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   Q3 : Week 8 – 3/25 – 3/29

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______________________________________________________                                                                 Jump to:  Monday Homework      3/25 – Monday – A Day – 2/3a Lab, 4

Main focus –                                                                                                                                                         
                                                  

a) To Complete calculations of Ksp Problems

b) To approximate the quadratic for very small Keq values (Ksp, Ka)

c) To calculate a Keq in a lab.

Period 2,3a:  

1.   Review Last nights Form, selected problems
2.   Model the Ksp Problem – Q vs. K       
 
Equilibrium 4 – Equilibrium constants 2.pdf
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Equlibrium 4 – Equilibrium constants 2 KEY.pdf
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3. Quadratic Catastrophe introduced
Equlibrium 3 – Equilibrium constants 1.pdf
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Equlibrium 3 – Equilibrium constants 1 key.pdf
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4. Quadratic Catastrophe avoided – (large and small Keq values)   

Circumventing the Quadratic Catastrophe – (large and small Keq values)   

     Approximating away the Catastrophe”.
 
Equilibrium 6a Small equilibrium constants.pdf
 
Equilibrium 6a Small equilibrium constants key.pdf
View Download
  
5. Solving for Ksp value, Common ion effect, Approximating away the Catastrophe with Ksp, 
 

Equlibrium 5 – Ksp.pdf

    
Equlibrium 5 – Ksp KEY 23-24.pdf
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6.  Lab 26 – Equilibrium of FeSCN
 
Lab 26 – Equilibrium of FeSCN.pdf
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Period 4:
 1 – 5 above.
 
 
This weekend we introduced a new Keq that we call Ksp when we look at ionization reactions:
 
                                                               AgCl(s)  <----->     Ag+(aq)   +    Cl(aq)
 
                                                         Keq or Ksp = [Ag+][Cl]
 
All ionization reactions have the same general reactions so all the equilibrium expressions look the same. Its called a solubility product (Ksp) instead of Keq because its Ksp depend on the product of the ions in solution. The more ions in solution at equilibrium the higher the Ksp thus the more soluble the salt is.  We can measure solubility based on equilibrium constants. We do not consider the AgCl (s) in the equilibrium expression because it is a solid, which has no dilution ability.

                                                                                                                                                   

 _____________________

TODAY’s NOTES: 

 

FROM the Ksp Notes section:

(SKILL # 2a) – Calculating the molar solubilities of individual ions ( 1 : 1 mole ratio)
 
Remember that Ksp is an equilibrium constant that has no units and is just the product of the molarities of the 
ions at equilibrium (saturated solution) at their respective exponents (coefficients).  Solubility is a measure of how many ions dissolve in certain volume of solution = molarity.
 
If we want the measure the actual solubility (molarity) of the Ca+2 or Mg+2 we will have to solve for them.
 
For instance the molar solubility of Ca+2  is solved the following way (SKILL # 2)
 
                                                 CaCO3 (s) <—>  Ca+2   +  CO3-2                       Ksp = 1.4 x 10-11
                                                 
                                                                        Ksp = [Ca+2][CO3-2]
 
We need an Equilibrium Ice Table to solve this problem since we need the value of [Ca+2] at equilibrium.
We have used stoichiometry tables already that use moles but now we are using equilibrium tables and they require molarity [   ] or partial pressures (   ) from equilibrium expressions!
 
                                                                    CaCO3 (s) <—>  Ca+2   +  CO3-2     
 
                                                    Initial :             I:                    0               0         we like to start from the zero position
                                                                                                                              and determine the change from this position
                                                    Change:          C:                 +x              +x       each ion is has 1 : 1 ratio; 
                                                                                                                                there is 1 Ca+2  for every 1 CO3-2
                                    Equilibrium:           E:               x               x       change from zero is just the change
                                                                                                                                        we need this value at equilibrium to place
                                                                                                                                         in our equilibrium expressions                                                
                                               
                                      Equilibrium expression = Ksp = [Ca+2][CO3-2]
 
                                                                   1.4 x 10-11  = [x] [x]     from the E :  part of the table
 
                                                                  1.4 x 10-11  = [x] 
 
                                                              1.4 x 10-11  = [x]  
                                                         
                                                    3.74 x 10-6 M = x = [Ca+2 ] = Molar solubility of calcium ions
                                                 
                                                    3.74 x 10-6 M = x = [CO3-2] = Molar solubility of calcium ions
 
              You should notice that x also = the molar solubility of the CO3-2   ion as well because the stoichiometric 
               ratio of the ions are 1 : 1  in the precipitation reaction.

