Today we jumped into acids and bases with both feet including Arrhenius' definition of acids and bases, the Bronsted-Lowry definition of acids and bases, and a couple "new" equilibrium constants (Ka, Kb, Kw) that let us assess the strength of an acid or a base. We just barely touched on pH, we'll get more into that on Monday.
I will also have your exams on Monday.
Tomorrow night (Saturday) is the last regular season home game for the Dragon basketball teams, come on out and watch our Dragons put a hurt on UM-Crookston. Women tip at 6pm, men at 8pm. Post season play starts next week....
Have a good weekend.
2009-02-27
2009-02-24
Questions...
Can you explain these two questions? Thanks!
9. Which of the following cannot be explained by LeChatelier’s Principle?
a. Removal of a gaseous product from a reaction drives the reaction to completion.
b. Any salt will dissolve in water, even though the amount might be extremely small.
c. The activation energy of a reverse chemical process is related to the activation energy and
ΔH of the forward process.
d. Less of a salt will dissolve in a solution that has a “common ion” than in pure water.
e. The equilibrium of an exothermic process will be more reactant-favored at higher
temperatures
"a" - removing a product will drive an equilibrium toward products, that's almost the definition of LeChatelier's Principle (I'll call it LCP from here on...)
"b" - Think about solubility in terms of a chemical reaction/equilibrium, salt + water yields ions in solution. Water acts as a reactant here, so adding more water will stress the equilibrium and shift it toward products.
"c" - This is purely kinetics, so it'd be tough to explain it with LCP.
"d" - Again, think about solubility as an equilibrium. NaCl(s) <-> Na+(aq) + Cl-(aq). If you use a "solvent" that already has a bunch of chloride ions dissolved in it, it will push the equilibrium left.
"e" - for an exothermic process, "heat" is a product of the reaction. If you add a product, the equilibrium will shift toward reactants.
14. You have found the following value in a table of equilibrium constants at 25ºC:
3 O2(g) 2 O3(g) Kc = 6.25x10-58
2 NO(g) + O2(g) 2 NO2(g) Kc = 2.25x1012
What is the equilibrium constant for the reaction:
NO(g) + O3(g) NO2(g) + O2(g)
a. 3.60x1069
b. 1.41x10-45
c. 2.03x1053
d. 6.00x1034
e. 9.38x10-52
How do we have to rearrange the two given reactions so they add up to the final reaction? Looks like the first rxn has to be reversed and multiplied by 0.5, so the "adjusted" equilibrium constant for that step is (1/Kc1)0.5. The second rxn has to be multiplied by 0.5, so the "adjusted" equilibrium constant is Kc20.5. Therefore, the equilibrium constant for the final rxn is: (1/Kc1)0.5 Kc20.5 = 6x1034. {Just a note in case you're having a bad math day, raising something to the 0.5 power is the same as taking the square root...}
I'll answer any other questions first thing in the morning.
9. Which of the following cannot be explained by LeChatelier’s Principle?
a. Removal of a gaseous product from a reaction drives the reaction to completion.
b. Any salt will dissolve in water, even though the amount might be extremely small.
c. The activation energy of a reverse chemical process is related to the activation energy and
ΔH of the forward process.
d. Less of a salt will dissolve in a solution that has a “common ion” than in pure water.
e. The equilibrium of an exothermic process will be more reactant-favored at higher
temperatures
"a" - removing a product will drive an equilibrium toward products, that's almost the definition of LeChatelier's Principle (I'll call it LCP from here on...)
"b" - Think about solubility in terms of a chemical reaction/equilibrium, salt + water yields ions in solution. Water acts as a reactant here, so adding more water will stress the equilibrium and shift it toward products.
"c" - This is purely kinetics, so it'd be tough to explain it with LCP.
"d" - Again, think about solubility as an equilibrium. NaCl(s) <-> Na+(aq) + Cl-(aq). If you use a "solvent" that already has a bunch of chloride ions dissolved in it, it will push the equilibrium left.
"e" - for an exothermic process, "heat" is a product of the reaction. If you add a product, the equilibrium will shift toward reactants.
14. You have found the following value in a table of equilibrium constants at 25ºC:
3 O2(g) 2 O3(g) Kc = 6.25x10-58
2 NO(g) + O2(g) 2 NO2(g) Kc = 2.25x1012
What is the equilibrium constant for the reaction:
NO(g) + O3(g) NO2(g) + O2(g)
a. 3.60x1069
b. 1.41x10-45
c. 2.03x1053
d. 6.00x1034
e. 9.38x10-52
How do we have to rearrange the two given reactions so they add up to the final reaction? Looks like the first rxn has to be reversed and multiplied by 0.5, so the "adjusted" equilibrium constant for that step is (1/Kc1)0.5. The second rxn has to be multiplied by 0.5, so the "adjusted" equilibrium constant is Kc20.5. Therefore, the equilibrium constant for the final rxn is: (1/Kc1)0.5 Kc20.5 = 6x1034. {Just a note in case you're having a bad math day, raising something to the 0.5 power is the same as taking the square root...}
I'll answer any other questions first thing in the morning.
