Wednesday and Thursday

4/2 and 4/2

As you enter the classroom

• Sit in your assigned seat• Get out your homework – we’re going to go

over the answers together• Reminder: 1) don’t turn your brains off, we are learning

today2) please ask questions today if you don’t

understand – see if your classmates can answer

Reminder

• The TRY 2 ticket for the Acid Base Test will be posted by Friday night. The Try 2 and Try 3 window is Monday-Wednesday. Check the website Friday evening for more information.

Homework Answers

3. Molecularity of elementary reactionsa. unib. bic. bid. bie. uni

4. a. There are two intermediates, B and Cb. There are 3 transition states (the peaks)c. The fast step is C D (the activation energy is

lowest)d. Endothermic (Products are higher in E than

reactants)

5. a. two peaks indicate 2 elementary reactionb. One intermediatec. The second step is rate-limiting (because the

activation energy is higher)d. exothermic

7. a. rate = k[H2O2][I]

rate = k[IO][H2O2]

b. 2H2O2 + I 2H2O + O2 + I

c. IO is an intermediated. rate = k[H2O2][I]

(I omitted charges because I was being lazy. I hope you included them)

16

17

18

Write down your homework (I won’t be doing the weebly)

• Packet questions: 9, 10, 15, 21• Please do your homework – remember that

you’re preparing for a mock AP and the real AP

Starting with the person who has longest hair

• Go around the table stating the three parts of collision theory

Take Notes (I won’t be posting this)

What affects the rate of a reaction?

There are a few…In your group, try to come up with at least 3

Share with the class…

You don’t have to write down fun facts…

1. Temperature of the system–Heat ‘em up speed ‘em up!–Rule of thumb: 10 K rise in temp usually

doubles the rate– Fun fact: refrigeration works to slow the rate

of biological reactions that cause spoiling

Temperature continued• At a higher temperature, a

larger fraction of the molecules are moving faster and thus meet the energy requirement for a successful reaction (have enough E to overcome the activation energy barrier)

• The Y axis is % of molecules with a given speed/KE

• Maxwell-Boltzman distribution:

2. Chemical nature of reactants– Inherent tendencies of

atoms/molecules/ions to undergo changes in chemical bonds–Fun fact: sodium exposed to air tarnishes

(loses electrons) almost instantly, potassium is slower and tarnishes (loses electrons) easier with moisture in the air

3. Ability of reactants to meet– Most reactions involve the collision of reactants– Homogeneous (reactants are in the same state, like both being

liquid or both being gases) vs heterogeneous (one reactant might be solid while the other is aqueous)

– Surface area! Increasing surface area increases the rate of a reaction

• Fun fact: liquid gas burns rapidly but gasoline vapor explodes

4. Concentrations of the reactants–Most reaction rates slow as the reactants are used

up– Fewer molecules mean few collisions mean few

reactions• This is the reason we study the INITIAL rate of a reaction

– Exception: if a reaction is zero order for a reactant, changing its concentration should not affect the rate

5. Presence of catalysts– Substances which increase the rates of

reactions without being used up– Important: they do this by creating a NEW

pathway that has a transition state with less energy; the activation energy of the new pathway is less

More on Catalysts• 3 classes of catalysts I want you to know about:1. Acid-base catalysts

Reactant either gains or loses a proton which changes the rate of the reaction

2. Surface catalysts(more on this on next slide)

3. Enzymes (more on this on next slide)

Surface catalyst• either a new reaction

intermediate is formed or the probability of successful collisions is modified, changing the rate of the reaction

• In the diagram, the surface helps reactants come together (both reactants sit on surface and are therefore more likely to find each other than running around in 3 dimensions)

Enzymes• some bind to a reactant in

a way that lowers Ea; others react with the reactant to form a new intermediate

• In the diagram, the enzyme helps with orientation to speed the reaction rate

With your face partner, answer the following: (groups of 3, just rotate around)

• Rally robin (go back and forth), starting with partner A, to answer the following:– What are 5 things that effect the rate of a reaction

• Partner B answers this question: what does the maxwell-boltman distribution show?

• Partner A answers this question: when does concentration have no effect on the rate?

• Starting with partner B, rally robin to answer this question: what are the 3 types of catalysts

Acid Base Try 1• You will now work as a team to do corrections to

the acid base test you took last week• Each team will get one fresh copy of the test• Alternate scribes between all 4 people– The purpose is learning, so you’re only jipping yourself if

you allow teammates to write answers that you don’t understand

– Answer all 3 parts for #4 and write two answers for #5c• When you’re done, hand in your individual tests and

the group test (2 separate stacks) – Start your HW if the bell hasn’t rung

You may use your notes. DO NOT HAVE A CELL PHONE OUT!!!

Try 2 ticket posted by Friday night. Retakes Monday-Wednesday.