 
(SKILL # 1b) – determining solubility by comparison on Ksp values (that have different mole ratios):
 
      We cannot evaluate the molar solubility from Ksp’s of salts that have different Ksp formulas that result from different the product ions that have different stoichiometric solubilities. For example: 
 
                                       CaCO3 (s) <—>  Ca+2   +  CO3-2                             Ksp = 1.4 x 10-11
 
                                        PbCl2 (s) <—>  Pb+2   +  2Cl-1                                 Ksp = 1.7 x 10-5
 
One might think that Pb+2  is more soluble at equilibrium and thus would have a greater molar solubility than Ca+2  but we must solve for Pb+2  first because of the different Ksp equations that result for precipitating reactions that have different number ions or coefficients.  
 

 
(SKILL # 2b) – Calculating the molar solubilities of individual ions (with 1:2 mole ratio)
Lets use are equilibrium ICE table to solve for the molar concentration of Pb+2  ions at equilibrium when PbCl2 (s) is added to water at a constant temperature.
 
                                                                                 PbCl2 (s) <—>  Pb+2   +  2Cl-1  
 
                                                    Initial :            I:                   0               0                 we like to start from the zero position
                                                                                                                                     and determine the change from this position
                                               Change:          C:              +x           +2x        each ion is has 1 : 2 ratio; 
                                                                                                                                     there is 1 Pb+2  for every 2 Cl-1
                                          Equilibrium:      E:               x              2x         change from zero is just the change
                                                                                                                                               we need this value at equilibrium to place
                                                                                                                                               in our equilibrium expressions      
                           *The +2x is only a stoichiometric value meaning that we double the x when we have a value that relates to it!
                                           
                                                             Equilibrium expression = Ksp = [Pb+2][Cl-1]2
 
                                                                   1.7 x 10-5  = [x] [2x]2     from the E :  part of the ICE table
 
                                                                   1.7 x 10-5  = [4x]
 
                                                             1.7 x 10-5    = ∛ [x]3      
                                          4
 
                                     1.62 x 10-2  [x]   = [Pb+2 ] = Molar solubility of lead ions
                                         
                                     3.24 x 10-2  [2x] = [Cl+2 ] = Molar solubility of chloride ions                       
 
*Notice a 1:1 ion salt like CaCO  always will have Ksp = [x][x] = [x]2
*Notice a 1:2 or 2:1 ion salt like PbCl2 (s) will Always have a  Ksp = [x][2x]2   = [4x]3                                        
    
There are other possibilities but these are the most common!  
 

 
-SO if we want to continue the comparison of  which salt is more soluble (free ions dissolved):
 
                                       CaCO3 (s) <—>  Ca+2   +  CO3-2                             Ksp = 1.4 x 10-11
                                        PbCl2 (s) <—>  Pb+2   +  2Cl-1                              Ksp = 1.7 x 10-5   
 
From skill 2a above:
 
 Equilibrium expression = Ksp = [Ca+2][CO3-2]
                         
                      1.4 x 10-11  = [x] [x]
                     
                    √1.4 x 10-11  = [x]  
                            
                  3.74 x 10-6 M = x = [Ca+2 ] = Molar solubility of calcium ions or CO3-2  ions
 
                                     
                                           PbCl2 (s) <—>  Pb+2   +  2Cl-1                              Ksp = 1.7 x 10-5   
 
From skill 2b above:

                       ∛ 1.7 x 10-5    = ∛ [x]3      
                                                             4
                                  1.62 x 10-2  [x]   = [Pb+2 ] = Molar solubility of lead ions
                                   
                                  3.24 x 10-2  [2x] = [Cl+2 ] = Molar solubility of chloride ions                       
 
  PbCl2 (s) (1.62 x 10-2  [Pb+2 ] ) then is more soluble than CaCO3  (s) , (3.74 x 10-6 M = x = [Ca+2 ] )
                            
                               The molar solubilities of the ions needed to be calculated to justify!
 
