any one got any gud ones????
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Chemistry and R.E revision websites watch
- Thread Starter
- 08-12-2003 17:15
- Thread Starter
(Original post by Saf!)
- 08-12-2003 17:18
any one got any gud ones????
i got exams tommorow and i'm stuck on revision!!!
(Original post by Saf!)
- 08-12-2003 17:23
ok, any websites........... even crappy ones
i got exams tommorow and i'm stuck on revision!!!
s-cool.co.uk (might be .com)
the bbc site, forgot the address.
Try doing a search on google.
- Thread Starter
(Original post by gemgems89)
- 08-12-2003 17:27
I'll give you a few, they aren't specifically R.E and chem, you might have already tried them too..here goes...
s-cool.co.uk (might be .com)
the bbc site, forgot the address.
Try doing a search on google.
thanx 4 remindin me!
- Thread Starter
- 08-12-2003 17:34
has anyone else got any gud uns!
i'm kinda despirate!
- Thread Starter
- 08-12-2003 17:53
(Original post by Saf!)
- 08-12-2003 17:58
thanx 4 remindin me!
educationindex.com/chem/ - found that as well, don't know where it's from though.
- 09-12-2003 10:28
not entirely sure if this is what you're looking for, but...
Acids and bases
Properties of acids
Typical strong acids are hydrochloric (HCl), sulphuric (H2SO4) and nitric (HNO3)
First defined by their action on various indicators
e.g litmus is red in acids and blue in alkali
Acids have typical reactions with:
Bases eg. CuO + H2SO4 CuSO4 + H2O
Metals eg. Mg + 2 HCl MgCl2 + H2
Carbonates eg. CuCO3 + H2SO4 CuCO3 + H2O + CO2
Acids are ionised by the action of water to release H+ ions in to solution. It is the H+ ion that is responsible for acid properties.
e.g. HCl (g) + water H+ (aq) + Cl—(aq)
Properties of bases
These neutralise acids, to form salts and water. Bases that are soluble in water are classified as alkalis eg. sodium carbonate (Na2CO3)
Typical strong alkalis are group I hydroxides (eg. NaOH) and group II (e.g. Ba(OH)2)
Alkalis turn litmus blue
Alkali are ionised by the action of water to release hydroxide OH—(aq) ions in to solution e.g. the hydrogen carbonate ion
NaHCO3 (s) + H2O Na+ (aq) + H2CO3(aq) + OH—(aq)
Strong and weak acids
Weak acids and alkalis are not fully ionised by the action of water – an equilibrium between ions and molecules results. They are classified as weak.
Examples of weak acids are carbonic acid and ethanoic acid
i.e. H2CO3 (aq) == HCO3— (aq) + H+ (aq)
Examples of weak alkalis are ammonia and ethylamine
i.e. NH3 (aq) + H2O == NH4+(aq) + OH—(aq)
This partial ionisation means that a solution of a weak acid is less acidic than a solution of a strong acid :
i.e. comparing a weak acid solution to that of a strong acid of identical concentration:
the weak cid will be less conductive,
the weak acid will react more slowly with metals and carbonates etc
the weak acid will have a higher pH
The pH scale
This is a measure of the acidity of a solution.
pH is defined as the negative log to the base 10 of the concentration of the hydrogen ion.
pH = - log10 H+ (aq)
Therefore a ten fold decrease in H+ in concentration produces an increase of 1 pH unit.
Two methods of measuring pH are :
1. using the Universal Indicator paper colour scale
2. using an electronic pH meter.
A number is obtained in both methods.
pH numbers between below 7 0 indicate an acidic solution
pH number 7 indicates a neutral solution
pH numbers between above 7 14 indicate an alkaline solution.
These solutions have the property of resisting changes in pH on addition of small amounts of strong acid or alkali. Blood is an example, maintaining a pH of close to 7.4.
Buffers are a mixture of solutions of either:
A weak acid and its salt ( eg. ethanoic acid and sodium ethanoate)
A weak alkali and its salt (eg. ammonia and ammonium chloride)
These are simple to make and test in the lab.
Using indicators such as methyl orange (strong acid/strong base) or phenolphthalein (weak acid/strong base) an accurate end point can be determined.
