Summary Notes (Mod. 1)Friday, June 10, 20051:37 PM 1.1 - Phases and Phase Diagrams

Transitionso Condensation: gas -> liquido Vaporization/evaporation: liquid -> gaso Fusion: solid -> liquido Freezing: liquid -> solido Sublimation: solid -> gaso Deposition: gas -> solid

Sublimation, fusion, and evaporation are all endothermic processes because energy is required to overcome the attractive intermolecular forces in the denser phase to spread them out in the less-dense phase (solid -> gas, for example)

Phase diagramso These are graphs of temperature vs. pressure to show where each phase is stableo Special notes:

At the "triple point", all 3 phases co-exist Beyond the critical point (Tc, Pc), neither gas nor liquid exists - we only have

something called super-critical fluid If the S-L line has a positive slope, it means that as we increase pressure, we

become solid. This further implies that solid is the densest phase. However, if the S-L line has a negative slope, we move into a liquid area as we increase pressure. As such, liquid is considered to be the densest phase (H2O is a good example of this.)

 1.2 - Clausius-Clapeyron Equation

The L-G line in a phase diagram tells us:o How the vapor pressure of a liquid changes with the temperature

Recall that vapor pressure is when you put a liquid in a closed container, and stuff starts evaporating into vapor until you reach an equilibrium point. At that point, the pressure exerted by the vapor is the vapor pressure.

o How the boiling temperature changes with external pressure The Clausius-Clapeyron equation is as follows:

   

Among other things, we can use it to deduce the normal boiling point of a substance. The normal boiling point is the boiling point under standard conditions - 1 atm of pressure - and so we can plug this into the equation to find out when liquid turns into gas under 1 atm of pressure.

 1.3 - Intermolecular Forces

The condensation of gas or the freezing of liquid occurs because as we lower the temperature, the kinetic energy of the molecules also decreases. If the kinetic energy dips low enough, there won't be enough energy to overcome the attractive intermolecular forces, and so the molecules cannot help but be drawn together.

Different substances have different physical properties for the following categories, depending on the strength of their intermolecular forces:

o Boiling point (high with strong attractive forces)o Melting point (high with strong attractive forces)o ΔHvap (high with strong attractive forces)o ΔHfus (high with strong attractive forces)o ΔHsub (high with strong attractive forces)o Vapor pressure (low with strong attractive forces)o Surface tension (high with strong attractive forces)o Viscosity (high with strong attractive forces)

There are many kinds of intermolecular forces which act between molecules:o Dipole-dipole forces

These occur between molecules with a permanent dipole moment - that is, molecules which are permanently polar due to one of the constituent atoms being more electronegative than the other(s)

o London dispersion forces

These occur between all types of molecules - regardless of whether they are polar or non-polar - because of the fact that electrons are in constant motion within a molecule

So at any given moment, it is possible that one end of the molecule could have more electrons than the other, thus making it a polar molecule

We use the term μinst to denote an instantaneous, temporary dipole moment created by the random movement of electrons, and the term μind to denote a molecule which has a temporary dipole moment because it is beside another molecule which has a μinst, and thus has had an attractive/repulsive effect on its electrons

The polarizability of a molecule measures how easily it can be affected by the μinst of other cells

Larger cells are more polarizable than smaller cells because their charge cloud are more diffuse, or spread out - and thus are more sensitive to the μinst of nearby molecules

o Hydrogen bonding forces These occur between cells containing N, O, or F When hydrogen is covalently bonded to N, O, or F, it is positive and is thus

attracted to lone pairs of electrons on other molecules - it's almost like a very strong dipole-dipole interaction

Strength comparison of various bonds…o Covalent and ionic bonds - REAL bonds - are the strongest. They are considered

"chemical bonding forces"o Intermolecular forces are easier to break

H bonds are stronger than dipole-dipole bonds and London Dispersion Forces Thusly, if you are ever asked to compare the intermolecular forces between X number of

substances, first consider:o Is there any hydrogen bonding going on? If so, how many hydrogen bonds can each

molecule form?o If that doesn't decide anything, consider dipole-dipole interactions - are some of the

substances extremely polar, and some others very non-polar?o Also, consider London Dispersion Forces - remember that bigger (and thusly heavier)

molecules are more susceptible to London Dispersion Forces because their charge clouds are diffuse

 1.4 - Heating Curves

Heating curves show us how the temperature varies with the amount of heat added We can obtain the following information from heating curves:

o The heat capacities of solid, liquid, and gas - how much temperature change can we induce with a given amount of heat added?

o The enthalpies of fusion and vaporization - how much heat do we have to add to fuse (melt) a substance? Or to vaporize (evaporate) it?

 1.5 - Introduction to Solids

2 main types of solids:o Crystalline - they have regular repeating patterns of molecules and have sharp, exact

melting points (because everything melts at the same time because the structure is constant everywhere)

o Amorphous - they have no regular repeating pattern and so they melt over a range of temperatures

There are different types of bonds which hold crystalline solids together in their crystal lattices:o Ionic bonds

These hold together positive and negative ions Examples: NaCl and NH4Cl

o Covalent bonds These hold together atoms by strong covalent bonds, like a "giant molecule" Examples: Diamond, SiO2

o Molecular bonds Here we have weak intermolecular forces holding molecules together Examples: H2O, S8, P4, CO2

o Metallic bonds These hold together metal cations with strong metallic bonds Examples: Cu, Zn, Al, etc.

