Module 3 (Periodicity) Flashcards

1
Q

What is the periodic table?

A

An arrangement of all the elements in order of increasing atomic number. Elements with similar chemical properties are found in the same group

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2
Q

What are the 4 blocks in the periodic table?

A
  1. S block
  2. D block
  3. P block
  4. F block
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3
Q

What is Periodicity?

A

Periodicity is a repeating pattern across different periods

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4
Q

Explain Atomic Radius

A

Atomic radii decrease as you move from left to right across a period because the increased number of protons creates more positive charge attraction for electrons which are in the same shell similar shielding.

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5
Q

What is Ionisation energy?

A

The energy required to remove an outer electron from an element

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6
Q

What is the equation for first ionisation?

A

H(g) -> H+(g) + e-

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7
Q

What factors affect ionisation energy?

A

There are three main factors
1. The attraction of the nucleus
(The more protons in the nucleus the greater the attraction)
2. The distance of the electrons from the nucleus
(The bigger the atom the further the outer electrons are from the nucleus and the
weaker the attraction to the nucleus)
3. Shielding of the attraction of the nucleus
(An electron in an outer shell is repelled by electrons in complete inner shells,
weakening the attraction of the nucleus)

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8
Q

Explain why there is a large increase from the 3rd and 4th ionisation energy in Aluminium

A

There is a large increase in ionisation energy as it is from an inner shell, so is under a stronger force of attraction from the nucleus and has less shielding

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9
Q

Why are successive ionisation energies always larger?

A

The second ionisation energy of an element is always bigger than the first ionisation energy. This is because the ion formed, is smaller than the atom and the proton-to-electron ratio in the 2+ ion is greater than in the 1+ ion. The attraction between the nucleus and electron is therefore stronger

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10
Q

Explain the trend in the first ionisation energy down group 1

A

The ionisation energy decreases down the group as the outer electron is further from the nucleus and experiences greater shielding from inner electrons so the attraction from the nucleus is weaker and therefore easier to overcome

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11
Q

Why does the ionisation energy increase across periods 2 and 3?

A
  1. The number of protons in the nucleus increases
  2. The outer electron is in the same shell with similar shielding
  3. The electrostatic attraction between the outer electron and the nucleus is greater
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12
Q

What are examples of an exception to the trend in ionisation energy?

A
  1. Boron and aluminium have lower first ionisation energies than beryllium and magnesium. This is because the outer electrons of boron and aluminium are in a p-orbital higher energy.
  2. Oxygen and sulfur have lower first ionisation energies than nitrogen and phosphorus, this is because the outer electrons of oxygen and sulfur have [aired electrons in a p-orbital, these repel each other, so one electron is more easily removed
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13
Q

Describe the bonding in metals

A

Metallic bonding is the electrostatic force of attraction between the positive metal ions and the delocalised electrons

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14
Q

What factors affect the strength of metallic bonds?

A

The three main factors that affect the strength of metallic bonding are:
1. Number of protons/ Strength of nuclear attraction.
(The more protons the stronger the bond)
2. Number of delocalised electrons per atom (the outer shell electrons are
delocalised)
The more delocalised electrons the stronger the bond
3. Size of ion.
(The smaller the ion, the stronger the bond.)

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15
Q

Describe the bonding and structure of diamond.

A

Tetrahedral arrangement of carbon atoms. 4 covalent bonds per atom

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16
Q

Describe the bonding and structure of graphite.

A

Planar arrangement of carbon atoms in layers. 3 covalent bonds per atom in each layer. 4th outer electron per atom is delocalised. Delocalised electrons between layers.

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17
Q

Properties of Giant covalent strucures:

A

Boiling and melting points: high- because of many strong covalent bonds in macromolecular structure. Take a lot of energy to break the many strong bonds

Solubility in water: insoluble insoluble

conductivity when solid:
- Diamond and sand: poor, because electrons can’t move (localised)
- graphite: good as free delocalised electrons between layers

conductivity when molten: poor

general description: solids

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18
Q

Properties of Metallic lattices:

A

Boiling and melting points: high- strong electrostatic forces between positive ions and sea of delocalised electrons

Solubility in water: insoluble

conductivity when solid: good, delocalised electrons can move through the structure

conductivity when molten: (good)

general description:
-shiny metal
-Malleable as the positive ions in the lattice are all identical. So the planes of ions can slide easily over one another
-attractive forces in the lattice are the same whichever ions are adjacent

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19
Q

Trends down Group 2:

A

Melting points decrease down the group. The metallic bonding weakens as the atomic size increases. The distance between the positive ions and delocalized electrons increases. Therefore the electrostatic attractive forces between the positive ions and the delocalized electrons
Atomic radius increases down the Group. As one goes down the group the atoms have more shells of electrons making the atom bigger
Group 2 metals all have the outer shell s2 electron configuration.

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20
Q

General reactions of Group 2 metals:

A

When the group 2 metals react, they lose their outer shell s2 electrons in redox reactions to form 2+ ions. The energy to remove these electrons is the first and second ionisation energies.

