Happy Christmas

Wishing you all a very Happy Christmas

Work out all the chemistry involved here. its simply copper in silver nitrate solution. However, to explain it properly you need the following topics: Redox, complex formation, bonding and electronic structure (d orbitals splitting etc.)

Enjoy yourself and work really hard to make sure that all the topics covered are known and understood.

Reox Reactions

The folowing links are useful to look at to help prepare for and understand the next topic.


IB syllabus: Redox Reactions

Chemguide Redox Reactions

IA links

Useful tips on IA's

Link to IB site for IA criteria

Revision note sites

There are lots of good links here. The first one seems a detailed great summary of the whole course. Take a look and tell me what you think.


Revision notes

Revision notes (whole syllabus)

IB site acid base

Student room notes

Test and IA dates

Please remember that your IA is due on monday and you will have a test on acids and bases on Thursday(dec 1st)

Acid/ Base Indicators

The choice of indicator depends on the type of acid / base titration that is being carried out. A strong base with a strong acid has a large pH range at the equivalence point (long vertical section of the graph). Therefore, many indicators would change colour at the equivalence point and would work. However, if there is a weak acid/ strong base or weak base/ strong acid, the pH change at the equivalence point is less and the pH at the equivalence point is not 7. Therefore, care needs to be taken with the choice of indicator, so that it changes colour at the equivalence point and not before or after it. Weak acid with weak bases, don't have clear distinct equivalence points and no indicator would work well. The following notes from chem guide, explain clearly how indicators work. They are simply weak acids where the acid and its conjugate base have different colours, and so the colour will be different in acidic or basic conditions. The indicator should be chosen so that the PkIn value is at or close to the pH at the equivalence point.

Indicators

Here are the acid/base colours that we saw of some indicators;

Buffer Practical Data

Drops of Acid     Water (pH)         Buffer ( PH)

0                              6.67                      10.15

1                              3.00                      10.02

2.                             2.57                       9.85

3                               2.39                      9.74

4                               2.17                      9.55

5                               2.08                      9.30

  

Note: It would have been better to have chosen a buffer with pH 7 so that it started at the same pH as water. However, this does clearly show the effectiveness of a buffer at keeping the pH relatively constant, compared to the massive change as seen when a  drop of acid is added to water

Drops of Alkali   Water (pH)         Buffer (pH)

0                                6.67                     10.15

1                               11.23                  10.31

2                               11.74                   10.46

3                               11.98                   10.66

4                               12.15                   10.76

5                               12.35                   11.06

Homework:

1. Put the data into an appropriate table

2. Plot appropraite graphs

3. Discuss the results. What does it illustrate?

4. The buffer (pH 10.01) conntained sodium hydrogen carbonate and sodium carbonate. Explain how it worked and write equations to illustrate your answer.

Notes on Acid / Base titrations and Buffer solutions

Acid/Base notes:

The equivalence point is when the acid base reaction is complete. For example if 25cm³ of 0.1M  of sodium hydroxide is added to 25cm³ of  0.1M of ethanoic acid the reaction will be complete. Adding more base will cause the pH to shoot up to the pH of the base.

The pH at the equivalence point is not necessarily 7. For strong bases with strong acids it is. This is because the salts formed are neutral and have no reaction with water. For example sodium chloride (NaCl) is a neutral salt and is formed from the reaction of sodium hydroxide with hydrochloric acid.

For weak acids and a strong base the pH at the equivalence point will be above 7, in other words it will be an alkaline solution.  This is because the salt that is formed is a conjugate base of the weak acid. For example the salt of the reaction with ethanoic acid and sodium hydroxide is sodium ethanoate (CH₃COONa). In solution the ethanoate ions (CH₃COO^-) act as a base, and accept a proton from water, therefore, releasing OH^- ions.

CH₃COO^-   +  H₂O  ↔  CH₃COOH  +  OH^-

For weak bases and strong acids the pH at the equivalence point will be acidic. This is because the salt formed is a conjugate acid of the weak base. For example the salt formed when a solution of ammonia reacts with hydrochloric acid is ammonium chloride (NH₄Cl). In solution the NH₄⁺ ions will donate a proton and therefore, the pH of the solution will be below 7.

