4.25 predict the effects of changing the pressure and temperature on the equilibrium position in reversible reactions

NOTE: It is important to remember in reversible reactions one reaction is endothermic and the other is exothermic.

The temperature and pressure of the reactants have a very strong effect on the position of the equilibrium. For example...

Temperature
If you increase the temperature - the endothermic reaction will increase to use up the heat
If you decrease the temperature - the exothermic reaction will increase to raise the heat

Pressure
If you increase the pressure - the reaction will produce the side of the equation that has the least moles (to work this out, just look at the ratio of moles on either side of the equation)
If you decrease the pressure - the reaction will produce the side of the equation that has the most amount of moles

4.24 understand the concept of dynamic equilibrium

Okay so this may need a few reads as it is a little complicated but this was the easiest way i could explain it... good luck :)

If a reversible reactions occurs (in a closed system), a dynamic equilibrium will be reached. All this means is that the relative % of reactants and products will reach a balance and stay there (NOTE: it is not always 50-50), the reactions will then continue to occur (basically the reaction takes place both ways, at the same time BUT there is no overall effect (one being made more than the other) as the relative % of reactants to products is balanced (remember :D) so the reactions kind of cancel each other out (as both reactions occur at the same rate)

NOTE: A closed system just means that none of the products/reactants can escape (this enables dynamic equilibrium to occur)

4.23 describe reversible reactants such as the dehydration of hydrated copper(II) sulfate and the effect of heat on ammonium chloride

Dehydration of hydrated copper(II) sulfate
Copper(II) sulfate is a white solid. When you added water to copper(II) sulfate it forms blue crystals, forming hydrated copper(II) sulfate. If heat this hydrated copper(II) sulfate it turns white as the water evaporates, forming (dehydrated) copper(II) sulfate. If you add water, it will turn blue again, if you heat it again it will turn white... etc

Ammonium chloride
Ammonium chloride is a white solid, when it is heated it breaks down into ammonia gas and hydrogen chloride gas. However, if you let these products (ammonia gas and hydrogen chloride) cool down, they will react with each other, forming ammonium chloride.

4.22 understand that some reactions are reversible and are indicated by the symbol ⇌ in equations

22. 22. ⇌ is the symbol for reversible reaction
23. 
    
24. NOTE: a reversible reaction is just a reaction where the products of the reaction can themselves react to produce the original reactants
    

4.19 understand the term activation energy and represent it on a reaction profile

The activation energy is just the minimum amount of energy the reactants needs for the reaction to start

4.21 explain that a catalyst speeds up a reaction by providing an alternative pathway with lower activation energy

Catalysts work by lowering the activation energy (the minimum energy required by reacting particles for the reaction to occur). This happens because a catalyst provides an alternative reaction pathway with a lower activation energy.

NOTE: with a catalyst, although the activation energy lowers, the enthalpy change (ΔH) is not affected.

4.18 describe the effects of changes in surface area of a solid, concentration of solutions, pressure of gases, temperature and the use of a catalyst on the rate of reaction

NOTE: this point is just about the changes that occur, if you want/need to understand why these changes occur, go to this point... 4.20


Changes in surface area
The bigger the surface area (to volume ratio) the faster the reaction. This is because there are more particles on the surface for the reactants to react with.  In a solid, to increase the surface area without decreasing the mass just break up the solid into smaller pieces.

Concentration of solutions
If a solution is very concentrated there are lots of particles for its volume (the particles are close together). Alternatively, if the concentration is very weak there are a very little amount of particles for its volume (the particles are very spaced out). 

Pressure of gases
This is very similar to concentration of solutions. If the gas is at high pressure there will be more particles squished into a certain space, more particles means a faster rate of reaction. Alternatively, low pressure results in little amount of particles meaning a slower rate of reaction.

Temperature
The hotter the reactants the faster the reaction. Alternatively, the colder the temperature of the reactants the slower the reaction rate. This is because the particles have very little energy. 

Catalyst
A catalyst works by giving the reactants a surface to 'stick' to. This will increase the rate of reaction.

4.20 explain the effects of changes in surface area of a solid, concentration of solutions, pressure of gases and temperature on the rate of reaction in terms of partial collision theory

Changes in surface area
The bigger the surface area (to volume ratio) the faster the reaction. This is because there are more particles on the surface for the reactants to react with, meaning more collisions with the particles meaning a faster reaction. In a solid, to increase the surface area without decreasing the mass just break up the solid into smaller pieces.

Concentration of solutions
If a solution is very concentrated there are lots of particles for its volume (the particles are close together). More particles means more collisions and therefore a faster reaction rate. Alternatively, if the concentration is very weak there are a very little amount of particles for its volume (the particles are very spaced out). This means collisions will be less frequent (are there are less particles to collide with) so the reaction is slower.

