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Reaction Kinetics – Concepts

1. Types of rate:
• Initial rate is change in concentration of reactants or product at time t = 0.
• Instantaneous rate is rate of reaction at any given time/instant.
• Average rate is total concentration of reactant used or total concentration of product formed over total time.

2. The minimum energy which colliding molecules must possess for successful collision/reaction is called the activation energy, Ea.

3. Rate law or rate equation is the mathematical relationship between the rate of a reaction and the concentration of the reactants in a reaction. E.g. A + B —> Products
The rate law is
Rate = k[A]^m[B]^n
where k is the rate constant
m is the order of reaction with respect to reactant A
n is the order of reaction with respect to reactant B
(m + n) is the overall order of the reaction
4. The order of reaction with respect to a particular reactant is the power to which the concentration of that reactant is raised in an experimentally determined rate equation / rate law.

5. The rate constant, k, is the proportionality constant in the experimentally determined rate law.

6. The half-life (t1/2 ) of a reaction is the time taken for the concentration of a reactant to fall to half its initial value. It is constant only for a first order reaction as it is independent of reactant concentrations.

7. A catalyst is a substance that increases the rate of a reaction by providing an alternative reaction pathway that has lower activation energy.

8. Biological catalysts such as enzymes are very selective in the reactions that they catalyze, and some are absolutely specific, operating for only one substance in only one reaction. For reactions that normally produce a pair of optical isomers (racemic mixture) when carried out in the lab, enzymes are able to selectively produce one optical isomer in the body.

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Organic Chemistry – Concepts

1. Empirical formula is the simplest formula that shows the ratio of each kind of atom in a molecule. e.g. C2H5 is the empirical formula for C4H10

2. Molecular formula shows the actual number of each kind of atoms in a molecule. e.g. C4H10

3. Structural formula shows how the atoms are connected to each other in a molecule. e.g. CH3CH2CH2CH3

4. Displayed/full formula shows all the bonds and relative placing of all the atoms in a molecule.

5. Homologous series are compounds have the same general formula and functional group and each homologue differs from its neighbor by a fixed group of atoms (e.g.–CH2). As we go down a homologous series, the chemical properties remain unchanged but there is a gradual change in physical properties. Examples of homologous series are alkanes, alkenes, alcohols…..

6. Structural isomerism refers to compounds with the same molecular formula but different structural formula. E.g. CH3COOCH3 and C2H5COOH

7. Stereoisomerism refers to compounds that have the same molecular formula but with different spatial arrangements.

• Geometric isomers have same carbon skeleton with double bonds restricting free rotation. For geometric isomerism to exist, there must be two different groups of atoms bonded to each side of the C=C bond.

• Optical isomers are non-superimposable mirror images of each other (enantiomers). Isomers have at least one chiral C atom, i.e. there are four different groups attached and have no plane of symmetry. An equal proportion of enantiomers forms a racemic mixture which is optically inactive.

8. The primary structure of a protein shows the exact order (or unique sequence) of the -amino acids held by peptide/amide linkages along the polypeptide chain. The primary structure determines what the protein is, how it folds and its function.

9. The secondary structure refers to the detailed configurations of the polypeptide chain. In a protein molecule, the long chain of amino acid units may be coiled into an -helix or folded into a -pleated sheet. Both structures are stabilized by hydrogen bonds between the N-H group of one amino acid residue and the C=O group of another along the main chain.

10. The tertiary structure of the protein refers to the overall 3-dimensional shape of the entire protein involving folding or coiling of the chains. It shows how protein molecules are arranged in relation to each other.

There are four types of R group interactions which hold the tertiary structure in its shape.

 van der Waals’ forces (induced dipole-induced dipole bonding) exist when non-polar R groups (e.g. alkyl or aryl groups) come close together. They are usually found on the inside of globular proteins where, because they are hydrophobic, they do not interfere with solubility.

 hydrogen bonding between polar groups (e.g.. –CH2OH, -COOH and –NH2 groups).

 ionic bonding eg. –COO-, -NH3+, and >NH2+.

 disulfide linkages eg. –SH or –CH2-S-S-CH2- groups.
Quaternary structure of proteins refers to the spatial arrangement of its protein subunits. It shows how the individually folded protein subunits are packed together to yield large structures. This only applies to proteins that contain two or more polypeptide chains. The individual polypeptide chains are called the subunits. E.g. haemoglobin contains 4 subunits, each containing a haem group.

