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...
2009-01-27
Another question...
Can you explain this? I don't understand the differences between the interactions.
3. Which of the following is true regarding intermolecular forces?
a. Hydrogen bonds are as strong as covalent bonds
b. On average, dipole-dipole interactions are weaker than induced dipole-induced dipole
c. Induced dipole-induced dipole interactions are stronger for large atoms/molecules
d. Hydrogen bonds occur whenever a molecule contains hydrogen
e. Hydrogen bonds are just a special kind of induced dipole-induced dipole interaction
Let's take these one at a time.
a. Hydrogen bonds are as strong as covalent bonds
Covalent bonds are the bonds that hold molecules together. They are MUCH stronger than any intermolecular force. If IMFs were stronger than covalent bonds, the IMFs would rip molecules apart.
b. On average, dipole-dipole interactions are weaker than induced dipole-induced dipole
Induced dipole-induced dipole interactions (or if you prefer, you can call them London forces, or dispersion forces, or London dispersion forces) are temporary dipoles than change over time. Although some induced dipoles can be very strong and some regular dipoles can be quite weak, an average London force will be weaker than an average dipole-dipole interaction.
c. Induced dipole-induced dipole interactions are stronger for large atoms/molecules
Induced dipoles form when electron clouds are deformed. The bigger an electron cloud, the easier it is to deform. This can be because it's part of a big atom (like I or Po), or because it's part of a big molecule (like long hydrocarbons). The more an electron cloud can deform, the stronger it's induced dipole will be.
d. Hydrogen bonds occur whenever a molecule contains hydrogen
The hydrogen involved in a hydrogen bond has to be polar, that is, it has to be part of a polar bond. A molecule of H2 contains hydrogen, but the H-H bond is not polar, so we wouldn't call its IMFs hydrogen bonds.
e. Hydrogen bonds are just a special kind of induced dipole-induced dipole interaction
H-bonds are a special/specific type of dipole-dipole interaction.
So it looks like the correct answer is "c".
I will not have access to the blog or email again until tomorrow morning, I will answer any questions I get overnight as soon as I get to my office tomorrow. {Usually by 7am}
3. Which of the following is true regarding intermolecular forces?
a. Hydrogen bonds are as strong as covalent bonds
b. On average, dipole-dipole interactions are weaker than induced dipole-induced dipole
c. Induced dipole-induced dipole interactions are stronger for large atoms/molecules
d. Hydrogen bonds occur whenever a molecule contains hydrogen
e. Hydrogen bonds are just a special kind of induced dipole-induced dipole interaction
Let's take these one at a time.
a. Hydrogen bonds are as strong as covalent bonds
Covalent bonds are the bonds that hold molecules together. They are MUCH stronger than any intermolecular force. If IMFs were stronger than covalent bonds, the IMFs would rip molecules apart.
b. On average, dipole-dipole interactions are weaker than induced dipole-induced dipole
Induced dipole-induced dipole interactions (or if you prefer, you can call them London forces, or dispersion forces, or London dispersion forces) are temporary dipoles than change over time. Although some induced dipoles can be very strong and some regular dipoles can be quite weak, an average London force will be weaker than an average dipole-dipole interaction.
c. Induced dipole-induced dipole interactions are stronger for large atoms/molecules
Induced dipoles form when electron clouds are deformed. The bigger an electron cloud, the easier it is to deform. This can be because it's part of a big atom (like I or Po), or because it's part of a big molecule (like long hydrocarbons). The more an electron cloud can deform, the stronger it's induced dipole will be.
d. Hydrogen bonds occur whenever a molecule contains hydrogen
The hydrogen involved in a hydrogen bond has to be polar, that is, it has to be part of a polar bond. A molecule of H2 contains hydrogen, but the H-H bond is not polar, so we wouldn't call its IMFs hydrogen bonds.
e. Hydrogen bonds are just a special kind of induced dipole-induced dipole interaction
H-bonds are a special/specific type of dipole-dipole interaction.
So it looks like the correct answer is "c".
I will not have access to the blog or email again until tomorrow morning, I will answer any questions I get overnight as soon as I get to my office tomorrow. {Usually by 7am}
Exam tomorrow...
Yesterday in class we (more or less) reviewed for the exam. We added normality to our list of concentrations, normality is closely related to molarity.
Since people didn't have any additional questions before the exam, we also started talking about kinetics, we'll dig deeper into rates starting Friday.