Friday

4/4/14Put desks into groups for class

Put groups back into rows at end of class

As you enter the classroom

• Sit in your assigned seat• Get out your homework – we’re going to go

over the answers together• Reminder: 1) don’t turn your brains off, we are learning

today2) please ask questions today if you don’t

understand – see if your classmates can answer

Reminder

• The TRY 2 ticket for the Acid Base Test will be posted by Friday night. The Try 2 and Try 3 window is Monday-Wednesday. Check the website Friday evening for more information.

Let’s go over homework

New stuff: Integrated Rate LawsA third method to determine the order of a reactant

(we already learned about reaction mechanisms and the method of initial rates)

You need a graphing calculator!!!

Graphs Can Give Us Orders

• If a reaction is zero order for a given reactant, the graph of [ ] vs time will be linear

• k=-slope

• If a reaction is first order for a given reactant, the graph of ln[ ] vs time will be linear

• k=-slope

Since the reaction is NOT zero order, the graph of [ ] vs time is NOT linear

• If a reaction is second order for a given reactant, the graph of 1/[ ] vs time will be linear

• k=slope

Since the reaction is NOT zero order, the graph of [ ] vs time is NOT linear

Since the reaction is NOT first order, the graph of ln[ ] vs time is NOT linear

Let’s create a summary chartorder Linear graph k=0 [ ] vs time -slope1 ln[ ] vs time -slope2 1/[ ] vs time slope

We’ll fill in the last column later

Data from an experiment on the reaction A B is graphed. Based on the experimental data, the rate law is most likely:

a. rate = k[A]0

b. rate = k[A]1

c. rate = k[A]2

d. Not enough information

Data from an experiment on the reaction A B is graphed. Based on the experimental data, the rate law is most likely:

a. rate = k[A]0

b. rate = k[A]1

c. rate = k[A]2

d. Not enough information

Instead of graphs, I could just give you data and you could use your calculator to do 3 linear regressions….get your calc

• Whichever R2 value is closest to one indicates the order

• First, turn your diagnostics on:– Go to catalog– Scroll to diagnosticON and hit enter until your

screen says DONE• Please help your teammates as we go through

this problem

Example problem – jot this info down before continuing to next slide

A reaction A B is studied in order to determine the order of the reaction.

a) Indicate the order of the reaction by writing the rate law

b) Calculate the rate constant

Data

time (seconds) [A]0 0.010001000 0.006251800 0.004762800 0.003703600 0.00313

Insert time into L1 of your calculator. Enter [A] into list 2.Continue to next slide when students have data entered.

• We need to compare time to [ ] to see if it’s zero order, but we also need to compare time to ln[ ] (first order) and time to 1/[ ]

• It’s all about staying organized:L1 = timeL2 = [ ]L3 = ln[ ] L4 = 1/[ ]

The quick trick to filling L3 is to scroll over and highlight L3. Then, type ln(L2). Hit enter and it will calculate everything for you.

For L4, highlight L4 and type in 1/L2.

Help your neighbor! Tell the sub when you’re ready

• We need to run 3 separate linear regressions, making note of the R2 value each time.

• Reminder on linear regressions:– Stat, Calc, #4– When this appears on your screen: LinReg(ax+b)

type in the red portion: LinReg(ax+b)L1,L2– Write down R2

• Do the linear regressions 2 more times (write down R2), but make them look like this:– LinReg(ax+b)L1,L3– LinReg(ax+b)L1,L4

• Tell the sub when you have the 3 R2 values, click on when the kids are ready

Here’s how I keep myself organized

Analyzing my results. Since the most linear set of data is time and 1/[ ], this indicates that the reaction is second order.

Therefore, the rate law is: rate = k[A]2

Now to solve for k….click on when the kids are ready.

Refer back to your chart• For a second order reaction, k=slope.• Make sure the linear regression for L1 and L4 is still

on your screen. If so, ‘a’ is the slope and is equal to k. k=0.0610

What units would it have? Rearrange the rate law to find out: k = rate/[A]2, so k has units of s-1·M-1

So k = 0.0610 s-1·M-1

We’ve now answered both questions (rate law and k)!

One last thing to add. These equations are from your yellow equation chart.

Get that out now and fill these in.order Linear graph k= Equation

0 [ ] vs time -slope1 ln[ ] vs time -slope ln[A]t – ln[A]0 = -kt

2 1/[ ] vs time slope Find the version for 1/[ ] on your sheet and write it here

[A]t means the concentration at time t (which is the time you would use for t).[A]0 means the initial concentration (which you can get from time 0 in the data table)

Let’s answer one last question about the example problem we’ve already been using

• What will the concentration of A be at 6215 seconds?

• Since it’s second order, use the equation your wrote down in your chart.– [A]t is the unknown

– [A]0 = 0.01000 (from time zero, look in your lists)– k=0.0610– t=6215– SOLVE for the unknown. I got: 0.00209 M

Begin your homework!

• Packet #28, 29, 36• Acid Base Try 2 and 3 Monday-Wednesday