Today, we took the allll the notes we neglected to take on Friday or Monday. The topics were collision theory, reaction mechanisms, reaction rate and factors affecting reaction rate. I give a short summary of each of them below, but you can find the real notes at http://GBS-Moodle.glenbrook225.org/moodle/mod/resource/view.php?id=86333
Collision Theory
When a reaction takes place, it is because molecules collide with each other at 1) the right speed and 2) the right orientation. The "activated complex" is this transition stage between reactants and products during a reaction. The "right speed" is otherwise none as the activation energy.
Reaction Mechanisms
A reaction does not happen all in one go; a series of steps makes a reaction happen. The "crash" happens, and then the old bonds break to form new bonds. The overall reaction, called the complex reaction, given on paper is just the end result, and many elementary or intermediate steps are taken to get there.
Reaction Rate
This is a measure of how quickly the reactants in a reaction become products, or how much reactant is used over time. We will calculate this in units of molarity (mol/liter) per any unit of time (seconds, minutes, etc). The elementary steps of the reaction all have different required activation energies, and the step with the highest activation energy will take the longest to complete. This is called the rate-determining step, because it will be the factor that affects the speed of the reaction.
Factors Affecting Reaction Rate
Factors include: an increase in temperature, an increase in concentration (molarity), an increase in surface area, and the addition of a catalyst. A catalyst lowers the activation energy of certain slow elementary steps and can also provide different paths for a reaction to take, making it go faster.
Homework:
Check the answers to the equilibrium packet assigned yesterday (http://GBS-Moodle.glenbrook225.org/moodle/mod/resource/view.php?id=86334)
Okay, so you might not be so excited about this. But hopefully this blog
can clear up any confusions you have about blending what we already
know about stoichometry with this unit's topic: solutions.
Today, we started class by turning in our Molarity Stiumlations packets, which we're on schedule to go over tomorrow. If you need one of these, they're in the Unit 4 Handouts folder!
Then, we started on the torture part of this unit: Solution stoich.
Here's a brief overview of the notes we took and the problems we did!
Disclaimer: Ms. Friedmann's notes, which she wrote in class today, are in the Unit 4 Notes folder
under 'Notes on Solution Stoichiometry'. Go check these out, as well.
First, we redefined stoichometry: looking at one "thing" in a reaction (either a reactant or a product) and figuring out how much of another "thing" in the same reaction will be used up or produced based on the measurement of "thing" 1.
This definition might seem a little extensive, but to remind everyone...
A stoichiometry set-up looks just like this!
Now, we all remember doing (a little bit too many of) these problems!
So, where do solutions come in?
Well, to solve stoichometry problems involving Molarity, we need new algorithms.
Our old algorithm looked like this:
We've used this algorithm to solve all of our stoichiometry problems to date. But now,
with solutions, we need a new algorithm involving molarity. The new algorithm:
Obviously, the new parts of this algorithm are:
Calculations using the volume to find other measurements. Use the molarity to convert from volume to moles, and vice versa.
Calculations using the molarity to find other measurements. Use the volume to convert from molarity to moles, and vice versa.
Let's do a problem:
This is the first problem on the worksheet entitled 'solution stoich', located in the Unit 4 Handouts folder.
What volume of 0.150 M AgNO3 is needed to react with 45.0 mL of 1.50 M CaCl2?
What is "thing 1"? CaCl2, because we are given its molarity and volume.
What is "thing 2"? AgNO3, because we are asked to calculate the volume of it needed to react.
There are two ways to start this problem, either with the molarity or the volume of CaCl2. We chose to start with the volume of CaCl2.
Since the volume is in mL, we must convert to L just by shifting the decimal 3 places to the left.
Knowing our first conversion factor must be 1.50 moles CaCl2/1L, because molarity is moles of solute over liters of solvent, we can begin to set this problem up.
0.045 L * 1.50 moles CaCl2/1L
Now, we need to convert from moles of CaCl2 to moles of AgNO3. To do that, we look at the balanced equation. There are 2 moles of AgNO3 for every 1 mole of CaCl2. So...
And that's it! The rest of this sheet, along with the two others recieved in class today, are for homework. Good luck, and feel free to ask questions in the comments section!