11. It is stabilized by the same R-group interactions that stabilise the tertiary structure.

12. Denaturation is the loss of biological activity of a native protein. When proteins are denatured, the secondary and tertiary structures are disrupted i.e. the R group interactions are broken or destroyed. Note that the primary structure remains unaffected.
Factors that can lead to denaturation include extremes in pH, temperature, ionic salts, heavy metal compounds, presence of organic solvents etc.

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TO Master to PERFECTION before A’levels (Part 1)

Standard Definitions (Don’t Memorize. But appreciate and understand why key terms are important)

– Relative atomic, isotopic, molecular and formula mass, based on the 12C scale (just give mathematical expression)

– Mole in terms of the Avogadro constant
– VSEPR (2 assumptions)
– Basic assumptions of the kinetic theory as applied to an ideal gas
– Standard enthalpies (11 of them)
– Hess’ Law
– Entropy
– Standard electrode potential and standard cell potential
– Dynamic Equilibrium, LCP
– Strong and weak acids and bases
– Kc, KP, Ka, Kb, Kw, KSP,pH etc. (m. expression)
– Rate of reaction; rate equation; order of reaction; rate constant; (m. expression)
– Half life of a reaction
– Rate-determining step
– Activation energy
– Catalysts
– Transition metal, ligands, complex, coordination number
– Proteins 1o,2o,3o structure, Denaturation

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FAQ – Group II

For Gp II metals (e.g. magnesium, calcium, barium)

1. Reactivity (with water or other substances) increases down the group

Explanation:
• Increase in shielding effect outweighs increase in nuclear charge down the group due to increase
in number of electron shells.
• Hence ionization energy decreases
• Metals are able to form ions more easily.

For Gp II ionic compounds (e.g. carbonates, nitrates and hydroxides),

2. Melting point decreases down the group

Explanation:
• Increase in cation size
• leads to decrease in magnitude of lattice energy.
• Hence, ionic bonds are weaker and more easily broken.

3. Thermal decomposition temperature increases down the group

Explanation:
• Increase in cation size,
• leads to decrease in charge density of cation.
• Hence, electron cloud of anion is less distorted and hence less easily decomposed.

Note: Quote ionic radius values from Data Booklet to explain this. Calculate charge density if necessary.

4. Solubility of sulfates decreases down the group (not in syllabus but
may still be tested)

Explanation:
• Increase in cation size
• leads to significant decrease in magnitude of hydration energy (as compared to slight decrease in magnitude of lattice energy)
• Hence, enthalpy of solution (= LE – ΔHhyd) is less exothermic.

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FAQ – Group VII

For Group VII halogens (e.g. Cl2, Br2, I2)

1. Volatility decreases down the group (i.e. boiling point increases)

Explanation:
• Increase in electron cloud size hence more easily polarised
• leads to stronger dispersion forces between molecules

2. Oxidising power decreases down the group

X2 + 2e ⇔ 2X−

Explanation:
• Decrease in effective nuclear charge, hence electron affinity decreases
• Hence less easily accepts electrons (i.e. less easily reduced)
• Eθ value decreases (Quote from Data Booklet)

Two important proofs of this are:
• Any halide (e.g. I−) can be displaced by the halogen (e.g. Br2) above it.
• Reaction with sodium thiosulfate (explain change in oxidation state of S)

3. Reactivity with H2 decreases down the group
X2 + H2 → 2HX

Explanation:
• Atomic size of X increases
• Hence H – X bond becomes longer and weaker
• Product formed is less and less stable, hence reactivity decreases.

Note: Quote bond energy values to explain this, NOT Eθ values

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FAQ – Gases

Gases

Difference between assumptions and conditions

• 2 assumptions of ideal gas
o Negligible intermolecular forces
o Negligible particle volume compared to volume of container

• 2 conditions at which a gas acts most ideally
o High temperature
o Low pressure

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Colour Summary

Flame Test Colors
Li Deep red
Na Yellow
K Violet
Mg Bright white
Ca Orange-red
Sr Red
Ba Green
Cu Blue-green
P Pale blue-green
S Blue
Fe Gold
Pb Blue-white
Zn Blue-green

Aqueous Ion Colors
Cu+ Green
Cu2+ Blue
[CuCl4]2- Yellow
Cu(NH3)4 2+ Dark Blue; produced when ammonia is added to Cu2+ solutions

Fe2+ yellow-green (depending on the anion)
Fe3+ orange-red (depending on the anion)
FeSCN]2+ Red-brown, Wine-red to dark orange