We classify crystalline solids by the geometry of the unit cell:o Cubic

o Trigonalo Tetragonalo Hexagonalo Monoclinico Triclinico Orthorhombic

 1.6 - Cubic Packing Arrangements

Simple cubic packingo Description: The spheres in each layer are lined up directly with the spheres in the layer

above and below ito Spheres per cell: 1 (1/8 of a sphere at each corner)o Edge length: 2 x ro Packing efficiency (volume of spheres in a cell/volume of cell): 0.5236

Body-centered cubic packingo Description: The spheres of one layer are placed in the holes between the spheres of the

layers above and below it. The contact between the spheres happens in the middle (body) of the unit cell.

o Spheres per cell: 2 (1/8 of a sphere at each corner, and 1 full sphere in the middle)o Edge length: 2.3ro Packing efficiency: 0.6802

Face-centered cubic packingo Description: The contacts between spheres happen on the face of the unit cell.o Spheres per cell: 4 (1/8 of a sphere at each corner, 1/2 of a sphere on each face)o Edge length: 2.2ro Packing efficiency: 0.7405

 1.7 - Closest-packed Structures

These are ways to pack atoms together such that the amount of unused space is minimal There are 2 main kinds of closest-packed structures:

o ABAB closest packing We start off with Layer 1, which is packed such that all the holes between atoms

are trigonal - that is, the hole is surrounded on any given plane by 3 atoms We place Layer 2 on top of Layer 1 such that Layer 2's atoms cover some of

Layer 2's trigonal holes are on top of atoms in Layer 1, and some are on top of holes in Layer 1 as well

Then we place Layer 3 in the exact same position as Layer 1 - its spheres and holes are on top of Layer 1's spheres and holes

Thus we call it "ABAB" closest packing because there are only 2 distinct positions for layers

The unit cells resulting from this closest packing structure can combine in groups of 3 to produce a hexagonal-faced prism

o ABCABC closest packing Here we are exactly the same as ABAB closest packing for Layers 1 and 2, so

the difference lies in how we place Layer 3 We place the spheres in Layer 3 over the locations where there is a hole in both

Layer 1 and Layer 2, so that all holes are now covered - thus, the position of Layer 3 is necessarily distinct from Layer 1 and 2, and so we call it "ABCABC" closest packing

The unit cells resulting from this closest packing structure can be divided into face-centered cubic unit cells

 1.9 - Ionic Solids and Interstitial Sites

Now we are going to talk about how different atoms are packed together, unlike before when we were talking about packing the same kinds of atoms together

For a lot of ionic solids, we have one of the ions forming a cubic lattice - like what we talked about before - and the other one occupying the holes in that lattice

There are a few different holes which we should be familiar with:o Trigonal hole

If we assemble 3 spheres such that they appear to each be a point on a triangle, there will be a hole between these 3

Therefore the co-ordination number (the number of spheres making up the hole) is 3

However, this hole is too small to worry about - nothing of practical interest will ever fit inside

o Tetrahedral hole This is when the spheres surrounding the hole make up the shape of a

tetrahedral molecule - imagine the locations of the four "corners" of a tetrahedral shape - that is where the spheres surrounding this hole are located

So the co-ordination number is 4 The ratio (as described under the "Octahedral hole" point) is 0.225

o Octahedral hole Here we have the spheres surrounding the hole in an octahedral shape - where

we have 4 spheres lying 2 x 2 in a plane, and then spheres directly above and below the hole in the middle of those original four

So the co-ordination is 6 The ratio of the radius of a sphere which could fit into the hole to the radius of

the spheres surrounding the whole is 0.414 So obviously, if we place a sphere into that hole whose radius is less

than 0.414 of the surrounding spheres, then it's not going to disturb the surrounding spheres. However, in nature, what often happens is that the sphere is a little too big for the hole, which means that the surrounding spheres are pushed apart and are no longer touching each other…

o Cubic hole The hole in the center of a normal 2 x 2 x 2 sphere

Thus the co-ordination number is 8 The ratio (as described under the "Octahedral hole" point) is 0.732

Consider the face-centered cubic cell:o Along each of the edges, there is 1/4 of an octahedral hole, as well as a full octahedral

hole right in the middle of the cello Each corner sphere forms a full tetrahedral hole with the half spheres on each of the 3

faces beside that cornero So all in all, we have 4 octahedral holes and 8 tetrahedral holes in the face-centered cell

In general, in a closest-packed structure containing N spheres, there are N octahedral holes and 2N tetrahedral holes

When we are dealing with binary ionic solids, remember that R- (the negative ion) usually makes up the crystal lattice, and R+ (the positive ion) fills in the holes

We can use this information to predict what kind of holes a given positive ion occupies, based on the size ratio between the positive and negative ions

o 0.225 < R+/R- < 0.414 || The positive ions occupy tetrahedral holeso 0.414 < R+/R- < 0.732 || The positive ions occupy octahedral holeso R+/R- > 0.732 || The positive ions occupy cubic holes

  Ionic crystal structures to remember:

o Sodium chloride Cl- ions form an fcc lattice, and Na+ ions occupy the octahedral sites

o Cesium chloride Cl- ions form a simple cubic lattice, and Cs+ ions occupy cubic holes in Cl-

latticeo Zinc blende (i.e. ZnS)

S2- ions form an fcc lattice, and Zn2+ ions occupy half the tetrahedral holes They have to be in half the tetrahedral holes instead of all the

octahedral holes because they are too big for the octahedral holes!o Fluorite (i.e. CaF2)

Ca2+ ions form an fcc lattice and F- ions occupy all of the tetrahedral siteso Antifluorite (i.e. Na2O)

O2- ions form fcc lattices and Na+ ions occupy all of the tetrahedral sites 

1.10 - The Born-Haber Cycle We can measure the stability of an ionic solid by what its lattice energy is…

o Lattice energy is the enthalpy change when gas-phase ions combine to form an ionic solid

Because we can't measure this enthalpy directly, what we often do is to break up this "solidification process" into separate steps that we know the enthalpies for…

