AP Chem unit 14 new - reaction mechanisms and rate od reaction laws PDF

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reaction mechanisms and rate od reaction laws ...

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AP Chem unit 14Kinetics: the rate at which a chemical process occurs Reaction Mechanism: exactly how the reaction occurs. Factors affecting rate: Catalyst, pressure/concentration, temperature, surface area, nature of reactants. Physical state of reactants:   

in order to react molecules must come in contact with each other, The more homogeneous the mixture of reactants, the faster the reactants can react. For heterogeneous mixtures, surface area plays a role, largter the surface area, the quicker the rate.

Concentration of Reactants: 

As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

Temperature: 

At higher temps, reactant molecules have more kinetic energy, move faster, and collide more often with greater energy.

Presence of a Catalyst: (lower activation energy, allow reactions to occur quickly)  

Speed up reactions by changing the mechanism of the reaction Catalyst are not consumed during the course of reaction.

Reaction Rates: 

Rates of reaction can be determined by monitoring the change in concentration of either reactants or products as a function of time.

Ex: AB is a hypothetical reaction. If we know that at 20s the concentration of A is .54 M, and at 40s the concentration is .30M, can we calculate average rate of reaction over this time interval?

The average rate of the reaction over each interval is the change in concentration divided by the change in time. Less reactants, less collisions. Note: that the average rate decreases as the reaction proceeds because there are fewer collisions between reactant molecules. If in [ ] = concentration usually in M. Use this equation to find rate of reaction when specific reactant or product is not specified: aA+bB=cC+dD. Lowercase letters are coefficients. At a certain time in a reaction, substance A is disappearing at a rate of 4.0 x 10-2 M/s, substance B is appearing at a rate of 2.0 x 10-2 M/s, and substance C is appearing at a rate of 6.0 x 10-2 M/s. Which of the following could be the stoichiometry for the reaction being studied?

How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction?

2O3(g)3O2(g)?

If the rate at which O2 appears, ∆[O2]/∆t is 6.0 x 10-5 M/s at a particular instant, at what rate is O3 disappearing at this same time, -∆[O3]/∆t?

Concentration and Rate: Rate law: Rate=K [A]a [B]b K= rate constant, A and B = concentrations, a and b = order of reaction (1st, 2nd, 3rd) Rate law shows the relationship between the reaction rate and the concentrations of reactions. The exponents tell the order of the reaction with respect to each reactant. Zero order rate of reaction is = to rate constant. Nothing changes in rate. First order Double concentration, Double rate Second order Double concentration, Quadruple rate Third order  Double concentration, 16 rate. The overall reaction can be found by adding the exponents on the reactants in the rate law. EX: The initial rate of reaction A+BC was measured for several different starting concentrations of A and B and the results are as follows: Determine the rate law for reaction:

Solve for rate constant: **Pick and experiment you want and plug in initial rate given in table.

Determine the rate of reaction when [A] = .050M and [B] =.100M

*when comparing experiments, make sure the alternate concentration has the same M, as the one youre looking at is changing!

Integrated Rate Laws: First Order: ln [A]t = − kt + ln [A]0  if the reaction is first order, a plot of ln[A] vs Time will yield a straight line and a slope that will be –K (neg slope). [A]0 is the initial concentration of A [A]t is the concentration of A at some time, t, during the course of the reaction. ** Sig figs come directly from concentration. Ex: The decomposition of a certain insecticide in water at 12∘C follows first-order kinetics with a rate constant of 1.45 yr -1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 107 g/cm3. Assume that the temperature of the lake is constant. a) What is the concentration of the insecticide on June 1 of the following year?

b) How long will it take for the insecticide concentration to decrease to 3.0 x 10-7 g/cm3 ?

Second Order Processes: 1/[A]t=Kt + 1/[A]0 If the reaction is second order reaction, a plot of 1/[A]t vs Time will yield a straight line with a slope of +K (positive slope). Zero Order process: in some reaction the rate is apparently independent of the reaction concentration. The rates of the zero order reactions do not vary with increasing nor decreasing reactant concentrations. This means that the rate of reaction is equal to the rate constant, k, of that reaction.

Half life- the time required for one half of a reactant ot react. Because [A] at t1/2 is one half of the original [A], [A]t = 0.5 [A]0 First order processes: .693/ K = t1/2 **Half-life does not depend on original concentration in first order!! Second order process: 1/ K [A]0 = t1/2 [A]0= initial concentration. Half-life for a first order reaction will decrease the concentration of the reactant by ½ in each of a regularly spaced time interval, each interval equal to ½. Half-life of a second order reaction depends on initial concentration of reactant. The lower the initial concentration, the lower the half-life.

To calculate rate constant from graph for first order reaction, fins slope, -k, but make slope positive. ** the larger the number for rate constant, the faster the reaction occurs. ** the ONLY thing that impacts K (rate constant) is TEMPERATURE. How does temperature increase the rate of a chemical reaction? 

Kinetic energy increases, collisions increase, chances for right collisions to occur and react increases, need right amount of energy as well for right collision to react.

Temperature and Rate: 

As temp increases, so does reaction rate. This is because K is temperature dependent.

Collision model:   

In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide w/ each other Molecules must collide with the correct orientation and with enough energy to casue bond breakage and formation.

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Activation energy: in other words, there is a minimum amount of energy required for reaction top occur: the activation energy, Ea+.

If more energy released than absorbed  exothermic If more energy absorbed than released  endothermic. **ONLY Catalyst can lower activation energy **the smaller the hill, the easier it is for the reaction to occur. Maxwell-Boltzmann Distributions:   

At higher temperatures, a larger population of molecules has higher energy. If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases.

Arrhenius Equuation: what you get from graph? Find slope to find activation energy. Slope = activation energy. ** look at activation energy when looking on rate of reaction, lowest greatest= slowestfastest.

TEST Q: In terms of molecules, why does rate of chemical reactions increase as you increase temp? 

As you increase temp, more molecules have the energy to overcome activation energy. Since more molecules can overcome activation energy, activation energy increases, allowing reaction rates to increase as well.

Reaction Mechanisms: the sequence of events that describes the actual process by which reactants become products.   

Reactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process. The molecularity of a process tells how many molecules are involved in the process.

Intermediate: Same molecule you cancel out in product and reaction: Spectator ions. Multistep Mechanisms: one of the steps will be slower than others. 

The overall reaction cannot occur faster than this slowest rate-determining step. ** The slow step determines how fast you will go.

Slow initial step: NO2 (g) + CO (g)  NO (g) + CO2 (g) •

The rate law for this reaction is found experimentally to be Rate = k [NO2]2

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A proposed mechanism for this reaction is

Step 1: NO2 + NO2  NO3 + NO (slow) Step 2: NO3 + CO  NO2 + CO2 (fast) •

The NO3 intermediate is consumed in the second step.

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As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. **only look at slow step to determine rate law. A+A= K[A]2 products so K[NO2]2..

1) Take out intermediate, then write overal equation.

2) From reactants in slow step, write rate law

Fast initial step: 2 NO (g) + Br2 (g)  2 NOBr (g) •

The rate law for this reaction is found to be Rate = k [NO]2 [Br2]

Step 1: NO + Br2

NOBr2

Step 2: NOBr2 + NO  2 NOBr

(fast)

rate of forward=rate of reverse

(slow)

** step 1 include both forward and reverse reactions The rate oc the overall reaction depends upon the slow step, The rate law for that step would be :...