Co2+ Pink
CoCl42- Blue (Co2+ with HCl will form a CoCl4 2- complex that is blue)

Cr3+ Violet (Cr(NO3)3 to Green (CrCl3)
CrO4 2- Yellow
Cr2O7 2- Orange

Ni2+ Green

Mn2+ Pink
MnO4 – Purple (Mn w/ +7 oxidation state is purple)
MnO4 2- Green

Pb3+ blue-green (Pb2+ and Pb4+ are colorless)

V2+ violet
V3+ blue-green

Ti(H2O)6 3+ Purple

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TO Master to PERFECTION before A’levels – Part 1

Standard Definitions (Don’t Memorize. But appreciate and understand why key terms are important)

– Relative atomic, isotopic, molecular and formula mass, based on the 12C scale (just give mathematical expression)
– Mole in terms of the Avogadro constant
– VSEPR (2 assumptions)
– Basic assumptions of the kinetic theory as applied to an ideal gas
– Standard enthalpies (11 of them)
– Hess’ Law
– Entropy
– Standard electrode potential and standard cell potential
– Dynamic Equilibrium, LCP
– Strong and weak acids and bases
– Kc, KP, Ka, Kb, Kw, KSP,pH etc. (m. expression)
– Rate of reaction; rate equation; order of reaction; rate constant; (m. expression)
– Half life of a reaction
– Rate-determining step
– Activation energy
– Catalysts
– Transition metal, ligands, complex, coordination number
– Proteins 1o,2o,3o structure, Denaturation

Standard Explanations (must be concise (save time), accurate and complete) – You must know this so well you have are absolutely confident of reproducing them under stressful exam conditions.

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TO Master to PERFECTION before A’levels – Part 2

1. Atomic Structure • Ionisation Energy (Trend across the period + 2 anomalies, Down the gp, successive IE,*TM)

Remark
• Predicting position from successive IE
Refer to AMS, Gases and Atomic Structure Revision Notes

2. Bonding
• Boiling point/melting point
• Volatility
• Electrical conductivity
• Solubility

3. Energetics (entropy)
• Discuss the effects on the entropy of a chemical system by the
following:
(i) change in temperature
(ii) change in phase
(iii) change in the number of particles (especially for gaseous
systems)
(iv) mixing of particles

Remark
• Predict the effect of temperature change on the spontaneity of a
reaction, given standard enthalpy and entropy changes disorderliness”/”ways to arrange particles” are key words.
Refer to entropy lect notes.

Using Gibbs equation ΔG=ΔHTΔS.
Make sure you are comfortable with putting your thoughts into words.

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Glossary of Terms

1. State
A concise answer with little or no supporting argument, e.g. a numerical answer that can be obtained ‘by inspection’ is required.

2. List
A number of points, generally each of one word, with no elaboration is required. Where a given number of points is specified, this should not be exceeded

3. Explain
Reasoning or some reference to theory is required (depending on the context)

4. Describe
State in words (using diagrams where appropriate) the main points of the topic. It is often used with reference either to particular phenomena (where answers should include reference to observations associated) or to particular experiments.

5. Discuss
A critical account of the points involved in the topic should be provided.

6. Outline
Be concise i.e. restrict the answer to giving the essentials only.

7. Predict
Make a logical connection between other pieces of information. Such information may be wholly given in the
question or may depend on answers extracted in an early part of the question.

8. Deduce
Used in a similar way as predict except that some supporting statement is required, e.g. reference to a law/principle, or the necessary reasoning is to be included in the answer

9. Comment
It is an open-ended instruction, inviting one to recall or infer points of interest relevant to the context of the question, taking account of the number of marks available.

10. Suggest
It is used in two contexts, i.e either to imply that there is no unique answer (e.g. in chemistry, two or more substances may satisfy the given conditions describing an ‘unknown’), or to imply that candidates are
expected to apply their general knowledge to a ‘novel’ situation, one that may be formally ‘not in syllabus’.

11. Find
Can be interpreted as calculate, measure, determine etc

12. Calculate
A numerical answer is required. In general working should be shown.
Note:The misuse of units and/or significant figures is liable to penalty.

13. Determine
It implies that the quantity cannot be measured directly but is obtained by calculation, substituting measured and known values of other quantities into a standard formula.

14. Sketch
When applied to graph work, the shapes and/or position o the curve need only be qualitatively correct but some quantitative aspects (e.g. passing through the origin, having an intercept at a particular value) may be looked for.

In diagrams, a simple and freehand drawing is acceptable but care should be taken over proportions and the clear exposition of important details.