Using a pH meter the pH changes during a titration can be followed.
- 09-12-2003 10:32
try the bitesize website.
this one helped me alot too.
- 09-12-2003 10:33
The nuclear atom
Chemistry uses the idea of 3 sub-atomic particles, forming all different atom types.
Protons = positively charged, of mass 1 unit, found in the nucleus
Neutrons = electrically neutral, mass of 1 unit, located in the nucleus.
Electrons = negatively charged, 1/1840 mass units, located in space surrounding nucleus
The numbers for each different atom type (element)
Atomic number (Z) = the number of protons in the nucleus
Mass number (A) = the number of protons + neutrons in the nucleus
7 mass number
3 atomic number
In neutral atoms, the number of protons must be equal to the number of electrons
Hence for Lithium above P = 3 N = 4 E = 3
Ions are formed by the loss or gain of electrons by an atom i.e. A and Z do not change.
Example Na Na+ O O 2-
P=11 P= 11 P= 6 P= 6
N=12 N= 12 N= 6 N= 6
E=11 E = 10 E= 6 E= 8
Atoms of the same element (same number protons) having different numbers of neutrons
35 37 1 2 3
Cl Cl H H H
17 17 1 1 1
p=17 p=17 p=1 p=1 p=1
n=18 n=20 n=0 n=1 n=2
e=17 e=17 e=1 e=1 e=1
These isotopes occur in different abundancies (% natural occurrence)
Abundancies Cl atoms 25% 37Cl 75% 35Cl
From which the Relative Atomic Mass is calculated
- a weighted average atomic mass compared to 1/12 of the mass of the 12C atom
Relative Atomic mass Cl = (25 x 37) + (75 x 35) = 35.5
- average mass of the Cl atom is 35.5 times more massive than 1/12 of a carbon-12 atom
other abundancies are 20Ne 90% 22Ne 10%
The Mass Spectrometer (Higher only)
The masses and abundancies for each isotope are found using a mass spectrometer.
The electrons in an atom exist at an energy level (shell) rather than in a particular place.
Electrons need particular amounts of energy (quanta) to occupy a particular energy level
Electrons cannot have intermediate amounts of energy
The evidence for this comes from emission spectra
When a compound like potassium chloride is heated in a Bunsen a particular colour light is emitted.
If this light is passed looked at through a spectroscope a series of lines of different colour are seen – of different frequencies
The energy of each separate frequency is given by
E = h f h= Plancks constant f= frequency
Hence hydrogen emits only certain energies of radiation.
These result from electron movements between specific energy levels (shells).
Energy is absorbed by the electron and it moves to a higher energy level.
It spontaneously returns to the lower energy level, emitting the difference in energy between the two.
As only certain energies are emitted, the levels must also have only certain energy values. These values converge.
Electron arrangement (configurations)
The Inert gases are very stable (almost totally un-reactive).
He 2 electrons
Ne 10 electrons (2 +8)
Ar 18 electrons (2+8+8)
This pattern led chemists to link these electron arrangements with chemical stability.
2 electrons in the 1st shell is stable – no more electrons will be added
8 electrons in the 2nd shell is stable – no more electrons will be added
8 electrons in the 3rd shell is stable – no more electrons will be added
Hence we write the electron arrangements for the 3 inert gases as:
He 2 Ne 2,8 Ar 2,8,8
For other elements, electrons are placed in the lowest energy shell, until it is complete.
Further electrons go in the next available shell.
Potassium 19K 2,8,8,1
Further evidence for the way electrons are arranged in energy levels with in atoms comes from ionisation energy data.
- 09-12-2003 10:37
Elements bond together to form compounds in two principal ways.
If the electro-negativity values of the elements are significantly different , as when a metal and a non-metal combine,
IONIC bonding results.
If the electro-negativity difference is small, as when two non-metals combine, or zero (as for elemental molecules),
COVALENT bonding results.
Electron transfer between atoms results in positive and negative ions being formed. These are then attracted electro-statically (opposite charges attract).
