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# AS and A Level: Physical Chemistry

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## The dynamic equilibrium and 'Le Chatelier Principle'

1. 1 Le Chatelier Principle states “If a system that is in dynamic equilibrium is subjected to a change, the position of equilibrium will shift to minimise that change”. Put simply- if you do something, the system will try and reverse what you’ve done!
2. 2 Adding a catalyst does not change the concentration, pressure or temperature of a system. Therefore adding a catalyst has no effect on the position of equilibrium.
3. 3 If you increase the concentration of one side of an equation, the position of equilibrium will shift to the other side (to try and return the concentration back to its original value).
4. 4 If you increase the pressure of the system, the position of equilibrium will shift to the side of the reaction with less moles of gas (to try and reduce the pressure back to its original value).
5. 5 If you increase the temperature the position of equilibrium will shift in the direction of the endothermic reaction (to try and reduce the temperature back to its original value).

## Top equations for acid / base chemistry (A level only)

1. 1 For a strong acid the acid concentration is equal to the H+ concentration. This is because strong acids fully dissociate their H+ ions. [acid] = [H+]
2. 2 For a weak acid, because they only partially dissociate their H+ ions, to find the H+ concentration we must use the following equation: [H+] = √ka[acid]
3. 3 For a buffer, we calculate the value of H+ by using: [H+] = Ka[acid] / [salt] (where Ka is the acid dissociation constant)
4. 4 For a strong base, we calculate the H+ value by using: [H+] = Kw / [base] (where Kw is the ionic product of water = 1 x 10-14)
5. 5 To convert [H+] into pH, we would use the equation: pH = -log[H+]

## Top tips for ionisation energy

1. 1 One of the factors which will affect ionisation energy is electron shielding. This is how many inner shell electrons an atom has. The more electron shielding, the lower the ionisation energy. Electron shielding stays the same across a period and increases down a group.
2. 2 The second factor affecting ionisation energy is the proton number / nuclear charge. The higher the nuclear charge the higher the ionisation energy. Nuclear charge increases across a period and down a group.
3. 3 The third factor affecting ionisation energy is the atomic radius (size of the atom). The higher the atomic radius the lower the ionisation energy. Atomic radius decreases across a period (as the increased number of protons pulls the electron shells closer) and increases down a group.
4. 4 All three of these factors combine to have an effect of increasing the ionisation energy as we go across a period (eg F has a higher ionisation energy than O)
5. 5 All three of these factors combine to have an effect of decreasing ionisation energy as we go down a group (eg K has a lower ionisation energy than Na)

• Marked by Teachers essays 14
• Peer Reviewed essays 19
1. ## Revision Notes. Substances Manufactured for use in Industries. Chemicals, alloys and polymers.

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Burning of these products will oxidise sulphur, S to form sulphur dioxide, SO2. 3. Sulphur dioxide, SO2, is an acidic gas. When it dissolves in rainwater, it forms sulphurous acid, H2SO3, and causes acid rain. 4. Sulphur trioxide, SO3, will also form when sulphur dioxide, SO2, reacts with oxygen, O2, gas in air. 5. When sulphur trioxide, SO3, dissolves in rainwater, sulphuric acid, H2SO4, is also formed causing acid rain. 6. The effects of acid rain on the environment are as follows: (a)

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2. ## Write an essay on electrode potentials.

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calculation of cell emf ; (b) prediction of reaction feasibility and limitations. ~ Sample Essay ~ An electrode potential is the difference in an potential between an electrode and its surrounding electrolyte. It is always referred to a "zero point" defined by the potential of a reference electrode (e.g. standard hydrogen electrode). Electrode potential is an important measurement in the realm of electrochemistry, and it is particularly useful in the prediction of the energetic feasibility of redox reactions as well as calculation of electromotive force (emf)

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3. ## Objective To find out the equilibrium constant, Kc, for the reaction below, using acid hydrolysis:

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Table 1 shows the amount of chemicals added: Reagent bottle's number 1A 1B 2 3 4 Volume of HCl(aq) added /cm3 5.0 5.0 5.0 5.0 5.0 Volume of CH3COOCH2CH3(l) added /cm3 - - 5.0 4.0 2.0 Volume of H2O(l) added /cm3 - - - 1.0 3.0 Total volume of mixture /cm3 5.0 10.0 4.> Record the total mass before and after each addition of chemical into Table 3. 5.> Set them aside for at least 48 hours and shake the bottles occasionally. Procedure -- Titration (48 hours later) 1.> Rinse and fill a burette with standardized NaOH solution. 2.> Pour the contents of bottle 1A into a conical flask and rinse the bottle with deionized water.

