Enter equilibrium

The steady state approximation applies to kinetic systems, but if we think about the nature of chemical reactions, we can say that all chemical reactions are microscopically reversible.  For the simple reaction A↔B

If we assume that the forward and reverse reaction both represent elementary steps, we can write rate law expressions for each direction:

Rate_forward = k_forward[A]₀

Rate_reverse = k_reverse[B]₀

After some time has passed, the rate of the forward reaction will be equal to the rate of the reverse reaction.

Rate_forward = Rate_reverse
k_forward[A]₀ = k_reverse[B]₀     

Rearranging to group constants and concentrations, we get the expression for the equilibrium constant.

K_eq = [products]_eq / [reactants]_eq  

If [products]_eq = [reactants]_eq, then K_eq = 1 and the reaction does not favor products or reactants.  If the equilibrium is product-favored, the the value of K_eq will be larger than 1; if the equilibrium is reactant-favored, the the value of K_eq will be less than 1 but still positive.

Mechanisms

The mechanism for a chemical reaction is described by the rate law.  Mechanisms describe the collisional events that allow a reaction to occur and must obey 3 rules.

1. A mechanism must be composed of elementary steps/reactions.
2. The elementary steps of a mechanism, when added together, must yield the overall reaction
3. The observed rate law for the overall reaction must be consistent with the rate law for the slowest step

Elementary steps are the collisions that occur in a reaction.  Thinking about the probability of a collision, we can simplify the picture a little bit because it is extremely unlikely that more than 2 particles will collide with the proper orientation and energy to react.  This means that all elementary steps are either unimolecular or bimolecular.  {Yes, termolecular elementary steps are possible, but they're rarely significant contributors so we will ignore them for now.}
Once we have this series of elementary steps, what do we do with them?  Because elementary steps describe collisions at the molecular level, we can write rate laws for each of the elementary steps based only upon the balanced reaction of the elementary step.  Since doubling the number of any molecule will double the probability of a collision (the rate), the elementary steps are first order with respect to each reacting molecule.
Rates are determined by the activation energy of a step or an overall process; the higher the activation energy, the slower the rate.  For a series of steps, whichever step has the highest activation energy will determine (or limit) the rate for the entire process, so it is known as the Rate Determining (or Limiting) Step, the RDS (or RLS).  With a little algebra, we can wrassle the rate law of the RDS into a form that looks like the observed rate law.  The rate law expression for the overall process shown below, A + B + D→E, is

Rate_obs = k_obs[A]₀^x[B]₀^y[D]₀^z

If the first step is RDS, we can write the rate law expression for the first step:

Rate₁=k₁[A]₀¹[B]₀^y

This rate law, looks just like the observed rate law if the reaction is first order with respect to [A] and [B], so if the first step is RDS, this is a pretty straight-forward problem with minimal algebraic wrasslin'.

For the second step RDS, we have to use a steady state approximation to make the rate law "consistent with" the observed rate law.  The rate law for the second step is:

Rate₂=k₂[C]₀¹[D]₀¹

But [C] does not appear in the overall observed rate law expression, so we have to figure out a way to make this expression "consistent with" the observed rate law for the overall process.  If most of "C+D" has quite a bit of energy, but not quite enough to get over the C+D→E hump, it probably has enough energy to go backwards and return to "A+B".  This means we can define another rate law, this time for the reverse of step 1, C+D→A+B:

Rate_-1=k_-1[C]₀¹[D]₀¹

As the rxn A+B→C+D proceeds, eventually we will build up a concentration of "C+D" that will remain essentially constant throughout the reaction.  This is the "steady state".  When we reach this steady state, the rate of step 1 going forward (Rate₁) will be equal to the rate of step 1 in reverse (Rate_-1):

Rate₁ = Rate_-1

Which means that:

k₁[A]₀¹[B]₀¹ = k_-1[C]₀¹[D]₀¹

Now we can solve for  [C]₀¹ :

[C]₀¹ = {k₁/k_-1}[A]₀¹[B]₀¹/[D]₀¹

Now, we can plug the expression for  [C]₀¹ into the rate law expression for step 2:

Rate₂ =  k₂({k₁/k_-1}[A]₀¹[B]₀¹/[D]₀¹)[D]₀¹ = k_combined[A]₀¹[B]₀¹  

Note: Since "k₁", "k_-1", and "k₂" are constants, we can just lump them together into a single constant, in this case labelled "k_combined".

So if the second step is slow, the observed rate law should be first order w.r.t. [A] and [B], zero order w.r.t. [D].  Note that this is the same as the observed rate law expression if the first step is slow, so how can we distinguish between these two mechanisms?  Typically there would have to be something else observable in the reaction.  If step 2 is slow, then during the course of the reaction, we might expect to see a measurable concentration of "C" appear when the steady state is established and then disappear when the reaction is complete.  We might also be able to analyse the value of "k_observed" to see a difference, but that's often a bit more challenging than detecting an intermediate in the reactions.

Yikes, that one got a little long.  Have a good weekend.

Logarithms

Logarithms used to be extremely important.  With modern calculators, logarithms aren't quite as critical as they used to be, but they're still very useful in some cases.  There are a few definitions and identities that will help us out a bit, the most basic ones are:

log (10^x) = 10^{log x} = x

ln (e^x) = e^{ln x} = x

The following work for either common logs (log, base 10) or natural logs (ln, base "e"):

log (AB) = log A + log B

log (A/B) = log A - log B

log (A^B) = B log A

A little practice will help lock these in.  Good luck.

Integrated Rate Laws and Activation Energy

There's a new bit of OWL assignment posted, make sure you take a look.

Rate laws can tell us a lot about a reaction, but a simple rate law doesn't do a great job of telling us how fast or slow a reaction really is.  In order to incorporate a time component, we can integrate the rate law expressions.  The integrated rate laws (IRLs) give us a way to monitor the way concentrations change over time.

0^th Order IRL → [A]_t = -kt + [A]₀

1^st Order IRL → ln[A]_t = -kt + ln[A]₀

2^nd Order IRL → {1/[A]_t} = -kt + {1/[A]₀}

IRLs can also be used to determine the order of a reaction with respect to a given reactant.  All of the IRLs listed above are equations of lines.  If we plot [A]_t vs. t and the result is a straight line, then the process must be 0^th order w.r.t. [A].  Likewise, a linear plot of ln[A]_t vs. t implies a 1^st order process, and a linear plot of {1/[A]_t} vs. t implies a 2^nd order process.

Why do reactions have the rates they do?  This is a function of the amount of energy required to make a reaction occur.  Recall from Collision Theory that collisions must occur and those collisions must be oriented and energetic.  The energy required to get a reaction started is the activation energy, E_a.  Activation energy is described by the Arrhenius equation:

k = A exp(-E_a/RT)

where:

k = rate law constant

A = frequency factor

E_a = activation energy

R = universal gas constant, 8.314 J/mol.K

T = temperature in units of Kelvin

Although this is an elegant little bit of mathematics, it's not the most useful form of the Arrhenius equation.  If 2 sets of conditions are known, we can set up a ratio of the Arrhenius equation for each run and ultimately find that:

ln(k₁/k₂) = (E_a/R)({1/T₂} - {1/T₁})

The comparative form works well, but it has a notable flaw.  We have to assume that both of the points of data that we have a quite good and accurate.  Hopefully this is the case, but it might not be, leading to error.  If we want to average out some of that error, we can do another transformation of the Arrhenius equation to form a line:

ln(k) = (-E_a/R)(1/T) + ln(A)

Friday we'll look more closely at activation energy and what it means in terms of reaction mechanisms.