So for example, the reaction Na + Cl -> NaCl…o First of all, consider what state the elements are normally in: Na(s) + Cl2(g)

We realize that Na is a solid (and will have to be sublimated) and Cl2 is a diatomic molecule, so the bond will have to be broken

o So now we have Na(g) + Cl(g)…But we need them to become ions as well! So…Na -> Na+ + e- (ionization) and Cl + e- -> Cl- (electron affinity)

o Then we use lattice energy - the energy resulting from the actual process which will combine Na+(g) and Cl-(g) into NaCl(s)

We also know the Δhoformation for this process - and from there we can figure out what the lattice energy (just that last step) is: ΔHo

f = ΔHsub + IE(1) + 1/2 DCl-Cl + EA(1) + ΔHlatticeo Don't forget that we only take half of the bond dissociation energy (D) for Cl2 because

we only want 1 atom!  Summary Notes (Mod. 2)Saturday, June 11, 20053:15 PM 2.1 - Chemical Kinetics and the Factors Affecting the Rates of Chemical Reactions

Chemical kinetics is the quantitative study of reaction rateso Our ultimate goal is to deduce information about the sequence of molecular processes

involved in the conversion of reactants into products, because usually there are multiple steps involved!

There are several factors which affect reaction rates:o Nature of the reactants

For example, ions in solution react rapidly because they are mobile and can exert strong forces on each other, even from far distances

But molecules with very strong bonds or complicated shapes will take longer to react

o Reactant concentration The more concentrated the reactants are, the more they will bump into each

other and therefore reacto Temperature

Increasing temperature will increase kinetic energy, which again will cause the molecules to move faster and collide more

o Catalysts These things lower the activation energy needed to get reactions started, and

so they happen quickero Physical state of the reactants

A lot of times, solids are slower to react because less surface area is in contact between two solids, so we have fewer molecules bumping into each other

 2.2 - Quantifying the Reaction Rate: Average and Instantaneous Rates

The rate of a chemical reaction tells us how much reactant is consumed (or how much product is produced) per unit of time

o We use the unit mol/L/s to measure this We can quantify the rate of reaction using two different types of rates:

o Average rate (measured over a length of time)o Instantaneous rate (measured at a specific point of time)

 2.3 - Rate Laws for Chemical Reactions

A rate law is a mathematical expression which shows us how rate depends on concentration. It is formed as follows:

  

We can use the "method of initial rates" to deduce the rate law for a reaction. This is when we carry out the same reaction multiple times, but we vary the concentrations of the reactants. Each time we run the experiment, we measure the "initial rate of reaction" - the instantaneous rate of reaction at t = 0, essentially.

o By comparing different runs of the experiment in terms of their reactant concentration and initial rate of reaction, we can deduce the orders of each reaction in the rate law

 2.4 - Integrated Rate Laws

Once we have the rate law for a given reaction, we can perform a calculus integration on it to find an expression which helps us to easily see how the concentrations of a reactant change as time goes by

The rate laws are as follows:       2.5 - Analysis of Concentration vs. Time Data

If we are smart, we can use the integrated laws on any random reaction and find what its order is and what the value of its rate constant (k) is! There are 2 ways to do this:

o Algebraic approach If we have information on a bunch of different instances in the experiment, each

of which have different starting and ending concentrations, and time changes, then we can sub each of these data sets into the general integrated rate laws for 0th order, 1st order, and 2nd order reactions…

Since k is (more or less) constant for a given reaction, the integrated rate law for the correct order of reaction will give the same values of k each time

o Graphical approach In the same way, if we have multiple data sets from different runs of the

experiment, we can graph them… To check for a 0th-order reaction, plot [A] vs. t

If the reaction really is 0-order, this will be linear with slope -k Linear because the slope (which is -k) shouldn't

change between different experimental trials if this is the right order

Slope is negative because [A] (the concentration) is decreasing

To check for a 1st-order reaction, plot ln[A] vs. t If the reaction really is 1-order, this will be linear with slope -k

Linear because the slope (which is -k) shouldn't change between different experimental trials if this is the right order

Slope is negative because [A] (the concentration) is decreasing and so ln[A] will as well

To check for a 2nd-order reaction, plot [A]-1 vs. t If the reaction really is 2-order, this will be linear with slope

+k Linear because the slope (which is -k) shouldn't

change between different experimental trials if this is the right order

Slope is positive because [A] is decreasing, yet we are plotting [A]-1

 2.6 - Half-life (t1/2)

t1/2 is the time it takes for the concentration of A to decrease to one-half its initial value Although we could of course use the integrated rate laws to calculate half-lifes, we can derive

equations which are even quicker. Here they are:  

Note that the half-life equation for 2nd-order reactions is NOT constant - it depends on how much reactant we are starting with!

 2.7 - Summary of 0, 1, and 2-Order Kinetics           

                 2.8 - Reaction Mechanisms

A reaction mechanism is the sequence of elementary processes which show how reactants are converted into products

o We can never know what the reaction mechanism for a particular reaction is for sure, but we can make a good guess if we compare the known rate law for the total reaction to the rate law implied by a proposed reaction mechanism

We can use collision theory to make predictions about the rates of the elementary processes…the key idea is that: rate of an elementary process is proportional to z x f x p

o Z = collision frequency, which is determined by the concentration of each reactanto F = the fraction of collisions with enough kinetic energy to lead to a reactiono P = the fraction of collisions with a favorable alignment for breaking the necessary

bonds and forming new ones The rate laws for elementary processes are easy to deduce - the order of each reactant is equal to

the number of its molecules involved in the process 

So how do we figure out the rate law for a reaction when we are given the reaction mechanism? There are 2 paths we can take…

o We calculate it based on the slowest step of the reaction…but we can only do this when we are told which step is the slowest!

o We make no assumptions about which steps are fast or slow - just pick one reactant or product and figure it out based on that

One very useful tool when doing this is the steady-state approximation! It states that we assume that the rate of producing an intermediate = rate of consuming an intermediate

o When we need to eliminate intermediates from a rate law, we can make substitutions based on the above approximation

o Using this assumption entails finding the rates for all the processes producing this intermediate, then setting that to equal the rates for all processes consuming this intermediate

 2.9 - How does Temperature Affect Reaction Rate?