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4. ## Stereochemistry of Butenedioic acid

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For example, an electrophile adds to the double bond, rotation becomes possible. Loss of the electrophile then regenerates the double bond. If rotation occurred in the intermediate, the result is the other isomer. The result of this kind of transformation is an equilibrium mixture of the cis and trans isomers. Frequently, the trans isomer is more stable, so the equilibrium mixture would contain more of the trans isomer. The greater the difference is in the stability of the isomers, the greater the concentration of the trans isomer will be at equilibrium.

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5. ## Electrochemical Cells

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This is measured by placing it with a standard hydrogen electrode. This is because for the half reaction that turns hydrogen ions into hydrogen gas, the standard electrode potential is 0.00V. If a half-cell has a positive standard electrode potential, then it means that it is more likely to receive than give away electrons, which would make it the reducing agent. The oxidising agent is the negative half-cell. For my investigation, I am using copper sulphate and zinc sulphate. When the solutions are 1mol dm-3, the redox reaction between them is: Zn(s)

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6. ## Electrochemistry Experiment. Objective: To investigate the effect of change in lead(II) ion concentration on the potential of the Pb2+(aq) &#8739;Pb(s) electrode

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electrode Introduction: This experiment investigates the e.m.f. of the cell: Cu(s) |Cu2+(aq) |Pb2+(aq)|Pb(s) Keeping the ion concentration in the copper electrode system constant(1M) and varying the ion concentration in the lead electrode system, the effect of change in lead(II) ion concentration on the potential of electrode as well as the Kc of the above reaction can be found. Chemicals: Copper foil x1, lead foil x1, 1M Cu2+ solution, 0.1M Pb2+ solution, saturated potassium nitrate solution Apparatus: 250 cm3 beakers, 50cm3 beakers multimeter, distilled water bottle, filter papers, electrical wires with electrode holders, forceps, 100ml volumetric flask,10ml pipette x2, dropper Procedure: 1.

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7. ## Ozone chemistry speech. Today, I am here to inform you that an ozone hole has been detected over Antarctica and the concentration of ozone will continue to decrease across the globe if the emission of ODCs, such as CFCs, does not reduce.

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USA: The CFCs' industry is worth 8 billion dollars per year. I understand. The chemical is ideal for many purposes as it is used as a solvent and cleaning agent in the electronic industry, a non-toxic propellant in aerosol cans, a refrigerant and is a blowing agent in producing Styrofoam and other plastics. However, CFCs account for 80% of ozone depletion so if we don't act soon, our children and grandchildren will have to pay the price. USA: And who's going to pay the price if we abide to this Montreal Protocol you are suggesting?

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8. ## Analysis of Two Brands of Commercial Bleaches

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Trial 1 2 Final burette reading / cm3 11.90 23.40 35.00 Initial burette reading / cm3 0.20 11.90 23.40 Volume of Na2S2O3 / cm3 11.70 11.50 11.60 Brand 2 Trade Name : LION Price : \$6.67per dm3 (\$10/1500mL) Trial 1 2 Final burette reading / cm3 14.60 28.70 42.80 Initial burette reading / cm3 0.50 14.60 28.70 Volume of Na2S2O3 / cm3 14.10 14.10 14.10 Questions 1. For each brand, calculate (a) the amount of the active ingredient available in g dm-3.

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9. ## The Development of the Periodic Table of the Elements

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An advantage of Lavoisier's work is that he distinguished between metals and non-metals but a disadvantage was that he included some compounds and mixtures as elements and substances that were neither, for example light and heat (which he called "caloric"). Nevertheless, this was recognised by the scientific community as a step forward in chemistry - he earned himself the nickname "the father of modern chemistry". Ultimately, his work marked the beginning of categorising the elements, which would prove incredibly useful later on.

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10. ## Redox titration

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+ 5Fe 2+ (aq) +8H+(aq) --> Mn2+(aq) +5Fe3+(aq)+4H2O(l) Between the titration of potassium manganate (VII) solution and ammonium iron (II) sulphate solution, The molarity of ammonium iron (II) Apparatus: Safety spectacles 25cm3 pipette Pipette filler 4 conical flasks 50cm3 burette Small funnel White tile Wash-bottle of distilled water Materials Solution A: ammonium Iron(II) sulphate solution 1M dilute sulphuric acid Solution B: potassium manganate (VII) Procedures: 1. The burette was filled with potassium manganate (VII) solution B 2. 25 cm3of the ammonium iron (II)

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11. ## Enthalpy of formation of calcium carbonate