NH₄⁺  ↔ NH₃ + H⁺

Or more truly:   NH₄⁺   + H₂0  ↔ NH₃  +  H₃O⁺

The buffer region in titrations:

When a strong base is added to a weak acid or a strong acid is added to a weak base, around the half equivalence point (if the equivalence point is when 25cm³ has been added, the half point is when 12.5cm³ has been added). At this point and around this point there is the unreacted weak acid and the salt formed which is the conjugate base. Or the unreacted base and the salt formed which is the conjugate acid. Let’s take the weak acid strong base example first:

 At the half equivalence point, half of the weak acid remains, E.g. CH₃COOH, in addition there is the same amount of salt in the solution. This means that there is the acid to react with small amounts of base that is added.  Therefore, pH remains relatively constant around the half equivalence zone. There is also the conjugate base which can react with small amounts of acid, if this was added. Buffer solutions can resist changes in pH upon small amounts of acid or alkali added.

For a weak base and a strong acid which is added, at the half equivalence point, there are equal amounts of the weak base and its salt (a conjugate acid). This means that the pH stays relatively constant around this area as there is some weak base to react with the added acid. It would also remain relatively constant if a small amount of base was added as there is the conjugate acid to react with it.

Buffer solutions

You have seen that a buffer solution is formed when a weak acid is partially reacted with a strong  base or a weak base is partially reacted with a strong acid. A buffer solution could also be formed by adding the salt of a weak acid to the weak acid or adding the salt of a weak base to the weak base. Effectively you end up with the same thing.

Weak acid buffer: E.g.  ethanoic acid and sodium ethanoate

The acid can react with small amounts of alkali added: CH₃COOH + OH^- → CH₃COO^-  + H₂O

The conjugate base can react with small amounts of acid added: CH₃COO^-  + H⁺  → CH₃COOH

Or CH₃COO^- + H₃O⁺  →  CH₃COOH  + H₂O

Note that the weak acid is even less ionised due to the presence of the conjugate base and the salt is therefore, pretty much totally from the salt. So we could put the following equation:

CH₃COOH  ↔  CH₃COO^-  +  H⁺

As:  Acid ↔ Salt + H⁺

Therfore Ka can be expressed as: Ka = [salt] [H⁺]/[acid]

[H⁺]= Ka [acid]/[salt]

Or pH = pKa + log [salt]/[acid]

And with this  equation you can see clearly that at the half equivalence point the pH = pKa. You can therefore, use this to obtain the pka value of a weak acid.

Now follow the same logic for a weak base and its salt.

The relationship between Ka and Kb for an acid base conjugate pair

Pick and example or just use HA and A^-  to represent a weak acid and its conjugate base, and then write the appropriate equations and see if you can work out the relationship between them. This can be useful if you are given one value and you need to use the other.

Buffer Solutions

Buffer Solutions

Titration Curves

Titration curves

Acid /Base equations

Acid and Base Equations

[H^+]   to  pH     = log (1/[H⁺])

pH to [H⁺]        =  10^-pH

kw = 1 x10^-14 at 25⁰C

[OH^-] to pOH   = log (1/[OH^-])

pOH to [OH^-]   = 10^-pOH

At 25⁰C  pH + pOH = 14

Ka = [H⁺] [A_-]/[HA]

Kb = [BH⁺] [OH^-]/[B]

Ka to pKa     = log (1/Ka)

pKa to Ka     = 10^-pKa

Kb to pKb    = log (1/Kb)

pKb to Kb     = 10^-kb

Ka x Kb = 1 x 10^-14   at 25^OC

PKa + pKb = 14 at 25^OC

pH of an acidic buffer =   PKa + log [salt]/[acid]

pOH of a basic buffer = pKb + log [salt]/[base]

Conjugate acid base pairs

Some things to remember:

Conjugate acid base pairs differ by H⁺. When an acid donates a proton it becomes its conjugate base. When a base accepts a proton it becomes it conjugate acid.  Some Substances can behave as an acid or base and they are called are amphoteric. Even an acid can be made to behave as a base if it is put with a stronger acid.

Complex ions

From our discussions today we could see the link between the Lewis theory on acids and bases and how it could be used to classify a complex formation as acid base. You can see the importance of definitions and the need for clarification in terms of which definition you are using. The following is a good link to complex ions and is a useful recap on this topic.

