Titrations I - Monoprotic/Monoprotic

When we're dealing with acid-base chemistry, we have to be able to determine the concentration of the acids and bases that we're using. This most often means performing a titration where one of the components is known. The absolutely 100% most important thing about titrations is to remember that titrations are stoichiometry problems, and they should be treated as such. There's nothing new and magical about titrations, they're just stoichiometry problems applied to a specific situation. The 4 steps to solve every stoichiometry problem are:

1) Write a balanced chemical equation

2) Calculate moles of one component (“moles of known”)

3) Convert moles of known to moles of interest using the ratios in the balanced chemical equation

4) Convert moles of interest to whatever you want to find

With acid-base titrations, it is most often convenient to monitor pH of the solution. To explore the pH-dependence of a titration, we can look at a system where we know all the concentrations. Let's say we are titrating 25.00mL of 0.85M nitrous acid, HNO₂(aq), with 0.85M KOH(aq). At the beginning, we have a solution of a weak acid, so we can calculate the initial pH using a K_a expression. Looking up the K_a of nitrous acid online gives a couple different values, let's say that it's around 5x10^-4 . K_a is just like every other equilibrium, so let's set up a table...

	HNO2(aq) +	H2O(l) ↔	H3O+(aq) +	NO2-1(aq)
[ ]initial	0.85M	XXXX	10-7 M	0 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.85 – x) M	XXXX	(10-7 + x) M	x M

Plugging in values:

That expression is solvable by the quadratic formula (and I encourage you to solve it for practice and to convince yourself that the things we're about to do are valid), but we might be able to simplify it if we make some assumptions about the size of “x”. Since this is a weak acid (K_a < 1), it's probably reasonable to assume that most of the HNO₂ molecules will still be intact when this system reaches equilibrium, so although we will lose some (Δ [ ] = -x), the value of “x” will probably be quite small compared to 0.85, so we can assume that (0.85 – x) ≈ 0.85. At the same time, although “x” is probably small, 10^-7 is really small, so there's a pretty good chance that (10^-7 + x) ≈ x. With these two assumptions (which we have to check later) in place, the K_a expression simplifies to:

x = 0.0206

Before we do anything else, we need to pause to check that the assumptions we made are indeed reasonable. 0.0206 is truly massive compared to 10^-7 so the second assumption is great. For the first assumption, it's a little closer; 0.0206 is certainly smaller than 0.85, but is it smaller enough?

(0.0206 / 0.85 ) *100 = 2.4%

Usually, the limit is around 5%, so we should be OK here as well. The pH of this solution at the beginning of the experiment should be:

pH = -log[H₃O⁺] = -log(0.0206) = 1.686

What happens to the pH when we start adding KOH(aq)? Well, KOH is a base, so the pH will go up, but how will it go up? We can calculate that. Starting with 25.00mL of 0.85M nitrous acid, let's add 5.00mL of 0.85M KOH(aq). To make it easier to keep track of everything, let's start by writing out a chemical equation and converting to moles:

HNO₂(aq) + KOH(aq) ↔ H₂O(l) + KNO₂(aq)

Mols HNO₂: (0.02500L)(0.85M) = 0.02125mols HNO₂

Mols KOH(aq): (0.00500L)(0.85M) = 0.00425mols KOH

We can assume that all of the OH^-1(aq) that is added will react with the H⁺/HNO₂ that is present in solution, so after this 5.00mL addition, we should have a solution that contains:

0.02125 – 0.00425 = 0.017mols HNO₂(aq)

This nitrous acid is now in (25.00 + 5.00 = 30.00mL) of solution {if we assume that the volumes are additive}, so the concentration of nitrous acid is:

0.017mols / 0.03000L = 0.5667M

And the concentration of nitrite ions is:

0.00425mols / 0.03000L = 0.1417M

We can calculate the pH by treating this like an equilibrium problem, just like above, setting up a table. The pH of the solution should be the same as if we had made a new solution by adding 0.017mols of HNO₂ and 0.00425mols of NO₂^-1 to enough water to make 30.00mL of solution:

	HNO2(aq) +	H2O(l) ↔	H3O+(aq) +	NO2-1(aq)
[ ]initial	0.5667M	XXXX	10-7 M	0.1417 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.5667 – x) M	XXXX	(10-7 + x) M	(0.1417 + x) M

