A number of people are still struggling with some of these ΔH manipulations, so let's look at a a bare-bones example. For the reaction A -> B, you have determined that ΔH = +10kJ/mol. If this is true, what's ΔH for the reaction B -> A? The difference in enthalpy between substance "A" and substance "B" is 10kJ/mol, the sign just tells you which is higher (or lower), therefore, ΔH for the reaction B -> A is the same magnitude but in the opposite direction. Similarly, what about the reaction 2A -> 2B? This reaction represents twice as much "action" as the original A -> B, so the energy required (it's endothermic) must be twice as much as the original. Summarizing:
A -> B ΔH=+10kJ/mol
B -> A ΔH=-10kJ/mol
2A -> 2B ΔH=+20kJ/mol
5B -> 5A ΔH=-50kJ/mol
Another question from a previous exam, I've copied the answer key below:
13. The specific heat capacity of gold is 0.128J/g•ºC and the specific heat capacity of iron is 0.449
J/g•ºC. You have heated a 51.294g block of iron to 49.318ºC and placed it on a gold block at
21.516ºC. When the system reaches thermal equilibrium, the temperature of the gold and
iron blocks are 34.468ºC. If the system is perfectly insulated, what was the mass of the gold
block in grams? (20pts each)
The assumption that the system is perfectly insulated implies that all of the energy lost by the iron block is
transferred to the gold block. Finding the energy lost by the iron block:
Eiron = (0.449 J/g•ºC)(51.294g Fe)(14.850ºC)
And the energy gained by the gold block:
Egold = (0.128 J/g•ºC)(x g Au)(12.952ºC)
Setting these energies equal:
(0.449 J/g•ºC)(51.294g Fe)(14.850ºC) = (0.128 J/g•ºC)(x g Au)(12.952ºC)
x = 206g Au
The question came up because this problem might appear to be missing a couple negative signs. Well, it's sort of missing a couple negative signs, so let me clarify (hopefully). The block of iron starts out hot and transfers heat to the gold block. That means that the temperature change for iron should be negative (34.468-49.318 = -14.850), which will give a negative value of Eiron. But look at how the answer is worded. By stating that the energy is lost by the iron block, the answer is implying a negative sign. In this case, the way I worded the answer takes care of the positive and negative signs because I was looking at the quantity of energy independent of the direction that energy was being transferred. If you leave the negatives in when you calculate Eiron, then you will need to include an additional negative sign when you set Eiron equal to Egold.
2008-10-28
Exam prep questions.....
A couple questions have come in on email:
How do we do a problem like this? i got it from the spring 2008 exam 3a.
You have determined that ΔHºreaction for the following reaction is –311.2kJ/mol.
3 Ca(OH)2(s) + 2 H3PO4(s) -> Ca3(PO4)2(s) + 6 H2O(l)
What is ΔHºreaction for the reaction:
2 Ca3(PO4)2(s) + 12 H2O(l)-> 6 Ca(OH)2(s) + 4 H3PO4(s)
a. 622.4 kJ/mol
b. 311.2 kJ/mol
c. 155.6 kJ/mol
d. -311.2 kJ/mol
e. -622.4 kJ/mol
This question is all about how the value of ΔHºreaction changes as a reaction is modified. The "new" reaction is the reverse of the original reaction, so the "new" process must be endothermic rather than exothermic (change the sign of ΔHºreaction). The "new" reaction is also balanced as a multiple of the first, so if 2x as much reaction occurs, then 2x the heat is required. The correct answer is "a".
Yes, I expect you to be able to draw these. "Qualitatively correct" means I'm not looking for perfectly measured jumps and drops, but if a step in a reaction is 20kJ/mol exothermic it should look significantly larger than a step that's 2 kJ/mol exothermic, and both of them should be steps down. Your text doesn't always exactly call them reaction coordinate diagrams, but the figures on p. 242, 260, and 264 are all reaction coordinate diagrams. Although they're formatted a little differently than the way I typically draw them, these diagrams are conveying the same type of information. For those of you who will be taking Chem 210, some of the concepts we talk about in the second semester will be significantly easier to visualize if you start thinking about reactions as they appear/occur in a reaction coordinate diagram.