Metal ions, typically form groups 1, 2 and 3, lose electrons to form + ve ions
Elements from group 6 and 7 typically gain electrons to form – ve ions
The number of electrons lost or gained is sufficient to achieve a stable electron structure - that is, like an inert gas
magnesium forms a +2 ion,
aluminium forms a +3 ion
sulphur forms a -2 ion
chlorine forms a –1 ion
These ion charges are typical of groups 1, 2, 3, 6 and 7.
Transition metals are less uniform, and can also form more than one type of ion
e.g. iron can form Fe2+ or Fe3+ ions in its compounds.
Ions combine to form neutral compounds. - the ion charges add up to zero
Hence the formula of simple ionic compounds can be deduced
e.g. magnesium chloride Mg = +2
Cl = - 1
Cl = - 1
Hence the formula is one magnesium ion and two chloride ions - MgCl2
The properties of ionic compounds
Typical properties include:
1. High melting point and boiling point. Low volatility.
3. Electrical conductance in solution and melt. Non-conductive as solid.
4. Soluble in water. Insoluble in non-polar solvents.
These can be predicted from the nature of ionic bonding.
A covalent bond is the force of attraction between two atoms, resulting for the sharing of a pair of electrons, one form each atom.
A stable electron arrangement for each atom results (inert gas structure)
Lewis structures for molecular (group) ions
Here an inert gas electron structure is achieved both by sharing and through the gaining of electrons.
Eg OH - HS- ClO- CO32- NO2- CN-
Covalent bond strength
The greater the number of bonds between two atoms, the greater the strength of bonding.
However the relationship is not directly proportional - accounting for the greater reactivity of the alkenes compared to that of the alkanes.
N.B. The greater the bond strength, the shorter the bond length.
Polarity of covalent bonds
When two atoms of different elements covalently bond, a pair of electrons is shared. However, as the two elements will have different electro-negativities the electron pairs are not positioned equi-distant between the two nuclei.
The electrons are more attracted to the chlorine (positioned to the right of the periodic table and therefore electronegative) than to the hydrogen.
The electron distribution within H-Cl bond is therefore unsymmetrical and the bond has an electrical dipole. Bonds with dipoles are described as polarised.
The shapes of covalent molecules and molecular ions
Using the the Valence Shell Electron Pair Repulsion Theory (VSEPRT) - Electron pairs repel each other, and adopt a position of minimum repulsion, as far apart as possible.
Some molecules contain valence shell electron pairs that are not involved in bonding. These are called ‘lone pairs’.
A lone pair is more repulsive than a bond pair.
N.B. Multiple bonds are treated as ONE centre of electron density.
- 09-12-2003 10:40
- are those in which the products are of lower enthalpy (chemical energy) than reactants. i.e. the products are more stable than the reactants.
The difference between the enthalpy of the reactants and that of the products is called the ‘enthalpy change’ or ‘heat of reaction’ - H
An exothermic reaction has a negative H value as the enthalpy of products is less than the enthalpy of reactants. The difference in energy is given out, generally as heat.
Therefore, in an exothermic reaction the temperature of the system is raised.
An endothermic reaction has a positive H value - the enthalpy of products is greater than the enthalpy of reactants. The difference is absorbed from the system.
Therefore in an endothermic reaction the temperature of the system decreases.
The reactants are more stable than the products (not just stable!)
Determining enthalpy changes
This is generally achieved by measuring mass and temperature change.
H = mass x temp. change x specific heat capacity
SHC (Cp) is the amount of energy needed to raise 1kg by 1o K - units kJ kg-1K-1
Each pure substance has its own Cp – found in data books.
For a given mass, H can be calculated.
e.g. 50ml of pure water at 25 C is heated by a candle flame. The temperature is raised to 31.9 C. How much energy has been absorbed by the water?
Enthalpy change H = - 50 x 6.9 x 4.18 = - 1.44 x 103 J
N.B. assuming the mass of 50ml of water to be 50g, and the specific heat capacity of the water to be 4.18 Jg-1K-1
- the enthalpy change of reaction produces a temperature change in the water present.
Hence measuring the mass and T of this water will provide the H of reaction.
e.g. 50 cm3 of 1.0M NaOH and 50 cm3 1.0M HCl were mixed (Cp water is 4.2Jg-1K-1)
A temperature rise of 6.8 oC was measured.