Complex ions

Strong and Weak Acids

Remember strong acids and Bases completely ionise and therefore the concentration of H⁺ or OH^- ions respectively can be easily seen and the pH calculated. PH = log 1/[H⁺] and

 Kw = [H⁺] [OH^-] = 1.0 x 10^-14 at 25⁰C (Note this increases at higher temperatures and decreases at lower temperatures). You should be able to explain why. This does not mean that hot water is acidic!! It’s neutral as [H⁺] = [OH^-], but the pH will be less than 7. PH 7 only means neutral at 25⁰C

Weak acids and Bases only partially ionise and therefore, you cannot use the initial concentration alone in order to determine the concentration of H⁺ or OH^- ions, you need to have the Ka or Kb values.

Weak/Strong Acids and Bases

The Ionic Product of Water

Acid Base Definitions

Acid Base Theories

Quadratic Equation

Having said that IB questions on equilibria don't usually involve the quadratic equation, I managed to include one on your test! Surprisingly it was only worth 2 marks. here's a link to help remind you about the quadratic equation.

The quadratic equation

Vapour Pressure and homework

Remember your homework is to finish the next sheet of equilibrium questions and to make some notes on liquid vapour equilibria. You should read through the whole chapter in your text book.

The following site on vapour pressure is worth looking at:

Vapour Pressure

Equilibia

Some useful sites:

Equilibria Introduction

La Chatalier's principle

Kc Equilibrium constants

The effect of changing conditions

Equilibium

CHEMICAL EQUILIBRIUM

Many reactions are not reversible and just occur in one direction. For example it would be rather difficult to get the combustion of petrol to go in the opposite direction! However, some reactions and physical processes are reversible i.e. can occur in the forward or reverse direction. When the products are never fully formed and the reactants never used up a state of equilibrium is reached. At this point the rate of the forward and back reactions is the same. For example if bromine is put in a gas jar, initially the liquid evaporates at a faster rate than the gas condenses and the gas gets observably thicker. However after a while the gas density and volume of liquid remain constant. This does not mean that nothing is happening but there is no observable change as the system has reached equilibrium. This means that the forward and back directions are occurring at the same rate. The system is said to be in a state of dynamic equilibrium and for this to occur there must be:

1. A closed system

2 .No observable change

3. Both directions occurring at an equal rate

4. The equilibrium can be achieved starting from either direction

This is rather like walking up the escalators at the same rate as they are travelling down, you are both moving but you remain in the same position. Of course this can occur at different points on the escalators, the same is true in chemistry i.e. you can have different positions of equilibrium. For example if some bromine is removed from a gas jar by connecting another vessel,  then the bromine will evaporate at a faster rate and a new position of equilibrium will be established. Every equilibrium system has an equilibrium constant Kc which is simply the concentration of the products divided by the concentration of the reactants raised to the power of their stoichiometric coefficients. Kc has the advantage over the position of equilibrium as it is only affected by temperature i.e. nothing else will affect the value of Kc.  If more of one of the reactants is added then the position of equilibrium will move towards the right i.e. the products and equilibrium will again be established. When the new concentrations of reactants and products are fed into the  equilibrium constant expression the value of Kc will be the same. If more product is added then the position of equilibrium will move to the left and equilibrium will again be established with the Kc value remaining constant. Only temperature affects Kc values and so they are useful to compare equilibrium systems. A large Kc value means that that the equilibrium is towards the products and a small one indicates that the equilibrium is towards the reactants.

Useful Chemistry site

You don't need to know all the salt chemistry but this site has some useful information:

General chemistry information site

Link to IB1 Blog

IB1 Blog

IB1 blog

Periodic Table

Link to periodic table

periodic table

Homework and Test Date

The practical assessment on rates of reaction should be complete by wed 5th Oct

Revise for test on rates of reaction on mon 3rd oct

Homework

Look at the kinetics power points, check that you understand all the aspects of the kinetics topic

Make sure that your investigation is properly planned.

Arrhenius calculations

The Arrhenius equation can be used to calculate the missing value of: A, K or Ea. Remember to convert Ea values to J/mol as they are usually in KJ/mol. Therefore you need to be consistent with the gas constant which is: 8.314 JK^-1 mol^-1. T must be in Kelvin. The other tip is that when you are calculating an Ea value, use the natural log form: lnK = lnA - Ea/RT.