                                                           

_____________________________________________________                                                                         jump to:  top                          3/25 – Monday’s Homework: –

I have Ksp Notes posted in 3rd quarter that go over every type of problem. 
Many past students have said they were very helpful.
WE are working in Skills 5, 6, 7, 8 from the Ksp- notes.

 

1. Please complete – Equilibrium 4 – Question 1 – Ksp problem explained/modeled – lecture below.

I did this in class with everyone.  Please look at the second question = Quadratic Catastrophe!!!

    

Equilibrium 4 – Equilibrium constants 2.pdf
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Equilibrium 4 – Equilibrium constants 2 KEY.pdf
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2. Complete  Equilibrium 5 – Ksp.pdf worksheet and/or with me with the lecture below:
We started this is class. Use my new Key !!!!!!!
If you feel that you are on your way feel free to complete without video lecture and review with the key below.  YOUR ARE TO COMPLETE LETTER D in the question. In the video I say not to but please complete it and review with Key.
 
Equilibrium 5 Lecture: see below 
 

Equlibrium 5 – Ksp.pdf

    
Equlibrium 5 – Ksp KEY 23-24.pdf
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3. Complete Equlibrium 5a – Ksp.pdf worksheet with me with lecture below:
 
Equilibrium 5a – Ksp continues.pdf
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Equilibrium 5a – Ksp continues key.pdf
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Ksp TEST – (10 point) AP Question –  1 question part 2 test – (20 minutes)   Tomorrow!

 

 1: Equilibrium – Question 1 Lecture: (for Equilibrium 4 – done in class today!) 

 2: Equilibrium 5 worksheet – from beginning.

 3: Equilibrium 5a worksheet review – whole video: 

________________________________________________________________________                                 Jump toTuesday Homework / TOP 3/26 – Tuesday – B Day – 2, 3b/4 Lab

Main focus –                                                                                                                                                         
                                                  

a) To complete a Ksp FRQ.

b) To Review the concepts of equilibrium and review Ksp related equilibrium problems.

Period 2/3a,4:

1. Hand back collected FRQ that was assigned last Thursday. Reviewed the energy shift question.
          Mentioned that the responses were similar to the posted AP key. Was I born 53 years ago?
 
2.  Reviewed part (d) from Equilibrium 5 worksheet – Temperature and Equilibrium
 
3. Quick General review of Ksp/equilibrium/Q vs Keq – used New Ksp notes
– Quadratic Catastrophe! Review! – When to approximate 
 
4.  Ksp TEST – (10 point) AP Question –  1 question part 2 test – (20 minutes)   
 
5.  Review of test and practice with the homework Ksp practice problem.
 

         _____________________

TODAY’s NOTES:

                                                      I have Ksp Notes posted in 3rd quarter that go over every type of Ksp problem. 
                                                      http://mrgrodskichemistry.com/ksp-notes/
  

 This is an excerpt for the notes on how to solve “which salt precipitates first”:

(SKILL # 6) Determining the which salt precipitates first
(SKILL # 7) Prevent the Quadratic Catastrophe by approximating.
 
                                                                                          
A beaker contains a 400. ml solution of .20 M of  Cl– and .10 M of I. If  a solution of Pb(NO3) was slowly added to the beaker, which salt precipitate’s first?  
 
The Ksp of  PbI is 1.4 x 10-8  .     The Ksp of  PbCl2 is 1.7 x 10-5
 
This precipitation reaction (net ion):    Pb+2 (aq)  +  2I–  (aq)   —> PbI (s) was what I did as demo in class:
 
We should realize the that precipitating reactions that will occur due to the soluble Pb(NO3) that will deliver the soluble Pb+2 ions are: (Based on soluble rules nitrates are always soluble!)
 
                                                                    PbI (s)   <—>  Pb+2    +      2I                   Ksp = 1.4 x 10-8
 
                                                              PbCl2 (s) <—>  Pb+2   +       2Cl-1              Ksp = 1.7 x 10-5
 
If both ions,  .20 M of  Cl   and .10 M of I, were dissolved with the same concentration then all we would have to do is look at the Ksp values to determine which salt would precipitate first because both salts have the same formula (same stoichiometric ratios of product ions that lead to the same Ksp formula).  The salt with the smaller Ksp value must precipitate first as it the most insoluble (produces the lest amount of ions before it reaches equilibrium).  Remember that Ksp = Keq and that when Keq < 1 reactants are more favored.  
 