Keep the questions coming, I will post answers to the blog as soon as I can.
How do we do a problem like this? i got it from the spring 2008 exam 3a.
You have determined that ΔHºreaction for the following reaction is –311.2kJ/mol.
3 Ca(OH)2(s) + 2 H3PO4(s) -> Ca3(PO4)2(s) + 6 H2O(l)
What is ΔHºreaction for the reaction:
2 Ca3(PO4)2(s) + 12 H2O(l)-> 6 Ca(OH)2(s) + 4 H3PO4(s)
a. 622.4 kJ/mol
b. 311.2 kJ/mol
c. 155.6 kJ/mol
d. -311.2 kJ/mol
e. -622.4 kJ/mol
This question is all about how the value of ΔHºreaction changes as a reaction is modified. The "new" reaction is the reverse of the original reaction, so the "new" process must be endothermic rather than exothermic (change the sign of ΔHºreaction). The "new" reaction is also balanced as a multiple of the first, so if 2x as much reaction occurs, then 2x the heat is required. The correct answer is "a".
I was doing some practice exams from previous years and are we
going to have to know how to do a qualitatively correct reaction coordinate
diagram for a stepwise process?
Yes, I expect you to be able to draw these. "Qualitatively correct" means I'm not looking for perfectly measured jumps and drops, but if a step in a reaction is 20kJ/mol exothermic it should look significantly larger than a step that's 2 kJ/mol exothermic, and both of them should be steps down. Your text doesn't always exactly call them reaction coordinate diagrams, but the figures on p. 242, 260, and 264 are all reaction coordinate diagrams. Although they're formatted a little differently than the way I typically draw them, these diagrams are conveying the same type of information. For those of you who will be taking Chem 210, some of the concepts we talk about in the second semester will be significantly easier to visualize if you start thinking about reactions as they appear/occur in a reaction coordinate diagram.
Keep the questions coming, I will post answers to the blog as soon as I can.
2008-10-27
Almost exam time....
Today we went over a few problems that people asked about in preparation for the exam on Wednesday. If you have questions, let me know and I will post answers on the blog.
If you're looking for sample exams, last fall and spring Exam #3 are probably a good place to start. There may be a couple questions on those exams from Chapter 7, so if a question seems beyond what we've covered in class it probably is. Any of the previous years' Exam #3 will have Thermodynamics, heat capacity and enthalpy problems, so check those out as well.
Test Review SI session will be Tuesday at 1:30 in LO 84
Good luck in your test prep and I'll see you all Wednesday if not sooner.
If you're looking for sample exams, last fall and spring Exam #3 are probably a good place to start. There may be a couple questions on those exams from Chapter 7, so if a question seems beyond what we've covered in class it probably is. Any of the previous years' Exam #3 will have Thermodynamics, heat capacity and enthalpy problems, so check those out as well.
Test Review SI session will be Tuesday at 1:30 in LO 84
Good luck in your test prep and I'll see you all Wednesday if not sooner.
2008-10-25
Net ionic equation and combustion reactions
Friday in class we examined how the enthalpy of a reaction can be calculated by using the net ionic equation rather than the full equation, with the result being the same in either case. This is because the net ionic equation is describing the chemistry that occurs without all the "fluff" of spectator ions. We also worked through an activity looking at the enthalpy of combustion for a variety of fuels. By critically evaluating the numerical data, we can determine which fuel is the "best" based upon whatever we consider most important (energy content, storage capacity, transportation issues)
2008-10-22
Coupled enthalpy problems and Hess' Law
Looks like I missed posting after Monday's class. Oop.
We're elbow-deep in enthalpy at this point, today we worked through another couple problems in which we calculated the enthalpy of reaction, and used the enthalpy/heat/energy liberated by a reaction to heat up a block of granite in a heat capacity problem. We also worked through an example of a multi-step reaction and the realtionship between the enthalpy of each step and the overall enthalpy of the net reaction. This is Hess' Law, and is basically an example of how state functions work: the path doesn't matter, just the starting and ending points.
A new Mastering Chemistry is posted, due Monday. Don't wait until the last minute, our next exam is next Wednesday. As I mentioned in class today, Exam #3 will only cover through Chapter 6, we'll roll Chapter 7 into the next exam.