Calculate the heat of neutralisation (per mole) of NaOH.
Total mass water = 100g
T = 6.8
The heat of reaction for the quantity of NaOH used is
= - 100 x 6.8 x 4.2 = - 2856 J
Number of moles NaOH reacting = (50/1000) x 1 = 0.050 moles
H for one mole of NaOH = - 2856 / 0.05 = 57.12 kJ per mole
Standard Enthalpy Change - H
This is the heat change for a reaction, for the number of moles as stated in the reaction equation, adjusted to standard conditions (101 325 Pa and 298 K).
e.g. 2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (l)
H for the above reaction is at standard conditions and for 2 moles of ethane – hence the unit of H is kJ per equation. The heat evolved by the combustion of 2 moles of ethane, as in the above equation, is 3120 kJ
However for fair comparisons between the combustion of ethane and of other substances quantity must be defined.
This leads to a general definition of Heat of Combustion as the heat produced on complete combustion of one mole of substance (at standard conditions)
Hence the combustion of ethane has a Hc value of -1560 KJmol-1
Standard Enthalpy change of Formation Hf
This is defined as
The enthalpy change on forming one mole of compound from its elements in their standard states (states at 25oC, 101.3 kPa)
Eg C(s) + 2 H2(g) CH4(g) H = Hf
Although the above reaction does not actually take place, the value Hf = -73kJmol-1 can be calculated from other data eg heat of combustion data (see Hess’s Law)
H may be expressed as kJ per mole of reactant e.g. heat of combustion Hc or
H may be expressed as kJ per mole of product e.g. heat of formation Hf
Bond Enthalpies and Molecular level theory
Any heat change can be explained by comparing the bonds in the reactants and products
i.e. 12 x C-H + 2 x C-C + 7 x O=O
absorbs less energy than is released by
forming all the bonds in the products
i.e. 8 x C=O + 12 x O-H
Hence, more energy is released than is absorbed and the above reaction is Exothermic.
The reverse would apply for endothermic reactions.
Each type of bond has a different energy value associated with it - the Bond Enthalpy
This is the average energy change on breaking/making one mole of a particular bond.
- obtained from combustion and spectroscopic data, and by deduction from this data.
- bond values vary slightly i.e. the C-C bond has different values in C2H6 and in C3H8
Typical values are:
kJ mol-1 bond kJmol-1
C - C 346 C - H 435
C = C 600 H - H 416
C = O 686 C - O 336
O = O 497 O - H 460
Another value that is useful is the enthalpy of atomisation of carbon (graphite)
C(s) C (g) H = +715 kJ mol-1
These values can be used to predict an enthalpy change for a given reaction
e.g. CH4 + 2 O2 CO2 + 2 H2O
C (g) + 4 H (g) + 4 O (g)
Bonds to break (endothermic + ve) Bonds made (exothermic – ve)
4 x C – H = 4 x 435 2 x C = O = 2 x 686
2 x O = O = 2 x 497 4 x O – H = 4 x 460
The total heat change is achieved by adding the two values (remembering the signs)
Therefore H = + 2734 + - 3212
H = - 478 kJ per mole
NB. the bond enthalpy values are per mole of bonds. Hence calculated value is per mole.
- 09-12-2003 10:41
Systems at equilibrium show no change in macroscopic properties – their properties appear to be unchanging with time.
e.g a cup resting on a table is at equilibrium
However, chemical equilibria are DYNAMIC – they appear to be static but this is the product of two opposite changes occurring at the same rate.
e.g. bromine (l) bromine (g)
For an equilibrium to be reached
The change must be reversible
The system must be closed (no particles escaping)
rate of forward reaction = rate of reverse reaction
- equilibrium can be attained starting from either the left or right hand side.
Another example is the shaking of aqueous iodine with an immiscible solvent (hexane)
I2 (aq) I2 (hexane)
The greater solubility of iodine in hexane will lead to a strong purple colour in the hexane layer and a pale brown colour in the aqueous phase, regardless of volume of solvent or amount of iodine. The same ratio of concentrations in the two phases will always result.