BUT this is NOT the case so we must calculate the equilibrium molar concentration of  Pb+2  for each of the precipitating reaction and compare [Pb+2]  to see which one is the smaller.  The smaller [Pb+2] will be the one that precipitates first because this concentration Pb+2 is the value that must be exceeded first to cause the Q > Ksp to cause precipitation.
The lower concentration will be reached first as you pour Pb+2 ions into the beaker.
 
So here we are calculating the molar concentration of I which is Skill 2b with a twist. The twist is that we are starting with .10 M before we add any Pb+2 .
 
                                                                    PbI (s)   <—>  Pb+2    +     2I
                                    
 
                                        Initial :            I:                                 0                .10                   The solution already has .10 M
                                                                                                                                                             of  I–  ions.
                                    Change:          C:                               +x                 +2x                 each ion is has 1 : 2 ratio; 
                                                                                                                                                              there is 1 Pb+2  for every 2 I-1
                              Equilibrium:      E:                                 x                 .10 + 2x         we are adding 2x as equilibrium 
                                                                                                                                                                        is reached                                                                                                                                                              
                                           Equilibrium expression = Ksp = [Pb+2][I-1]2
 
                                                                                       Ksp = [x][.10 + 2x]2
 
                                                                                       Ksp = 4x3  +  .4x2  +  .01x    (CUBIC Equation!)
 
                                                            The Quadratic Catastrophe!
 
So eliminate the need to solve for these three roots of the cubic equation by using the idea that because the Ksp is so small that the reaction will move only a very small amount forward to reach equilibrium. Thus x is very tiny as compared to the .10 M that was initially added.  We eliminate the Quadratic Catastrophe by approximating the 2x away (this is skill 7): 
 
                                                                                         Ksp = [x][.10 + 2x]2
 
                                                                             Ksp = [x][.10]2
 
                                                                  1.4 x 10-8 = [x][.10]2
                                                                  1.4 x 10-6 = [x] = [Pb+2 
  
This all means the when molarity of Pb+2  is greater than  1.4 x 10-6  PbI starts Precipitating!  Or Better Yet it means that the value of Pb+2  if increased from 1.4 x 10-6  will cause Q to become bigger than K(sp) and precipitation BEGINs (reverse reaction becomes more spontaneous!
 
Okay so lets calculate the [Pb+2 when  PbCl2 starts precipitating:
 
                                                                 PbCl2 (s) <—>  Pb+2   +     2Cl-1                  Ksp = 1.7 x 10-5
                                  
                                        Initial :            I:                                          0             .20                  The solution already has .20 M
                                                                                                                                                                of  I–  ions.
                                    Change:          C:                                        +x           +2x                 each ion is has 1 : 2 ratio; 
                                                                                                                                                                there is 1 Pb+2  for every 2 Cl-1
                               Equilibrium:      E:                                        x            20 + 2x           we are adding 2x as equilibrium 
                                                                                                                                                                             is reached .                                                                                                                                                            
                                           Equilibrium expression = Ksp = [Pb+2][Cl-1]2
 
                                                  Approximate away the quadratic or cubic formula!
                                                                        
                                                                                      Ksp = [x][.20 + 2x]2
 
                                                                                      Ksp = [x][.20]2
 
                                                                     1.7 x 10-5 = [x][.20]2
                                                             
                                                                  4.25 x 10-4 = [x] = [Pb+2 
 
                   This all means the when molarity of Pb+2  is greater than  4.25 x 10-4  PbCl2  starts Precipitating!
 
Compare the molar solubilities of the [Pb+2 from both calculations: 
 
When we pour Pb(NO3) ( delivering  Pb+2 ions) into the beaker which Threshold [Pb+2] is reached first?
 
                                                          From PbI2                               From PbCl2
                                            [Pb+2 = 1.4 x 10-6                     [Pb+2] =    4.25 x 10-4 
 
PbI  precipitates first because the concentration of Pb+2 ions is smaller and thus equilibrium is reached first.
 