Questions...
A couple email questions....
Someone asked about #5 and #9 from the problem set that was handed out on the day I missed class. All the questions on that problem set are from previous problem sets, so you should be able to find answer keys posted on my web page under Previous Gen Chem II.
For #5, we will ultimately need to plug into the comparative form of the Arrhenius equation, so we need to wrestle the information given into rate law constants and temperatures. The rate law expression is given, so we can plug in the room temperature rate and the given concentrations to find "kroom". We can also plug in to get "kunknown". With the given activation energy and the appropriate value of R (8.314 J/mol.K), you can plug in all of the values in the comparative Arrhenius equation and solve for Tunknown. Make sure you pay attention to units, Ea is given in kJ and R uses J, you have to convert one of them before you solve for Tunknown or you'll have a random factor of 1000 floating around in your problem.
For #9, we're looking at integrated rate law problems. In all cases, we're using the 1st order IRL because the problem tells us that the reaction is 1st order w.r.t. Cl2(g). For all parts, plug in what you know and solve for the variable.
Other questions, let me know and I'll post answers to the blog.....
Someone asked about #5 and #9 from the problem set that was handed out on the day I missed class. All the questions on that problem set are from previous problem sets, so you should be able to find answer keys posted on my web page under Previous Gen Chem II.
For #5, we will ultimately need to plug into the comparative form of the Arrhenius equation, so we need to wrestle the information given into rate law constants and temperatures. The rate law expression is given, so we can plug in the room temperature rate and the given concentrations to find "kroom". We can also plug in to get "kunknown". With the given activation energy and the appropriate value of R (8.314 J/mol.K), you can plug in all of the values in the comparative Arrhenius equation and solve for Tunknown. Make sure you pay attention to units, Ea is given in kJ and R uses J, you have to convert one of them before you solve for Tunknown or you'll have a random factor of 1000 floating around in your problem.
For #9, we're looking at integrated rate law problems. In all cases, we're using the 1st order IRL because the problem tells us that the reaction is 1st order w.r.t. Cl2(g). For all parts, plug in what you know and solve for the variable.
Other questions, let me know and I'll post answers to the blog.....
2009-02-21
Manipulating equilibrium...
On Friday, we talked about how changes in the way a chemical equation is written affect the form and value of the equilibrium constant. A few notable cases:
1. If the chemical equation is reversed, the value of K is inverted. If A<->B has equilibrium constant Kf, then B<->A has equilibrium constant 1/Kf.
2. If a chemical equation is multiplied by some constant, the value of the equilibrium constant is raised to that power. If A<->B has equilibrium constant K, then 4A<->4B has equilibrium constant K4.
3. If two (or more) chemical equations are added together, the value of the equilibrium constant is the product of the equilibrium constants for all the individual reactions. If A<->B has equilibrium constant K1, and B<->C has equilibrium constant K2, then A<->C has equilibrium constant Knet = (K1)(K2).
There's a new Mastering Chemistry assignment posted, due Tuesday. Also, we have an exam on Wednesday, so if you have questions let me know or bring them to class on Monday, we will have time to review.
Don't forget that tonight is Go Pink with the Dragons. The women's basketball game starts at 6pm and they will be raising money for the Roger Maris Cancer Center.
See you all on Monday.
1. If the chemical equation is reversed, the value of K is inverted. If A<->B has equilibrium constant Kf, then B<->A has equilibrium constant 1/Kf.
2. If a chemical equation is multiplied by some constant, the value of the equilibrium constant is raised to that power. If A<->B has equilibrium constant K, then 4A<->4B has equilibrium constant K4.
3. If two (or more) chemical equations are added together, the value of the equilibrium constant is the product of the equilibrium constants for all the individual reactions. If A<->B has equilibrium constant K1, and B<->C has equilibrium constant K2, then A<->C has equilibrium constant Knet = (K1)(K2).
There's a new Mastering Chemistry assignment posted, due Tuesday. Also, we have an exam on Wednesday, so if you have questions let me know or bring them to class on Monday, we will have time to review.
Don't forget that tonight is Go Pink with the Dragons. The women's basketball game starts at 6pm and they will be raising money for the Roger Maris Cancer Center.
See you all on Monday.