Rate Laws

We can calculate the instantaneous rate for any point in a concentrations vs. time experiment, but the only really unique and important one is the initial instantaneous rate.  The initial rate of a reaction is described by a rate law.  The initial rate of a chemical reaction is proportional to the initial concentration of all the reactants raised to some power.  To remove the proportionality, we can add a constant, the rate law constant.

Rate₀ = k [reactant]₀^x

We can write a rate law expression for any chemical reaction as long as we know the reactants.  For example, for the generic reaction:

aA + bB → cC + dD

The rate law expression is

Rate₀ = k[A]₀^x[B]₀^y

If we think about this rate law expression, there seem to be a LOT of variables present.  We can determine the value of a number of those variables if we design our experiments thoughtfully.  Let's look at a specific example.
For the reaction of NO₂(g) with Cl₂(g), we have performed the following experiment: [NO₂]₀ = 1.228M, [Cl₂]₀ = 1.316M, Rate₀ = 2.881x10^{-3 M}/_min.  The rate law expression for this reaction is:

Rate₀ = k [NO₂]₀^x[Cl₂]₀^y

Notice that we don't really need to know the products or the balanced chemical equation to write out the rate law expression.  That doesn't mean we don't have to practice balancing chemical equation, keep on practicing!!  Plugging the numbers in to the rate law expression:

(2.881x10^{-3 M}/_min) = k (1.228M)^x(1.316M)^y

That's still 3 variables, so we need (mathematically) more equations to help us solve them.  We could just randomly start mixing reactants together, but if we're deliberate in our planning, we can make our jobs a little easier.  As good scientists, we try to change one 1 variable at a time whenever we're doing an experiment.  Why?  Because then if the result changes, we know it has to be caused by the variable we changed.  In this case, we can set up another experiment, and let's change the initial concentration of NO₂ but hold the initial concentration of Cl₂ constant.  For our second run: [NO₂]₀ = 2.456M, [Cl₂]₀ = 1.316M, Rate₀ = 1.152x10^{-2 M}/_min.  Plugging in to the rate law expression again, we get:

(1.152x10^{-2 M}/_min) = k (2.456M)^x(1.316M)^y

Let's get a little mathematical here... if these two equalities are valid (which they are), then their ratio is also a valid equality.  Bam.  Cancelling out everything that can cancel, we're left with the simplified expression:

(1/4) = (1/2)^x

x = 2

Therefore, the reaction is second order with respect to [NO₂]₀.

Kinetics, at an average rate...

Kinetics is the study of the rates and mechanisms/pathways of chemical reactions.  Most of kinetics can be understood by probability and the 3 points of Collision Theory: For a chemical reaction to occur: 1) Collisions between reactant particles must occur; 2) The collisions must be energetic enough to allow reaction; 3) The colliding particles must be oriented in a way that gives the desired reaction.  The probabilities is Collision Theory are affected by changes in temperature and concentration/pressure.

Rates can be expressed in a number of ways, but for most chemical systems, the rate is equal to the change in concentration over the change in time.  When that change in time is (relatively) long, we have an average rate.  Average rates can be either in terms of consumption of a reactant or production of a product.  The mathematical formality of "change in concentration" requires a negative sign on rates of consumption.  Rates of consumption or production are related to the stoichiometric coefficients in the balances chemical equation that is being observed and can be unified as a rate of reaction.

A significant disadvantage of average rates is that they change depending upon the time period being measured.  To more accurately estimate the rate of a reaction, we should use smaller time periods.  Taken to the extreme, we can calculate an instantaneous rate at any point during a reaction.  Of all the instantaneous rates for a given reaction, the only important or unique instantaneous rate is the very first one, the initial instantanous rate of the reaction.

On Friday, we'll continue with kinetics.  You will very likely NOT get your exams back on Friday.  A number of things came up today (and continue tomorrow) that will make it difficult to get exams back on Friday.  I will have them done over the weekend and will return them first thing Monday morning.  Sorry for the delay.

Nature of solutions

We took a little step away from colligative properties today to make sure we're all familiar with some of the terminology of solutions.  Solubility is not a "yes/no" question.  Is potassium nitrate soluble?  Yes, all potassium salts are soluble and all nitrate salts are soluble, so potassium nitrate must be soluble.  If we take 1 gram of KNO3(s) and add it to 1L of water, it will dissolve.  What if we take 1 kilogram of KNO3(s) and add it to 1mL of water?  Will it dissolve?  Yes, but will ALL of it dissolve?  I think not.  If we start with 1L of water and slowly add KNO3(s), at some point the excess KNO3(s) will no longer dissolve because the solution has become saturated.  A saturated solution represents the maximum amount of a solute that can dissolve in a given amount of solvent.  Sometimes, a solution can become supersaturated when "extra" solute is dissolved; supersaturated solutions are unstable and will form precipitate if a nucleation site is present.  This can be a speck of dust, or a scratch in the glass of the container, or a seed crystal of the substance.

We also looked at the temperature dependence of solubility.  For solids dissolving in liquids, heating the solution will usually allow more solute to dissolve.  This is one way to make supersaturated solutions; a hot saturated solution is allowed to cool in the absence of nucleation sites.  For gases dissolved in liquids, the situation is reversed; cold solvent is usually able to hold more dissolved gas than hot solvent because the gas solute particles in the hot solution have more kinetic energy and are more likely to escape from the solution.

Returning to colligative properties, we began to discuss the most important colligative property in biological systems, osmosis.  When two solutions of differ in concentration are separated by a semipermeable membrane, solvent tends to flow from the less concentrated side to the more concentrated side.  What's a semipermeable membrane?  Cell walls.  Skin.  LOTS of biological things are semipermeable membranes and control function by regulating osmosis.  Awesome.

There's new OWL posted, and don't forget to take the pre-lab quiz before 8:00am tomorrow.

Freezing Point Depression and Boiling Point Elevation

Continuing with colligative properties, if the presences of a solute affects the vapor pressure of a solution, then it must also affect the boiling point.  Liquids "boil" when the vapor pressure of the liquid is equal to the atmospheric pressure on that liquid.  If a solute is added to a solvent, the vapor pressure is depressed, so the temperature must be increased further to increase the vapor pressure to the point that it is equal to the atmospheric pressure.  Boiling point elevation is determined by the equation:

ΔT_bp = k_bpe•m•i

Where m = molality = (mols solute) / (kg solvent)

i = van't Hoff factor

The van't Hoff factor, sometimes denoted as "n", describes the number of particles formed when a solute dissolved.  The presence of a solute also affect the freezing behaviour of a solvent by changing the energetics of the system.  For a solvent to freeze when solute is present, the average kinetic energy of the particles must be slowed even more (temperature must be lower) than for a pure solvent.  Freezing point depression uses essentially the same equation, although with different values for the constant.