Making similar assumptions, the expression simplifies to:

x = 0.002000

Checking assumptions again, they are all valid, so:

pH = -log[H₃O⁺] = -log(0.002000) = 2.699

We can continue adding 5.00mL portions of the KOH(aq) and calculating to get the values shown in the table:

mL KOH(aq) added	pH
0.00	1.686
5.00	2.699
10.00	3.125
15.00	3.477
20.00	3.903

But what happens when 25.00mL of the KOH(aq) solution is added? At that point, the mols of OH^-1 that have been added is exactly equal to the mols of acid that were present in the original solution. This is an equivalence point in the titration. If we look at the balanced chemical equation for the process,

HNO₂(aq) + KOH(aq) ↔ H₂O(l) + KNO₂(aq)

The solution we have produced by titration is the exact same solution that we would have produced if we had simply dissolved 0.02125mols of KNO₂ in enough water to make 50.00mL of solution. Since NO₂^-1(aq) is the conjugate base of nitrous acid, we can think about it being involved in a K_b equilibrium with water, K_b(NO₂^-1) = 2x10^-11. Set up a table...

	NO2-1(aq) +	H2O(l) ↔	OH-1(aq) +	HNO2(aq)
[ ]initial	0.425M	XXXX	10-7 M	0 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.425 – x) M	XXXX	(10-7 + x) M	x M

Again, we can make assumptions similar to the above examples to get x = 2.92x10^-6 = [OH^-1].

pOH = -log[OH^-1] = -log(2.92x10^-6) = 5.535

pH = 14 - pOH = 8.465

If we keep adding KOH(aq), the mixture will continue to get more basic, eventually leveling off as the concentration of excess hydroxide reaches a limit. {Sounds like a fascinating calculus problem, the concentration should asymptotically approach 0.85...} If we plot this pH data, we get a titration curve like the one shown below.

Measuring the Acidity of a Solution

It is often useful and necessary to measure the acidity or basicity of a solution. There are a number of ways this could be described or reported, but in the context of the Bronsted-Lowry definitions of acids and bases, it is probably most convenient to monitor the concentration of H⁺(aq). The [H⁺] in a solution can be a very small number. Although modern pocket calculators can readily handle very small numbers, it's a little easier to describe these small concentrations using pH. pH expresses very small concentrations without having to use scientific notation and will be useful for a number of quantities.

pH = -log[H₃O⁺]

Reversing and re-solving that expression:

[H₃O⁺] = 10^-pH

Why do we follow [H⁺] or [H₃O⁺]? We know from the K_w expression that [H₃O⁺] and [OH^-] are related. Depending upon the type of problem, sometimes it's easier to think in terms of [OH^-], but [OH^-] is also usually a very small number, so it's useful to define an analogous quantity, pOH:

pOH = -log[OH^-]

Reversing and re-solving that expression:

[OH^-] = 10^-pOH

pH and pOH are related by K_w and we can derive that expression from K_w:

K_w = [H₃O⁺][OH^-]

-log(K_w) = -log([H₃O⁺][OH^-])

If we generalize that “-log” can be replaced by “p”, and remember that log AB = log A + log B, then:

pK_w = -log [H₃O⁺] + (-log[OH^-]) = pH + pOH

At 25°C, K_w = 10^-14, so pK_w = 14.

Strengths of Acids and Bases

If an acid is an H⁺-donor, then the stronger an acid is, the more effectively it will be able to donate H⁺. If that acid is in aqueous solution, we can think of “strength” by the following equilibrium for the generic acid, HA(aq):

HA(aq) + H₂O(l) ↔ H₃O⁺(aq) + A^-(aq)

In equilibrium terms, the stronger an acid is, the more product-favored the equilibrium will be. Since this equilibrium expression can be applied to any acid, and acids are an important and diverse class of compounds, we define this equilibrium as an acid dissociation equilibrium and call its corresponding equilibrium constant the acid dissociation constant, or K_a.:

Similarly, we can think of the strength of a base using the equilibrium for the generic base, B(aq):

B(aq) + H₂O(l) ↔ OH^-(aq) + BH⁺(aq)

With its corresponding base dissociation constant, Kb:

Looking at the K_a equilibrium equation, the water that appears on the reactant side is accepting H⁺ to become H₃O⁺(aq). If water is accepting a proton, it is acting as a Bronsted-Lowry base. Considering the reverse reaction, H₃O⁺(aq) is a proton-donor so it is acting as an acid, while A^-(aq) is accepting a proton as B-L base. These acids and bases are not independent of each other, they are conjugate acid-base pairs. A^-(aq) is the conjugate base of HA(aq), and HA(aq) is the conjugate acid of A^-(aq); H₂O(aq) is the conjugate base of H₃O⁺(aq), and H₃O⁺(aq) is the conjugate acid of H₂O(aq). A conjugate acid-base pair are related by the addition or removal of a single H⁺. The same relationships can be described for the K_b equilibrium expression.

In the K_a equilibrium, water is acting as a base, while in the K_b equilibrium, water is acting as an acid. So what is water, an acid or a base? The answer is BOTH! The acid or base behavior of a substance is dependent upon its environment because acid and base are relative terms. In the case of water, if the water molecules are interacting with something that is more acidic than water, then water will act as a base. Likewise, if the water molecules are interacting with something that is more basic than water, the water will act as an acid. This brings up an interesting question: is water an acid or a base when it's not interacting with any other substance? Consider the following equilibrium:

H₂O(l) + H₂O(l) ↔ H₃O⁺(aq) + OH^-(aq)

Once again, water is acting as both an acid and a base. This process is called autoionization. Since water is such an important substance for life on Earth, this equilibrium also has a specific letter assigned to it, K_w, the autoionization constant for water:

Like almost all equilibria, K_w is dependent upon temperature. At 25°C, K_w = 10^-14. For pure water, this means that at 25°C, [H₃O⁺] = [OH^-] = 10^-7M. If this equilibrium constant only applied to pure water, it would be interesting but of limited use. Fortunately, it can be applied to any relatively dilute aqueous solution to understand the relative amounts of hydronium and hydroxide ions present in the solution. Consider an acid, HA(aq), and its conjugate base, A^-(aq), both interacting with water:

HA(aq) + H₂O(l) ↔ H₃O⁺(aq) + A^-(aq)

A^-(aq) + H₂O(l) ↔ OH^-(aq) + HA(aq)

If these equilibrium equations are added together, the result is the K_w equilibrium equation. When two (or more) sequential equilibria are added together, the equilibrium constant for the overall process is the product of the equilibrium constants for the individual steps. This means that for any conjugate acid – conjugate base pair, K_a x K_b = K_w. This also implies a general relationship, the stronger an acid is, the weaker its conjugate base, and vice versa.

This has been a lot of acid and base information wrapped up in a big discussion of equilibrium. K_a, K_b, and K_w all describe specific systems, but the most important thing to remember is that at their core, these are all equilibrium constants. They behave like every other equilibrium constant, they follow all the same rules as every other equilibrium constant, and they can be manipulated just like any other equilibrium constant. The only thing special about them is that they refer to a specific type of chemical equation.

Acids and Bases - Definitions

A lot of the behaviors and properties of acids and bases can be described by equilibrium.  Before we get into that, we need to establish some definitions.  In Gen Chem 1, we tried to recognize acids by looking for H⁺ and bases by looking for OH^-.  That's a good start, and is the basis for the Arrhenius definitions of acids and bases:

Acid = H⁺-donor

Base = OH^--donor

These very brief definitions work, but we quickly run into a problem.  Aqueous ammonia is a base.  Aqueous ammonia {NH₃(aq)} does not have a OH^- to donate.  Arg.  Fortunately, there's a qualifier there, this is aqueous ammonia, and if there's water around, we can use it:

NH₃(aq) + H₂O(l)  <=>  NH₄⁺(aq) + OH^-(aq)

So NH₃(aq) can be forced to fit the simple definition above by using water.  In fact, a more proper and complete version of the Arrhenius definitions of acids and bases is:

Acid = a substance that, when dissolved in water, increases the concentration of H⁺(aq)

Base = a substance that, when dissolved in water, increases the concentration of OH^-(aq)

These definitions are a little better, but they still seem a bit restrictive.  To make definitions that are a little more general, let's look at that ammonia equation again.  In that equilibrium, ammonia is "increasing the concentration of OH^-(aq)" by removing H⁺ from water.  If we're trying to understand the function of acids and bases, it might be nice to follow one consistent thing around, so maybe a "better" definition of acids and bases could be:

Acid = H⁺-donor

Base = H⁺-acceptor

These are the Bronsted-Lowry definitions of acids and bases, and they are the definitions we will use most often in Gen Chem 2.  It's important to note that the Bronsted-Lowry definitions and the Arrhenius definitions are (and must be) consistent with each other.  It wouldn't be very helpful if 1 definition called something a base while the other definition called that exact same substance an acid.