Arrhenius Equation Calculations

The Arrhenius Expression

The rate constant is only constant for a fixed temperature. It increases exponentially as the temperature increases. Arrhenius came up with the following relationship:

k= A exp ^-Ea/RT       

This is often shown in logarithmic form so that it is the form of y=mx + c form and a graph of lnk against 1/T can be plotted which has a slope of -Ea/R.

Look at page 123 in The Chemistry Course Companion (Geoff Neuss) and the following Chemguide link is useful.

The Arrhenius Equation

Possible Graphs For Rate Experiments

Often concentration of a reactant against time is plotted or rate against the concentration of a reactant is plotted. You need to know and understand how each of the graphs would look like for: 0, 1st and 2nd order with respect to reactants concentration. Don't get confused between the two graphs; you need to look carefully at the axis.

See if you can plot and explain each of the possibilities?

Rates Investigation and Homework

To find the value of k, you just to insert the values from a particular experiment making sure that the concentrations are raised to the appropriate powers according to the order with respect to each reactant. Then just solve for k. To work out the units you simply do the same but with all the units inserted into the rate expression.

Rate investigation.

Please have the details of the investigation that you want to carry out ready. You must know how you are going to follow the rate, what your independent and dependent variables are and how you are going to control the other varaibles. I need to make sure that all the apparatus is ready, so please email your list of apparatus and chemicals.

The Rate Expression and the link with Mechanisms

The Rate Expression and Orders of Reaction

Simplistically one would imagine that doubling the concentration of a reactant would double the rate of a reaction. This seems logical; there will be double the number of collisions. However, although this is often the case it is not always true. Indeed, sometimes doubling the concentration of a reactant could have no affect on the rate and sometimes it will increase the rate by more than double, for example the rate could be quadrupled.

How can this be so? The answer lies in the fact that reactions usually take place by a series of steps and one of the steps (the slowest one (rate determining step RDS)) will control the rate. Only reactants involved in this step will affect the rate.  For reactants that are not involved in the RDS their concentration will have no affect on the rate. The order with respect to this reactant will be zero and this reactant will not appear in the rate expression. If a reactant is involved once in the RDS then doubling its concentration will cause the rate to double. It is said to be first order with respect to this reactant. The rate expression would be rate ≈ [conc] or Rate = k [conc]. It could be the case that a reactant is involved twice in the RDS, in which case, doubling its concentration will cause the rate to go quadruple i.e. go up by 4. The rate expression would be: Rate ≈[conc]²  or  Rate = k[conc]². K = the rate constant and the powers are the orders for each of the reactants. The overall order is simply the addition of all of the powers in the rate expression.

For example for the rate expression:  Rate = k  [BrO₃^-] [Br^-] [H⁺]²

What is the order with respect to each reactant?

What is the overall order?

What units will k have in this rate expression.  Note that the units of k will vary depending on the expression.

 

Remember that the rate expression can only be determined experimentally by following the rate and changing one reactant at a time. The stoichiometry of a reaction tells you nothing about the rate expression or possible mechanism.

Deducing  a Mechanism from the Rate Expression

This can be quite complex. However, simplistically if a reactant has zero order then it is not in the RDS but must be in a faster step. If a reactant is first order then it is directly or indirectly involved once in the RDS. If a reactant is second order then it’s either directly or indirectly involved twice in the RDS. Indirectly means that it could be involved with producing an intermediate, which is then involved in the RDS.

It can be fun proposing mechanisms that fit with the rate expression. There may well be several possibilities.

 The following links are useful:

Orders and Mechanisms

 Virtual Experiment

 Orders of reaction

Homework

Write about the uses of catalysts. Please cover, The Haber process, The Contact process, catalytic convertors in car exhausts, plus others of your choice. You should also be ready to discuss orders of reaction, the rate expression and how this links to mechanisms.

Full Syllabus

Here's the link to the pdf file of the detailed syllabus. look at the S.Land H.L sections and check that you understand each part of the topic as we go through the syllabus. Ask about any parts that you are not clear about.