A few questions have come in:
How do we do a problem like this? (from Spring 2005 exam)
10. What is the mol fraction of sugar (C6H12O6) in a saturated aqueous sugar solution at 25ºC?
(Solubility of sugar in water at 25ºC = 211.4 g/100.mL)
a. 0.1744
b. 0.6789
c. 11.73
d. 2.114
e. 0.2112
This is largely just a unit conversion problem. We're looking for mol fraction of sugar in water, so we need to know (mols of sugar), (mols of water), and (total mols). Looking at that solubility, we can convert 211.4g of sugar to mols (211.4g / 180.155g/mol = 1.1734mols sugar), and we can convert 100.mL of water to mols (100.g / 18.015g/mol = 5.5509mols water). The mol fraction of sugar is (1.1734mols sugar / (1.1734+5.5509)total mols = 0.1745) so the answer should be "a". Just a quick note on test strategy here, you know by definition that the only possible values of mol fraction are between 0 and 1; therefore, you know that "c" and "d" must not be correct. Even if you had to guess, eliminating those two impossible answers increases your odds from 20% to 33%.
Hi, for the Spring 2008 exam you gave, I was hoping you could post how to do number six regarding vapor pressure on the blog.
6. If each of the following solids is added to 500.0mL of water, which will change the vapor pressure the most?
a. 1.2mols sugar
b. 0.4mols calcium phosphate
c. 0.6mols sodium chloride
d. 0.5mols calcium nitrate
e. 0.7mols ammonium phosphate
We worked through this one in class on Monday. Since all of these solutes are being dissolved in the same amount of water, the one that provides the most solute particles will affect the vapor pressure the most. Sugar is a molecular solute, so 1.2mols of sugar will result in 1.2 mols of solute particles in solution. Calcium phosphate will break into 5 pieces, so 0.4mols of Ca3(PO4)2 will result in 5(0.4) = 2.0mols of particles in solution. Sodium chloride = 2 particles, 2(0.6) = 1.2mols particles. Calcium nitrate = 3 particles, 3(0.5) = 1.5mols particles. Ammonium phosphate = 4 particles, 4(0.7) = 2.8mols particles. Since ammonium phosphate results in the most solute particles, it is expected to affect the vapor pressure the most.
Other questions, let me know.
Since people didn't have any additional questions before the exam, we also started talking about kinetics, we'll dig deeper into rates starting Friday.
A few questions have come in:
How do we do a problem like this? (from Spring 2005 exam)
10. What is the mol fraction of sugar (C6H12O6) in a saturated aqueous sugar solution at 25ºC?
(Solubility of sugar in water at 25ºC = 211.4 g/100.mL)
a. 0.1744
b. 0.6789
c. 11.73
d. 2.114
e. 0.2112
This is largely just a unit conversion problem. We're looking for mol fraction of sugar in water, so we need to know (mols of sugar), (mols of water), and (total mols). Looking at that solubility, we can convert 211.4g of sugar to mols (211.4g / 180.155g/mol = 1.1734mols sugar), and we can convert 100.mL of water to mols (100.g / 18.015g/mol = 5.5509mols water). The mol fraction of sugar is (1.1734mols sugar / (1.1734+5.5509)total mols = 0.1745) so the answer should be "a". Just a quick note on test strategy here, you know by definition that the only possible values of mol fraction are between 0 and 1; therefore, you know that "c" and "d" must not be correct. Even if you had to guess, eliminating those two impossible answers increases your odds from 20% to 33%.
Hi, for the Spring 2008 exam you gave, I was hoping you could post how to do number six regarding vapor pressure on the blog.
6. If each of the following solids is added to 500.0mL of water, which will change the vapor pressure the most?
a. 1.2mols sugar
b. 0.4mols calcium phosphate
c. 0.6mols sodium chloride
d. 0.5mols calcium nitrate
e. 0.7mols ammonium phosphate
We worked through this one in class on Monday. Since all of these solutes are being dissolved in the same amount of water, the one that provides the most solute particles will affect the vapor pressure the most. Sugar is a molecular solute, so 1.2mols of sugar will result in 1.2 mols of solute particles in solution. Calcium phosphate will break into 5 pieces, so 0.4mols of Ca3(PO4)2 will result in 5(0.4) = 2.0mols of particles in solution. Sodium chloride = 2 particles, 2(0.6) = 1.2mols particles. Calcium nitrate = 3 particles, 3(0.5) = 1.5mols particles. Ammonium phosphate = 4 particles, 4(0.7) = 2.8mols particles. Since ammonium phosphate results in the most solute particles, it is expected to affect the vapor pressure the most.
Other questions, let me know.
2009-01-23
Perhaps the most important force in biological systems...
Today we talked about osmosis, osmotic pressure, colloids and surfactants. These are all extremely important concepts in biological systems, so they should be pretty important to all of us biological organisms.
There is a new Mastering Chemistry assignment, due Tuesday.
Exam is next Wednesday. Old exams are posted on my MSUM web page (www.mnstate.edu/bodwin), if you haven't looked at them yet you might want to take a peek.
Have a good weekend, if you have any questions feel free to email me, I will answer all questions to the blog.
There is a new Mastering Chemistry assignment, due Tuesday.
Exam is next Wednesday. Old exams are posted on my MSUM web page (www.mnstate.edu/bodwin), if you haven't looked at them yet you might want to take a peek.