Since we're dealing with definitions here, what should we call "H⁺(aq)"?  "H plus" works, but we're also going to run across a few other descriptions.  If we think about the subatomic particles present in "H⁺(aq)", there's 1 proton (the atomic number of hydrogen, all hydrogens have to have 1 and only 1 proton), the "+1" charge means that the single electron of a hydrogen atom has been removed, and in the most common isotope of hydrogen there are zero neutrons.  This means that, in terms of subatomic particles, "H⁺(aq)" is just a proton, and that is what it is very often called in discussions of acids and bases.  Acids are "proton donors" and bases are "proton acceptors.
If we think in the other direction, a proton is a pretty concentrated little lump of positive charge.  If this little lump of positive charge is floating around in a polar solvent like water, it's probably going to attract (or be attracted to) the negative end of the water dipole.  We can write a chemical equation to describe this:

H⁺(aq) + H₂O(l)  <=>  H₃O⁺(aq)

"H₃O⁺(aq)" is called a "hydronium ion".  {For practice, draw Lewis structures of all the species in that equation.}
So when we're talking about acids and bases, we're likely to encounter a number of different descriptions of the same thing.  To keep them straight, remember that:

"proton" = H⁺(aq) = H₃O⁺(aq)

Now that we have a few definitions in place, what else can we do to set the table for acids and bases...

Class start delayed tomorrow

The University has delayed the start of classes tomorrow (Feb. 29th) until 10:30am.  That means we will not meet at 8:30.  Enjoy the extra sleep.

Solubility. {AGAIN?!?!}

When we talked about solubility, we said it was a dynamic process.  "Dynamic process" is super-secret science code for equilibrium.  If we want to think about solubility in its most generic terms, it is the reaction:

"ionic solid" ↔ "component ions"

Since pure solids do not appear in equilibrium constant expressions, the equilibrium constant for this type of reaction is just the product of the concentrations of the component ions, and is called a solubility product constant, with the symbol K_sp.  Although K_sp refers to a specific type of reaction, it's just another equilibrium constant so it follows all the same rules and has the same meaning as any other equilibrium constant.
For a specific example, let's consider the "insoluble" salt calcium sulfate.  Writing a K_sp-type chemical equation for calcium sulfate:

CaSO₄(s) ↔ Ca²⁺(aq) + SO₄^2-(aq)

K_sp = [Ca²⁺]_eq[SO₄^2-]_eq

"Insoluble" salts have reactant-favored K_sp's,
If we think about making a precipitate rather than dissolving a salt, we might think of a reaction like:

Ca(NO₃)₂(aq) + K₂SO₄(aq) ↔ CaSO₄(s) + 2 KNO₃(aq)

Equilibrium is really all about the chemical processes that are happening, not all the random fluff that might also be included like catalysts or spectator ions.  This means that we can always think about the net-ionic equation for a process at equilibrium rather than the full-molecular or full-ionic equation.  In fact, we can often completely eliminate terms from the equilibrium constant expression by using a net-ionic equation.  For the calcium sulfate equilibrium above, the full-ionic and net-ionic equation is:

Ca²⁺(aq) + 2 NO₃^-1(aq) + 2 K⁺(aq) + SO₄^2-(aq)  ↔  CaSO₄(s) + 2 K⁺(aq) + 2 NO₃^-1(aq)

Ca²⁺(aq) + SO₄^2-(aq)  ↔  CaSO₄(s)

That's the reverse of the K_sp equation for dissolving calcium sulfate, so we can manipulate the equilibrium constant expression and value to use in this situation.