Full syllabus

Types of Catalyst

A useful link with some diagrams illustrating heterogeneous catalysis.

Types of Catalyst

Homework

Please look at the link to rates of reaction and make sure you understand the following: The collision theory, factors that affect the rate of reaction [temperature, concentration, surface area, catalysts and pressure (for gaseous systems)]. In addition you should be able to sketch enthalpy level diagrams showing the activation energy for exothermic and endothermic reactions. You should also be able to sketch the Maxwell-Boltzmann distribution curve for two different temperatures and be able it to use it to explain how the use of catalyst and increasing the temperature causes an increase in the rate of reaction.


Chem guide rates link Rates of reaction

You also need to think of an experiment that you can investigate a factor that you think will affect the rate. you need to think about all the variables and how you will measure the rate. Please look at the link to practical work to remind yourself of the criteria. This could possibly be used for all three skill areas.

Explaining the colour in complex ions

This chemguide site is worth looking at:

Colour in complexes

Hybrid Orbital Shapes

This is a great site which visually shows you the formation of the hybrid orbital shapes.

Hybrid Orbital Shapes

Homework

Please go through the rest of p2 and have ready any questions. Also please look at the kinetics chapter, you could also look at some good websites e.g chem guide.

Rates of reaction

Notes from going through the exam

Remember ionic oxides are basic (if they dissolve) and covalent oxides are acidic. One of the gases which leads to acid rain is sulphur dioxide which is a covalent oxide.

Ionic oxides contain the oxide ion which reacts with water to form the hydroxide ion.


The following chem guide link is good: Period Three Oxides

Why is SO3 a solid at room temp?

One of the questions stated that sulphur trioxide was a solid at room temp. This may seem strange as one would presume that there are just weak forces of attraction ( Van Der Waal's)  between the molecules.
The following Chem Guide link is useful:Chem Guide

Power points (knockhardy) www.knockhardy.org.uk/ppoints.htm

This site has some very useful powerpoints which will help with your revision and can be a useful tool.

Powerpoints

Practical Assessment

Practical (Internal) Assessment 

Criterion	marks available	marks
Design	two grades out of 6 are to be presented	12 marks
Data collection and processing	two grades out of 6 are to be presented	12 marks
Conclusions and Evaluation	two grades out of 6 are to be presented	12 marks
Manipulative skills	1 grade out of 6 will be presented	6 marks
Personal skills	1 grade out of 6 will be presented	6 marks
	Total =	48 marks

Every year a sample of the internal assment work is sent by the teacher in charge for external moderation. In this process a third part (the moderator) tries to decide whether the teacher has applied the criteria correctly in his/her assessment. More about moderation.

Final grades

The final grade awarded for an IB subject is from 1-7. The schools receive a breakdown of the grade achieved in each part of the exam.

The requirements to achieve the IB diploma are fairly complex and may be found here: IB diploma award requirements

Homework

Homework is to look through the rest of the MC questions and have questions ready about anything that you want me to go through.

Double and triple bonds

Double bond C=C

   

                                                                   Sp² Hybrid orbitals overlap

                                                                    Forming sigma bond

                                 Unhybridised p orbitals overlap

                                   To form a pi bond. Note that the

                                                                  pi Bond has a lobe above and

                                      below the Sigma bond. However,

                               it still Constitutes One bond.

                                        The C=C double bond is Therefore,

                              one sigma and one pi bond.

        

                 Triple bond C=_C

                                                                                  pz and py unhybridised

                                                                               orbitals overlap to form

                                                                                two pi bonds. 

 sp hybrid orbitals overlap to form                      The triple bond is therefore

 a sigma bond.                                                       Two pi bonds and one sigma bond.

Shapes of Hybrid Orbitals

Shapes of Hybrid Orbitals

Note: The hybridisation not only explains observed shapes but it also explains the expansion of the octet which sometimes occurs. For example SF₆  involves d²sp³  hybridisation, therefore, explaining the six bonding pairs.

Interactive Website

Please look at this site all the time. There is lots of great stuff and you can follow what we are doing and click on the related links. If you exchanged your facebook time for IB interactive chemistry time I'd be very happy.


IB Interactive Syllabus

Welcome back

I hope that you have all had fantastic holiday's and are eager to get back into the chemistry class.