2009-02-18
Equilibrium in the semi-dark
Today we looked at a couple additional equilibrium problems and behavior including:
The similarities, differences and conversion between concentration-based equilibrium constants (Kc) and pressure-based equilibrium constants (Kp). They're related by the ideal gas law!
How do we determine if a reaction is at equilibrium? The reaction quotient has the same mathematical form as the equilibrium constant and can tell us where a reaction is relative to equilibrium.
What happens when we push around a system that's at equilibrium? If experiences stress, and it squirms around until it relieves that stress. This is LeChatelier's Principle and it can be used to explain chemical equilibria, human behavior, economic trends, population dynamics, and a whole lot more.
There's a new Mastering Chemistry assignment posted, due Monday. There's also a MC due Friday, so keep up with them.
This weekend the MSUM women's basketball team is going pink to put a full court press on breast cancer. T-shirts are available at the bookstore, and the 21st-ranked Dragons would love to see a mountain of pink in the stands at their game against Concordia (St. Paul) on Saturday. This is one of the last home games of the regular season, so come and cheer the Dragons into the post season.
http://msumdragons.com/sports/2009/1/20/WBB_0120091718.aspx?id=82
See you all on Friday, the lights should be working again in SL104 by then....
The similarities, differences and conversion between concentration-based equilibrium constants (Kc) and pressure-based equilibrium constants (Kp). They're related by the ideal gas law!
How do we determine if a reaction is at equilibrium? The reaction quotient has the same mathematical form as the equilibrium constant and can tell us where a reaction is relative to equilibrium.
What happens when we push around a system that's at equilibrium? If experiences stress, and it squirms around until it relieves that stress. This is LeChatelier's Principle and it can be used to explain chemical equilibria, human behavior, economic trends, population dynamics, and a whole lot more.
There's a new Mastering Chemistry assignment posted, due Monday. There's also a MC due Friday, so keep up with them.
This weekend the MSUM women's basketball team is going pink to put a full court press on breast cancer. T-shirts are available at the bookstore, and the 21st-ranked Dragons would love to see a mountain of pink in the stands at their game against Concordia (St. Paul) on Saturday. This is one of the last home games of the regular season, so come and cheer the Dragons into the post season.
http://msumdragons.com/sports/2009/1/20/WBB_0120091718.aspx?id=82
See you all on Friday, the lights should be working again in SL104 by then....
2009-02-16
Equilibrium calculations
Today we went through a few more equilibrium calculations and looked at a couple ways equilibria can be simplified. First, the "concentrations" of pure solids and liquids do not appear in equilibrium constants because their concentrations are constant (essentially) throughout the reaction.
Next we looked at assumptions that can simplify the math involved in equilibrium calculations. In many problems, the numerical value of "x" {the change in concentration when a reaction goes to equilibrium} is so small that it can be ignored in places where it's added to or subtracted from another term. For example, if the term "1.328 - 3x" appears in your equilibrium calculation, you should examine just how big you expect "x" to be. If the value of "x" turns out to be 0.00000000168, then it really doesn't affect the term "1.328 - 3x", and that term can be simplified to "1.328". Knowing when to make these approximations takes a little practice, so always check your assumptions after you finish working through a problem.
FInally, we looked at the difference between using a "full molecular" chemical equation and a "net ionic" chemical equation to determine the equilibrium constant. In many ways, the net ionic equation will always give you a more correct picture of the actual chemistry that takes place, but as long as the equilibrium constant you calculate is consistent with the chemical equation you use, I don't mind which you choose. But I strongly encourage you to think about reactions as net ionic reactions whenever possible...
There's a new Mastering Chemistry assignment, due Friday.
Next we looked at assumptions that can simplify the math involved in equilibrium calculations. In many problems, the numerical value of "x" {the change in concentration when a reaction goes to equilibrium} is so small that it can be ignored in places where it's added to or subtracted from another term. For example, if the term "1.328 - 3x" appears in your equilibrium calculation, you should examine just how big you expect "x" to be. If the value of "x" turns out to be 0.00000000168, then it really doesn't affect the term "1.328 - 3x", and that term can be simplified to "1.328". Knowing when to make these approximations takes a little practice, so always check your assumptions after you finish working through a problem.
FInally, we looked at the difference between using a "full molecular" chemical equation and a "net ionic" chemical equation to determine the equilibrium constant. In many ways, the net ionic equation will always give you a more correct picture of the actual chemistry that takes place, but as long as the equilibrium constant you calculate is consistent with the chemical equation you use, I don't mind which you choose. But I strongly encourage you to think about reactions as net ionic reactions whenever possible...
There's a new Mastering Chemistry assignment, due Friday.