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and at temperature of 25 oC (298 K).Moreover, the solution should have unit activity(1mol dm-3 ). 3. Write the equation for the formation of calcium carbonate under standard conditions. (Call this Equation 1) Ca(s) + C(s) + 3/2 O2(g) --> CaCO3(s) 4. Write an ionic equation for the reaction taken place. (Call this Equation 2) Ca(s) + 2H+(aq.) ? Ca2+(aq.) + H2(g) 5. Assuming (a) the solution in the plastic beaker has the same specific heat capacity as water, i.e., 4.2 kJg-1K-1 and (b)

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12. ## Nuclear Fusion as energy provider

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Radioactive decay is a passive action, while fission is active. For radioactive decay, the atom is unstable; while the nuclei in the process of fission absorb a neutron, then oscillate to become unstable. Moreover, the product of radioactive decay is only an atom of other element; while the products of fission are 3 neutrons and 2 different elements. Hydrogen and helium are by far the most abundant element, which is 89% and 11% respectively [1]. The process of making the simple elements (like lithium, etc.) is called nucleogenesis. Hydrogen acts as a producer, which is the start of the nucleogenesis.

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13. ## Titration Lab Report

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On each of the four printed graphs, draw a horizontal line from a pH value of 9 on the vertical axis to its intersection with the titration curve. In which trial(s) does this line intersect the nearly vertical region of the curve? In which trial(s) does this line miss the nearly vertical region of the curve? For Trials 1 and 3, the horizontal line from pH 9 intersects the S curve. For Trials 2 and 4, because the equivalence point is lower than pH9, they nearly miss the vertical region of the curve.

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14. ## Scientists and Theories of Atomic Structure

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He also said that these indivisible pieces (atoms) were infinite ? always moving and capable of moving together. However his theory was not widely accepted and forgotten about for over the next 2000 years. Later in the early 1800s an English chemist named John Dalton performed experiments which led to the acceptance of the ideas of atoms. Dalton?s theory consisted of 5 main points, he deduced that: 1. All Elements are composed of extremely small particles called atoms. 2. Atoms are indivisible and indestructible particles, which was along the lines of what Democritus had said but it was developed further.

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15. ## Gas Behaviours and the Weather

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Air pressure also relates to the density of the air and the height above the Earth's surface from where it is being measured (see Passante, 2006). The lines on the weather map (Fig. 1) represent the variations in pressure from region to region shown in 'isobars'. Temperature is a measurement of the speed of the molecules' movement in a substance. The more energy the molecules have, the faster they move and the higher the temperature is. When measuring the temperature of the air, it is the speed of the molecules that is being measured (Bureau of Meteorology, 2010).

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16. ## Spectroscopy

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The Electromagnetic Spectrum White light (what we call visible or optical light) can be split up into its constituent colours easily and with a familiar result - the rainbow. All we have to do is use a slit to focus a narrow beam of the light at a prism. This set-up is actually a basic spectrometer. The resultant rainbow is really a continuous spectrum that shows us the different energies of light (from red to blue) present in visible light.

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17. ## Atmospheric Pollution

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The output pollutants produced by coal fired stations are, sulphur oxides, nitrogen oxides and carbon dioxide. Sulphur is found in coal. When coal burns the sulphur compounds are oxidised so convert into sulphur oxide gases. Nitrogen oxide emissions can be produces in two ways. Firstly, most fuels contain compounds of nitrogen as it is formed by proteins. When nitrogen combusts, the nitrogen compounded are oxidised giving us NOx gases. The other way is that at high temperatures of combustion atmospheric nitrogen and oxygen combine to produce thermal NOx gases. Carbon dioxide is formed by the carbon in the coal reacting with the oxygen, making carbon dioxide when combusted.

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18. ## How does aspirin work?

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Salicin was concerted into Salicylic acid by the Italian chemist Piria in 1839. It was synthesised by a process discovered by Kolbe and Lautemann in 1860 which led to the introduction of Salicylic acid and sodium salicylate (forerunners of aspirin) for treatment of fever and arthiritis. However, these compounds were toxic to the stomach and caused diarrhoea and vomiting. German chemist Felix Hoffmann was set the task by Arthur Eichengrun of Friedrich Bayer & Co in 1893, to find a less toxic alternative.