ΔT_fp = k_fpd•m•i

For the problems we looked at in class, the answers are below.  The problem was "23.381g of “salt” dissolved in 500.00mL of water, what is the freezing/boiling point?":

Compound	m	i	ΔTfp	Tfp	ΔTbp	Tbp
KNO3	0.462522	2	1.72	-1.72	0.48	100.48
Na3PO4	0.331763	4	2.47	-2.47	0.69	100.69
Mg(ClO3)2	0.244565	3	1.36	-1.36	0.38	100.38
(NH4)2SO4	0.353884	3	1.97	-1.97	0.55	100.55
CaCl2	0.421340	3	2.35	-2.35	0.66	100.66

Southwestern Advantage

Many of you may have had a short presentation/survey in some of your classes from representatives of Southwestern Advantage during which they talked about a summer "internship" opportunity.  I did not have them come to our Gen Chem class because I knew that they came to the Bio class that most of you are in at 9:30 and I didn't think you all needed to have a second dose of their sales pitch.

It has come to my attention that the representatives of Southwestern Advantage may have been a bit overzealous in their tactics when strong-arming their way into some classrooms.  A message was sent to faculty from the Career Development Center :

It has come to our attention that representatives from Southwestern Advantage are visiting your classes to explain their Internship/ Employment Program and to request that a survey be completed by your students. We also understand that they are stating their visit has been authorized by The Career Development Center and/or our Director, Greg Toutges. This is not the case.

During their pitch in class, I have also noticed that the recruiter is very reluctant to describe exactly what this "excellent internship and independent business opportunity" is.  Southwestern Advantage is a door-to-door bookselling business.  They sell books and "study systems" that are intended to help pre-college students with their studies.  I have not seen these products, so I don't know whether they're good or not, but doing a quick web search leads me to believe that the products sold by Southwestern Advantage are quite expensive, and given the wealth of FREE information and tutorials available online, I personally would never pay the prices I saw mentioned even if my child was struggling.

Southwestern Advantage uses an independent contractor/seller model, so although the recruiter very likely spoke about earning $8000 during the summer (a number he used when talking to me), that number may not be realistic, and may require 10-15 hour days, 7 days a week for weeks at a time.  In addition, there will be living expenses that you would not incur while living at home and working for minimum wage, so if you are considering exploring a summer job with Southwestern Advantage, make sure you really analyze the numbers they provide, although I would expect that they will offer very few concrete details until you have signed a contract.  Sales can be a very good career for some people, but it's not for everyone.  Set up a spreadsheet to calculate income and expenses to compare your various summer option before you are coerced into signing a contract.

A few students have also mentioned that the Southwestern Advantage recruiter was asking for Dragon ID#'s and social security numbers.  I did not attend the presentation in any classes this year, but if the recruiter was really asking for this type of information, I would be EXTREMELY suspicious of their intent, or at the very least their tactics.  In addition, when Lucas Odegard, the "Corporate Recruiter" who probably talked to your class, met with me about coming into Gen Chem, he consistently referred to all of you as "kids".  Like it or not, you are not "kids", you are adults.  If Mr. Odegard considers you all to be "kids", it seems to me that he has a profound lack of respect for all of you, and merely sees you as another resource or product that he may be able to use to make money.

I am sure that there are pre-college students who have benefited from Southwestern Advantage's products, and I am sure that there are college students who have earned good money selling these products, but I have been extremely unimpressed with the tactics that have been used by Southwestern Advantage on our campus.  If you choose to explore this opportunity, please make sure you are fully informed and are not taken in by a slick and predatory sales pitch.

Colligative Properties

"Solubility" is not a yes/no answer, it's always a matter of degree.  Although we may describe something as "soluble", there obviously must be a limit to that solubility: NaCl is "soluble" in water, but you can't dissolve a bucket of NaCl in a glass of water.  Similarly, if we look back at the solubility rules we used in Gen Chem I, just because something is described as "insoluble" doesn't mean that none of it will dissolve, it just means that very little will dissolve.  It's an important distinction and we'll re-visit it in a few weeks.

When a solute is added to a solvent, the properties of that solvent are affected.  A colligative property is one that is dependent upon the number of solute particles present in the solution and not necessarily the identity of those solute particles.  If 0.1mol of sucrose and 0.1mol of fructose (0.2mols of solute particles) are dissolved in 10.0L of water, the colligative properties of the solution will change by the same amount as if 0.2mols of sucrose alone is dissolved in 10.0L of water.  We looked at vapor pressure depression as a colligative property; the presence of a solute decreases the vapor pressure of a solution due to solvent-solute interactions and surface blocking.

Next week we'll look at more colligative properties.  Have a good weekend.

The solution to solution is solution

When a volatile liquid is in an open container, it evaporates.  When a volatile liquid is in a closed container, it builds up vapor pressure.  Vapor pressure is the function of the temperature of the system (an indication of the average kinetic energy of the particles) and the intermolecular forces holding the particles together in the liquid state.  It represents a dynamic system, where the rate of liquid particles vaporizing to the gas phase is exactly equal to the rate of gas particles condensing to the liquid phase.

The properties of pure liquids are fascinating, but they become even more interesting when we add another component to the system.  A solution is a homogeneous mixture of two or more components.  The major component(s) is/are the solvent(s); the minor component(s) is/are the solute(s).  When a solute dissolves in a solvent, solvent-solvent and solute-solute interactions must be broken (requires energy) and solvent-solute interactions must form (releases energy).  If the energy released is greater than the energy required, a solution forms.

We also reviewed molarity and stoichiometry problems.  All stoichiometry problems follow the same 4 steps:
1. Write a balanced chemical equation
2. Convert the quantity of the known compound(s) to moles
3. Using the mol-mol ratio in the balanced chemical equation, convert moles of known substance to moles of interest
4. Convert moles of interest to whatever you're looking for

Wow, big day.  See you in lab tomorrow.

Friday - Solids and liquids

On Friday we looked at some similarities, differences, and various properties of solids and liquids.  The behaviour of many solids can be explained by comparing the relative magnitude of intramaterial intermolecular forces to the intermaterial intermolecular forces that would allow the solids to melt, sublime, or otherwise break apart.  When looking at the properties and behaviour of liquids, these comparative intermolecular forces become even more important because liquid is an intermediate state between solids and gases.

Surface tension and capillary action are due to relative liquid-liquid, liquid-atmosphere, liquid-surface IMFs.  Viscosity and volatility are due to the relative magnitude of IMFs compared to the average kinetic energy of a sample.

Sorry I didn't get this posted sooner, it kind of slipped away from me.  Enjoy the rest of your weekend.

Transition to transitions (!!)

Today we looked at Dalton's Law of Partial Pressures and did a little bit of derivation and/or proof of where this Law comes from.  {English note: ending a sentence with a preposition is not great, perhaps that should have been "proof of whence this Law came"}  An ideal gas follows all gas laws exactly and does not violate Kinetic Molecular Theory of Gases; when real gases are studied, there are deviations, especially at high pressure and low temperature.  Some of these deviations are phase changes.  The behaviour of a sample as energy is added or removed can be visualized as a series of heat capacity and enthalpy events in a heating/cooling curve.  If heating/cooling curves are observed at multiple pressures, a phase diagram can be constructed.

Don't forget to get signed up for OWL and look at the currently posted assignments.  And as mentioned below, Chem 210L labs will not meet this week.

Chem 210L this week

I just realized that I forgot to mention in class yesterday that Chem 210L labs WILL NOT MEET THIS WEEK.  We will get rolling in lab next week, January 19th.  There will be a pre-lab quiz for next week's lab, so make sure you keep an eye on your email for a message that the lab info is posted and ready in D2L.