Have a good weekend, if you have any questions feel free to email me, I will answer all questions to the blog.
2009-01-22
Solution behavior and colligative properties
In class Wednesday we reviewed concentration units and worked through a vapor pressure depression/boiling point elevation/freezing point depression problem using potassium bromide in water. We also talked a little bit about how temperature affects solubility of solids or gases in liquid solvents.
There is a new Mastering Chemistry assignment posted, due Monday.
Exam 1 has been moved up to next Wednesday, January 28th. Please make note of this and remind anyone else who is registered for this class.
See you all on Friday...
There is a new Mastering Chemistry assignment posted, due Monday.
Exam 1 has been moved up to next Wednesday, January 28th. Please make note of this and remind anyone else who is registered for this class.
See you all on Friday...
2009-01-17
Phase diagrams and solution
On Friday we took the 2-dimensional heating/cooling curves we talked about on Wednesday and translated them into a phase diagram that relates pressure and temperature to the states of matter for a substance. Although understanding how pure substances behave is important, the majority of "matter" we encounter is not pure substance but various mixtures, so we started talking about solutions and how addition of a solute can change physcial properties. These colligative properties depend upon the number of solute particles present in the solution, not the identity of those solute particles.
See you all on Wednesday. (or Tuesday if you're in my lab section...)
See you all on Wednesday. (or Tuesday if you're in my lab section...)
2009-01-14
Liquids and phase changes
Today we talked about some properties of liquids that can be explained using intermolecular forces including viscosity, surface tension, capillary action and vapor pressure. We then moved to phase transitions and looked at how a heating/cooling curve could be used to understand how much energy in transferred as heat is added or removed from a substance.
There is a new Mastering Chemistry assignment posted, due Monday. My web page is back up so you should be able to access all the information found there, please let me know if there are broken links or missing files.
See you all on Friday.
There is a new Mastering Chemistry assignment posted, due Monday. My web page is back up so you should be able to access all the information found there, please let me know if there are broken links or missing files.
See you all on Friday.
2009-01-12
First day...
Today we ran through the syllabus and began looking at the interaction between energy and matter as related to states of matter.
Mastering Chemistry Course ID = C210SP09JB
If anyone has tried to look at my MSUM web page, I had a slight problem this morning and managed to destroy my page. I've gotten a few bits of info back up, but it will likely take a day (or 2 or 3) to get everything in full working order. Sorry about that.
Mastering Chemistry Course ID = C210SP09JB
If anyone has tried to look at my MSUM web page, I had a slight problem this morning and managed to destroy my page. I've gotten a few bits of info back up, but it will likely take a day (or 2 or 3) to get everything in full working order. Sorry about that.
2009-01-07
Welcome to Chem 210
Welcome to Dr. Bodwin's blog for the spring semester. Check back for class announcements and other info.
2008-12-16
Grades are in...
I hope everyone made it through the lovely weather we experienced Sunday and yesterday. I have submitted grades to the Registrar, they should be visible within a day. The average on the final exam was around 140/200. If you would like to see your final exam, stop by my office. I do not return final exams, but if you'd like to take a look at it you are welcome to do so.
I hope everyone has a great break and I look forward to seeing many of you back in class next semester. Don't forget about Lewis and VSEPR structures, we'll need them to get started in January....
I hope everyone has a great break and I look forward to seeing many of you back in class next semester. Don't forget about Lewis and VSEPR structures, we'll need them to get started in January....
2008-12-11
Questions...
A few questions came in overnight.....
just like exam #4, are there going to be an "explanation" section about trends in the periodic table or is it going to be "circle which one is bigger/smaller, has higher first ionization energy..."?
Yes, there will be questions where you will be expected to explain your answer.
could you give a brief explanation of endothermic vs exothermic in reaction...i remember i got a couple of those wrong. it seems like it should be straight forward but it's not always!
Endothermic reactions require heat, exothermic reactions liberate heat. The "trick" question here usually involves water freezing. To make water freeze, you must remove heat from the system, so freezing water (or any substance) is an exothermic process, even though it "feels cold" if you were to hold it in your hand.
Other questions, let me know. See you in a few hours...
just like exam #4, are there going to be an "explanation" section about trends in the periodic table or is it going to be "circle which one is bigger/smaller, has higher first ionization energy..."?
Yes, there will be questions where you will be expected to explain your answer.
could you give a brief explanation of endothermic vs exothermic in reaction...i remember i got a couple of those wrong. it seems like it should be straight forward but it's not always!
Endothermic reactions require heat, exothermic reactions liberate heat. The "trick" question here usually involves water freezing. To make water freeze, you must remove heat from the system, so freezing water (or any substance) is an exothermic process, even though it "feels cold" if you were to hold it in your hand.
Other questions, let me know. See you in a few hours...
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