For the iodine phase equilibrium above, it is found that
iodine in hexane
iodine in water
The position of equilibrium
From experimental data, there is a general law for all one phase equilibria, such as
a A + b B c C + d D
It is found that
Cc D d
= Kc the equilibrium constant
A a B b
The value of Kc is maintained on changes to concentration, pressure and amounts, but is altered by changes to temperature.
If the position of equilibrium lies well to the right hand side
the value of Kc is very large
If the position of equilibrium lies to the left hand side
the value of Kc is very small
The units of Kc depend on the Kc expression itself, and varies for different equilibria.
Le Chatelier’s Principle.
An equilibrium will respond to imposed changes by shifting to left or right, so altering concentration values to maintain Kc. In other words, an equilibrium shifts position to counter any external changes imposed.
Consider the gas phase equilibrium:
A (g) + B(g) == C(g) H = - 120kJ mole-1
• raising temperature will cause the equilibrium to shift to the left,
• so the concentration of C decreases whilst those of A and B increase
• because a shift to the left is in the endothermic direction,
• hence the original increase in temperature is countered
• the value of Kc is decreased
• increasing the pressure of the reaction system causes a shift to the right
• because there are less moles of gas on the right hand side
• less gas particles exert less pressure (by collision with the container walls)
• hence a shift to the right counters the original increase in pressure
• the value of Kc is maintained by this adjustment
Industrial processes have to consider position of equilibrium, rate and costs.
Consider the industrial Haber Process
N2 (g) + 3 H2 (g) == 2 NH3 (g) H = - 92kJmol-1
If the equilibrium is subjected to:
high pressure - the equilibrium moves to the right, reducing the number of moles of gas, and, hence reducing pressure
low temperatures - the equilibrium moves in the exothermic direction (to the right) to increase the temperature.
Therefore to achieve maximum yield of ammonia
high pressures and low temperatures are indicated.
at low temperatures the rate of attainment of equilibrium is very slow
at high pressures , the costs of gaseous containment are high
Therefore in practice a compromise between yield, rate and cost is used
Industrial conditions are about: 550oC and 250 atms pressure.
To increase the rate of attainment of equilibrium further a catalyst is used
Industrial catalyst is : finely divided iron
The catalyst does not affect the position of equilibrium (nor yield) but does decrease the time taken for equilibrium to be reached – hence saving man-hours and energy.
Uses of ammonia
Ammonia: the good, the bad and the really ugly.
In the 20th century German Chemist Fritz Haber began to look for ways to combat the shortage of nitrogen-rich fertilisers. Haber invented a process in which he combined nitrogen from the air with hydrogen to form ammonia, which can then be used as a basis for the manufacture of fertilisers. The production of ammonia then became known as the Haber process.
The Haber process is used to create materials that are used in our everyday lives. The use of ammonia as a base for the manufacture of fertilisers has helped farmers to grow their crops in rich soil, increasing their quality and quantity of sales. Pharmaceutical companies use ammonia in the production of drugs such as anti-malaria and vitamins. Ammonia is a key component in cleaning agents used very frequently in your home. The Haber process plays a large role in refrigeration and air-conditioning. Also playing a role in the production of fibres and plastics.
The Haber process is used to create medicines that are misused and sold as illict drugs today. The addiction to a few of these drugs can account for a large amount of crime that occurs in our society at the moment.
THE REALLY UGLY.
The Haber process can be used to inflict pain and death among people. Fritz Haber and the Haber process he had engineered became a great interest to the German chemical industry shortly before the start of World War I. Within the Haber process, Fritz Haber had manufactured various nitrogen compounds that were needed for the production of bombs and poisonous gases.
In 1911 Fritz Haber was y invited to run the Kaiser Wilhelm Institute of Chemistry and Electrochemistry in Berlin. The German army overlooked his Jewish origin early in the war because of his large contributions to the war effort.
Fritz Haber’s wife, Clara,was very unhappy with Haber’s process begin used for the manufacture of explosives and poison gas. As soon as the German army began to use chlorine gas, manufactured by her husband’s process, on the French she killed herself.
Haber became a very unhappy man, in 1933 the Nazi regime forced Haber to resign from the Kaiser Wilhelm Insitute of Chemistry and Electrichemistry because of his refusal to dismiss his Jewish staff. Haber remained firm in his belief that “I have always employed my staff on the basis of their abilities and not because of their origin!”