States of Matter - Gases

As we begin looking at states of matter, we start with gases.  Gases are convenient to study because many real gases behave in a very theoretically "correct" manner, meaning that their behaviour can be understood and explained using the Kinetic Molecular Theory of Gases: 1)gas particles are very much smaller than the space between the particles; 2) gas particles move randomly; 3) except during collisions, attractive and repulsive forces between particles are negligible when compared to the kinetic energy of the gas particles; 4) collisions are elastic; 5) the average E_kin of the particles in a sample of gas is proportional to the absolute temperature of that sample.  KMToG can be used to explain a number of gas laws including Avogadro's (V ∝ n), Boyle's (V ∝ 1/P), Charles' (V ∝ T), and the Ideal Gas Law (PV=nRT).

Next time, we'll wrap up gases and move on to liquids and solid as well as the phase changes between them.

Chem 210 - Spring 2012 - IT HAS ARRIVED!!

The semester has started, I hope everyone had a restful break and is ready to dive in.  Remember to get signed up in OWL and take a look at the first assignments.  I will try to have smaller assignments more often, so there should be something active in OWL just about any time you log in.  Make sure you work on those assignments early.

I'm going to try a little experiment this semester, I will be tweeting class topics using the hashtag #GenChem2012.  If you're a Twitter user, you know what that means.  If you're not, don't worry.  I'll also be posting class info here on the blog.  And, of course, I'll be giving info IN CLASS.

Final Exam

The final exams are graded, I'll enter course grades first thing tomorrow morning.  If you'd like to see your final exam, you can stop by and look at it but you can't keep it.  You should be able to see your grade in eServices some time tomorrow or Friday at the latest.  Have a great break and I'll see you next semester.

Final exam...

Less than 1 hour until the final exam starts...

Final exam info

You will have the same info on the front of your final exam as was on Exam 4.

Finding exams and keys...

A few people have been having trouble finding the exams and keys I posted yesterday.  I suspect you may be running into a cached page problem?  If you're using Chrome, try opening my webpage in an incognito window; other browsers have similar features, but I don't know what they're called.

web.mnstate.edu/bodwin
In the left panel, click on "Chem 150" under "Fall 2011"
The new page should open in the right panel.  Scroll down and all the exams and keys should be there.

Exams and keys posted

All the exams from this semester are posted, all the keys except Exam 4 are posted.  I might get to those today, but we went through most of Exam 4 in class just a few days ago.  Let me know if you have any questions.

VSEPR

Valence Shell Electron Pair Repulsion Theory!
VSEPR is the theory used to predict molecular shapes.  Because each region of electron density (lone pair, single bond, double bond, triple bond) is negatively charged, the regions of electron density repel one another as much as possible.  This repulsion dictates the shape of the molecule or polyatomic ion.  If you're having trouble visualizing these 3-dimensional shapes, try the PhET simulation we looked at in class:
http://phet.colorado.edu/en/simulation/molecule-shapes

Have a good weekend and don't forget to look at the OWL assignments that are currently posted.

Lewis Structures

As with everything, your ability to understand and draw Lewis Structures depends upon 3 equally important things.  #1 - Practice drawing Lewis Structures.  #2 - Practice drawing Lewis Structures some more.  #3 - Most importantly, everyone needs to practice drawing Lewis Structures.  We'll work through some more examples in class and (hopefully) do some practicing in class, but you really really really really need to practice them yourself.  I've posted the lab info for the experiment we're doing after break, take a peek for some more practice using Lewis Structures.

On Friday, VSEPR.  What's VSEPR?  Come to class on Friday...

Lab exam

A few people have asked about the lab exam you will have this week, specifically how to study/prepare for it.  To give everyone the same info, here's the reply I sent to someone who asked:
--------
Hopefully you're already prepared. ;)  I might review some of the techniques and procedures we've consistently used throughout the semester; things like error handling, graphing, different types of glassware, etc.  Because this is more of a techniques and procedures exam, it's not necessarily something that can be studied for.  As I've said, the exam is not going to be a bunch of experiment-specific detail (What was room temperature for the Al + HCl experiment?, What color was the nickel solution in the Clandestine Lab experiment?, etc).
--------
Let me know if you have any further questions, I'm not sure how much more detail I can go into about the lab exam, but you can always ask.

Electron configurations, etc

We've been looking at electron configurations and what we can do/predict with them.  Sizes, charges, stability, magnetism.  Don't forget about the OWL assignments that are posted.

All the nitty gritty of the electron world...

Monday and today we've been exploring the world of the electron a bit more.  The vast majority of chemistry is really a study of the electron: where are they, why are they there, where do they move, when do they move, how fast do they move.  On Monday, we looked at quantum numbers as a way to address electrons, but writing out explicit quantum numbers can be a bit ponderous, so today we looked at a shorthand way to express quantum numbers with electron configurations.  Electron configurations describe the energy levels and orbitals that are occupied (or might be occupied) in an atom or ion and provide a very useful tool for studying electrons.  Practice them.

Light!!

We've been looking at the nature of light for the last 2 days (as well as getting exams back and going through problems) and have just gotten to the point of using that light to explore the structure of atoms. Next week, the fun will be beyond measure.  On Monday, I promise there will be fire.

Have a great weekend.  Volleyball has their final home games of the year tonight and tomorrow, and football has their last home game tomorrow.

Exam tonight...

Exam 3 is tonight so we spent most of class reviewing.  On little bit that we (sort of) added...  When calculating the heat of reaction for an aqueous process, it is sometimes easier to use the net ionic equation.  There aren't always tabulated values available for every soluble salt, but the ions can be found for most elements.

See you tonight, 6pm in SL104.

emailed question...

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Hello Dr.Bodwin,

I have been studying up on the Exams that you have on your page and some of them have some things that we didn't go over in class and I was just wondering if those things would still be on our exam. The things that I am wondering about is like the quantum numbers, electron configuration, and the wavelength problems. 
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We'll be getting to that material after this exam, don't worry about it for now.

Old exam keys...

Looks like I haven't had time to put a key together for a while...

http://msumgenchem.blogspot.com/2010/10/old-exam-keys.html

Question

Question from email:
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Dr. Bodwin,

I am wondering if the test on Monday will have material from our last exam and the new material covered since then or just the enthalpy??
Thanks

-----------------
A few people have asked me this question and the best answer I can give is "yes".  Although there will not be any questions that are strictly "exam 2 questions" on this exam, you will have to know how to do things from exam 2 to answer exam 3 questions.  If you are trying to calculate the heat liberated or absorbed by a reaction, you will have to be able to write a balanced equation.  If you're trying to write a balanced equation, you will have to be able to write balanced formulas.  Many enthalpy problems are the same as all the stoichiometry problems we looked at for exam 2, the only difference is that instead of calculating grams or molarity or volume, you'll be calculating heat.  You may need to determine the limiting reagent, or percent yield, just like any other stoichiometry problem.

Other questions, let me know.

Heat, heat, heat...

Today we went through some more enthalpy/heat transfer problems.  We also went through a problem that demonstrated Hess' Law; for a multi-step process, the sum of the enthalpies for all the steps should equal the enthalpy of the whole process.  We've actually been using Hess' Law the whole time we've been looking at enthalpy, but we didn't formally call it Hess' Law.