Pressure of gases
This is very similar to concentration of solutions. If the gas is at high pressure there will be more particles squished into a certain space, more particles mean more collisions which mean a faster rate of reaction. Alternatively, low pressure results in little amount of particles in a certain space meaning less collisions (as there are not as many particles to collide with) meaning a slower rate of reaction.

Temperature
The hotter the reactants the faster the reaction. This is because the particles have more energy due to the heat. Because they have more energy this means they will more faster and therefore collide more. Alternatively, the colder the temperature of the reactants the slower the reaction rate. This is because the particles have very little energy. 


NOTE: faster collisions are only increased by temperature, all other ways to increase reaction rate increase the amount of particles to collide with.

4.17 describe experiments to investigate the effects of changes in surface area of a solid, concentration of solutions, temperature and the use of a catalyst on the rate of reaction

Changes in surface area

A change in surface area will affect the rate of reaction as there will be more/less surface area, meaning more/less collisions with more/less particles.

Method
- Measure out 50g of large pieces of marble chips and put them in a conical flask
- Add 50ml of dilute HCl
- Straight after you add the acid, put a bung in the top and attach it to a delivery tube attached to a gas syringe and start a stopwatch
- For 5 minutes, and at 30 second intervals, record the volume of gas collected in the syringe.
- Plot your results on a graph with time (the independent variable) as x and volume of gas collected (the dependant variable) as y.
- Repeat this experiment with smaller marble chips (still the same amount of mass... 50g).
- Now repeat again with powdered marble chips
- Plot these results on the same graph for easy comparison

Conclusion
Should all go well, you should conclude that the experiment using powdered marble chips produced the most amount of CO2 in the same amount of time This is because an increased surface area causes more collisions, therefore the rate of reaction is faster, therefore more gas is produced.

NOTE: The gas given off is CO2, remember this is a small scale version of how to make CO2 in a laboratory, if you are unsure of how this is done, here it is... 2.20  :)



Changes in concentration

A change in concentration will affect the rate of reaction as there will be more particles to collide, so the reaction will be completed quicker

Method
- Measure 50g of magnesium metal and put it in a conical flask
- Put the flask on scales
- Add 50ml of least concentrated HCl (do not remove flask from scales)
- Record the mass of the flask with magnesium and HCl then immediately start a stopwatch
- Record the mass of the experiment for 5 minutes, at 30 second intervals
- Plot your results on a graph with time (the independent variable) as x and mass loss (the dependant variable) as y. NOTE: the mass will decrease as this reaction produces hydrogen gas, which 'floats' away.
- Repeat this experiment with different concentrations of HCl (ensure to keep the same mass of magnesium...50g and volume of acid...50ml) and plot your results on the same graph

Conclusion

You should find that the higher the concentration the steeper the graph (the quicker the reaction).

Changes in temperature

The higher the temperature the faster the reaction, this is because the particles have more energy from the heat.

Method

- Draw an X (or any clear shape) on a piece of paper and put a conical flask on top

- Add 50ml of sodium thiosulphate and 50ml of HCl and immediately start a stopwatch

- Time how long it takes for the X to 'disappear' (NOTE: it will become not visible as this reaction produced a yellow precipitate of sulfur which cloudy the flask so you can't see the X, just time how long it takes until you can't see the X).

- repeat the experiment but with the two solutions at a higher temperature (using a water bath to heat the two solutions before adding them together).

- Repeat 5 times, increasing the heat of the solutions each time.

- Plot your result in a table for easy comparison.

Conclusion

This reaction should show that the higher the temperature, the quicker the X disappears, therefore the quicker the reaction.

Using a catalyst (decomposition of hydrogen peroxide)

The use of a catalyst will increase the rate of reaction.

Method

- Add 50ml of hydrogen peroxide to a conical flask
- Put a bung on top and attach it to a delivery tube attached to a gas syringe
- Start a stopwatch
- Time how much gas is collected (in the gas syringe) in 10 minutes at 30 second intervals (probably VERY little as the reaction naturally is very slow)
- Plot your results on a graph with time (the independent variable) as x and volume of gas collected (the dependant variable) as y.
- Repeat experiment, but add in a bit of manganese (IV) oxide (this is a catalyst)
- Repeat this experiment with smaller marble chips (still the same amount of mass... 50g).
- Now repeat again with powdered marble chips
- Plot these results on the same graph for easy comparison

Incase you were wondering, the equation for this reaction is...

