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Flashcards in Topic 4 Deck (23):
1

Describe the use of the indicators litmus, phenolphthalein and methyl orange to distinguish between acidic and alkaline solutions

Litmus:
In acid, turns red
In neutral, turns purple
In alkali, turns blue

Phenolphthalein:
In acid and neutral, turns colourless
In alkali, turns pink

Methyl Orange:
In acid, turns red
In neutral, turns orange
In alkali, turns yellow

Universal Indicator:
In acid, turns red
In neutral, turns green
In alkali, turns blue

2

Understand how the pH scale, from 0–14, can be used to classify solutions as strongly acidic, weakly acidic, neutral, weakly alkaline or strongly alkaline

The pH scale is used to measure acidity and alkalinity

Solutions with a pH less than 7 are acidic. 0 being a really strong acid and 6 being a very weak acid.

Solutions with a pH of 7 are neutral

Solutions with a pH greater than 7 are alkaline. 14 being a really strong alkali and 8 being a very weak alkali.

3

Describe the use of universal indicator to measure the approximate pH value of a solution

Red, Orange, Yellow - 1,2,3 = Strong Acid
Lime, Green - 4 - 6 = Weak Acid
Green - 7 = Neutral
Turquoise, Blue - 8 - 11 = Weak Akali
Purple - 12,13,14 = Strong Alkali

4

Define acids as sources of hydrogen ions, H+, and alkalis as sources of hydroxide ions, OH¯

An acid is a substance that dissolves in water to produce hydrogen ions (H+)

An alkali is a substance that dissolves in water to produce hydroxide ions (OH-)

5

Predict the products of reactions between dilute hydrochloric, nitric and sulfuric acids; and metals, metal oxides and metal carbonates (excluding the reactions between nitric acid and metals)

Metals:
Metal + Acid -> Salt - Hydrogen
A metal only reacts with an acid if it is MORE reactive than hydrogen

- Metal + Hydrochloric Acid -> Metal Chloride Salt + Hydrogen
- Metal + Sulphuric Acid -> Metal Sulphate + Hydrogen

Metal Oxides/Hydroxides:
Metal Oxide/Hydroxide + Acid -> Salt + Water
The reaction will also heat up a bit

- Metal Oxide/Hydroxide + Hydrochloric Acid -> Metal Chloride Salt + Water
- Metal Oxide/Hydroxide + Nitric Acid -> Metal Nitrate Salt + water
- Metal Oxide/Hydroxide + Sulphuric Acid -> Metal -Sulphate + water

Metal Carbonates:
Metal (Hydrogen) Carbonate + Acid -> Salt + Water + Carbon Dioxide
Reaction will fizz (as CO2 is being produced)

- Metal (Hydrogen) Carbonate + Hydrochloric Acid -> Metal Chloride Salt + Water + Carbon Dioxide
- Metal (Hydrogen) Carbonate + Nitric Acid -> Metal Nitrate Salt + Water + Carbon Dioxide
- Metal (Hydrogen) Carbonate + Sulphuric Acid -> Metal Sulphate + Water + Carbon Dioxide

6

Understand the general rules for predicting the solubility of salts in water

- All common sodium, potassium and ammonium salts are soluble
- All nitrates are soluble
- Common chlorides are soluble, except silver chloride
- Common sulfates are soluble, except those of barium and calcium
- Common carbonates are insoluble, except those of sodium, potassium and ammonium

7

Describe experiments to prepare soluble salts from acids

Acid + Insoluble Base:
- Put some dilute acid into a beaker and heat it using a bunsen burner. Do not let it boil
- Add the insoluble base, a little at a time, to the warm dilute acid and stir until the base is in excess
- Filter the mixture into an evaporating basin to remove the excess base
- Leave the filtrate in a warm place so the water evaporates and crystals form
- Remove the crystals and dry them on filter paper

Acid + Soluble Base (Alkali):
- Put an aqueous solution of the alkali into a conical flask and add a suitable indicator (litmus or methyl orange)
- Add dilute acid from a burette until the indicator just changes colour
- Add powdered charcoal and shake the mixture to remove the colour of the indicator
- Filter to remove the charcoal and then obtain crystals from the filtrate in the usual manner

8

Describe experiments to prepare insoluble salts using precipitation reactions

- Add dilute sodium chloride solution to dilute silver nitrate solution until no more precipitate forms
- Filter to collect residue
- Wash the residue with cold distilled water
- Leave the residue to dry on filter paper

eg. Sodium Chloride + Silver Nitrate -> Sodium Chloride + Sodium Nitrate

9

Describe experiments to carry out acid-alkali titrations.

A titration is a method of finding out exactly the volume of one solution that is required to react with a given volume of another solution.

Titrations are commonly used to find out the volume of acid required to react exactly with a given volume of an alkali:

- Using a pipette, put 25.0cm3 of an alkali solution into a conical flask
- Add a few drops of an indicator (methyl orange)
- Put the acid in a pipette and note the initial reading
- Add the acid to the alkali until the indicator changes colour
- Note the final reading of acid in the burette
- Subtract the initial reading to obtain the volume of acid added. This is the volume required to neutralise the 25.0cm3 of the alkali.

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: Energy is given out when bonds are made, and taken in when bonds are destroyed

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

Exam Definition: An exothermic reaction is one which gives out energy to the surroundings, usually in the form of heat and usually show 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 is required to break old bonds is greater than the energy released when new bonds are formed.