We're elbow-deep in enthalpy at this point, today we worked through another couple problems in which we calculated the enthalpy of reaction, and used the enthalpy/heat/energy liberated by a reaction to heat up a block of granite in a heat capacity problem. We also worked through an example of a multi-step reaction and the realtionship between the enthalpy of each step and the overall enthalpy of the net reaction. This is Hess' Law, and is basically an example of how state functions work: the path doesn't matter, just the starting and ending points.
A new Mastering Chemistry is posted, due Monday. Don't wait until the last minute, our next exam is next Wednesday. As I mentioned in class today, Exam #3 will only cover through Chapter 6, we'll roll Chapter 7 into the next exam.
2008-10-17
Coupled systems and heat of reaction
Today we worked through a coupled systems problem in which a warm block of lead transferred energy to a cooler block of silver. In working through the algebra, I made an error somewhere and came up with the wrong answer, so let's try to correct it here. The set-up was OK, so we got to the following equation:
(0.128 J/g.degC)(17.89g)(Tfinal-42.73degC) = -(0.235 J/g.degC)(23.13g)(Tfinal-15.42degC)
This is set up correctly, now we have to work through the algebra:
(Tfinal-42.73degC) = -{(0.235 J/g.degC)(23.13g)}/{(0.128 J/g.degC)(17.89g)}(Tfinal-15.42degC)
Turning the great big ugly fraction into a number, we get:
(Tfinal-42.73degC) = -(2.3737)(Tfinal-15.42degC)
Distributing the -2.3737:
(Tfinal-42.73degC) = -(2.3737Tfinal)+(2.3737)(15.42degC)
Moving the Tfinal terms to one side and the numerical terms to the other:
Tfinal + 2.3737Tfinal = (2.3737)(15.42degC) + 42.73degC
Factoring out Tfinal and solving:
Tfinal = {(2.3737)(15.42degC) + 42.73degC}/{1 + 2.3737} = 23.5degC
There is a new Mastering Chemistry posted, due Tuesday. Have a great weekend.
(0.128 J/g.degC)(17.89g)(Tfinal-42.73degC) = -(0.235 J/g.degC)(23.13g)(Tfinal-15.42degC)
This is set up correctly, now we have to work through the algebra:
(Tfinal-42.73degC) = -{(0.235 J/g.degC)(23.13g)}/{(0.128 J/g.degC)(17.89g)}(Tfinal-15.42degC)
Turning the great big ugly fraction into a number, we get:
(Tfinal-42.73degC) = -(2.3737)(Tfinal-15.42degC)
Distributing the -2.3737:
(Tfinal-42.73degC) = -(2.3737Tfinal)+(2.3737)(15.42degC)
Moving the Tfinal terms to one side and the numerical terms to the other:
Tfinal + 2.3737Tfinal = (2.3737)(15.42degC) + 42.73degC
Factoring out Tfinal and solving:
Tfinal = {(2.3737)(15.42degC) + 42.73degC}/{1 + 2.3737} = 23.5degC
There is a new Mastering Chemistry posted, due Tuesday. Have a great weekend.
2008-10-15
Internal energy and heat capacity
Today we talked about internal energy and how keeping track of the work done on or by the system and the heat transferred into or out of the system can help us follow the change in internal energy of a system. In order to relate heat transfer and temperature change, we explored the concept of heat capacity, especially the more generally applicable specific heat capacity and molar hear capacity. We'll take a closer look at some heat capacity problems on Friday and begin to look at the factors that influence heat transfer in chemical reactions.
2008-10-10
Exam returned and Thermodynamics
Today in class exams were returned. If you were not in class, I will bring exams with me next Wednesday. The average was a little lower than last time. I also took a closer look at Mastering Chemistry scores and there are quite a few people who are just not doing Mastering Chemistry assignments. Try to keep up on these, MC is around 13% of your grade.
For anyone in my lab sections, hand-ins are due next Wednesday by 12:30pm for all of my sections (Tuesday 9am, Wednesday 11:30am, Thursday 9am). Chem 150L will not meet next week due to the Fall Breather.
Have a great weekend and I'll see you all on Wednesday.