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21
Q

The trend of ionisation energy down the group:

A

The first and second ionisation energies decrease down the group. The outermost electrons are held more weakly because they are successively further from the nucleus in additional shells. In addition, the outer shell electrons become more shielded from the attraction of the nucleus by the repulsive force of inner shell electrons

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22
Q

The trend of reactivity down group 2:

A

Reactivity of group 2 metals increases down the group (The reactivity increases down the group. As the atomic radii increase there is more shielding. The nuclear attraction decreases and it is easier to remove outer electrons. Cations form more easily.)

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23
Q

Reactions with oxygen:

A

The group 2 metals will burn in oxygen.
e.g.
Mg burns with a bright white flame
2Mg + O2 -> 2MgO
MgO is a white solid with a high melting point due to its ionic bonding
Mg will also react slowly with oxygen without a flame.
Mg ribbon will often have a thin layer of magnesium oxide on it formed by reaction with oxygen
2Mg + O2 -> 2MgO
This needs to be cleaned off with emery paper before doing reactions with Mg ribbon
If testing for reaction rates with Mg and acid, an uncleaned Mg ribbon would give a false result because both the Mg and MgO would react but at different rates.
Mg + 2HCl -> MgCl2 + H2
MgO + 2HCl -> MgCl2 + H2O

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24
Q

Reactions with water:

A

Magnesium reacts in steam to produce magnesium oxide and hydrogen. The Mg would burn with a bright white flame
Mg (s) + H2O (g) -> MgO (s) + H2(g)
Mg will also react with warm water, giving a different magnesium hydroxide product
Mg + 2 H2O -> Mg(OH)2 + H2
This is a much slower reaction than the reaction with steam and there is no flame
The other group 2 metals will react with cold water with increasing vigour down the group to form hydroxides
Ca + 2 H2O (l) -> Ca(OH)2(aq) + H2(g)
Sr + 2 H2O (l) -> Sr(OH)2(aq) + H2(g)
Ba + 2 H2O (l) -> Ba(OH)2(aq) + H2(g)
The hydroxides produced make the water alkaline
One would observe:
fizzing, (more vigorous down group)
the metal dissolving, (faster down group)
the solution heating (more down group)
and with calcium, a white precipitate appears (less precipitate forms down group)

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25
Q

Reactions with acids:

A

The group 2 metals will react with acids with increasing vigour down the group to form a salt and hydrogen
Ca + 2HCl (aq) -> CaCl2(aq) + H2(g)
Sr + 2 HNO3(aq) -> Sr(NO3)2(aq) + H2(g)
Mg + H2SO4(aq) -> MgSO4(aq) + H2(g)
If barium metal is reacted with sulfuric acid it will only react slowly as the insoluble barium sulfate produced will cover the surface of the metal and act as a barrier to further attack. Ba + H2SO4 -> BaSO4 + H2 The same effect will happen to a lesser extent with metals going up the group as the solubility increases. The same effect does not happen with other acids like hydrochloric or nitric as they form soluble group 2 salts.

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26
Q

The action of water on oxides of elements in Group 2

A

The group 2 oxides react with water to form hydroxides of varying solubility
CaO (s) + H2O (l) -> Ca(OH)2(aq) pH 12

Group 2 oxides are basic as the oxide ions accept H+ ions to become hydroxide ions in these reactions

MgO (s) + H2O (l) -> Mg(OH)2(s) pH 9Mg(OH)2
is only slightly soluble in water so fewer free OHions are produced and so lower pH

Calcium hydroxide is reasonably soluble in water. It is used in agriculture to neutralise acidic soils. If too much calcium hydroxide is added to the soil, excess will result in soils becoming too alkaline to sustain crop

Magnesium hydroxide is classed as partially soluble in water.
A suspension of magnesium hydroxide in water will appear slightly alkaline (pH 9) so some hydroxide ions must therefore have been produced by a very slight dissolving. Magnesium hydroxide is used in medicine (in suspension as milk of magnesia) to neutralise excess acid in the stomach and to treat constipation.
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
It is safe to use as it so weakly alkaline.

An aqueous solution of calcium hydroxide is called lime water and can be used as a test for carbon dioxide. The limewater turns cloudy as white calcium carbonate is produced.
Ca(OH)2 (aq) + CO2 (g) -> CaCO3 (s) + H2O(l)

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27
Q

What are Halogens?

A

Group 7 elements, all of which exist as diatomic molecules:
Fluorine (F2): very pale yellow gas. It is highly reactive
Chlorine (Cl2): greenish, reactive gas, poisonous in high concentrations
Bromine (Br2): red liquid, that gives off dense brown/orange poisonous fumes
Iodine (I2): shiny grey solid sublimes to purple gas.

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28
Q

Trend in melting point and boiling point

A

Increase down the group
As the molecules become larger they have more electrons and so have larger induced dipole-dipole forces (London forces) between the molecules. As the intermolecular forces get larger more energy has to be put into breaking these intermolecular forces. This increases the melting and boiling points

29
Q

Electronic configuration.