Simplifying Approximations:
There are a couple approximations or assumptions that can make many of our equilibrium calculations a little easier to deal with.  These assumptions are usually valid for problems where the equilibrium is either quite strongly reactant-favored or quite strongly product-favored.  For very reactant-favored equilibria, we can assume that the change in concentration of reactants is small enough to be negligible.  For strongly product-favored equilibria, we can often treat the problem more like a limiting reactant/theoretical yield problem.  These approximations can be the only way to solve some of the higher order polynomials that come up fairly often in equilibrium calculations.

Correction from class:
As I said a few days ago, equilibrium problems are the places I am most like to get myself in trouble when I make things up during class.  I had a little "oops" today in class with the initial concentrations of the lead and sulfate stock solutions.  The set-up should have been 50.0mL of 1.0M solutions.  All the concentrations (and moles) in the equilibrium tables were fine if this stock concentration is used.

Manipulating equilibrium constants

What happens when we change the written form of a chemical reaction that describes an equilibrium?  This is usually necessary when we're looking at reaction mechanisms or other stepwise processes.

Multiplying a reaction:
Consider the reaction 2A↔B with K=4.  We can multiply that reaction by 3 to get the reaction 6A↔3B.  Fundamentally, this is the same reaction and the same process, but what about the equilibrium constant?  Writing out the equilibrium constants for both reactions:

So if K = 4, K' = 4³ = 64

Changing direction:
If we reverse a chemical reaction, we once again have to adjust the equilibrium constant.  Changing 2A↔B to B↔2A means that we are changing the identity of reactants and product.  This means that the concentrations that used to be in the numerator of the equilibrium constant are now in the denominator and vice versa.  Mathematically, this has the effect of inverting the value of the equilibrium constant.  If K = 4 for 2A↔B, then K"=1/4 for B↔2A.

Multipart/Multistep:
For a multistep process, the equilibrium position of each step will affect the overall equilibrium of the system.  This is in contrast to kinetics where the rate law of the overall process relies only upon the rate law of the slowest step.  For a stepwise process, the equilibrium constant for the overall process is equal to the product of the individual steps.

Manipulating Equilibrium

Because equilibrium is a dynamic, thermodynamic process, there are a number of things we can do to "fiddle with" equilibrium systems.  Perhaps the most significant conceptual tool we have at our disposal is LeChatelier's Principle.  {Side note: To the francophones in the audience, I believe the "a" in "LeChatelier" is supposed to have a little hat on it...}
When a system at equilibrium experiences a stress, the position of that equilibrium will shift to relieve the stress (as much as is possible).
What's stress?  The most common type of stress we're likely to see in chemical reactions is adding or removing a reactant or product.  Let's think about the simple equilibrium reaction 2A↔B.  The equilibrium constant is:

If we add a little extra "A" to the reaction after equilibrium has been established (a stress), the reaction will have to shift toward the products to re-establish equilibrium.  If we look at the reaction quotient, Q, for a reaction, we can evaluate whether a system is at equilibrium, and if it is not, what direction the system has to shift to reach equilibrium.  The reaction quotient has the same mathematical form as the equilibrium constant; if Q if less than K, then there are too many reactants in the system and it must shift toward products.  If Q is greater than K, then there are too many products and the system must shift toward reactants.

Kinetics Experiments - The Nuts and Bolts

Last week in lab we performed a kinetics experiment in which we observed the iodination of acetone. For those of you who may be curious, there were a number of things that went into the planning of that experiment that are required to make it work as smoothly as it did. There is a certain art to experimental design, but it is all firmly rooted in the science that is being observed. For the kinetics of the iodination of acetone, the experimental design considerations are (in my opinion) fascinating, largely because they all make very good sense and can be understood with a Gen-Chem-level of knowledge.