Often during an elementary process, there is even a sub-step within the process where we get an "activated complex" (made with the reactants) that quickly turns into the products of the elementary process

o The most important thing about this activated complex is that it is short-lived and high-energy…the kinetic energy from the collision (and thus the start of the reaction) is converted to potential energy such that this activated complex has a lot of energy

o So exactly how much energy do we need to input in order to get to this high energy activated complex state? The amount is called activation energy, and it is denoted by Ea

Now think about this…if we have a reversible reaction, there is an activation energy to go from the reactants to the activated complex (and then, spontaneously, to the products) and also a different activation energy to go from the products back to the activated complex (and then spontaneously to the reactants)…

o Think about the difference between the two activation energies (we'll denote them with Ea

forward and Eareverse)…The overall enthalpy change for the reaction is ΔH = Ea

forward - Eareverse

The equation relating temperature of a reaction with the rate constant for the reaction is as follows:

  

2.10 - Catalysts A catalyst provides a new mechanism with a lower activation energy, which means that a

greater fraction of collisions lead to reactions, and so reactions proceed more rapidly 2.11 - One Final Point

Thinking about activation energies again, we should note that the step with the highest activation energy is not necessarily the rate-determining step!

Consider a situation where the first step (although having the highest activation energy) is a reversible step, and so the activation complex is much more likely to go backwards than to keep going to completion…that means that very little product from the first step will be produced!

So the reaction which takes this product and makes another product - the overall product of the reaction - is the rate-determining step!

   Summary Notes (Mod. 3)Sunday, June 12, 20054:04 PM 3.1 - Chemical Equilibrium is a Balance of Reaction Rates

We need to remember that all reactions are reversible, meaning that the products can turn back into reactants!

Eventually we reach an equilibrium point, which is when the rate of reactants -> products (known as the "forward" reaction) and the rate of products -> reactants (known as the "reverse" reaction)

o At the beginning of the reaction, we expect the rateforward to be large and the ratebackward to be small (because there is a lot of reactant and no product), but then as the reaction proceeds they grow closer to each other

o When the rates are equal, we say that a state of dynamic equilibrium has been reached, which means that there is still motion at a molecular level (hence the word "dynamic"), but that there is an overall balance in the concentrations of the reactants and products (hence the word "equilibrium")

 3.2 - Law of Chemical Equilibrium

We use a function called the reaction quotient (denoted by Qc) to describe the relationship between the concentration of the products and that of the reactants. It looks like this:

  

If we happen to measure this relationship when the reaction is at equilibrium, we receive a special value of Qc named Kc - the equilibrium constant. Here is how we derive it:

  

o A very important thing to remember about the equilibrium constant Kc is that its value does not change, even if the starting concentrations of the reactants in a reaction do

However, the composition of a system at equilibrium will change depending on the starting concentrations - but just know that this is different from the equilibrium constant!

We can learn a lot of information from the instantaneous value of Qc - that is, the relationship between the concentration of products and the concentrations of reactants at the beginning of the reaction, before anything has happened - when we compare it with Kc, which is just Qc at equilibrium

o When Q < K, the reaction must go in the forward direction to reach equilibriumo When Q > K, the reactions must go in the reverse direction to reach equilibriumo When Q = K, there is no "net" reaction

Summary Notes (Mod. 3)Monday, July 11, 20056:20 PM 3.1 - Chemical Equilibrium is a Balance of Reaction Rates

We need to remember that all reactions are reversible, meaning that the products can turn back into reactants!

Eventually we reach an equilibrium point, which is when the rate of reactants -> products (known as the "forward" reaction) and the rate of products -> reactants (known as the "reverse" reaction)

o At the beginning of the reaction, we expect the rateforward to be large and the ratebackward to be small (because there is a lot of reactant and no product), but then as the reaction proceeds they grow closer to each other

o When the rates are equal, we say that a state of dynamic equilibrium has been reached, which means that there is still motion at a molecular level (hence the word "dynamic"), but that there is an overall balance in the concentrations of the reactants and products (hence the word "equilibrium")

 3.2 - Law of Chemical Equilibrium

We use a function called the reaction quotient (denoted by Qc) to describe the relationship between the concentration of the products and that of the reactants. It looks like this:

  

If we happen to measure this relationship when the reaction is at equilibrium, we receive a special value of Qc named Kc - the equilibrium constant. Here is how we derive it:

  

o A very important thing to remember about the equilibrium constant Kc is that its value does not change, even if the starting concentrations of the reactants in a reaction do

However, the composition of a system at equilibrium will change depending on the starting concentrations - but just know that this is different from the equilibrium constant!