If you have questions, email me.  I have a few other things going on this weekend, but I will do my best to post answers to the blog ASAP.  If you need to take Monday's exam at an alternate time and have not yet talked to me, please check in either by email or in person before Monday.

And most importantly, take a break or two over the weekend.  The weather is supposed to be quite nice, so take a little walk around the block for a study break.  Volleyball is home tonight and tomorrow and football is home this weekend.

Enthalpy

Wednesday in class we went through a couple more enthalpy problems/calculations.  Enthalpy of formation values are tabulated and refer to the heat transfer when 1 mol of the substance is produced from its standard state elements.  The magnitude of that heat transfer is the same whether a substance is being formed from its elements or the elements are being formed from the substance, only the direction of the heat transfer (and therefore the sign of {delta}H) changes.  This is the key to calculating enthalpy of reaction using tabulated enthalpy of formation values.

Tonight at 6pm in HA113, Tri-Beta will be hosting research night.  Faculty from Biosciences, Chemistry, and Physics will give brief descriptions of their research and be available for questions.  If you're interested in doing research, this is a good opportunity to see a variety of the projects taking place on campus.

New OWL

Oh, and there are new OWL assignments posted.  Enjoy.

Enthalpy - the heat of a process

Today we linked heat capacity to the larger idea of enthalpy.  Enthalpy is the heat transfer associated with a chemical reaction or physical process.  We'll work through a few more examples on Wednesday.

The Exam

We spent most of today going over the exam.  Exam 3 will cover a LOT of the same material, so make sure that you use your performance on Exam 2 to guide you toward the areas you need to study more.  Having graded the exam, I can identify polyatomic ions as the biggest problem most people had.  There's no real trick for polyatomic ions, you just have to memorize them.  As with anything (music, language, sports, etc), the more you practice the more automatic a thing becomes.  When you write out the formula for nitrate 100 times while you're studying, you will tend to remember the formula for nitrate.

We also did a quick heat capacity problem at the end of class, we'll get more into that next week.  Keep an eye on the schedule, Exam 3 is a week from Monday, so it's coming up quickly.  Have a good weekend.

Volleyball tonight - Diggin for a Cure!

Tonight the MSUM volleyball team will be Diggin for a Cure to raise funds for and awareness of cancer research and treatment.  Come out and support Dragon Volleyball as they crush the Golden Eagles of UMinn-Crookston.

Exam Results

Exam 2 is graded, I'll give it back in class tomorrow.  The scores were not good.  A few people did quite well, but many did very poorly leading to an average just under 50%.  We'll spend some time talking about this tomorrow in class.

With renewed vigor...

Today in class we started looking at thermochemistry, the ways that heat and chemical processes interact.  This is a small part of the larger field of thermodynamics.  We looked at some of the foundational energy things (types, units, transfer) and just got to heat capacity.  On Friday we'll get exams back and dig a little deeper into heat capacity.

Exam 2 results

OK, not exactly results, I haven't graded the exam yet, but it seems like a lot of people struggled with this exam.  This is typically the most challenging material in Gen Chem I so it's not unusual for this exam to have lower scores.  I've posted a poll, let me know if you've identified specific problems in your approach to class.  The poll is 100% anonymous and is not monitored or moderated in any way, so it's not definitive scientific data, but if there's a consistent problem identified I might be able to do some things to help.

EXAM TONIGHT

Don't forget that we are in a different room and a different building for the exam this evening.  We will be in CB111, the Center for Business.

Chem Club Tutoring Schedule

I've posted the Chem Club Tutoring schedule in a panel to the left (<-- that way <--).  Take advantage of this service and let me (or them) know what's working or not working with the schedule or tutors.

Monday's in-class problem

I'll get answers posted for Monday's in-class problem some time today or tomorrow.

Lab Hand-In

Yes, there is a hand in assignments due for last week's lab.  The assignment should be turned in to the mailboxes outside HA103.  Make sure you put your assignment in the right box (by Lab Assistants) or you will not get credit for the assignment.

OWL question - Roots

There have been a few people asking about the "Roots" question in OWL and I think the confusion may be coming from the way it's worded.  It sounds as if you will get a very large or very small number as an answer that will require scientific notation to answer, but the 3-4 times I've tried the question, my correct answer has always been easily expressed without scientific notation, answers like "4.293" and "0.271".

Simplest help = just evaluate the mathematical expression given in the problem and type in the answer you get.

First day!

Welcome to Fall 2011! Check here for class info, answers to emailed questions, and other randomness.

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For question #9 on exam 3 from this summer, how does the balanced equation that you give on the answer key match up with the stoichiometry that is also given? In the equation there are 2NaOH(aq) but in the stoich problem under it, there are 2mol KOH. I guess I am not understanding that relationship.

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Typo. It should be KOH in the equation.

One more...

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I have a question on #5b on the practice test. You posted the answer as

H3AsO4(aq) + 2 KOH(aq)  2 H2O(l) + K3HAsO4(aq)
(0.02500L H3AsO4(aq)) (0.127 M H3AsO4(aq)) (2mol KOH / 1mol H3AsO4) ( 1/0.03868L KOH(aq)) = 0.164M KOH(aq)

I am confused at where the .127 M H3AsO4 came from.

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The 0.127M came from the first part of the question. In 5b, you're using the arsenic acid solution you determined the concentration of in 5a to titrate a new KOH(aq) solution that has an unknown concentration.

Email questions...

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1. For question #14 on exam 1 from this summer, how is the "i" value 3? Also, do you have to add 100 to get the bp for these types of problems, but if it were a fp question would you subtract the value from zero?
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When potassium sulfate dissolves in water, it forms 2 potassium ions and 1 sulfate ion for a total of 3 particles. Remember, when you're calculating these numbers, you are most often calculating a change in freezing point or boiling point. Boiling point is elevated in solution, so the change you calculate is above the boiling point of the pure solvent; freezing point is the opposite direction.

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2. For question #18 on exam 1 from this summer, I understand that the order of [CH3I] is 1st and that [F2] is 0. But, if a concentration is 0 order, does that make it not part of the rate law expression? I just don't see why the [F2] isn't part of the problem after you say that its 0 order on the answer key to the test. Say if it was 1st or 2nd order, how would the problem be different?
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We could explicitly include a "[F2]^0" term in the rest of the problem, but any number raised to the zero power is equal to 1. Since we're only multiplying and dividing, including an extra term that's the equivalent of "1" will not affect the answer. If it was 1st or 2nd order, we would have to include that term. This would impact the units of "k" as well as changing the numerical value.

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3. How many questions will be on the exam? and how much time will we have on the test? Will it mostly be problems to work through or will there be some multiple choice as well?

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30-40 questions, you will have the full 2 hour 10 minute class period. This exam will have a slightly higher proportion of multiple choice questions than most of your previous exams, maybe over half multiple choice. Although there are more questions, the questions will be similar to the type of questions you've seen on previous exams.

Email me any other questions, I'll be checking in throughout the rest of the day and this evening. I will plan to be in my office (HA407H) by 7am tomorrow, if you have other questions you can stop in early. Good luck.

Questions

A couple email questions...

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Hello professor Bodwin,
I have two questions about the material for exam four. When assigning oxidation numbers, would diatomic ions have a charge of 0, like I2? And on problem set #10 number 3, for the second reaction, on the answer key you have that 3 e- needed to be added to both sides of the reduction half-rxn. I am confused why it is added to both sides and not just the reactant side.