15. Construct
Often used in relation to chemical equations where one is expected to write a balanced chemical equation,not by factual recall but by analogy or by using information in the question.

16. Compare
Both similarities and differences between things or concepts should be provided.

17. Classify
Group things based on common characteristics.

18. Recognise
Often used to identify facts, characteristics or concepts that are critical (relevant/appropriate) o the understanding of the situation, event, process or phenomenon.

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Power Revision

A Level Chemistry – 2 hrs Each Lesson

1. Atoms Molecules and Stoichiometry – 2 lessons
2. Chemical Bonding – 2 lessons
3. Chemical Energetics- 2 lessons
4. Reaction Kinetics – 2 lessons
5. Chemical Equilibrium – 2 lessons
6. Ionic Equilibrium – 2 lessons
7. Introduction Organic/Alkanes/Alkenes – 2 lesssons
8. Arenes – 1 lessons
9. Halogen Derivatives – 2 lesons
10. Hydroxy Compounds – 2 lessons
11. Carbonyl Compounds – 2 lessons
12. Carboxylic Acids and Derivatives – 2 lessons

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Atoms, Molecules and Stoichiometry

1 Define the following terms:

(a) Isotopes are atoms of the same element with the same number of
protons but different number of neutrons.

(b) Relative isotopic mass is the number of times the isotope is heavier than 1/12 the mass of an atom of carbon-12.

(c) Relative atomic mass of an element is the average mass of its atoms in the isotopic mixture, compared to 1/12 the mass of an atom of carbon-
12.Symbol: Ar

(d) Relative molecular mass of a molecule is the average mass of its
molecule compared to 1/12 the mass of an atom of carbon-12.
Symbol: Mr

(e) Avogadro’s Law states that equal volumes of all gases under the same
conditions of temperature and pressure contain the same number of
molecules/atoms (valid for ideal gases or gases under ideal-like
conditions).

(f) Molecular formula gives the actual number of the number of atoms of
each element present in the compound.

2 Concept : Limiting Reagent and Dilution

The process of obtaining iodine from oil field brines involves the following three step process. In the first reaction, 250 g of sodium iodide (NaI) is reacted with 340 g of silver nitrate (AgNO3).

NaI + AgNO3 —> AgI + NaNO3
2AgI + Fe —> FeI2 + 2Ag
2FeI2 + 3Cl2 —> 2FeCl3 + 2I2

(a) Determine the limiting reagent in the first reaction.
No of mol NaI available = +25023 127 = 1.6667
No of mol AgNO3 available = ++340108 14 48 = 2.00
NaI ≡ AgNO3 ⇒ NaI is the limiting reagent

(b) Calculate the mass of iodine crystals that can be obtained from the whole process.
NaI ≡ AgI ≡ ½ FeI2 ≡ ½ I2
No mol of I2 obtained = ½ ( 1.6667 ) = 0.83335
Mass of I2 obtained = 0.83335 x 2(127) = 212 g

(c) The iodine crystals obtained was then dissolved in 50.0 m3 of organic solvent, trichloromethane. Calculate the concentration of iodine solution in g dm-3.

Concentration of I2 = 50000212 = 0.00423 g dm-3

(d) In a further experiment, a certain volume of organic solvent, trichloromethane was added to lower the concentration of the iodine solution calculated in (c) to 1.40 X 10-5 mol dm-3. Calculate the volume of trichloromethane added to achieve this effect.

CoVo = CdVd ⇒ (0.00423254)(50000) = (1.40 x 10-5) (Vd)
Vd = 59480 dm3
∴ Volume of trichloromethane = 59480 – 50000 ≈ 9.48 x 103 dm3

(e) State the assumption made in determining your answers in parts (c) and (d).

Iodine is soluble in organic solvent / trichloromethane

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Atoms, Molecules and Stoichiometry Part 2

3. (a) Define the term mole.

The mole is defined as the amount of substance that contains the same
number of particles as there are atoms in 12 g of pure carbon-12.

(b) Why is the phrase “the mass of one mole of oxygen” ambiguous?

It is because the statement can either refer to one mole of oxygen atoms or one mole of oxygen molecules (O2). The statement can also refer to 16O or 18O isotopes.

(c) A meteorological balloon of 2 m diameter has a volume of 4.19 m3. It floats since it is given an upthrust equal to the mass of air it displaces.