We can learn a lot of information from the instantaneous value of Qc - that is, the relationship between the concentration of products and the concentrations of reactants at the beginning of the reaction, before anything has happened - when we compare it with Kc, which is just Qc at equilibrium

o When Q < K, the reaction must go in the forward direction to reach equilibriumo When Q > K, the reactions must go in the reverse direction to reach equilibriumo When Q = K, there is no "net" reaction

 3.3 - Rules and Conventions for Qc and Kc

Qc and Kc are always evaluated in mol/L! We never include the concentrations of pure solids (s), pure liquids (l), or liquid solvents (often

H2O) in our Qc expression Sometimes when we are calculating equilibriums, we can use the 5% rule: we say that C - x = C

if the value of x turns out to be no more than 5% of C Special feature: Method of Successive Approximations

o Sometimes we need to figure out the value of x, but the quadratic formula is too messy and the 5% rule is not satisfied, so we can't approximate it

o So we use the "method of successive approximations", which is essentially a recursive technique we use to find x

o We take the equation in question and isolate some function of x (let's say x, or x2 - usually the numerator) on one side, then solve for it firstly by making the C - x = C approximation

o We will get some value of x which is too large of a percentage of C to use, but what we then do is re-calculate x using this value (instead of the approximation value)

o We continue doing this until we get a similar value of x 2 times in a row - then we are sufficiently close to the desired value

 3.5 - Gas-phase Equilibria

Sometimes it is desirable to find an equilibrium based not on concentrations of substances, but on the partial pressures of them (i.e. when our reactants and products are gases)

o We use Kp to denote this, and it is [product of partial pressures of products] / [product of partial pressures of reactants]

o Kp is related to Kc by the following equation: Kp = Kc(RT)Δngas

 3.6 - Manipulating Reactions and Equilibrium Constants

If we play around with reactions, we have to adjust the equilibrium constants accordingly!o If we reverse a reaction, then invert ko If we multiply a reaction by "n", then we raise k to the power "n"o If we add 2 reactions together, then we multiply the k's together

 

3.7 - Le Chatelier's Principle A guy named Le Chatelier came up with a way for qualitatively describing the ways that different

actions can affect a system:o "When a system at equilibrium is disturbed, the equilibrium shifts in the direction that

partially counteracts the disturbance." So what does that mean? Essentially, if a system is at equilibrium and we do something which

affects the equilibrium, the system automatically shifts so as to try and recover its previous state (although it can never do so fully)

o However, note that it is very important that these "self-adjustments" only happen when the change affects the value of Qc

So, what are the consequences of this rule?o If the volume of the reaction vessel decreases: this means that we are increasing

the pressure of the system…and since we want to maintain the same pressure, the equilibrium will shift (either right or left) so that fewer moles of gas are produced

o If a quantity of some substance is removed: the equilibrium will shift such that we produce more of that substance

o If the temperature is increased: this is going to increase the energy of the system (since there will be more kinetic energy, and so forth) and so since we want to maintain the energy of the system, we are going to shift in a direction where energy is consumed, rather than produced (i.e. we will go in the endothermic direction)

o A small amount of some solid is removed: nothing will happen, because solids are not in the equilibrium quotient, so the equilibrium is not disturbed!

o A catalyst is added to the reaction mixture: nothing will happen, because catalysts merely affect the speed of a reaction - they don't change the equilibrium!

 3.8 - Temperature Dependence of K

Remember in Module 2, how there were rate constants for every reaction, and they were affected by what the temperature was? In much the same way, equilibrium constants (Kc) are affected by temperature

o A very important note is that we are talking about the thermodynamic equilibrium here - which means that we express non-gases in concentrations, but gases in partial pressures!

The temperature and thermodynamic equilibrium interaction according to this equation:  Summary Notes (Mod. 4)Monday, July 11, 20056:20 PM 4.1 - Acids, Bases, and Conjugate Acid-Base Pairs

According to the Bronsted-Lowry definition:o Acid = proton (H+) donoro Base = proton (H+) acceptor

Conjugate acid-base pairs are pairs of acids and bases which can be converted to each other via a proton transfer reaction

o Proton transfers happen when one molecule has a lone electron pair (negative charge) which is attracted to a proton on another molecule (positive charge)

o The proton gets taken away from its molecule and is now part of a new one According to the Lewis definition:

o Acid = electron pair acceptoro Base = electron pair donor

The Lewis definition encompasses all the Bronsted-Lowry acids and bases, and also covers reactions where proton transfers don't happen

o Consider the description of a proton transfer above to understand the wording of the Lewis definition: when a base with a free electron pair attracts and ultimately steals a proton from another molecule, the base now no longer has a free electron pair! Instead, it is bonded to a proton where the free pair used to be. Essentially, the base has donated an electron pair.

o And the acid which the proton was removed from now is less one proton, and has an electron pair there in its place! Essentially, the acid has accepted an electron pair.

 4.2 - Self-ionization of Water and the Ion Product of Water

Water can self-ionize, which means that all by itself, it splits into H3O+ and OH- molecules:

o 2H2O (l) <--> H3O+ (aq) + OH- (aq) The equilibrium constant for this reaction is denoted Kw, and we have Kw = [H3O+]eq[OH-]eq (a very

important relationship) 4.3 - The pH Scale

Important formulas/relationships:o pH = -log[H3O+]o pOH = -log[OH-]o pH + pOH = pKw = -logKw = 14

 4.4 - Ionization of Acids and Bases in Water

Water molecules are amphiprotic, which means that they can accept or donate protons (i.e. act as an acid or a base)

o Which one they do depends on the nature of the substances which are dissolved in the water

Acids protonate water molecules, which produces H3O+ ions (thus all acids must have at least H+ present)

Bases de-protonate water molecules, which produces OH- ions (thus all bases must have a lone electron pair to bond with this new proton)

When it comes down to it, the strength of an acid or base is determined by the equilibrium constant for its ionization reaction

o If it is large (>> 1), that means that the acid/base fully ionizes, so it is very strongo If it is small (<< 1), then not much of the acid/base ionizes, and therefore it is weak

The equilibrium constant for ionization reactions is called an ionization constant, and is represented by Ka (for acids) or Kb (for bases)