 2H₂O_2(aq) ---> 2H₂O_(l) + O_{2 (g)}

NOTE: The gas given off is oxygen, remember this is a small scale version of how to make oxygen in a laboratory, if you are unsure of how this is done, here it is... 2.18 :)

4.16 use average bond energies to calculate the enthalpy change during a simple chemical reaction

Each type of bond (for example O-O or C-H) has a particular bond energy... we will be given these in the exam so don't worry about learning them. These energies can be used to calculate enthalpy change, for example...

Using bond energies, calculate the enthalpy change for the following reaction
H2 + Cl2 ---> 2HCl

Bond energies...
H-H: +436 kJ/mol
Cl-Cl: +242 kJ/mol
H-Cl: +431kJ/mol

1. Work out what bonds are broken & the energy made... 

1 mole of H-H is broken and 1 mole of Cl-Cl is broken. Therefore, +436 + +242 = +678 kJ/mol is required to break the bonds in this reaction

2. Work out what bonds are being made & the energy released...

Forming 2 moles of H-Cl bonds. This released 2 x +431 = 862kJ/mol


3. Use the formula 'ΔH = total energy absorbed to break bonds - total energy released in making bonds' to find out the enthalpy change

ΔH  = 678 - 862 = -184 kJ/mol


4. ΔH  is negative which means the reaction must be exothermic

4.15 understand that the breaking of bonds is endothermic and hat the making of bonds is exothermic

Not much to say here... when you make a bond it is an exothermic reaction, when you break a bond it is an endothermic reaction.

Energy must be supplied to break existing bonds, so bond breaking is endothermic. However, energy is released when new bonds are formed, so bond formation is exothermic.

4.14 represent exothermic and endothermic reactions on a simple energy level diagram

In exothermic reactions
This energy level diagram shows an exothermic reaction, we can tell this because the products are at a lower energy level than the reactants. The difference in height (ΔH) represents the energy given out in the reaction. ΔH is negative here because the reaction is giving out energy (as it is an exothermic reaction). The activation energy represents the energy needed to break the old bonds.


In endothermic reactions

This energy level diagram shown an endothermic reaction, we can tell this because the products are at a higher energy level than the reactants. The difference in height (ΔH) represents the energy taken. ΔH is positive because the reaction is taking in energy (because is an endothermic reaction).

Basis of notes source: CGP

4.13 understand the use of ΔH to represent enthalpy change for exothermic and endothermic reactions

The enthalpy change is the overall change in energy in a reaction, it is symbolised by ΔH and its unit is kJ/mol, as it is the amount of energy in kilojoules per mole of reactant. It can be positive or negative, if the reaction is exothermic, the enthalpy change is negative because the reaction gives out energy, if the reaction is endothermic, the enthalpy change value is positive because the reaction takes in energy.

4.12 Calculate molar enthalpy change from heat energy change

Okay so you have calculated the amount of energy produced , this can be used to work out the molar enthalpy change (this is basically the enthalpy change given out by one mole of the reactant).

NOTE:

To calculate the molar enthalpy change, you need to know the equations moles = mass / Mr (if unsure of this equation, click here) and molar enthalpy change = energy produced / moles. For example...


0.9g of methylated spirit produces 6510J of heat energy, work out the heat produced per mole. (The Mr of methylated spirit is 44.6)

- First, work out the amount of energy transferred...

We know that 6510J of heat energy was produced, this means 6510J of energy was transferred... this needs to be converted into kJ (as the unit or enthalpy change is kJ/mol)... 6.510kJ

- Next, find out how many moles of fuel produced this heat...

moles = mass / Mr

= 0.9 / 44.6

= 0.02 moles

- Now, divide the amount of heat energy produced by the amount of moles...


6.510 / 0.02 = 325.5 kJ/mol

The end (:

4.11 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralisation in which heat energy changes can be calculated from measured temperature changes

NOTE: THIS IS SO CONFUSING I DON'T UNDERSTAND WHATS GOING ON the only example/explanation for this I could find was in the CGP revision guide but I still don't understand it so will defiantly be coming back to this post to edit it after i have finished the spec, if anyone knows how to do it or has any info or anything please comment it and I will add it to the post!

Combustion
This is basically the same method as 2.32 of the biology spec but with fuel not food... to measure the amount of energy produced when a fuel is burnt, just burn the fuel and use the flame to heat up some water...

- Put 50g of water into a copper can (because copper is a very good conductor of heat)
- Record the temperature of the water
- Weigh the spirit burner and lid (the spirit burner contains the fuel)
- Place the spirit burner underneath the copper can and light its wick.
- Stir constantly until the water reaches about 50ºC
- Put the flame out using the burner lid
- Record the final temperature of the water
- Weigh the spirit burner and lid again
- calculate the enthalpy change

Your setup should look something like this...