The Haber process, like many things today, can be used for good, bad and evil. Without the Haber process many everyday objects like refrigeration, medicine and plastic would not exist. However if the Haber process did not exist, the human lives that were lost during the wars following the invention of poison gas and the attempted extermination of the Jews might not have occurred. When Haber received the Nobel prize many people were outraged that he should be honoured for creating a process for the manufacture of explosives and poison gas.
The Contact Process
A multi-staged process producing sulphuric acid on an industrial scale:
Sulphur is combusted in air
S (s) + O2 (g) SO2 (g)
Sulphur dioxide is then reacted with oxygen in an equilibrium reaction
2 SO2 (g) + O2 (g) == 2 SO3 (g) H = -196kJmol-1
The industrial conditions employed are
(the equilibrium mixture is cooled, to lesson the effect of increased temp on the amount of SO3 formed)
close to atmospheric pressure
(the cost of increasing pressure is not justified by the small effect on yield)
a catalyst of vanadium V oxide pellets is used
(at low cost the rate of attainment of equilibrium is increased)
Un-reacted equilibrium gases are recycled through further converter (catalyst) beds until a conversion to SO3 of about 98% is achieved
The sulphur trioxide, SO3 , produce is then dissolved in 98% sulphuric acid to form Oleum.
SO3 (g) + H2SO4 (l) H2S2O7 (l)
The Oleum is then diluted to give an increased volume of 98% H2SO4
H2S2O7 (l) + H2O (l) 2 H2SO4 (l)
N.B. The direct reaction between SO3 and water to give H2SO4 is very exothermic and hazardous acidic fumes are produced. Hence the preferred method of Oleum formation.
- 09-12-2003 10:44
What is rate of reaction?
We define rate of reaction as the change of amount (in moles or as mol.dm-3) of a particular reactant or product per unit time (secs, mins etc.)
i.e. change in amount (or concentration) of a substance
The units of rate are therefore generally mol.dm-3 s-1
Measuring amount of reactant/product should be carried out at regular time intervals, appropriate to the rate of reaction. If a time interval is too long, then only data on the average rate between time intervals will be obtained.
End-point methods of timing a rate may be appropriate, as long as it is appreciated that an average rate over the time interval is being reflected in the time taken.
There are several ways to follow the reaction between oxalate and manganate VII ions
5 C2O42— (aq) + 2 MnO4—(aq) + 16 H+ (aq) 10 CO2 (g) + 2 Mn2+(aq) + 8 H2O (l)
This is based on the assumption that
for a reaction to occur particles must collide.
However, the kinetic theory of gases calculates a very high collision frequency – indicating that all reactions should be instantaneous if every collision led to a reaction.
only collisions with a minimum energy (activation energy) will react.
The Maxwell-Boltmann distribution of molecular energies indicates that only a few particles in any system have high enough energy to react on collision. Hence most collisions are unsuccessful (result in rebound).
Factors which alter the frequency of collision, or the numbers of collisions with enough energy to react, will affect the rate of reaction. These are:
Concentration of reactants
The more randomly moving particles in a given volume, the more collisions that occur per second,. Hence there are more successful collisions per second simply because the totals number of collisions is increased. The proportion of unsuccessful collision remains the same.
Particle size (and available surface area.)
The larger the surface area available for collision, the more chance there is of a collision. Hence there are more successful collisions per second simply because the total number of collisions is increased. The proportion of unsuccessful collision remains the same.
Reducing particle size (for the same total mass) increases the available surface area, and hence tends to increase rate of reaction.
Increasing temperature increases the average kinetic energy of the particles, and, hence, increases the proportion of particles with more than Ea as described in the MB distributions below.
A small increase in total collisions per second also occurs.
The proportion of collisions that will have Ea is greatly increased, hence there are many more successful collisions per second, although the total number of collision per second is only slightly increased.
Catalysts increase rate of reaction by providing an alternative reaction pathway of lower minimum energy, Ea.
Hence more particles possess this new lower energy requirement for reaction. Hence the number of possible successful collisions is greatly increased.