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In diatomic molecules like I2, each iodine has an oxidation number of 0. This is iodine in its neutral, uncombined {with other elements} form. All the elements that are diatomic molecules (H2, N2, O2, halogens) are oxidation number zero when they are their uncombined diatomic molecule.

On problem set #10... oops, that's a mistake. I was copying and pasting reactions and I must have forgotten to delete some of those electrons. It should be:

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Reduction half-rxn: 2( 3 e^- + Cr³⁺(aq) ó Cr(s) )

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Let me know if there are more questions.

PS6 and E2 keys...

Blanks and keys are posted on my mnstate.edu page.

I won't be in my office this afternoon, I have an appointment off campus. Email if you have any questions.

Error in PS#05 Key

There was an error in the key I posted yesterday for PS#05, it should be corrected in the version that is online now. In the first problem, there are 3 gas particles on the reactant side forming 2 gas particles on the product side. The error said 2 and 2. Sorry about that, please let me know when you find errors (or things you might think are errors) so I can correct them.

PS#05 and key posted...

On my mnstate.edu page.

If you have questions, I'll be in my office (HA407H) tomorrow morning, probably until the early afternoon. You can also email, I'll answer questions here.

Have a good weekend.

Keys posted

The keys from Exam 1 and Problem Set #4 are posted on my mnstate.edu page.

http://www.mnstate.edu/bodwin/

And for today's problem set...

http://www.mnstate.edu/bodwin/courses/genchem2x/c210nps2k.pdf

Problem Set 1...

Here's the key for today's problem set.

http://www.mnstate.edu/bodwin/courses/genchem2x/c210nps1k.pdf

Make sure you try to work through the problem before checking the key, if you just read through the key you will think you understand the problems, but when faced with a blank page you might not know where to start.

----Question----

On the Spring 2010 exam, for number 17, it says to redo experiment 3 at 16.31 degrees, the rate is 7.53x10-4. How do you calculate the new k for this problem. The answer is 7.17x10-5

----Answer----

The new k is calculated from the concentrations used in experiment 3 and the new rate. When the temperature of a reaction changes, the mechanism doesn't change (over small temperature changes) so the rate law expression is the same, you can plug everything in and use the same orders you determined at the original temperature.

----Question----

I'm having trouble with #14 on the Spring 2008 exam. How do you solve it?

14. You have found the following value in a table of equilibrium constants at 25ºC:

Cu2+(aq) + 4 NH3(aq) 􀀧 [Cu(NH3)4]2+(aq) Kc = 1.7x10-13

What is the equilibrium constant for the reaction:

2 [Cu(NH3)4]2+(aq) 􀀧 2 Cu2+(aq) + 8 NH3(aq)

----Answer----

This is similar to the lead bromide question... To get from the first equilibrium reaction to the second, we have to: 1)reverse the reaction; 2)multiply by 2. To convert the equilibrium constant, that means we have to: 1)take the inverse; 2)raise it to the 2nd power. So:

K{new} = ( 1 / (1.7E-13))^2 = 3.46E25

----Question----

17. You have found the following value in a table of equilibrium constants at 25ºC:

PbBr2(s) 􀀧 Pb+2(aq) + 2 Br-(aq) Kc = 6.29x10-6

What is the equilibrium constant for the reaction:

½ Pb+2(aq) + Br-(aq) 􀀧 ½ PbBr2(s)

----Answer----

To get the second reaction, the first reaction must be: 1)reversed; 2)multiplied by 0.5. Therefore, the equilibrium constant must be: 1)inverted; 2)raised to the 0.5 power. So the "new" equilibrium constant value is:

( 1 / 6.29E-6 )^0.5 = 399

----Question-----

On the exam -summer 2007 (exam 2) #9, I wasn't sure how you got the second grams (16.246 g / 331.208) because in the periodic table it said pb is 207.2, so what do I have to do to get it to 331.208?

----Answer----

The source of the lead(II) is lead(II) nitrate. When you're looking at equilibrium problems (or kinetics problems, or any stoichiometry problems) you can usually just use the net ionic equation, but you need to include spectator ions when you're weighing out reactants.

----Question-----

On exam - fall 2004 (Exam 1), I wasn't sure how to calculate #12. For #14 I converted it to grams then to the concentration but I don't seem to be getting the right answer either.

----Answer----

That's an old exam... #12 has an error, so there is not correct answer listed, but here's how to do it. The rate of oxygen consumption is:

0.433mols/8.5oL/min = 0.0509 M/min oxygen consumed

From the balanced chemical equation, for every 5 mols of oxygen consumed in the reaction 4 moles of ammonia is consumed, so the rate of ammonia consumption is:

(0.0509 M/min oxygen)(4 moles ammonia / 5 moles oxygen) = 0.0408M/min

Last minute questions...

----Question----

I can't figure out the ppm question when it asks what is the concentration in ppm of a solution made by dissolving 14.18mg of dinitrotoluene (182.13g/mol) in 4.00L of water. I thought you took .01481g/182.13 than you took that answer divided by 4kg and than take that times 1 million but its not working out for me.

----Answer----

"ppm" is (by convention) usually thought of as a mass-mass unit, so it's part of the mass fraction family.

ppm (dinitrotoluene, DNT) = [ {mass DNT} / {total mass of sample} ] * 10^6 =

[ {0.01418g DNT} / {4000g + 0.01418g} ] * 10^6 = 3.55ppm DNT

NOTE: because the volume of water only known to 3 sig figs and is so large compared to the mass of DNT, the denominator simplifies to just 4000g.

----Question----

You don't need to know molar mass of a substance to find what: molarity, molality, normality, mole fraction, or mass percent?

----Answer----

The only reason you would need the molar mass would be to calculate moles of the substance, so anything that includes moles will require the molar mass. Because mass percent is just a ratio of masses, you can calculate it without knowing the molar mass of the substances involved. This is related to the previous problem above.

Questions...

A few email questions came in. I may be able to answer a few more of these tonight, but by about 8 or 9pm I'll be offline until morning, so if you have questions, get them in sooner rather than later.

--Question--------------

I just had a question for the freezing point depression problems..for example on the Spring 2009 exam 1b #13, as far as the calculations where do you get the #mols particles/mols ? In that problem it says (2 mols particles/mol LiNO3) how do you get that part?

--Answer--------------

Quite a few people are having trouble with this. The number of particles per solute (the van't Hoff Factor) is a measure of how many pieces each formula unit of solute breaks into when it dissolves. For molecular solutes, the solute formula is a single piece in solution; for example, when a sugar molecule dissolves in water, it's still a single sugar molecule, it doesn't break down into individual carbon, hydrogen and oxygen atoms. When ionic solutes dissolve in water, they break down into the ions that make up the formula. In the LiNO3 example given above, when LiNO3 dissolves in water it does NOT float around as LiNO3 units in solution, it breaks down into lithium ions in solution and nitrate ions in solution, so for every 1 LiNO3 unit, there are 2 particles in solution.

--Question--------------

I am not really understanding the formula for the ppm problems, for ex. (Winter 2006, Exam 1), you have times 10^6, how did you get to the sixth power?

--Answer--------------

10^6 is a million. When you're converting a fraction (mass fraction or volume fraction) to ppm {or ppm(v)}, you have to multiply by a million, 10^6.

