appearance of B

 Average rate can also be written in terms of A:  This is the rate of disappearance of A (equal to B only negative)  Average Rates can only be positive

 Start with one mole of A at time zero, measure amounts of A and B at given time intervals

 Data for Reaction A  B

 When mole ratios of equations are not 1:1  For the reaction: aA + bB  cC + dD

 Example ◦ If the rate of decomposition of N 2 O 5 in a reaction vessel is 4.2 x 10-7 M/s, what is the rate of appearance of NO 2 and O 2 2 N 2 O 5(g)  4 NO 2(g) + O 2(g)

 Consider the reaction between butyl chloride and water: C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) +HCl (aq)

 Using the curve created from the data, we can determine the instantaneous rate for any given point on the curve  Recall: slope is rise over run!

 Analogy: Distance between Fall River and Norton is 29.7 mi along a certain route. It takes Mr. N 30 minutes to get to school. His average rate is 59.4 mph.  But at t = 15, Mr. N’s instantaneous rate is 95 mph, and at t = 1 Mr. N’s instantaneous rate is 25 mph.

 Increasing concentration of reactants gives increasing rate  Decreasing rates of reactions over time is typical ◦ Due to decreasing concentration of reactants

 Rates of a reaction can be related to concentrations with a rate constant (k)  For reaction:  Rate laws are defined by reactant (not product) concentrations aA + bB  cC + dD

 For the rate law expression:  The overall order of reaction is the sum of the powers (x + y) ◦ However, rate with respect to [A] is only x

 In most rate laws reaction orders are 0, 1, or 2 ◦ Can be fractional or negative at times ◦ Most commonly 1 or 2  Reaction orders are determined experimentally, and do not necessarily relate to coefficients of a balanced equation

 Example: What is the overall order of reaction for the reaction below? CHCl 3(g) + Cl 2(g)  CCl 4(g) + HCl (g) Rate= k [CHCl 3 ][Cl 2 ] 1/2 A.) ½ B.) 2 C.) 3/2 D.) 2/2

 Zero order for a reactant means concentration changes have no effect on reaction rate ◦ Example: Drinking  1 st order means concentration changes give proportional changes in reaction rate ◦ Double the concentration, double the rate

 2 nd order rate law, increasing in concentration results in a rate increase equal to the concentration increased to the second power ◦ Example:  Double conc. = 2 2 = 4 (rate increase)  Triple conc. = 3 2 = 9 (rate increase)

 The units for the rate constant depend on the order of the rate law Z = overall order of reaction

Rate LawOverall OrderUnits of Rate Constant (k) Rate = kZeroM/s (M s -1 ) Rate = k [A]First1/s (s -1 ) Rate = k [A][B]Second1/M s (M -1 s -1 ) Rate = k [A] 2 [B]Third1/M 2 s (M -2 s -1 )

 What is the unit for the rate constant for the reaction below? CHCl 3(g) + Cl 2(g)  CCl 4(g) + HCl (g) Rate= k [CHCl 3 ][Cl 2 ] 1/2 A.) M ½ /s B.) M/s C.) M 2 /s D.) M -1/2 /s

 A particular reaction was found to depend on the concentration of the hydrogen ion. The initial rates varied as a function of [H + ] as follows: [H + ] (M) Initial rate (M/s) 6.4x x x10 -7  What is the order of the reaction in [H + ]  A.) 1  B.) 2  C.) -1  D.) -1/2

Rate Data for the Reaction Between F 2 and ClO 2 [F 2 ] (M)[ClO 2 ] (M)Initial Rate (M/s) x x x What is the rate law expression for the reaction?

 Rate law tells how rate changes with changing concentrations at a particular temperature  We can derive equations that can give us the concentrations of reactants or products at any time during a reaction (instantaneous) ◦ These are known as integrated rate laws

Rate = k  Using calculus, the integrated rate law is:  [A] t is concentration of reactant at time t  [A] 0 is initial concentration of reactant

 This has the same form as the general equation for a straight line  Graphically, the slope is equal to -k

 Separate equations can be derived relating to time required for reactants to decrease to half of initial concentration (aka half-life or t 1/2 )  When t = t 1/2, [A] t is half of [A] 0 ([A] t =[A] 0 /2)

Rate = k [A]  Using calculus, the integrated rate law becomes:  This equation is of the general equation for a straight line (like before)

 Note that only the second graph is used so that the slope can be determined  Also, y-intercept is ln [A] 0

 Example 2N 2 O 5(g)  4NO 2(g) + O 2(g)  The decomposition of dinitrogen pentoxide is a 1 st order reaction with a rate constant of 5.1 x s -1 at 45ºC. If initial conc. is 0.25M, what is the concentration after 3.2 min.?