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19. ## Analysis of Oxygen Absorber. How can the oxygen absorber absorb oxygen in the food package? What is the composition of the oxygen absorber? In this activity, we will plan and carry out an investigation to find out the chemical nature of the oxygen absorb

Besides, oxidation of food can cause the colour change of food. If the colour of foods changes, people will have a lower longing to have these food. Using oxygen absorber can prevent the change of colour. When people discovered that oxygen can be the main cause of degeneration of food, scientists started to find out the solutions. This caused the fast development of oxygen absorber in the 20th century. Scientists combined two simple ideas: "Irons absorbs oxygen when it rust." and "Prevent oxidation by eliminating oxygen.", and then iron powder package was first acted as oxygen absorber.

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20. ## Analysis of Two Commercial Bleaches

first, then one was washed by potassium iodide solution, one was washed by sulphuric acid), 1 pipette filler, 2 250cm3 volumetric flasks (Washed by deionized water only), 1 conical flask (Washed by deionized water only), 1 burette (Washed by deionized water first, then washed by standard thiosulphate solution), 1 washing bottle (Filled with deionized water), and 1 dropper Reagents used: "Kao" Bleach, "Clorox" Bleach, 1.0 M potassium iodide solution (KI), 1.0 M sulphuric acid (H2SO4), Standard thiosulphate solution(S2O32-) with conc.

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21. ## Explain what is meant by the following terms: i) Oxidising agent: ii) Displacement reaction: iii) Electronegativity

Though their ability and strength as oxidising agent alters as the group is descended- more on this later. Oxidising agents become reduced in the process of a reaction. Take the following reaction as our example: 2 Na + Cl2 = 2 Na Cl 2 Na + Cl2 = 2 Na Cl 0 0 +1 -1 0 = neutral oxidation state, uncharged. +1 = the oxidation state has risen, Na has become oxidised -1 = the oxidation state has reduced, Cl has become reduced Therefore, because chlorine has been reduced, it has accepted electrons from the sodium, it is said to

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22. ## Quantitative Determination of Food Colouring in Jelly Crystals using UV/Vis spectroscopy

The absorption in the visible range affects the perceived colour of chemicals used in food products involved. In this region of the electromagnetic spectrum, molecules undergo electronic transitions from the ground state to the excited state The Beer-Lambert law states that the absorbance of a solution is directly proportional to the concentration of the solution and the path length. Thus, for a fixed path length, UV/VIS spectroscopy can be used to determine the concentration of a solution. It is necessary to know how quickly the absorbance changes with concentration. Therefore the equation used is A=Ecl where A is the absorbance, E is the molar extinction coefficient, c is the sample concentrations in moles/litre and l is the length of light path through the sample.

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23. ## Objective:-To prepare a buffer solution and observe the properties of a buffer

On the other hand, if a small amount of base is added to the solution, CH3COOH(aq) will react with the base, giving out salt i.e. CH3COONa. This shifts the equilibrium of CH3COONa (aq) CH3COO- (aq) + Na+ (aq) to the right. The increase of [CH3COO-] does not contribute to any change in pH. Therefore, the pH of the solution remains almost constant. Procedures Preparation of Solution A(Buffer) 1. The pH meter was calibrated with a buffer of pH 10. 2. 25 cm3 of 0.1M NaOH was measured with a measuring cylinder and poured into a 100 cm3 beaker.

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24. ## Analysis of 2 commercial brands of bleaching solution & a determination of the best buy

4. Titrate this mixture against the standard sodium thiosulphate solution provided. Add starch indicator when the solution become pale yellow. 5. Record your results and calculate the molarity of the orginal bleach solution. 6. Determine which bleach has the lowest price per mole of actual bleach. Results : Brand A = ___________KAO_____________ Brand B = _________Best Buy____________ Volume = ___________1.5L______________ Volume = ____________2L_______________ Price = _______\$12.9 / Bottle____________ Price = ________\$10.9 / Bottle____________ Mole = _________0.753 Mol_____________ Mole = ___________1.152 Mol ___________ \$/mole = _______\$17.13 / Mol ___________ \$/mole = __________\$9.46 / Mol__________ Best buy is ________ Best Buy___________ Titration of Brand A against the standard sodium thiosulphate solution Titration 1 2 3 4 Final Burette Reading (ml)

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25. ## Analysis of Commercial Vitamin C Tablets

Procedures : 1. Accurately weigh out 0.6 to 0.7g of potassium iodate(V). Record the mass.. 2. Dissolve this in deionized water and make up to 250 cm3 in a volumetric flask. 3. Use such iodate(V) solution to standardize the given thiosulphate solution as follows : (a) Pipette 25.0 cm3 of KIO3(aq) into a conical flask, add to it about 5 cm3 of 1 M KI(aq) followed by about 10 cm3 of 0.5 M H2SO4(aq) , then immediately titrate with 0.05 M Na2S2O3(aq)

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