Exam Definition: 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.

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

1. Measure the required amount of each substance
2. Record the temperature of any liquids before mixing
3. Mix the reactants in a polystyrene cup
4. Measure the temperature
5. However much heat has gone up or down is the calorimetry of the reaction

12

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

The enthalpy change is the over 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 enthalphy change is negative because the reaction gives out energy. If the reaction is endothermic, the enthalpy change is positive because reaction takes in energy.

13

Represent exothermic and endothermic reactions on a simple energy level diagram

What is needed:
y-axis = energy, x-axis = progress of reaction
Reactants, Activation Energy, Products and Overall Energy Change

In Exothermic Reactions:
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. The activation energy represents the energy needed to break the old bonds.

In Endothermic Reactions:
The products are at a higher energy level than the reactants. The different in height (ΔH) represents the energy taken. ΔH is positive because the reaction is taking in energy.

14

Understand that the breaking of bonds is endothermic and that the making of bonds is exothermic

During chemical reactions, the bonds in the reactants must be broken, and new ones formed to make the products.

Breaking bonds needs energy and therefore is described as ENDOTHERMIC.

Energy is released when new bonds are made and therefore is described as EXOTHERMIC.

The reaction is EXOTHERMIC because the energy NEEDED to break the bonds is LESS than the energy RELEASED in making new bonds.

If a reaction is ENDOTHERMIC then the energy NEEDED to break the bonds is MORE than the energy released in making new bonds.

15

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 a 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 stop watch
- Plot your results on a graph with time (independent variable) as X and volume of gas collected (dependent variable) as Y
- Repeat the experiment with smaller marble chips (same amount of mass 50g)
- Now repeat again with powdered marble chips
- Plot these results on the same graph for easy comparison.

Conclusion:
Powdered marble chips produced the most amount of CO2 in the same amount of time. This is because an increased SA causes more collisions, therefore the rate of reaction is faster, therefore more gas is produced.

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
- 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 results on graph with time on x and mass loss on y. Mass will decrease as the reaction produces hydrogen gas which 'floats' away.
- Repeat experiment with different concentrations of HCl, ensure the mass of magnesium 50 g and volume of acid 50 ml is the same and plot results on the same graph.

Conclusion:
The higher the concentration, the steeper the graph (the quicker the reaction).

Changes in Temp:
The higher the temp, the faster the reaction, this is because the particles have more energy from heat.

Method:
- Draw an X on a piece of paper and put a conical flash 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'/become not visisble
- Repeat experiment with two solutions at higher temp (using a water bath to heat the two solutions before adding them together).
- Repeat 5 times ,increasing heat everytime
- Plot results in a table for easy comparison.

Conclusion:
Should show that higher the temp, the quicker X disappears, therefore 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 gas syringe) in 10 mins at 30 seconds intervals (probably little as reaction is slow)
- Plot results on a graph with time as X and gas collected as Y.
- Repeated experiment but add a bit of manganese (IV) oxide (catalyst)
- Now repeat with powdered marble chips
- Plot these results on the same graph for easy comparison.

2(H2O2)(aq) -> 2(H2O)(l) + 02(g)

16

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 a reaction

Changes in SA:
The bigger the SA (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 (particles are close together). Alternatively, if the concentration is very weak there are a very little amount of particles for its volume (particles are spaced out).

Pressure of Gases:
This is similar to concentration of solutions. If the gas is at a 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 temp, the reacts are slower and so is the reaction rate. This is because the particles have little energy.

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

17

Understand the term activation energy and represent it on a reaction profile

The activation energy is the minimum amount of energy required for a reaction to occur.

18

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

Surface Area:
There is more surface that is in contact, so it is more likely that there will be successful collisions that lead to a reaction/faster rate of reaction.

Concentration/Pressure:
There's more particles so, again, it's more likely that there will be successful collisions that lead to a reaction/ faster rate of reaction.

Temperature:
Particles are moving faster, so they collide more frequently, because of this, there is going to be more successful collisions which increase the rate of reaction.

Catalyst:
Catalysts make the reaction require less activation energy, providing an alternate 'route' for the reaction.

19

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

It speeds up a reaction by providing an alternate pathway, so lower activation energy is required.

20

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

⇌ is the symbol for a reversible reaction.

A reversible reaction is just a reaction where the products of the reaction can themselves react to produce the original reactants.

21

Describe reversible reactions 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 add water to it, blue crystals will be formed, forming hydrated copper (II) sulfate. If this is heated, the hydrated copper (II) sulfate turns white as the water evaporates forming (dehydrated) copper (II) sulfate.

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 cool down, they will react with each other, forming ammonium chloride.

22

Understand the concept of dynamic equilibrium
Equalibrium:
A state in which opposing forces or influences are balanced

Dynamic Equilibrium:
Reversible reactions that take place at the same rate, in both directions (reaction hasn't stopped).

If reversible reactions happen in a closed system (meaning that none of the reactants or product can escape), equilibrium will always be reached.

The position of equilibrium depends on: temp, pressure of reacting mixture (this can change to give more products and less reactants).

A catalyst doesn't alter the position of equilibrium, it only makes the reaction reach equilibrium faster.

23

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 on reaction is endothermic and the other is exothermic.

The temp and pressure of the reactants have a very strong effect on the position of the equilibrium. e.g.:

Temperature:
If you increase the temp - the endothermic reaction will increase to use up the heat
If you decrease the temp - 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.