There are only 6 strong acids:o HI // HBr // HCl // H2SO4 // HNO3 // HClO4

Note 1: HI, HBr, and HCl all are from Group 17, and are binary acids (just H+ and another atom)

Note 2: H2SO4, HNO3, and HClO4 are oxoacids, which means that their structure is such that the proton which they donate is bonded to an oxygen

Note 3: H2SO4 (sulfuric acid) is a special case because it ionizes twice (there are 2 protons)

There are 7 strong bases:o LiOH // NaOH // KOH // Mg(OH)2 // Ca(OH)2 // Sr(OH)2 // Ba(OH)2

Note 1: LiOH, NaOH, and KOH are from Group 1 and are VERY soluble in water This means that NaOH (s) will go 100% to NaOH (aq), which will go

100% to Na+ and OH- Note 2: Mg(OH)2, Ca(OH)2, Sr(OH)2, Ba(OH)2 are from Group 2 and are not

totally soluble in water This means that Mg(OH)2 (s) will not go 100% to Mg(OH)2 (aq),

although Mg(OH)2 (aq) will go 100% to Mg+ and OH-

o Also note that ions such as H-, O2-, and S2- can act as bases because they accept a proton from water

Understandably, these will be strong bases because they don't need to ionize - they are already ions! (Recall that the strength of an acid/base is determined by how much it ionizes…)

However, note that these ions will never exist by themselves…we usually get them as part of compounds such as NaH, Na2O, and Na2S

 4.5 - Molecular Structure and Acid Strength

In very general terms, it is easiest to ionize an acid when the H-A bond (the bond between the proton and the rest of the molecule) is long and polar…or in other words, acids are stronger when the H-A bond is long and polar

o Why long? Because long bonds are weaker bonds (think about how triple bonds are much closer than single bonds), and so the proton can break away more easily!

o Why polar? Because the more polar the bond is, the more charge already exists on the proton…and so it is more strongly attracted by the negatively charged lone electron pair on the base

For binary acids:o Acid strength increases as you go left to right across a period because the

electronegativity of the element to which H is bonded increases, making the bond more polar

o Acid strength increases as you go down a group because the bond length increases For oxoacids:

o We don't really care about bond lengths here, because (by definition) the proton is always bonded to an oxygen

o But we do care about the polarity, which is affected by electronegative atoms and/or electron-withdrawing groups (so the more there are, the more polar the bond will be, and therefore the acid will be stronger)

Special feature: What about HF?o The reason this is interesting is because it fits the profile of a strong acid - it is in Group

17, and F is fairly electronegative…the only beef could be that the bond length might not be super long, but still there is a huge drop-off between the ionization constants of HF and HCl, for example

o The reason why HF is so weak is that when it ionizes, the H+ and F- ions (for whatever reason) stay together to form an ion pair…and so the H+ ions can't do their thing and make the solution acidic!

o We end up with the following equilibrium constants: Ka = [F-][H3O+] / [HF] = relatively small Kion pair = [(H3O+)(F-)] / [HF] = relatively large

 4.6 - Polyprotic Acids

Polyprotic acids have two or more hydrogen atoms…and they ionize sequentially, in separate reactions

o Usually, the ionization constants decrease dramatically from one reaction to the next…this is because as we ionize, the resulting acid is negatively charged, and therefore attracts the proton to itself more strongly…so it is harder to tear the proton away!

o When we're trying to determine the pH of weak polyprotic acids, we just have to consider the first reaction - even though H3O+ is produced in subsequent reactions - because a) recall that the ionization constant for the second reaction is going to be super-small anyway, and b) since it is a weak polyprotic acid, not much of the second acid is produced from the first reaction anyway!

Of course, sulfuric acid is an exception because it is a strong acid… 4.7 - Systematic Treatment of Equilibrium

Sometimes we want to know the exact concentrations of substances without making any simplifying approximations…but it's hard when we have only 4 equilibrium equations (let's say)…so we use 2 more types of equations:

o Material balance equation (MBE): The initial concentration of some acid equals the concentration of that acid and all the other acids after ionization

o Charge balance equation (CBE): Positive charge equals negative charge…so just count up all the ions - positive equals negative

Don't forget H3O+ and OH-! Don't forget to use coefficients for the ions which have a non-unitary charge!

 4.8 - Conjugate Acid-Base Pairs Revisited

When an acid (HA) loses a proton, it is converted into its conjugate base (A-)o The strength of A- as a base is characterized by its Kb value…and it is inversely related to

the acid strength of HAo We know this from the following relationship: Kw = Ka x Kb

The conjugate base of a weak acid is a weak base The conjugate base of a strong acid is a very weak base

  Summary Notes (Mod. 4 cont'd)Tuesday, July 12, 20059:43 AM 4.9 - Salt Solutions

OK, essentially a salt solution is an ionic compound - meaning that it consists of two ions!o And this is important for acids & bases because the ions could have acidic or basic

character, thus affecting the pH! Many salts are composed of a metal cation (i.e. positive ion) and an atomic or molecule anion (i.e.

negative ion) - that is, the anion is either an atom of some element or a molecule of some substance (there may be more than one element)

Interesting fact: Metal ions with a charge of +1 or +2 usually don't affect the pH of a solution, but when we get metal ions with a charge of +3 or more, they do! And this is because on a molecular

level, each individual metal cation is surrounded by some number of water molecules (this is called the "primary hydration sphere", because the positive charge on the cation attracts the negatively charged oxygen atom of the water molecules (well…the SEMI-negatively charged oxygen atom -- remember that oxygen is more electronegative than hydrogen, so in a bond we're probably going to see a slightly negative nature in the oxygen)

o And then, the aluminum ion will pull electron density away from the hydrogen even MORE…and so now that bond is really polar and it's easy for the H+ ion to leave and form H3O+ with other water molecules NOT in the primary hydration sphere, thus affecting the pH!