Catalysts are thought to participate in the reaction but to be regenerated in the final steps of the reaction pathway. Hence catalysts are not used up the reactions they catalyse.
A step – wise theory of reaction kinetics is required to explain catalysis. The catalyst provides an alternative pathway (series of steps) that requires less energy to reach the activated state.
Step-wise reaction pathways
Reactions may take place via more than one chemical change – in a series of steps.
The reaction sequence is
A B C
In any sequence of steps, it is the step with the largest Ea that is the slowest step.
It is the slowest step which determines overall rate. The other steps often have very low Ea values, and are fast.
Hence, in this example, step
is the ‘the rate determining step’
- 09-12-2003 10:45
The millions of different carbon based compounds are organized into
Homologous series contain molecules which:
1. share a general formula - e.g. CnH2n+2 nearest neighbors differ by -CH2 -
2. share chemical properties – all contain the same functional group e.g. >C=O
3. exhibit a trend in boiling point - increases with increasing number of carbons
Aliphatic alkanes (straight-chain alkanes)
CH4 C2H6 C3H8
The general formula is CnH2n+2
The next members of the series are, butane, pentane and hexane.
For a given molecular formula, more than one carbon chain can exist = structural isomerism
Naming alkanes follows this pattern:
• name the longest carbon chain and number each carbon.
• name and count (di, tri) the side chains, and position with the lowest appropriate number
• name in sequence of: numbers ; side-chains (in alphabetical) ; longest chain
• punctuation: separate numbers by comas; separate letters and numbers by a dash.
Example 2,3 - dimethylbutane
The next members of the series is pent –1-ene and its isomers
For a given molecular formula the double bond can be positioned differently.
Reactions of the alkanes
Alkanes tend to be inert. The C –H and C - C bond enthalpies are high, and breaking them in the process of forming new bonds with other atoms provides a high energy barrier to reaction. However, alkanes do undergo combustion in air.
Combustion of hydrocarbons
Hydrocarbons react with oxygen (combust) in an exothermic reaction.
In a plentiful supply of oxygen the products are always carbon dioxide and water
eg. C3H8 + 5 O2 3 CO2 + 4 H2O
eg. C2H4 + 4 O2 2 CO2 + 2 H2O
The CO2 produced has been linked to global warming.
However if the supply of oxygen is limited then carbon monoxide or carbon may form.
eg. CnHx + limited O2 H2O + mix of CO and C
This limited oxidation produces less heat per mole of hydrocarbon.
The carbon monoxide produced is highly poisonous. Carbon particles are associated with respiratory diseases.
Other Functional Groups
The many oxygen or nitrogen containing organic molecules are also grouped in to homologous series.
Aldehydes eg. Ethanal CH3CHO
Ketones eg. Propanone CH3COCH3
carboxylic acid e.g ethanoic acid CH3COOH
Alcohols eg. ethanol CH3CH2OH
Amides eg. ethanamide CH3CONH2
Amines eg.ethyl amine CH3CH2NH2
Esters eg. methyl ethanoate CH3COOCH3
Halogenoalkane eg. bromoethane CH3CH2Br
- 09-12-2003 10:46
Oxidation and Reduction
Consider the following oxidation of magnesium:
2Mg + O2 2 MgO
writing two half-equations for this process
Mg - 2e— Mg2+ oxidation
O + 2e— O2— reduction
If we compare this to a similar reaction
Fe + S FeS
Writing two half equations
Fe - 2e— Fe2+
S + 2e— S2—
The two reactions are identical in nature – involving the loss and gain of electrons.
Hence oxidation and reduction are defined in terms of electron transfer
Oxidation is loss of electrons
Reduction is gain of electrons
Not all reactions involve ion formation and the complete transfer of electrons
eg. C + O2 CO2 ( -O = C+ = O - )
Here Carbon has partially lost the 4 electrons forming the 4 polar covalent bonds, and each oxygen has partially gained 2 electrons from within the polar covalent bonds.
Hence carbon is said to be oxidised
oxygen is said to be reduced.