3/26 – Tuesday’s Homework: – 

 

1. Review the PRACTICE KSP Test we took in class with the keys below if you did not do so in class.                       I will have your test graded and returned by email for this evening. Please Review with the Key below. 

Today’s Test Keys 
KSP FRQ TEST 2001 – Grodski & College Board Keys.pdf
View Download
 
KSP TEST 2001 – blank.pdf
View Download
 
You will have a 2 question (40 minute) part 2 TEST tonight.  A good review would be the following AP questions. 
 
2.  Please complete Equilibrium 5b – Ksp practice.pdf  and review with key below.
      I made a lecture for this problem below!
 
***In Question (b) ii  of Equilibrium 5b – Ksp practice.pdf I did not finish the sentence: 
Calculate the minimum concentration of F(aq)  necessary to initiate precipitation of the salt selected in part (b)(i).
 
Ksp Practice: Another Ksp problem!
Equilibrium 5b – Ksp practice.pdf
View Download
 
Equilibrium 5b – Ksp practice key (AP Central key).pdf
View Download 
 
3.  I will be be having a Zoom meeting Tonight to review the homework or Test 1. The Link is below.

       This is an optional activity and you do not have to attend for and grade.

Mr. Grodski is inviting you to a scheduled Zoom meeting.

Topic: Ksp Review 2
Time: Mar 26, 2024 07:00 PM Eastern Time (US and Canada)

Join Zoom Meeting
https://us02web.zoom.us/j/87434515439?pwd=MXJBY3dxb0QzY29hTUhMYnpnYnM5Zz09

Meeting ID: 874 3451 5439
Passcode: Svz6wU

 

 

 2: Equilibrium 5b worksheet review – whole video: 

____________________________________________________________________________                    Jump to: Wednesday Homework / top 3/27 – Wednesday A Day – 2/3a Lab, 4

Main focus –                                                                                                                                                         
                                                 
a) To Review the measure of solubility through Ksp’s. 
                                                                     
b) To introduce Rate Law 
 
c) To determine the order of a reaction through experimental evidence AND                                determine the best mechanism for an overall reaction.

Period 2,3a:

1. Take the Ksp Test 2!

2.  Review the Ksp Test 2 and start the FRQ homework.

Period :4

1. Take the Ksp Test 2!

RATE LAW Begins: – did not do!!!  we will start this next week!

LAST NEW TOPIC: Rate Law – path function from Reactants to Products
 
3. Intro to Rate Law – Last New topic! We are very close to being done with the entire course!
 

     a) effective collisions  – collision theory

     b)  factors that effect RATE (speed) of reactions (increase the frequency of effective collisions) 

           i) RATE Law deals with the initial rate of a (spontaneous) reaction BASED on the amount of concentration

Rate Law does not measure spontaneity and spontaneity does not measure Rate Law!

    c) Determine the Rate Law based experimental evidence     

            i) eyeball method vs calculating 

  d) rate Law predicts the best mechanism by predicting the elementary step that is rate determining.

              The slowest step in a mechanism is Rate Determining Step (RDS) !

i) Coefficients = exponents ONLY for single step reactions = elementary reactions = single step reactions

             Coefficients becomes orders for Reactants in the RATE LAW EQUATION for ELEMENTARY REACTION ONLY !

            ii) Catalysts vs Intermediates.

iii) Determining the mechanism ( a group of elementary steps that overall = the overall reaction) that best represents                          the experimentally determined RATE Law.

Classwork modeling:

Rate Law Rumble 1819.pdf
View Download
 
Rate Law Rumble key .pdf
 
Kinetics 1 Rate Law .pdf
View Download
 
Kinetics 1 – Rate Law Key.pdf
         
 

                                                                                 

 

Rate Law Intro Lecture 1:

Rate Law Intro Lecture 2:

   _____________________

TODAY’s NOTES: 

a) You are to read my notes on RATE LAW that is posted in the 3rd quarter:  
                                    
                                                                           http://mrgrodskichemistry.com/rate-law-notes/      
 
*You are to read my notes up to and including skill 5 and skill 7!  
Complete RATE LAW Skills from the NOTES:
Skill #1 – Identify the orders of the reaction from the experimental evidence – Eyeball method
Skill #2 – Determining catalysts and intermediates in mechanisms
Skill #3 – Determining Rate Law from mechanisms or evaluating the best mechanism
Skill #4 – Determine the rate law constant with correct Units
Skill #5 – Calculate the Rate for the initial concentrations that are not in the experiment.
Skill #7 – Determine the Orders by math (when the eyeball method can’t be used.) – NON-EYEBALL METHOD!