--Question--------------

I wanted to make sure I did my work right for the mol fraction questions. If the question asks, what is the mol fraction of sugar in a saturated aqueous sugar solution at 25 C. (solubility of sugar in water is 211.4 g/ 100 mL.

Would it be 211.4 - 180.1548 (total grams of sugar) / 180.1548 = .1744

--Answer--------------

This is essentially a unit conversion problem. The given solubility means that 211.4g of sugar will dissolve in 100mL of water. The mols of sugar is:

211.4g / 180.1548g/mol = 1.173mols sugar

Mols of water in the system:

(100mL)(1g/1mL) / 18.015g/mol = 5.55mols water

So the mol fraction of sugar in this solution is:

(mols of sugar) / (total mols) = (1.173mols sugar) / (1.173mols sugar + 5.55mols water) = 0.1745

--Question--------------

I know we did our lab on molar mass today, but I am not sure how to calculate question #16 on exam 1a (spring 2009). It has atm added to it.

16. A newly discovered protein has been isolated from seeds of a tropical plant and needs to be characterized. A total of 0.137g of this protein was dissolved in enough water to produce 2.00mL of solution. At 31.68°C the osmotic pressure produced by the solution was 0.134atm. What is the molar mass of the protein? (20pts)

--Answer--------------

In lab we were looking at one colligative property (freezing point depression) to determine the molar mass of an unknown. This problem is looking at a different colligative property (osmotic pressure) to determine the mass of an unknown. The expression for osmotic pressure is:

P = MRTi

OK, we have to make an assumption here, we're going to assume that the protein is a single particle in solution, which makes the van't Hoff Factor (i) equal to 1. After that, we can start plugging in the info from the problem.

0.134atm = M(0.08206 L.atm/mol.K)(31.68+273.15K)(1)

M = 0.0053569 mols protein/L solution

We've made 2.00mL of solution so:

(0.0053569 mols protein/L solution)(0.00200L solution) = 0.000010714mols of protein

So the molar mass of the protein is:

(0.137g protein)/(0.000010714mols of protein) = 12800g/mol

Exam #1 next Friday

We've gone through Chapters 10 (Gases) and 11(Solids and Liquids), we'll look at Chapter 15 (Solutions) next. The first exam is next Friday, let me know when you have questions, I'll post answers here.

Questions...

From email...
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On Exam 4, Fall 2007, you have 2 answers highlighted for number 7. I don't understand how Li and P can both be the smallest; can we circle more than one answer on the exam?
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Depending upon your explanation, I would have accepted either of those answers. (That's one of the reasons more recent exams have these comparisons as short answer questions.) If you look at the electron configuration, P has more than a full shell of additional electrons, so it might seem like Li is the smaller atom, but because atomic size decreases left-to-right across the Periodic Table and P is much farther right than Li, P might be smaller. Looking at the actual data (Figure 7.22 in your textbook, page 307), P has a radius of 110pm and Li has a radius of 157pm, so it looks like in this case the left-right trend makes more of a difference than the up-down trend.

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On Exam 4, Fall 2006, you circled A for question number 7 for being the largest ions. I thought the largest ion was the lowest negative charge, and the smallest ion was a positive charge. I am not sure how to figure out what ion is the largest?
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It's not only a matter of charge, we also have to look at the size of the parent atom. For a given element, the higher the charge the smaller the ionic radius and the lower the charge the larger the atomic radius, so, for example, Ge⁴⁺ is smaller than Ge²⁺ which is smaller than Ge which is smaller than Ge^2-. Within a row, this trend is pretty reliable, but as we move far up or down the P.T. things can change. Fr⁺¹ is smaller than Fr, and F^-1 is larger than F, but Fr⁺¹ is much larger than F^-1 because the parent Fr atom is SO huge compared to the parent F atom that the change in size when they form ions doesn't make up for the original difference in size. Looking at the ions in this question, F^-1, Li⁺¹ and Al⁺³ are all definitely small, so it comes down to comparing Pb²⁺ and Br^-1. Pb²⁺ has over a full shell of additional electrons, but it's a cation. Looking at the atomic radii, Pb = 180pm and Br = 115pm, so a Pb atom is bigger than a Br atom, but a Pb²⁺ ion should be smaller than 180pm and a Br^-1 ion should be larger than 115pm. How much smaller and larger will determine the answer to this question... I accepted either answer for this question, but looking at real data, Pb²⁺ has a radius of around 140pm and Br^-1 has a radius of around 180pm, so the correct answer should be Br^-1.

----------------------------------------------
Have we talked about number 11 and 16 on the Fall 2006 exam or is it just something we should study and know?
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We have addressed these, but maybe not in exactly these terms. #11 is based upon forming stable electron configurations, so being able to write a correct electron configuration and then adding or removing electrons to give full shells, full subshells, or half-full subshells will demonstrate which ions are (relatively) stable. #16 is an application of VSEPR, lone pairs are more "sterically demanding" than bonding pairs so the repulsion in each of these molecules will affect the bond angle. We talked about this comparing methane, ammonia and water bond angles in class.

----------------------------------------------
Then I also had a question on the most polar bonds. Is the most polar the furthest apart on the periodic table?
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In general, yes, but... The polarity of a bond is a function of the difference in electronegativity of the elements involved, so it's better to look at it from that perspective, with fluorine being the most electronegative. For example, if we're comparing a P-Cl bond to a Si-Cl bond, the Si-Cl bond is a little more polar. If instead we compare a C-S bond to an O-S bond, the O-S bond is quite polar while the C-S bond is barely polar at all, even though C and S are farther apart on the P.T. than O and S.

BTW, atomic and ionic radii numbers came from your textbook and from WebElements.com, it's an interesting website if you haven't checked it out. It's much more information-based than explanation-based, but it's a handy one to keep in mind.

Lewis structures...

We are getting in to Lewis structures and looking at how the electrons are distributed in ionic and molecular/covalent substances. Lewis structures are all about practice. The rules I typically use for Lewis Structures are a little bit different from those listed in the book, so:

Lewis Structures – electron counting method

1. Add up total valence electrons in the molecule or ion

2. Draw a skeleton structure using all single bonds (usually the least electronegative atom is central, hydrogen is NEVER the central atom, some structures have multiple “central” atoms)

3. Fill the octet of all peripheral atoms (hydrogen exception…)

4. Place any extra electrons on the central atom, pair up if possible

5. Check formal charge (find missing or extra electrons…)

6. Minimize formal charge distribution (if possible) by forming multiple bonds (resonance?)

7. Check formal charge

My favorite thing about Lewis Structures is that there are a couple places in the "rules" where you can check yourself and find mistakes early without going through the entire process. Formal charge is an extremely useful tool (perhaps even more useful for those of you who will have to take organic chemistry at some point...) so make sure you're comfortable calculating formal charge.

There's a new OWL assignment posted, due next Wednesday.

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:

------

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

------

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:
------
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

----

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.

Titrations are stoichiometry problems

Today we talked about titrations. Titrations are (usually) used to determine the concentration of an acid or base in solution, although they can be used in a LOT of places other than acid-base reactions. Treat them the same way you approach all stoichiometry problems.

The extra SI session next week will be next Monday at 6:30 in BR269. We have an exam next Wednesday.

Catch up...

Yikes, it's been a while...

Since the last exam, we've been looking at stoichiometry and various classes and types of reactions. This is chapter 4 & 5 material from the textbook. The heart of every stoichiometry problem is the mol-to-mol conversion made using the coefficients from the balanced chemical equation. The rest of the problem is all about getting into and out of moles.