Carbon in CO2 is given the oxidation number of + 4
Oxygen in CO2 is given the oxidation number of - 2
The rules for assigning oxidation number are :
1. The oxidation number of elements is 0
2. In neutral molecules the sum of the oxidation numbers is 0. In ions, the sum is equal to the ion charge.
3. In a compound, the more electronegative atom will have a negative oxidation number.
4. The oxidation number of group I elements in their compounds is +1
5. The oxidation number of group II elements in their compounds is +2
6. The oxidation number of H in its compounds is +1 (except in hydrides, -1)
7. The oxidation number of oxygen in compounds is – 2 (except in peroxides, -1)
Oxidation number is used to decide what, if anything, in a reaction has been oxidised or reduced.
An increase in oxidation number is oxidation
A decrease in oxidation number is reduction
Often the oxidation number of an element which has more than one possible oxidation state is listed in the name of a compound :
e.g. iron II chloride FeCl2 and iron III chloride FeCl3
Roman numerals are used to avoid confusion with ionic charge
Oxidising and reducing agents
Oxidising agents are the particle responsible for oxidising something else.
An oxidising agent is itself reduced in a redox reaction.
Reducing agents are the particles responsible for reducing something else
A reducing agent is itself oxidised in a redox reaction.
Reactivity can be thought of as relative strength of oxidising or reducing power.
This can be applied to the displacement reaction of metals
e.g Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)
• Zinc has been oxidised in preference to the copper (Zn Zn 2+)
• Zn is more reactive than copper
• Zn is a stronger reducing agent than copper (Cu2+ Cu)
This idea can be applied to the displacement reactions of the halogens
e.g. Cl2 (aq) + 2 Br— (aq) 2Cl—(aq) + Br2 (aq)
• chlorine has been reduced (Cl2 2Cl- )in preference to bromine, which has been oxidised.
• chlorine is a stronger oxidising agent than bromine
• chlorine is more reactive than bromine
The more reactive an element the more likely it is to displace another less reactive element from its compounds.
Hence we can predict that Zn will reduce Pb2+ to Pb metal, i.e. Zn will displace Pb.
Pb2+ + Zn Pb + Zn2+
Pb2+ + 2e- Pb red
Zn Zn2++ 2e- ox
In a voltaic cell the two parts of a spontaneous redox reaction are separated, and the electron transfer between them carried through a wire.
This flow of electrons can be made to do work, such as light an electric bulb.
The Daniel cell:
The spontaneous reaction between zinc and copper ions is split in to two half cells.
Zn2+ + 2e- Zn
Cu2+ + 2e- Cu
As Zn is the more reactive, it is the stronger reducing agent
Hence Cu 2+ is reduced to Cu, and Zn is itself oxidised to Zn2+
These two changes occur at the two separate electrodes
Anode Zn - 2e— Zn2+
Cathode Cu2+ + 2e— Cu
The flow of electrons between the electrodes carries a voltage of 1.1V (at standard conditions). The greater the difference in the reactivity of the metals inolved, the greater the voltage.
In general, reductions always occur at the cathode, oxidations at the anode.
N.B. electrons flow through the connecting wires from the oxidation to the reduction.
An electrolysis cell is typically:
The molten electrolyte is a heated ionic compound e.g. NaCl . In the solid state these ions are fixed in place, but when molten (or aqueous) these ions are free to move and are attracted to the electrode of opposite charge.
Electrode reactions (for molten sodium chloride (NaCl))
These two half reactions are an oxidation and a reduction.
Anode (+) 2Cl— Cl2 + 2e—
Cathode (-) 2 Na+ + 2 e— 2 Na
Overall reaction equation
2 NaCl 2 Na + Cl2
This redox reaction does not happen spontaneously at standard conditions. Hence the electrical power source is needed, to drive the electrolytic decomposition forward.
Electrolysis can also be used to cover one metal with another – as in copper plating.
The main difference to the cell for decomposition is in the use of an active electrode.
As before, these two half reactions are an oxidation and a reduction.
Anode (+) Cu (s) Cu2+ (aq) + 2 e—
Cathode (-) Cu 2+ (aq) + 2e— Cu (s)
At the copper anode, copper metal dissolves.
At the metal cathode, copper metal forms on the surface of the electrode.
The net effect is to transfer copper of the anode on to the surface of the metal cathode.
i.e. The metal cathode has been electroplated (copper plated)