2009-02-13
Equilibrium
Today we transitioned fully into equilibrium and looked at what the equilibrium constant can tell us about a system at equilibrium. We worked though a problem that let us calculate the value of the equilibrium constant using initial concentrations, and then calculated equilibrium concentrations when we knew the value of the equilibrium constant.
Remember, equilibrium is a dynamic process, so it doesn't matter if we start with all reactants, all products, or some mixture of both, the system will always settle into conditions that give the same equilibrium constant. (As long as temperature and pressure remain constant...)
I looked over some Mastering Chemistry problems, and I didn't see any that I liked right now. A bunch of them will be much better after class on Monday, so I'll wait until then to post a new assignment.
Have a great weekend. If you're looking for something to do, the MSUM Symphony Orchestra is performing a program called "Symphonic Romance" Saturday night at 7:30pm in the Hansen Theatre. A number of your fellow Chem 210-ers will be part of the performance, so come out and show your support.
See you all Monday.
Remember, equilibrium is a dynamic process, so it doesn't matter if we start with all reactants, all products, or some mixture of both, the system will always settle into conditions that give the same equilibrium constant. (As long as temperature and pressure remain constant...)
I looked over some Mastering Chemistry problems, and I didn't see any that I liked right now. A bunch of them will be much better after class on Monday, so I'll wait until then to post a new assignment.
Have a great weekend. If you're looking for something to do, the MSUM Symphony Orchestra is performing a program called "Symphonic Romance" Saturday night at 7:30pm in the Hansen Theatre. A number of your fellow Chem 210-ers will be part of the performance, so come out and show your support.
See you all Monday.
2009-02-11
Kinetics to Equilibrium to cancelled class
On Monday we talked about reaction mechanisms and looked at how the overall rate of a reaction is determined by the slowest step in the mechanism. (Not too surprisingly, this step is called the "Rate Determining Step" or the "Rate Limiting Step", and is often noted/labeled "RDS".) Our discussion of reaction mechanisms led us to the idea that if a reaction can proceed in the forward direction, it might also be able to move in the reverse direction, establishing a condition called equilibrium.
For those of you who were in my lab yesterday this may come as no shock, but I have chosen to stay home today and try to finish recovering from this nasty little cold bug that has invaded my sinuses. I will have a problem set printed out for class today that should provide some good practice in Kinetics, I encourage you all to work on it together, but class is (technically) cancelled today.
Don't forget the current Mastering Chemistry assignment, due tomorrow. See you all on Friday.
For those of you who were in my lab yesterday this may come as no shock, but I have chosen to stay home today and try to finish recovering from this nasty little cold bug that has invaded my sinuses. I will have a problem set printed out for class today that should provide some good practice in Kinetics, I encourage you all to work on it together, but class is (technically) cancelled today.
Don't forget the current Mastering Chemistry assignment, due tomorrow. See you all on Friday.
2009-02-06
Activation energy....
Today we worked through a kinetics/rate law/activation energy problem. We will dive into reaction mechanisms on Monday...
There's a new Mastering Chemistry assignment, due Thursday.
Basketball is at home this weekend if you're looking for something to do. Tonight our teams take on Southwest MN State, tomorrow it's MSU Mankato. Women tip at 6, men at 8.
There's a new Mastering Chemistry assignment, due Thursday.
Basketball is at home this weekend if you're looking for something to do. Tonight our teams take on Southwest MN State, tomorrow it's MSU Mankato. Women tip at 6, men at 8.
2009-02-04
Kinetics goes BOOM...
Looks like I missed a couple days...
Since the last exam, we've been looking at Kinetics and trying to understand rates of chemical reactions. That has included average rates over (relatively) long time periods, and rate laws which describe the instantaneous initial rate of a reaction. By doing a little calculus, rate law expressions can be converted to integrated rate laws which show the relationship between concentration and time. Today we got to the idea of activation energy and looked at an example of a reaction that doesn't occur until a spark is added. Activation energy is found by using the Arrhenius equation in one of its forms ("standard", comparative or linear).
The new Mastering Chemistry Assignment is posted, due Monday. There will be another MC assignment posted on Friday, so don't wait to get started on this one...
Since the last exam, we've been looking at Kinetics and trying to understand rates of chemical reactions. That has included average rates over (relatively) long time periods, and rate laws which describe the instantaneous initial rate of a reaction. By doing a little calculus, rate law expressions can be converted to integrated rate laws which show the relationship between concentration and time. Today we got to the idea of activation energy and looked at an example of a reaction that doesn't occur until a spark is added. Activation energy is found by using the Arrhenius equation in one of its forms ("standard", comparative or linear).
The new Mastering Chemistry Assignment is posted, due Monday. There will be another MC assignment posted on Friday, so don't wait to get started on this one...
Subscribe to:
Posts (Atom)