Old exam keys

I'm not going to be able to post regular keys to Fall 2008 and Fall 2009 exam 3, so to let you check yourself, here are the answers to the "a" forms...

Fall 2008, Exam 3a

1 = d; 2 = b; 3 = c; 4 = c; 5 = endo,exo,exo,endo; 6 = f; 7 = d; 8 = c; 9 = c; 10 = -143kJ/mol; 11 = 127g; 12 = -39.79kJ/mol

Fall 2009, Exam 3a

1 = d; 2 = b; 3 = c; 4 = endo,exo,exo,endo; 5 = +1184.0kJ/mol; 6 = 35.42kJ; 7 = 1692kJ; 8 = +3810kJ/mol; 9 = -2375.2kJ/mol; 10 = 81.4g; 11 = +6.71kJ/mol

A few people are having problems solving problems like Winter 2006, Exam 3, #13. It sets up as:

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(0.84 J/g•ºC)(2.95x103 g)(Tfinal – (-77.91ºC)) = - (1.015 J/g•ºC)(10.00x103 g)(Tfinal – (20.00ºC))

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Solving this is algebra. I know the units all work out so let me work through this solution just using the numbers...

(0.84)(2950)(x - (-77.91)) = -(1.015)(10000)(x-20.00)

Let's combine all the numerical terms...

(2478)(x + 77.91) = (-10150)(x-20.00)

Distribute the numerical terms...

2478x + 193061 = (-10150)x + 203000

Moving all the "x" terms to one side and all the numerical terms to the other side...

(2478+10150)x = (203000-193061)

12628x = 9939

x = 0.787 = Tfinal

Another...
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How do you figure out problems like #7 on the Fall of '07 exam 3b?
7. Rust (Fe₂O₃) can be converted to iron by the following reaction:
2 Fe₂O₃(s) --> 4 Fe(s) + 3 O₂(g)
What is ΔHºreaction for this process? (ΔHfº = -824.2^kJ/_mol for Fe₂O₃.)
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This is an enthalpy of reaction problem, although it might seem like there's not enough info given. The products here are both uncombined elements in their standard states, so their standard enthalpy is zero. That makes the enthalpy of this reaction:
2(824.2^kJ/_mol) + 4(0^kJ/_mol) + 3(0^kJ/_mol) = 1648.4kJ (or ^kJ/_mol, or ^kJ/_{mol rxn}, or ^kJ/_rxn, see the discussion of this below...)

"Moles of Reaction"

Email question:
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I know you answered this question in class on Friday but I'm still uncertain on how you get mols of rxn? Can you give me a hypothetical on the equation we did on Friday from exam 3a from Fall '08? In this it's 1 mol rxn per 2 mols of haxane.
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OK, this one always causes some trouble, largely because I try to tie together the units that we typically see on enthalpy and the balanced equations. It might be easier to think of it simply as "reaction" rather than "moles of reaction", so we could look at something like:
2 H₃PO₄(aq) + 3 Ca(OH)₂(aq) --> Ca₃(PO₄)₂(s) + 6 H₂O(l)
Calculating {delta}H_rxn for this process, we get...
2(1288.3^kJ/_mol) + 3(542.8^kJ/_mol) + 6(230.02^kJ/_mol) + 1(-4120.8^kJ/_mol) + 6(-285.8^kJ/_mol) = -250.48^kJ/_mol
{Note: These are numbers I pulled from a table similar to the one in your textbook. Change the sign on reactants because these are being consumed in the reaction, not formed.}
Since {delta}H is negative, this reaction is exothermic, but what exactly are those units? Remember when I ran through one of the first enthalpy problems in class I used very complete and expanded units, let's just look at the first term here. The units on the enthalpy of formation for phosphoric acid are "kilojoules per mole of phosphoric acid formed". If we want to properly add these terms together, they have to have the same unit, so we have to convert/relate "moles of phosphoric acid" into something that is consistent throughout this problem. That's where the units on the "2" become important, even through they are often left off. That "2" comes from the balanced chemical equation and is really "2 moles of phosphoric acid per balanced chemical equation" or "2 moles of phosphoric acid per reaction". OK, so we do that for every term in the problem and then add them together to get the final answer with units of "kilojoules per balanced chemical equation" or "kilojoules per reaction" and everything is great... except that these enthalpies of reaction are often reported with units of "kJ/mol". Mole of what? In the above reaction, we could say it's per mole of calcium phosphate because for each "reaction" there is 1 mole of calcium phosphate formed, but that seems to limit us to problems that only deal with calcium phosphate. Here comes the magic unit "mol of reaction". Using the terminology above, we can say that one "balanced chemical equation" is one "mol of reaction".
Many (most? maybe all?) textbooks get around this problem by being quite explicit in the way they present thermochemical reactions, in fact your textbook has a section in Chapter 6 called "Thermochemical Expressions" {Sec. 6.5, p 230) that gives a nice example. If the {delta}H for a chemical equation is shown right next to the balanced chemical equation to which it refers, it can be implied that the {delta}H is valid only for the exact balanced equation shown, so the "per mol" or "per mol rxn" is often omitted and {delta}H is just reported with units of "kJ".