 

 

I have worked really hard to write a complete chapter in RATE Law that I hopefully is clear and logical and breaks down every skill you need to learn.
RATE LAW looks daunting at first but trust me everyone loves rate law once you let it sink in.  PLEASE do this before you view todays class lectures.  The Lectures will make MUCH more sense.
b) Rate Law Lecture 1 is what I did today, keeping in mind what you read in the Rate Law Notes.

 

 _____________________________________________________                                                                                jump to:  top               3/27 – Wednesday Homework:

1.  You are to complete the FRQ that was given out today in class and Review with the key!

I will post the key soon.

This above is the only homework for tonight and the last amount till Tuesday night. Enjoy!

 

 

________________________________________________________________________                            Jump to: Thursday Homework / top 3/28 – Thursday – B Day – 2, 3b/4 Lab 

Main focus –                                                                                                                                                         
                                                  

 I will update..

    a) To Calculate the orders of reaction (when eye ball method does not work)

    b) To use stoichiometry in Rate Law questions.

     c) To determine the Rate Law through Fast Equilibrium problems.

 

Period 2/3a,4:

 

1. Review of orders determined through RATE experiments:

      a) help us build an equation that will measure initial rate or instantaneous rates of reactions

      b) Help us understand how a reaction occurs either by a single step or a multiple step process = mechanism

2.  Review the Fast Equilibrium elementary step that helps determine the Rate Law formula for the overall reaction.

3. Review of Rate Law Rumble worksheet.

 

4. Review the homework problems – appropriate mechanisms for Rate Laws –

                       also skill #6 – stoichiometry in rate law 

5. When you need to go beyond the eyeball method – Calculating the orders. 

* Eyeball method vs. Mathematical method (NON-EYEBALL Method) of finding the orders –skill #7

Kinetics 1 Rate Law .pdf
View Download
 
Kinetics 1 – Rate Law Key.pdf

 

6. catalyst  – Demo 

7. Complete 1999B of Kinetics 2 worksheets:

Kinetics 2 – rate law.pdf
View Download
 
Kinetics 2 – rate law KEYp .pdf
View Download  

 

 

Rate Law Presentation:

_______________________

TODAY’s NOTES:

 
RATE orders –
 
                  R = [x]0  =   zero order    =    no effect on the rate of a reaction
 
                       R = [x]1   =   1st order       =    Rate changes exactly as the [x] changes
 
                       R = [x]2  =    2nd order     =    Rate change is the square of the [x] change
 
                       R = [x]3  =    3rd order      =    Rate change is the cube of the [x] change
 
                        R = [x]1/2 =  1/2 order       =    Rate change is the square root of the [x] change
 
                        R = [x]-1  =  -1 order         =     Rate change is the inverse of the [x] change
 
 
 
Complete RATE LAW Skills from the NOTES:  http://mrgrodskichemistry.com/rate-law-notes/      
Skill #1 – Identify the orders of the reaction from the experimental evidence – Eyeball method
Skill #2 – Determining catalysts and intermediates in mechanisms
Skill #3 – Determining Rate Law from mechanisms or evaluating the best mechanism
Skill #4 – Determine the rate law constant with correct Units
Skill #5 – Calculate the Rate for the initial concentrations that are not in the experiment.
Skill #6 – Determine the Stoichiometric loss of reactants to gain of products
Skill #7 – Solve for the Rate Law orders without EYEBALL method
Skill #8 – Solve for the Rate Law orders with Graphs 
Skill #9 – Solve for time, concentration, or using integrated Rate Law formulas

 

Worksheets worked on so far.