NOTE: the draught shield is just there to ensure no/little heat escapes

Dissolving, displacement and neutralisation reactions

4.9 describe experiments to carry out acid-alkali titrations

By doing a titration you are able to find out exactly how much acid is needed to neutralise a certain amount of alkali, exactly how much alkali is needed to neutralise a certain amount of acid.

Method

- Add 25cm³ of alkali to a conical flask (with a pipette and pipette filler)
- Add a few drops of phenolphthalein indicator
- Fill a burette with your acid (NOTE: ensure you have the burette below eye level incase the acid sprays out etc)
- Add the acid to the alkali a bit at a time (using a burette). NOTE: regularly swirl the conical flask to ensure the acid is evenly distributed throughout the alkaline
- Stop adding acid as soon as the solution in the conical flask changes colour (with phenolphthalein, it will go colourless) - this means that the alkali has been neutralised.
- Using the burette, record the amount of acid used to neutralise the alkali.
- Repeat a few times to avoid anomalies etc.



If you are asked to find out the amount in moles of acid needed to neutralise an alkali (or vice versa), just do the same as a mole calculation. For example...

It takes 30cm³ of sulphuric acid of an unknown concentration to neutralise 25cm³ of sodium hydroxide with a concentration of 0.1 moles per dm³. Find the concentration of the sulphuric acid.

- First, work out how many moles of the 'known' substance (sodium hydroxide) you have...

moles = concentration x volume

= 0.1 x (25/1000)

= 0.0025 moles

- Now, write down the equation to work out the ratio of acid : alkali...

2NaOH + H2SO4 ---> Na2SO4 + 2H2O

This means there are two moles of sodium hydroxide to every 1 mole of sulphuric acid.

- This means we need to half the amount of moles of sodium hydroxide to get the amount of moles of sulphuric acid...

0.0025/2 = 0.00125

- Now we know the amount of moles and volume of the acid, so we can work out the concentration...

concentration = moles / volume

= 0.00125 / (30/1000)

= 0.0417 moles per dm³

4.10 understand that chemical reactions in which heat energy is given out are described as exothermic and those in which heat energy is taken in are endothermic

NOTE: before you get your head round endothermic and exothermic, it might be a good idea to understand that energy is given out when bonds are made, and taken in when bonds are destroyed...


If a reaction gives out heat, it is said to be exothermic and the energy released in bond formation is greater than the energy used in breaking bonds.

Definition for exams: An exothermic reaction is one which gives out energy to the surroundings, usually in the form of heat and usually shown by a rise in temperature.


If a reaction is endothermic, it takes in heat from its surroundings during the reaction process. This is because the energy required to break old bonds is greater than the energy released when new bonds are formed.

Definition for exams: An endothermic reaction is one which takes in energy from the surroundings, usually in the form of heat and usually shown by a fall in temperature.

4.8 describe experiments to prepare insoluble salts using precipitation reactions

You can use a precipitation reaction. To do this just pick two solutions that contains the ions you need and add them together. For example, to make barium sulfate (insoluble) you need to add together one solution containing barium ions and another containing sulphate ions. To do this, you cam mix barium chloride with sulphuric acid...

barium chloride + sulphuric acid ---> barium sulphate + hydrochloric acid

BaCl_2(aq) + H₂SO_4(aq) ---> BaSO_4(s) + 2HCl_(aq)

4.7 describe experiments to prepare soluble salts from acids

You need to pick an insoluble base, and the acid of the salt you want to make (sorry that was a bit confusing, I couldn't really find another way to word it). Basically, if you was to make copper nitrate, mix nitric acid and copper carbonate (remember all carbonates are insoluble except sodium, potassium and ammonium carbonate)...

CuCO_3(s) + 2HNO_3(aq) ---> Cu(NO₃)_2(aq) + CO_2(g) +  H₂O_(l)


All you need to do is add the metal oxide, carbonate or hydroxide (aka the insoluble base) to the acid. The base will react with the acid and dissolve.  

NOTE: This is a neutralisation reaction, you will know when the reaction is over as this is when all the acid has been neutralised (this is when the excess solid will sink to the bottom of the flask and not react anymore)

Next, filter the solution to get rid of the undissolved base and evaporate the water to leave pure salt crystals.

4.6 understand the general rules for predicting the solubility of salts in water

Some salts are soluble whilst others are insoluble, we may get a question where we are given a salt and we have to determine whether is is soluble or insoluble, these are the rules we must learn is order to be able to do this...

- Sodium salts are soluble
- Potassium  salts are soluble
- Ammonium salts are soluble
- All sulphates except barium sulphate and calcium sulphate are soluble
- All chlorides except silver chloride are soluble
- All carbonates except sodium, potassium and ammonium carbonates are insoluble (this is because all sodium, potassium and ammonium salts are soluble remember)