OK, after that LONG explanation, let's try a shorter answer. You can think of "mol of rxn" simply as "rxn". Each time the reaction happens once (as balanced), the calculated heat is liberated or consumed. The reason I tend to use the "mol of rxn" label is because it naturally leads to the question "What reaction?" which means that every enthalpy you calculate MUST be related to a specific balanced chemical equation.

Other questions, let me know...

A couple notes...

#1. If you look at old exams from a couple years ago, you might see some questions about photon energy and deBroglie wavelengths. These are topics from the next chapter and will not be included on this exam.

#2. I may post some answer keys for last year's exams before Monday, but I can't guarantee that I'll have time. I will try.

#3. As I mentioned in class, I misplaced my regular Casio calculator. If you happened to slip it in your bag when you visited my office I'd love to get it back. Thanks.

Email any questions, I'll answer them here on the blog. Have a good weekend, if you're looking for a study break the soccer team is home tomorrow (noon, v. Wayne State) and Sunday (1pm, v. Augustana). This may be the last weekend with nice weather, try to enjoy it. It might snow Tuesday or Wednesday night...

In-class problem

I think a number of people were confused at how to approach the problem we did in class on Wednesday. The problems was:

"fuel" reacts with oxygen to produce carbon dioxide gas and water gas. If 10.00g of "fuel" is burned in excess oxygen and all of the energy is transferred to 5.00L of water initially at 11.24degC, what is the final temperature of the water?

I'll pick a fuel none of you had, benzene, C6H6(l). This is a coupled-systems problem, the benzene will burn to produce/liberate heat in an enthalpy process, then the 5.00L sample of water will absorb the heat in a heat capacity process. Start with a balanced equation:
2 C₆H₆(l) + 15 O₂(g) --> 12 CO₂(g) + 6 H₂O(g)
Now calculate the {delta}H for the reaction from the standard enthalpies of formation found in the table in the back of your book.
2(-49.03^kJ/_mol) + 15 (-0^kJ/_mol) + 12(-393.509^kJ/_mol) + 6(-241.818^kJ/_mol) = -6271.08^kJ/_{mol rxn}
NOTE: change sign on reactants, don't change sign on products.
10.00g of benzene does not represent a "mol of reaction", so we need to scale that number to the amount of fuel being used:
(10.00g C₆H₆) / (78.113 ^g/_mol) = 0.1280mols benzene
(0.1280 mols C₆H₆) (1 mol rxn / 2 mols C₆H₆) = 0.06401 mols rxn
(0.06401 mols rxn) (6271.08^kJ/_{mol rxn}) = 401.4kJ of energy released by the reaction

OK, now the heat capacity part of the problem. I'm putting 401400J of heat into this 5.00L sample of water and changing its temperature. Use the units on heat capacity to set up the problem correctly:
(401400J) (1 g.degC / 4.184J) (1 / 5000g) = 19.19degC
This is the change in temperature, so the final temperature of the water must be (11.24+19.19)degC = 30.43degC

Other questions, let me know...

This week and new OWL

This (short) week we have started Chapter 6, Thermochemistry. Thermochemistry is all about heat: where it is, where it's moving, how much is moving, what it does when it moves, etc. As with everything, these problems become MUCH easier with some practice, so make sure you work some examples. To help you with that, there's a new OWL assignment posted. I've included a few more simulations and other rather visual modules in this set of assignments, if you're struggling with some of these energy issues it will help to work through some of these, be sure to take advantage of these opportunities rather than just clicking through to get the right answer.