Calculate:
(i) the mass of hydrogen in the balloon,

No of moles of hydrogen gas = 4190 / 23 = 182.2
Mass of hydrogen = 18.2 x 2 = 364 g

(ii) the mass of air it displaces,

Volume of air displaced = 4190 dm^3
There should be 182.2 mol of air.
Mass of air displaced = 182.2 x 29 = 5280 g

(iii) the load the balloon can carry for it just to lift off from the ground.

Upthrust = mass of air displaced = mass of hydrogen + load
Load = 5283.8 – 364. = 4919.4 = 4.92 kg

4. The reaction of silicon tetrachloride with moist ethoxyethane produces two oxochlorides with the formulae Si2OCl6 and Si3O2Cl8. When 0.10 g of one of these oxochlorides completely reacted with water, all of its chlorine was converted into chloride ions, and produced 0.303 g of silver chloride precipitate when an excess of aqueous silver nitrate was added.
Deduce the identity of the oxochloride.

AgCl Ξ Cl
No of mol of Cl in 0.303 g of AgCl
= 0.303/(108+35.5)
= 2.11 x 10-3

Relative molecular mass of Si2OCl6
= 2×28.1+16+6×35.5
= 285.2

Relative molecular mass of Si3O2Cl8
= 3×28.1+2×16+8X35.5
= 400.3

No of mole of Cl in 0.1g of Si2OCl6
= (0.1/285.2) x 6
= 2.10 x 10-3

No of mole of Cl in 0.1g of Si3O2Cl8
= (0.1/400.3) x 6
= 2.00 x 10-3

The oxochloride is Si2OCl6.

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Atoms, Molecules and Stoichiometry Part 3

Concept: Determining Formula of compound

A metal hydroxide, M(OH)n, is one of the products formed in an electrochemical cell used to power golf trolleys. 50.0 cm3 of solution containing 0.028 mol of M(OH)n in 1 dm3 requires 21.00 cm3 of sulfuric acid for complete neutralisation. The sulfuric acid contains 0.2 g of hydrogen ions in 1 dm3.

(a) Calculate the number of moles of sulfuric acid that will react with 1 mole of M(OH)n.

H2SO4 ≡ 2H+

No. of moles of H+ in 1 dm3 = 0.2/1.0 = 0.200

[H2SO4] = 0.2002= 0.100 mol dm-3

No. of moles of H2SO4 in 21.10 cm3 = (21.00×0.100)/1000 = 2.10 x 10-3

No. of moles of M(OH)n in 50.0 cm3 = (50.00×0.028)/1000 = 1.40 x 10-3

No. of moles of H2SO4/No. of moles of M(OH)n
= 2.10 x 10-3/1.40 x 10-3
= 1.5

1 mol of M(OH)n reacts with 1.50 mol of H2SO4

(b) Hence, determine the value of n.

2M(OH)n (aq) + nH2SO4 (aq) → M2(SO4)n (aq) + 2nH2O (l)
Mole ratio: 2M(OH)n ≡ nH2SO4
M(OH)n ≡ (n/2)H2SO4
n/2 = 3/2
n = 3

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Atoms, Molecules and Stoichiometry

Practice Question (15 min)

The alums are a series of double salts formed when a cation, with a charge of +1, having a large radius and a cation, with a charge of +3, having a small radius combined with sulfate ions.

Ammonium chromium alum has the formula of (NH4)aCr(SO4)b.xH2O. 1.28 g of the salt was dissolved in 100 cm3 of 0.0500 mol dm-3 ammonium chloride solution and the solution was divided into two equal portions.

To one portion was added an excess of sodium hydroxide and the mixture was boiled. The ammonia that was evolved neutralized 25.60 cm3 of 0.150 mol dm-3 nitric acid.

(a) Write a balanced ionic equation for the reaction between sodium hydroxide and the ammonium ions in solution. [1]

(b) Calculate the number of moles of NH3 from the alum. [2]

To the other portion, an excess of zinc was added which reduced the Cr3+(aq) to Cr2+(aq). The mixture was then filtered and the filtrate was titrated with acidified potassium dichromate(VI). It was found that 22.35 cm3 of 0.0100 mol dm-3 acidified potassium dichromate(VI) was required for the titration.

(c) Write a balanced ionic equation for the reaction between Cr2+ ions and acidified potassium dichromate(VI). [1]

(d) Calculate the number of moles of chromium ions from the alum. [1]

(e) Use your answer to (b) and (d), calculate the values of a and b. [1]

(f) Hence, find the relative formula mass of ammonium chromium alum and the value of x. [2]

(g) Calculate the percentage of chromium in the sample. [2]

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