  So when we are trying to figure out the pH of salt solutions, we should split the salt into its ions

and figure out the acidic/basic character of the ions When we are looking at salts derived from polyprotic acids though, we must be careful!

Because…the ion can either be oxidized or reduced - i.e. we must consider both its Ka and Kb value! Whichever is higher will tell us if the result is acidic or basic

 4.10 - Common Ion Effect

OK…think about the ionization of a weak acid, HA: HA (aq) + H2O (l) <--> A- (aq) + H3O+ (aq)o Now think about what would happen if we added A- or H3O+…the equilibrium will go way

to the left and the amount of A-/H3O+ which actually came from the ionization of HA is reduced a lot!

 4.11 - Acid-Base Neutralization Reactions

We can have all sorts of different neutralization reactions: strong acid/strong base, strong acid/weak base, weak acid/strong base, weak acid/weak base

o We assume that ANY neutralization reaction where either the acid or the base is strong will go 100% to completion…Here is the proof:

  Also, here is how to figure out the Kneutr:

o And the KEY idea is this: since we know that the reaction will go to completion, we can just calculate how much was left over (i.e. which reactant was in excess), and then figure out what pH that will give us (not forgetting to take into account the contribution to the pH made by the salt…)

 4.12 - Buffer Solutions

OK, firstly: a buffer solution contains appreciable amounts of a weak acid and its conjugate base (or the other way around)

o The idea here is that since they are conjugates, they will not neutralize each other, but will rather exist in equilibrium

And a buffer's big ability is that it can absorb of strong acid/strong base, but not change appreciably in pH (as long as it's not TOO much acid/base…)

o And this is because it neutralizes the strong acid/base! A buffer will turn strong acid into weak acid (less of an effect on pH), and strong base into weak

base: 

Here are two ways of calculating the pH of a buffer solution: 

Note that when we talk about concentrations in a buffer, we can just use the starting concentrations for each substance, because they are pretty much the same as the equilibrium concentrations. Why? Because of the common ion effect!

  If we want to PREPARE buffer solutions and we're trying to figure out how much of each thing to

put, there are a few things we can do:o Do a straight calculation to figure out how much of the weak acid and conjugate base (or

vice versa) we will needo Or figure out how much of the weak acid and strong base (which will NEUTRALIZE some

of it, producing a salt which is made up of the conjugate base) 

Buffer capacity is the number of moles of strong base that must be added to 1.0 L of the solution to raise pH by 1 unit

o So when we have a buffer and we add something, we have to figure out how that affects the buffer ratio!

Which neutralizes it, acid or base? And what is the effect of this neutralization? Particularly, the salt?

  

Tricky Things…Saturday, July 30, 200512:13 AM Module 3

When I work with equilibrium problems and I'm given the equilibrium of concentration of something, it doesn't always equal x! (i.e. Example on Page 3-7)

Again with equilibrium problems, don't forget what we have to do when a substance has a non-unitary coefficient!

o If affects how much ("x") it loses or gains…o And also it must be raised to the appropriate power in the Qc expression!

When solving for x in equilibrium problems, don't forget to square root both sides if that's what you want to do!

Justify your damn assumptions! When you are using the Van't Hoff equation…don't forget to use joules! Not kJ!

o And also…use the thermodynamic equilibrium constant! Whenever you're given a Kc to apply to a reaction, make sure it's the exact same reaction, and

that you don't have to manipulate Kc! Get your mol and mol/L straight!

Summary Notes (Mod. 6)Sunday, July 31, 20054:18 PM 6.1 - Electrochemical Cells

OK…now we're going to talk about electron transfer reactions - i.e. redox reactions, where electrons are lost by one substance and gained by another

Think about a zinc rod in an aqueous solution of copper nitrateo You will observe that a copper coating forms SPONTANEOUSLY on the rod…why?o Because this reaction happens: Zn (s) + Cu2+ (aq) --> Zn2+ (aq) + Cu (s)

So here we have solid zinc (i.e. the stuff which was on the rod) being oxidized and going into solution

But then we have copper cation in solution being reduced and turning to solid, and adding itself to the rod (this is where we get the coating from!)

o There are actually two half reactions going on here: Oxidation: Zn (s) --> Zn2+ (aq) + 2e- Reduction: Cu2+ (aq) + 2e- --> Cu (s) Overall: Zn (s) + Cu2+ (aq) --> Zn2+ (aq) + Cu (s)

OK, so that's cool…but now the big concept we have to realize is that electron transfer reactions like this one can happen even when the two reactants aren't in direct contact!

  In electrochemical cells, we have two reactants (just like Zn and Cu which we just considered!)

which are physically separated, but electrically connected by a conducting materialo It is made up of two "half-cells" and a salt bridge

It looks like this:   

So notice firstly that there is a platinum wire (which can carry electrons) going from one reactant (zinc) to the other (copper)

So think about the first reaction we talked about, and how that could happen here…o On the zinc side, the oxidation half reaction will happen and the electrode (that's what

we call the block of zinc metal) will decrease in mass, and we will have Zn2+ and 2e- in solution

o But instead of going into solution, the electrons will travel through the wire to the copper electrode! And what's going to happen there? We have Cu2+ (from the solution) and all these extra electrons, so a reduction is going to happen and we will get Cu (s) added onto the copper electrode - thus the electrode is going to increase in mass!