1. Relative concentrations of the reactants. Some of you may have noticed that the concentration of the iodine was very much less than the concentration of the other reagents. This was not an accident. First, since we were observing the color of the solution and that color was due to the iodine, the concentration of iodine had to be high enough to be easily observable but low enough to react in a reasonable amount of time. In addition to this very practical consideration, there is a chemical reason for the concentrations used. The rate of a chemical reaction is dependent upon the concentration of the reactants. As “good” scientists, we always want to design our experiments in such a way that only one variable is changing. If, for example, the concentration of iodine was 1M and the concentration of acetone was also 1M, then as the iodine concentration changed (and affected the rate of the reaction), the acetone concentration would also change and would also affect the rate of the reaction. Although this data could be mathematically interpreted, it would be much more convenient if only one of the concentrations was changing during the reaction. Is this possible? Of course not. Iodine cannot disappear unless it reacts with acetone (in this reaction). BUT, we can make it seem like the acetone concentration is not changing by making the initial concentration of acetone SO high, that the change will be very small, indeed “negligible”, compared to the change in the iodine concentration. To put some numbers to it, if the initial concentration of iodine is 5mM and the initial concentration of acetone is 1.000M and the reaction is allowed to proceed, when all of the iodine has reacted, the final concentration of acetone would be 0.995M. Yes, the concentration has changed, but it has changed by only 0.5%. This means that the change in rate caused by the change in concentration of acetone can be ignored. The reaction (and reaction rate) will observe the rate law order with respect to iodine only.
2. Limits of the Spec-20s. When we did the experiment, we said that the sample did not have to be put in the spectrometer until most of the color had faded. If anyone put their samples in the Spec-20 immediately, you might have noticed that the Spec-20 was unable to read the absorbance of the solution until most of the iodine color faded. This can be explained by thinking about the nature of “absorbance”. Absorbance is a logarithmic scale related to percent transmittance. An absorbance of “0” is equivalent to a percent transmittance of 100%. If the percent transmittance drops to 10%, the absorbance is “1”, which means 90% of the light that is shining on the sample is being absorbed. If the absorbance climbs to “2”, it means that only 1% of the incident light is getting through the sample (99% is absorbed). An absorbance of “3” means only 0.1% of the light is getting through. As we can see, increasing absorbance ver quickly decreases the amount of light getting through the sample. This brings us to the limits of the detector we are using. It can pretty easily discern the difference between 5% of the light being absorbed and 85% of the light being absorbed, but it's less reliable when trying to distinguish 97.8% absorbed and 98.5% absorbed. It is usually best practice to design experiments so the absorbances used are ~1.5 or less. If the absorbance is too high, the data will either get very “noisy”, or will just be abnormally low, making trends that should be linear appear curved.
3. Aggregation of solutes. This probably wouldn't have been a big problem in our iodination of acetone experiment, but Beer's Law assumes that the solute particles are separate and independent in solution. If the concentration of colored solute is relatively high, it's possible that the colored solute particles will start to interact in a way that will make them absorb differently than if they were truly separate and independent in solution. This is often solved by addressing the instrument limits mentioned in #2; if the sample is diluted sufficiently bring to the absorbance down below ~1.5, it is usually dilute enough to avoid interactions between solute particles.

There are other considerations, but these are the big ones that often come up in experimental design for kinetics experiments and any experiment where we are observing color using a spectrometer. For most of our experiments, we set things up behind the scenes to minimize or eliminate these problems, but I hope that some of you were curious enough to wonder why the experiment was set up the way it was.

See you in the morning.

Equilibrium is Organization!!

The key to equilibrium problems is almost always figuring out the best way to organize the data presented in the problem or experiment.  This is most often done by putting together a table.  Consider the following problem.

0.80mols NF₃(g) and 0.60mols O₂(g) are combined in a 1.00L vessel and allowed to reach equilibrium. At equilibrium, the concentration of NF₃(g) is found to be 0.30M. What is the value of the equilibrium constant for this reaction?
Write a balanced chemical equation:

2 NF₃(g) + O₂(g) ↔ 2 NO(g) + 3 F₂(g)

Set up a table to organize the data from the problem:

	2 NF3(g) +	O2(g) ↔	2 NO(g) +	3 F2(g)
[ ]initial	0.80 M	0.60 M	0 M	0 M
Δ[ ]	- 2x	- x	+ 2x M	+ 3x M
[ ]equil	(0.80-2x) M	(0.60-x) M	2x M	3x M

Write an expression for the equilibrium constant:

Plug in values from the table:

Determine "x" from the problem.  We are told that [NF3]eq=0.30M, and we see in the table that [NF3]eq=(0.80-2x), so:

0.30 = 0.80-2x

x = 0.25

Plugging in:

K = 3.4

This equilibrium is (very slightly) product-favored.

Practice, practice, practice...

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:
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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).
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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. 
----------

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

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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/