Here's an awesome Khan Academy video on solution stoichiometry if you're still confused:
We started off the day checking in our homework from yesterday, the Calculating Molar Mass worksheet. The sheet can be found in the Unit 2 Handouts folder. Answers are as follows:
CaCO3 is ionic and is named calcium carbonate. It's molar mass is 100.09 g.
N2O6 is molecular and is named dinitrogen hexoxide. It's molar mass is 124.02 g.
Na2SO4 is ionic and is named sodium sulfate. It's molar mass is 142.05 g.
C8H18 is molecular and is named octane. It's molar mass is 114.26 g.
Fe3(PO4)2 is ionic and is named Iron (II) Phosphate. It's molar mass is 357.49 g.
Some things to remember about calculating molar mass:
Be careful with a polyatomic ion in parentheses, like (SO4)2. Since there are 2 atoms of SO4, there are 2 sulfur and 8 oxygen.
The masses of each element and your answer will always go to two decimal places.
The notes worksheet we filled out (titled Mole 1) can be found in the Unit 2 Handouts section.
First of all, we defined a mole. A mole is a unit similar to a dozen. Whilst a dozen contains 12 items... A mole contains 6.022×1023 items!
Then, we practiced a few molar mass problems. The answer to 2) is 44.01 g. The answer to 3) is 261.35 g.
It's been duly noted that sig figs don't come into play during molar mass problems, rather, every answer is rounded
to two decimal places.
Then, we started the real lesson: Mole conversions!
We wrote down a diagram to help us remember the conversion factors between moles, grams and particles.
The important thing to remember is: MOLES ARE THE CENTER OF THE UNIVERSE!!
So, moles go in the middle of all conversions, as this picture shows.
As you can all see, the conversion factor between mass (shown here as grams) and moles is the
molar massand the conversion factor between moles and particles (shown here as molecules)
is Avogadro's number. We will be using these conversion factors to solve these problems!
The first problem gave us the number 2.0 x 10-3 g of SnO2 and asked us to find how many moles were present in the sample.
First off, we need to name SnO2. It's correct name is Tin (IV) oxide. Next, we list what we have: 2.0 x 10-3 g of SnO2 Because we know g (grams) is a unit of mass, and we are trying to convert to moles, the conversion factor we are using is molar mass. We set up our equation. 2.0 x 10-3 g of SnO2 * 1 mol / ? g = our answer But how do we know how many grams to put on the bottom of our fraction? It's our conversion factor, the molar mass. We find the molar mass of SnO2 to be 150.71 g. Now, our equation reads: 2.0 x 10-3 g of SnO2 * 1 mol of SnO2 / 150.71 g of SnO2 = our answer Using our prior knowledge of dimensional analysis, we can solve this. Our answer is 1.3 x 10-5 mol of SnO2.
We completed another problem in class, 5a, using the same conversion factor but going from moles to grams and not the other way around. The answer to this one is 370 g of Li2CO3.
Our next problem, 6, was a bit trickier. Given 0.0908 g of nickel (II) chloride, find the number of molecules in the sample.
Molecules = particles. So, looking back at our diagram, we need to convert twice: once from grams of nickel (II) chloride to moles, and once from moles to particles.
First, we need to name nickel (II) chloride. It's name is NiCl2.
Then, we establish what we do know: 0.0908 g of NiCl2
Our first conversion will be from mass (grams) to moles, so we will use the molar mass to convert.
The molar mass of NiCl2 is 129.59. So the first conversion is...
0.0908 g of NiCl2 * 1 mol / 129.59 g
Now that we have converted from grams to moles, we now convert from moles to particles.
To convert from moles to particles, you use Avogadro's number. So, our full equation is...
0.0908 g of NiCl2 * 1 mol / 129.59 g * 6.02 x 1023 / 1 mol = our answer Canceling out the correct units and multiplying across, the answer is 4.22 x 1020.
Ms. Friedmann let us know that she's busy making lesson plans and hasn't had enough time to grade our quizzes or labs yet. Don't worry, Ms. Friedmann, we all still love you!
She also briefly discussed that only covalent (molecular) compounds make true molecules. The things that ionic compounds make are called formula units, although the terms are used interchangeably.
A tricky little thing about ionic compounds: although a compound like MgSO4 is obviously ionic, the bonding between the S and the 4 O's in the sulfate ion is molecular.
And last but not least... thanks for the laughs today KG. Or should I say Scotty :)