                                  Rate Law Rumble 1819.pdf
                                   View Download

                                         RATE LAW SKILLs (in your notes) :  #1, #2, #3, #4, #5, #7

                           

                                   Kinetics 2 – rate law.pdf – (1984B, 1991B)
                                   View Download

                                   RATE LAW SKILLs (in your notes) :  #1, #4, #7 

Rate Laws from Mechanisms (Optional) Lecture:

4: NONEYEBALL Method of getting the order of reaction: SKILL 7 in the notes

3: Fast Equilibrium Lecture:

___________________________________________________                                                                                jump to:  top                     3/28 – Thursday Homework: –

                                                                                                                                                                                                                               

 THERE IS NO HOMEWORK for the mini spring break!

 

I will update..

1.  You are to read the notes to review skills 1 – 5 (ALL Classes)

 

2.   Please complete the first Rate Law Rumble Worksheet with Rate Law lecture 2 posted above.

     Please use the RATE LAW Lecture 2 above to complete.

This Question contains: RATE LAW SKILLs (in your notes) :  #1, #2, #3, #4, #5, #7 
 
Rate Law Rumble 1819.pdf
View Download
 
Rate Law Rumble key .pdf

NOT for Homework below this.. I will update.

3.   Please view the Fast Equilibrium Lecture below and follow along with me following worksheet.                      

Kinetics mechanisms – fast equilibrium.pdf
View Download
 
Kinetics mechanisms – fast equilibrium KEY.pdf
View Download

4. Please complete the first 2 pages of the Kinetics 2 worksheet  (1984B, 1991B) and review with the key.             (ALL Classes)

*If you can use the eyeball method to determine the Rate Law then do so!!!!!

SKIP question C of 1984B

This question contains: RATE LAW SKILLs (in your notes) :  #1, #4#7

You might (period 3/4 saw this but not period 2) need me to show you how calculate the orders from the experiment (IF THE EYEBALL METHOD – (SKILL 1) will not work)
Please view the NON- EYEBALL METHOD (SKILL 7) video below:
 
Kinetics 2 – rate law.pdf
View Download
 
Kinetics 2 – rate law KEYp .pdf
View Download

 

Integrated Rate Law Formulas:

 Please follow these instructions. If you do these out of order you will not learn an reinforce the concepts that I want you to review.
 
1. Please watch the Integrated rate law derivation lesson below on the derivation of the 0th, 1st, and 2nd Order Integrated  Rate Law formula. 
 
These formulas help us utilize time with changes in our reactant concentrations!  
These formulas are integrated formulas (Calculus) and they are needed to link “real time” with actual concentrations (area under curve).  Remember that Rate Law is the Rate from INITIAL CONCENTRATIONS of reactants. That Rate that we have been calculating for changes the instant the chemical reaction starts so it really becomes useless AFTER some time period after time zero.  They utilize concepts of derivative and integrals which many of you have NOT had in math YET! You do not need to use these concepts to answer the following questions, just the formulas:
 
How long does it take for the reaction to be completed?
How long does it take for the concentration of my reactant to reach a certain value?
Given the time, how much of the reactant remains?
What is the shelf life of the chemical? (how long does it take to decompose?)
 
2.  Read the notes on the new INTEGRATED RATE LAW Formulas – Skill 8 and Skill 9
 
 
3. Complete the Form that goes along the video (this will be in auto-reply)
 
4. Complete the following worksheet and review with the keys: 
Give yourself 40 minutes, these are 2 part 2 questions, then grade. The goal is 7 out of 10.
 

2005B 2009 Kinetics quiz 2013 with college board and grodski keys.pdf

 
These formulas are USED WITH TIME!! They are used 2 ways.
 
1: They can be used to identify the Rate Order By how they are graphed. For instance if you graph a 1st order reaction using the ln[A]t  (ln of concentration of reactant at some time period) in the y axis and the t (time)  in the x axis it will be linear with a negative slope.  If it is not linear with a negative slope then the reactant is not 1st order. 
 
See the reference table below. 
 
2: Once you identify the whether the RATE ORDER is 1 or 2 use the correct integrated rate law equation to solve either the time (t), initial concentration [A]0 , concentration of reactant after some time[A]t ,
or the rate constant, k.
 
What this lecture describes is the mathematics  (Calculus ) needed to solve for or Link Time with remaining reactants.  These important derivations will derive our Integrated Rate Law Formulas that are given to us in our reference tables.   They help us identify the order (or exponents) of our reactants in our Rate Law Equation just by graphing the changes of concentrations of the reactants in time during a chemical reaction.   
 