Our next exam will cover ONLY chapter 6 and is just over a week away (October 25th). Do not wait until the last minute to study. I expect a significantly better average on Exam 3, but that won't happen if you don't start working on this material this weekend.

If you need a study break, there's a football game AND a volleyball game at home today (Saturday). What happens to the kinetic energy of a wide receiver and a linebacker when they collide? When the libero gets an awesome dig and the volleyball goes straight up in the air, how are the kinetic and potential energy of the ball changing? What kind of energy is represented by the glucose molecules in the energy drinks that the players drink during the game? See? There are even fun energy problems to consider when you're watching sports...

More questions...

A couple question...

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I was just wondering about the oxidation number. example: PH3 P has 3+ and H has 1- so all you need to do is subract the two or do i have the method wrong. thanks
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I'd start from the other direction on this, hydrogen is almost always oxidation number +1, so if there are 3 hydrogens at +1, and the molecule is neutral overall, then the phosphorus must be oxidation number -3.

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I was doing one of the "Conceptual Exercise" problems and can't figure out how they came up with the answer that they did. It is 5.4 on page 175 part B. I came up with H^+ + OH^- yields H_2_0. It looks like that is the answer but then it has another equation for an answer as well. I know I am forgetting something simple but I can't figure out what it is and it's driving me nuts. Please help me out and thanks in advance.
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These are a bunch of acid-base net ionic equations, so I'll address them all.

HCl(aq) + KOH(aq) --> H₂O(l) + KCl(aq)
H⁺(aq) + Cl^-(aq) + K⁺(aq) + OH^-(aq) --> H₂O(l) + K⁺(aq) + Cl^-(aq)
Chloride ions and potassium ions don't change, so they are spectators and the net ionic equation is:
H⁺(aq) + OH^-(aq) --> H₂O(l)

H₂SO₄(aq) + Ba(OH)₂(aq) --> 2 H₂O(l) + BaSO₄(s)
This one should look familiar...
2 H⁺(aq) + SO₄^2-(aq) + Ba²⁺(aq) + 2 OH^-(aq) --> 2 H₂O(l) + BaSO₄(s)
Since water is a molecule and barium sulfate is a precipitate, these both stay together in the full ionic equation, so the net ionic equation doesn't have any spectator ions here...
2 H⁺(aq) + SO₄^2-(aq) + Ba²⁺(aq) + 2 OH^-(aq) --> 2 H₂O(l) + BaSO₄(s)

CH₃COOH(aq) + NaOH(aq) --> H₂O(l) + NaCH₃COO(aq)
You could also represent acetic acid and acetate ions as CH₃CO₂H/CH₃CO₂^- or HC₂H₃O₂/C₂H₃O₂^-. Acetic acid is a weak acid, so it should not be split up in the ionic equation:
CH₃COOH(aq) + Na⁺(aq) + OH^-(aq) --> H₂O(l) + Na⁺(aq) + CH₃COO^-(aq)
The sodium ion is a spectator, so the net ionic is:
CH₃COOH(aq) + OH^-(aq) --> H₂O(l) + CH₃COO^-(aq)

Other questions, let me know, I'll be checking email all evening. I'll also post answers first thing in the morning. Good luck...

From email...

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On the exam, will you write the chemical formula as words or shorthand?
for example, hydroiodic acid as opposed to HI

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Yes. You will most likely see both, depending upon the goal of the problem. In some cases I want to see if you can write balanced formulas and balanced reactions, so I will give you the name of the chemical and expect you to write the correct formula. In other cases, I want to see if you can do the stoichiometry from a given chemical reaction so I'll use a chemical formula. In "special" cases, there are only a few names I expect you to know. You should know the names and formulas of all strong acids, ammonia, etc.

Almost exam time...

We've finished up chapters 4 and 5, exam next Wednesday. Let me know if you have any questions or anything specific you'd like to review on Monday in class. I've posted problem set #4 and the answer key on my mnstate.edu page, let me know if there are problems.

Have a good weekend and good luck preparing for the exam.