In this example, we say that the zinc electrode is the anode (the electrode where reduction occurs) and the copper electrode is the cathode (where reduction occurs)

  We can classify electrochemical cells as:

o Galvanic (aka voltaic) cell - if they produce energy (electron movement) due to the spontaneous redox reaction which occurs!

o Electrolytic cell - if the redox reaction is NOT spontaneous 

Here is how we describe the components of an electrochemical cell in shorthand notation:   

And lastly…we need the "salt bridge" in our electrochemical cell so that we don't get a charge build-up on either half cell…

o Think about it! If the salt bridge isn't there, then as the half-reactions happen in each cell, we are going to get a build-up of either positive or negative charge!

In the zinc cell, the electrode is in a solution of Zn(NO3)2, and as the half-reaction occurs, we are getting more and more Zn2+ ions in this solution…and so there is lots of positive charge!

In the copper cell, we have an aqueous solution of Cu(NO3)2, but Cu2+ ions are disappearing and we're just left with negative stuff - (NO3)2!

o So what does the salt bridge do for us? Well, it is made up of positive and negative ions (say, KNO3) and so the positive ions can go into the copper solution to neutralize all the negative charge, and the negative ions can go into the zinc solution to neutralize all the positive charge!

 6.2 - Cell Potential

OK…here's the deal…when we talk about "cell potential", we are referring to how much work can be done for each unit of electric charge that goes through that wire between the anode and the cathode!

So the formula is like this: 

And the cell potential depends on a bunch of stuff…o The oxidizing or reducing strength of each half-cell (as it shouldn't surprise you, some

substances are better at doing the half-cell/redox reaction thing than others!)o The conditions under which the cell operates (the temperature, pressure, etc.)

  Alright, our next objective is to come up with a way to find the oxidizing/reducing strength of

different half cells (i.e. different substances)…but how can we do this, since it's all relative?o Well firstly, we will say that we are measuring stuff at certain conditions (25o C, 1 atm, 1

mol/L)o Also, we are going to compare possible half-cells to an accepted "reference" half-cell -

this is the "standard hydrogen electron" (SHE) 

6.3 - Standard Hydrogen Electrode and Standard Electrode Potentials Alright…it's kinda hard to find hydrogen as a metal…so if we want to look at half-reactions

involving hydrogen, we are going to put an inert metal (platinum) into the aqueous solution and in a tube above it, put H2 gas…what we're going to see is an equilibrium established between the H2 molecules and H3O+ molecules in the aqueous solution

o So there are obviously half-reactions for the SHE, and they differ based on whether the SHE is the cathode or the anode in any given system

Alright…now this is important…each half-reaction has a cell potential…what is a cell potential exactly?

o It is the tendency for a REDUCTION process to occur at an electrodeo So if we put that in the perspective of a half-cell, we're talking about how well an

electrode can attract electrons…o So if the right-hand electrode attracts electrons (i.e. is reduced) very strongly and the

left-hand electrode doesn't really attract electrons too strongly, then there will be a big-time electron flow to the right, and the voltage will be great!

However, we can't measure this potential directly!o But what we can do is…if we set up a system where two half-reactions are occurring, and

electrons are going across a wire, we can put a voltage meter on the wire and figure out what the total voltage is!

o And so…if we set up systems with the SHE and some other half-cell, then we just say (for reference's sake) that the potential of the SHE is 0, then we can figure out what the cell potential of the other half-cell is!

So, FOR EXAMPLE:o If we hook up a system such that a SHE is on the left and a zinc half-cell is on the right,

then measure the flow of electrons and see that it is -0.763 V, what does that mean?

o It means that the electrons are flowing to the left more than they are flowing to the right! What does that tell us about how good each electrode is at getting reduced? It tells us that the SHE is BETTER at getting reduced than the Zn is…but if we say that the SHE's potential is 0, then the potential of the Zn electrode must be -0.763

And, following…how do we calculate the potential of a full electrochemical cell in general?o Eocell = Eo(right) - Eo(left)

Why? Think about it…if the right half-cell is stronger than the left one, than the electron flow will be to the right…and thus the potential of the cell will be positive! According to the formula, does this make sense? Yes!

Note that this is ONLY when we put anode on the left and cathode on the right! 

Yeah, so there's this Table of Standard Reduction Potentials with the potentials for a bunch of half-cells…a couple things to note about it:

o All the reactions are written as reduction reactions! So basically, the larger the Eo, the more likely that this reaction will occur…

And the smaller the Eo, the LESS likely that this reaction will occur…and the MORE likely that the OPPOSITE reaction will occur (i.e. oxidation)

o The Eo is listed is what the voltage would be if it was connected to the SHE 

6.4 - Using Standard Reduction Potentials to Calculate Eocell

OK…here are some rules for how we can manipulate the Eo for half-reactions:o If we reverse a half-reaction, then we reverse the sign…because now we are talking

about what the OXIDATION potential for the reaction is! Because if we reverse a reduction reaction, now there is an oxidation reaction happening!

o If we multiply a half-reaction for "n", NOTHING HAPPENS! Why? Because remember, we are measuring in volts, which is a ratio between the work done and amount of charge transferred! But if we multiply the half-reaction, both the numerator and denominator increase, and so the ratio stays the same!

o If we add 2 half-reactions together, just add the E0 values (we will do a lot of this…) 

6.5 - Oxidizing and Reducing Agents Alright…let's review:

o An oxidizing agent is something that causes something itself to be oxidized…thus it is a substance which is easily reduced (so on the Table, the best oxidizers are at the top…do you understand why?)

o An a reducing agent causes something else to be reduced…therefore a good reducing agent is easily oxidized (so the best reducers are the bottom of the Table…)

 6.6 - The Nernst Equation

We use the Nernst equation to calculate the cell potential under different concentrations of each substance:

  

A few special notes:o Q is the thermodynamic reaction quotiento We can apply this to the overall reaction, or to half-reactions individually