For anyone in my lab sections, hand-ins are due next Wednesday by 12:30pm for all of my sections (Tuesday 9am, Wednesday 11:30am, Thursday 9am). Chem 150L will not meet next week due to the Fall Breather.
Have a great weekend and I'll see you all on Wednesday.
2008-10-06
Acid strength
One of the multiple choice questions brings up the idea of strong vs. weak acids and bases:
Which of the following is the strongest acid?
a.KOH (aq)
b.HClO4 (aq)
c.HC2H3O2 (aq)
d.H20 (aq)
e.NH3 (aq)
Strong acids (and bases) are those which ionize/dissolve completely when added to water. Let's approach this question by first identifying each substance listed. KOH is a strong base, so it's definitely not the strongest acid. H2O is water, doesn't seem like a strong acid. NH3 is ammonia, it's a weak base. That leaves us with HClO4 and HC2H3O2. Perchloric acid is a strong acid, HC2H3O2 is vinegar, a weak acid. By eliminating the bases and water, you turn this from a 20% chance of random guessing, to a 50:50 chance.
Check Table 4.2 in the text, the acids and bases listed in black are all "strong", those listed in pink are weak. For our purposes, any acid or base that is not listed in that table as being "strong" should be considered weak.
Other questions, let me know...
Which of the following is the strongest acid?
a.KOH (aq)
b.HClO4 (aq)
c.HC2H3O2 (aq)
d.H20 (aq)
e.NH3 (aq)
Strong acids (and bases) are those which ionize/dissolve completely when added to water. Let's approach this question by first identifying each substance listed. KOH is a strong base, so it's definitely not the strongest acid. H2O is water, doesn't seem like a strong acid. NH3 is ammonia, it's a weak base. That leaves us with HClO4 and HC2H3O2. Perchloric acid is a strong acid, HC2H3O2 is vinegar, a weak acid. By eliminating the bases and water, you turn this from a 20% chance of random guessing, to a 50:50 chance.
Check Table 4.2 in the text, the acids and bases listed in black are all "strong", those listed in pink are weak. For our purposes, any acid or base that is not listed in that table as being "strong" should be considered weak.
Other questions, let me know...
Review day....
Today we worked through a few review problems for the exam on Wednesday and talked about the density of gases for a minute or two.
SI sessions: Because of the exam, there will not be an SI session this Wednesday or Thursday. SI will resume next week.
Exam issues: If you choose to use a calculator, remember that graphing/programmable calculators are not allowed.
Feel free to email me any questions, I will posts answers to the blog. Good luck in your preparation.
SI sessions: Because of the exam, there will not be an SI session this Wednesday or Thursday. SI will resume next week.
Exam issues: If you choose to use a calculator, remember that graphing/programmable calculators are not allowed.
Feel free to email me any questions, I will posts answers to the blog. Good luck in your preparation.
2008-10-03
Gas stoichiometry and mixtures
We worked through a couple stoichiometry problems, including one that required use of the ideal gas law to determine the volume of gas produced by a reaction. These are the same stoichiometry problems that we've been working on for the last couple weeks, they just use different bits and pieces of info to get into and out of moles. We also talked about pressure and how the atomic understanding of "pressure" helps us understand mixtures of gases including Dalton's Law of Partial Pressures.
There is a new Mastering Chemistry assignment, due Tuesday. Exam #2 is next Wednesday. Have a good weekend.
There is a new Mastering Chemistry assignment, due Tuesday. Exam #2 is next Wednesday. Have a good weekend.
2008-10-01
Kinetic Molecular Theory and "Ideal" gases
Today we explored an explanation of the gas laws we talked about on Monday, the Kinetic Molecular Theory of Gases. Gas laws are based upon observation and tell us what happened or will happen; kinetic molecular theory attempts to explain why those observations happen. We had a new problem set in class today as well, it's posted on my web page.
I did not put up a new Mastering Chemistry assignment today, but there will almost certainly be a new one of Friday. Your next exam is next Wednesday, be sure to keep up on those MC assignments and take a look at the old exams on my web page.
I did not put up a new Mastering Chemistry assignment today, but there will almost certainly be a new one of Friday. Your next exam is next Wednesday, be sure to keep up on those MC assignments and take a look at the old exams on my web page.
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