 Example #2 2N 2 O 5(g)  4NO 2(g) + O 2(g) The decomposition of dinitrogen pentoxide is a 1 st order reaction with a rate constant of 5.1 x s -1 at 45ºC. How long will it take for the concentration of N 2 O 5 to decrease from 0.25M to 0.15M?

 For first order reactions:  Note it is independent of concentration!  This is used to describe radioactive decay and elimination of medications from the body

Rate = k [A] 2  Using calculus, the integrated rate law becomes:  Just like the previous two, this equation is of the general equation for a straight line

 Unlike first order, second order does depend on initial concentrations:

Ways to find Rate Constant or Reaction Order Conc. Vs. Time (Graphically) k = slope Rates vs. Conc. (Exp. Data/Rate Laws)Half-life expressions

 Most reactions increase in rate with increasing temperature  This is due to an increase in the rate constant with increasing temperature

 Minimum amount of energy required to initiate a chemical reaction ◦ Varies from reaction to reaction  This is the kinetic energy required by colliding molecules in order to begin a reaction ◦ Remember, even with sufficient KE, orientation is still important

 Activation energy must be enough to overcome initial resistance for a reaction to take place

 Diagram can be used to determine if reaction is exothermic (- ∆H) or endothermic (+∆H)  Activated complex (or transition state) is the arrangement of atoms at the peak of the E a barrier ◦ Unstable and only appears briefly

 The relationship between rate and temperature was non-linear  Reaction rate obeyed an equation based on 3 factors: 1.Fraction of molecules that possess E a 2.# of collisions per second 3.Fraction of collisions with proper orientation

 k = the rate constant  R = gas constant (8.314 J/mol*K)  T = Absolute temperature (K)  E a = the activation energy  A = frequency factor ◦ A is mostly constant with variations in temperature

 Taking the natural log of both sides gives a formula in straight line form:  Graph of ln k versus 1/T will be a straight line with a slope of –E a /R and a y-intercept of ln A

 In order to compare different rates at different temperatures the equation can be rearranged:

 Example ◦ The rate constant of a first order reaction is 3.46 x 10-2 s-1 at 298 K. What is the rate constant at 350 K if the activation energy is 50.2 kJ/mol? k 1 = 3.46 x 10-2 s -1 k 2 = ? T 1 = 298 KT 2 = 350 K

 Example

k 2 = s -1

 The process by which a reaction is broken down into multiple step reactions ◦ Chemical equations only show beginning and ending substances ◦ Can show in detail bond breaking and forming and structural changes that occur during a reaction

 Elementary steps ◦ Processes that occur in a single event or step, are elementary processes.  Can determine rate laws from elementary steps, unlike overall reactions ◦ Particles collide with sufficient energy and proper orientation for reaction to occur ◦ These are the small step reactions in which an overall reaction occurs

 Often times chemical reactions are a result of multiple steps not shown by the overall equation  The above rxn below 225 ° C occurs as 2 elementary steps

 1 st step ◦ 2 NO 2 molecules collide  2 nd step ◦ The NO 3 then collides with CO and transfers an O  The elementary steps must add to result in the overall chemical equation

 Every reaction is made up of a series of elementary steps  Rate laws reflect the relative speeds of these steps  The rate law of an elementary step is directly related to its molecularity ◦ This is the number of molecules that participate as reactants

 Unimolecular = 1 st order (A  product) ◦ Rate =k[A]  Bimolecular = 2 nd order (A+B  prod) ◦ Rate= k[A][B]

 Most reactions involve multiple steps  Often one step is much slower than the other  The Rate Determining Step (RDS) is the slowest step in the reaction  The slowest step of a multi-step reaction determines the overall rate of reaction!

 Good things to know/determine: ◦ Steps must add up to overall reaction ◦ Identify Catalyst  Consumed at first, regenerated later (not in overall) ◦ Identify Intermediates  Produced and then consumed later ◦ Each step has a rate law  Depends on number of reactants ◦ Rate Determining Step => slowest step

 Example: ◦ What is the rate law of the following multi-step reaction? slow fast

 As a general rule catalysts change the rate of reaction by lowering the E a  Usually this is done by giving completely different mechanism for the reaction ◦ This lowers the overall Ea