We're currently in the middle of reaction types, we've talked about precipitations and molecule-forming reactions so far. We'll finish up acids and bases on Monday and move on to everyone's favorite reaction type, redox!

Have a good weekend, the weather's supposed to be nice. If you need a study break, the soccer team is at home this weekend on both Saturday and Sunday at 1pm. What could be better than a beautiful fall day and a little soccer?

Another question...

From email...
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Just wondering if you could tell me what I am doing wrong for #8 on last years chem 150 test. The question is "What is the formula weight of nickel(II) nitrate?"
Heres what I did:
Nickel is 58.69 and there are 2 so I took (2)58.69. Nitrate is NO3 so Nitrogen is 14.01 and Oxygen is 16.00 and there are 3 so it would be 14.01+3(16). This leads me to:
2(58.69)+14.01+3(16) which gives me 179.39. This is the wrong answer...on the answer key the answer is 182.70. Just wondering what I did wrong. Thanks!
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A lot of people trip up on this one. Remember, when there's a roman numeral after a metal, that tells you the charge of the metal cation, it does not tell you how many of that cation are in the balanced formula. For this one, the nickel has a +2 charge. Nitrate has a charge of -1, so to balance the charge of the formula, we need two NO₃^-1 for each Ni⁺², Ni(NO₃)₂, so the formula weight of nickel(II) nitrate should be:
(58.69g/mol) + 2(14.007g/mol) + 6(15.999g/mol) = 182.70g/mol

Other questions, let me know...

Exam questions...

Already a couple questions...

---Question----------------------------------------------------------------

Is the empirical formula just the smallest whole number ratio? What if it came out C=1.5 H=2.5 O=3 or something like that? Is that still the empirical formula? or would the empirical formula be C=3 H=5 O=6, and then work from there to get your molecular formulas?!

---Answer----------------------------------------------------------------

Yes, the empirical formula is the smallest whole number ratio, so in your example the most correct way to report the empirical formula would be C₃H₅O₆. The molecular formula would be some multiple of that and you'd have to be given more information in the problem to determine the correct molecular formula.

---Question----------------------------------------------------------------

For percent composition questions with multiple elements, will be expected to have the elements in the correct order in the final answer? For example: KMnO₄ instead of say MnKO₄....Or will the main concern be that we achieved the correct amount of each element?

---Answer----------------------------------------------------------------

Quite a few people have asked me about this and in general the order doesn't matter. The only place your should pay attention to the order and groupings is in the formulas of ionic compounds and polyatomic ions. When writing the formula for an ionic compound it is usually best accepted practice to list the cation first, followed by the anion, and you should always write polyatomic ions as their common formula is written. In your example, since permanganate is a polyatomic ion, it should always be written together as "MnO₄^-". Since this is an ionic compound, the cation {potassium ion} should also be written first, so this should be written KMnO₄. That's not a result of it being a percent composition problem, that's the naming convention for ionic compounds.

If you have other questions, let me know, I'll post answers to the blog ASAP.

Suggested problems from the text...

I've posted some suggested problems from the textbook on my web page. These are not required problems, they will not be collected, they will not be graded, they are merely problems that I think are good ones from the end of each chapter if you'd like some additional practice.

Almost exam time...

Sara was unable to get a room, so there will not be an extra SI session tonight.

Next Monday is your first exam, be sure to look at the old exams on my webpage to get an idea of what to expect. Also, make sure that if you intend to use a calculator on the exam it cannot be a graphing/programmable calculator, or a cell phone calculator, or an iPod/iPad calculator, or any other networked or interactive "calculator experience". Sit with at least 1 open chair between everyone.

Today we worked through another percent composition question, this time using percent composition to determine the formula of a salt that someone carelessly neglected to label completely.

If you have questions you'd like to go over in class on Friday, let me know in advance or just bring them to class and we'll go through as many as we have time for.

The cusp of the Labor Day weekend...

There is (was?) a problem with the time in OWL, it has been set to east coast time. I've contacted a few people at OWL about this and it should be fixed soon if it's not already corrected. If it's not corrected and you happen to be doing your OWL assignment between 11:00pm and 11:59pm, please continue to do the assignment even if you get a "past due" message. When the clock is corrected, the system will automatically re-grade your assignment.

There's new OWL posted, due next Friday.

Lab hand-ins turned in by 3:00pm on Mondays get a bonus point, BUT next Monday is Labor Day, so the university will not be open (at least not completely). Therefore, for this assignment only you will get a bonus point as long as your assignment is in by noon Tuesday. The regular deadline, Wednesday at 3:00pm, is still in place.

Today in class we looked at ionic formulas again and checked out how to calculate percent composition from a chemical formula, and a chemical formula from percent composition. One question a few people had about the example I did was "Where did that 78g/mol number come from?" That's a number that would have to be given in the problem for you to use, it's not something you would calculate in that specific example.

If you have any input about the SI schedule, please let Sara know. If none of the current times fit in your schedule, or if some other time would work better in your schedule, she is open to adjusting the SI times to better serve as many student as possible. Keep in mind that Sara is a student, so she also has a lot of time commitments for classes and study time and other activities (and she probably likes to sleep an hour or two every night...).

Enjoy your 3-day weekend, be safe and I'll see you on Wednesday.

Announcements and ionic compounds

Today we had LOTS of announcements, so:

Dr. Wallert announced the bi-weekly (that's every other week, not twice a week...) Biochemistry and Biotechnology Seminar Series. There are some very interesting talks and topics planned, so keep them in mind. There are poster hanging around the building, look for the ear of corn with some funny looking kernels for more details.

Tonight is the first meeting of the Chemistry and Biochemistry Club. SL104, 7pm. CBC is a great opportunity to get involved on campus and in the community, and gives you an opportunity to get to know other chemistry and biochemistry majors. There will be pizza and soda...

Keep an eye on the schedule, our first exam is coming up on September 13th...that's less than 2 weeks away. It will cover Chapters 1-3. Don't forget about the calculator rules: no graphing calculators, no sharing calculators, no cell phones, no iPods/iPads. If it's more complex than a basic $15 scientific calculator, it's probably not allowed. If you're not sure, check with me BEFORE the exam.

Today in class we looked at ionic compounds. You will be required to know any polyatomic ion listed in Table 3.7, page 88 of the textbook.

SI Information

There was an error in the room posted for SI, sorry for the confusion. The correct rooms are now posted.

We are pretty much through chapters 1 & 2, getting a good start on chapter 3. I will post some suggested problems from the textbook as soon as I have a chance to look through them. These will not be required problems, but are good examples if you'd like a little more practice.

Welcome to Fall 2010!

Welcome to the Fall semester! I'm looking forward to a great semester. There is already an OWL assignment (actually 2) posted, take a look at them as soon as possible. The first assignment is sort of an introduction to the OWL system. The second assignment is a math review. The math review assignments use applied examples, but if you look at the problems carefully you'll see that you really don't have to understand anything about pH or radioactive decay to do the problems, they're just algebra problems that you have to plug values into and solve. They're also good practice using your calculator.

The direct link to initially register in OWL is:

http://www.cengage.com/owl/

Once you're registered, you can use the link in the upper left corner of this page to get to the login page. Good luck and let me know if you're having any problems.

See you Wednesday.