Here they are in our reference tables:

It is usually very helpful to utilize these equations when you arrange them into a linear equation.

  1st Order:                   X   –>  Y  
 
                                      R = K [X]1    
                                
                             ln[A]t – ln[A]0 =  -kt
 
                             ln[A]t  =   -kt   +    ln[A]0
                                 y      =    mx  +      b
                         (negative slope)

 

 2nd Order:            X  +   X  –>  X2                                    
 
                               1   –         =  kt
                              [A]t    [A]0
 
                                =  kt   +    1
                                [A]t                  [A]0   
 
                            y     = mx  +   B
                                                 (positive slope)
 
 
 

                                                                                                                                   

  .

1: Integrated Rate Law formulas LESSON:

  
3 : Integrated Rate Laws Formulas Form: To be posted…..

C

Derivations of the 0th, 1st, and 2nd order Screencast lecture – optional lecture

Chenmistry Calculus (optional)- Why we need the differential Rate Laws

Catalyzed decomposition of hydrogen peroxide demo – lowering the Ea with a mechanism!

__________________________________________________________________                                   Jump to: Friday Homework / top  

3/31 – Friday – B Day – 2, 3b/4 Lab NOT UPDATED BEYOND THIS

I am out today and I am not happy but I feel like I got hit by a truck. I am so close to finishing the course with you guys!

Main focus –                                                                                                                                                         
                                                  

    a) To review a frq (2022)

  

Period 2,3b/4:  

1.  Please print out page 9 to page 14 , question 2 and 3 from this test.  
2. complete question 2 and 3.
3. review with the key and hand it in to the crate.  I understand that period 1 will not get as far as period 3,4.
4. IF printing is a problem then just use a piece of paper. 

 

2022 AP Chem FRQ:

https://secure-media.collegeboard.org/apc/ap22-frq-chemistry.pdf

 

2022 AP Chem FRQ College Board Scoring guideline:

https://apcentral.collegeboard.org/media/pdf/ap22-sg-chemistry.pdf

‘ 

 

_____________________

TODAY’s NOTES: 

Reference Table:
         This is the first order integrated formula.

        This is the second order integrated formula.

There is no zero order integrated formulas on the AP.

This is the Arrhenius equation that ties activation energy ( Ea) and temperature to the rate law constant. 

 It is usually very helpful to utilize these equations when you arrange them into a linear equation.

                                                          1st Order:                   X   –>  Y  
 
                                                                                          R = K [X]1    
                                
                                                                                   ln[A]t – ln[A]0 =  -kt
 
                                                                                 ln[A]t  =   -kt   +    ln[A]0
                                                                                      y     =    mx  +      b
                                                                                                   (negative slope)

 

                                                            2nd Order:                X  +   X  –>  X2
 
                                                                                                   –       =  kt
                                                                                               [A]t    [A]0
 
                                                                                                    =  kt   +    1
                                                                                               [A]t                [A]0   
 
                                                                                                y     = mx  +   B
                                                                                                         (positive slope)
 
 

Mr. Grodski Solutions – 2022

_________________________________________________                                                                                              jump to:  top

3/31 – Weekend Homework: –

1.  Please review Thursday nights FRQ’s with the video below. I would of reviewed it in class today.

2005B 2009 Kinetics quiz 2013 with college board and grodski keys.pdf

 

 

2. Complete a short 5 question Rate law MC Form.
 
Catalysts and Activation Energy (Ea = Activation Energy)
The Arrhenius equation calculations are not on the AP but the concepts of how the k (Rate law Constant, like the equilibrium constant) is related to temperature.
 
3.  Please watch the Lecture on the derivation of the half – life formulas and lecture on the Activation Energy, Ea.
 
4. Complete the form below that is based on the 2 lectures.
 
5. Complete a 2 question FRQ (rate law and KSP).
 
Kinetics (rate law),Ksp Free Response, 2004B, 2009 FRQ .pdf
View Download 
 

 

 

  
2 : Rate Law MC Questions Review Form.

C>

3 : Half Life Derivations of the oth, 1st and 2nd order

3 : Activation Energy (Ea) Lecture.

  
4: Half Life / Arrhenius/ Ea Form 21-22

C>