A

All group 7 elements have the outer shell s2p5 electron configuration. They will often react by gaining one electron in redox reactions to form 1– ions

30
Q

The displacement reactions of halide ions by halogens.:

A

A more reactive halogen will displace a halogen that has a lower reactivity from one of its compounds
The reactivity of the halogens decreases down the group as the atoms get bigger with more shielding so they less easily attract and accept electrons. They therefore form -1 ions less easily down the group
Chlorine will displace both bromide and iodide ions; bromine will displace iodide ions

potassium chloride (aq):
-plus chlorine: Very pale green solution, no reaction
-plus bromine: Yellow solution, no reaction
-plus iodine: Brown solution, no reaction

potassium bromide (aq):
-plus chlorine: Yellow solution, Cl has displaced Br
-plus bromine Yellow solution, no reaction
-plus iodine: Brown solution, no reaction

potassium iodide (aq):
-plus chlorine: Brown solution, Cl has displaced I
-plus bromine: Brown Solution, Br has displaced I
-plus iodine: Brown Solution, no reaction

The colour of the solution in the test tube shows which free halogen is present in the solution. Chlorine =very pale green solution (often colourless), Bromine = yellow solution Iodine = brown solution (sometimes black solid present)

31
Q

Observations if an organic solvent is added:

A

potassium chloride (aq):
-plus chlorine: colourless, no reaction
-plus bromine: yellow, no reaction
-plus iodine: purple, no reaction

potassium bromide (aq):
-plus chlorine: yellow, Cl has displaced Br
-plus bromine: yellow, no reaction
-plus iodine: purple, no reaction

potassium iodide (aq):
-plus chlorine: purple, Cl has displaced I
-plus bromine: purple, Br has displaced I
-plus iodine: purple, no reaction

The colour of the organic solvent layer in the test tube shows which free halogen is present in the solution.
Chlorine = colourless
Bromine = yellow

32
Q

Explanation of reactivity (Halogen):

A

Chlorine is more reactive than bromine because it will gain an electron and form a negative ion more easily than bromine. This is because an atom of chlorine is smaller than bromine and the outermost shell of chlorine is less
shielded than bromine so the electron to be gained is attracted more strongly to the nucleus in chlorine than bromine.

Cl2(aq) + 2Br – (aq) -> 2Cl – (aq) + Br2(aq)
Cl2(aq) + 2I – (aq) -> 2Cl – (aq) + I2(aq)
Br2(aq) + 2I – (aq) -> 2Br – (aq) + I2(aq)

33
Q

The disproportionation reactions of chlorine:

A

Disproportionation is the name for a reaction where an element simultaneously oxidises and reduces.

Chlorine with water:

Cl2(g) + H2O(l) -> HClO(aq) + HCl (aq)
Chlorine is both simultaneously reducing and oxidising. It changes from 0 in Cl2 to -1 in HCl and +1 in HClO
If some universal indicator is added to the solution it will first turn red due to the acidity of both reaction products. It will then turn colourless as the HClO bleaches the colour.

Chlorine is used in water treatment to kill bacteria. It has been used to treat drinking water and the water in
swimming pools. The benefits to the health of water treatment by chlorine its killing of bacteria outweigh its
risks of toxic effects and possible risks from the formation of chlorinated hydrocarbons. Reaction of chlorine with cold dilute NaOH solution:
Cl2, (and Br2, I2) in aqueous solutions will react with cold sodium hydroxide. The chlorine is reacting by disproportionation. The colour of the halogen solution will fade to colourless

Cl2(aq) + 2NaOH(aq) -> NaCl (aq) + NaClO (aq) + H2O(l)

The mixture of NaCl and NaClO (sodium chlorate (I)) is used as Bleach and to disinfect/ kill bacteria

If the hot sodium hydroxide is used a different disproportionation reaction occurs forming sodium chlorate(V)
3Cl2 + 6NaOH → NaClO3 + 5NaCl + 3H2O

34
Q

The reactions of halide ions with silver nitrate:

A

This reaction is used as a test to identify which halide ion is present. The test solution is made acidic with nitric
acid, and then Silver nitrate solution is added dropwise.

The role of nitric acid is to react with any carbonates present to prevent the formation of the precipitate Ag2CO3. This would mask the desired observations

2HNO3 + Na2CO3 -> 2 NaNO3 + H2O + CO2 Fluorides produce no precipitate
Chlorides produce a white precipitate:
Ag+(aq) + Cl- (aq) -> AgCl(s)
Bromides produce a cream precipitate:
Ag+(aq) + Br- (aq) -> AgBr(s)
Iodides produce a pale yellow precipitate:
Ag+(aq) + I- (aq) -> AgI(s)

The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar:
Silver chloride dissolves in dilute ammonia to form a complex ion
AgCl(s) + 2NH3(aq) -> [Ag(NH3)2] + (aq) + Cl- (aq)
Colourless solution

Silver bromide dissolves in concentrated ammonia to form a complex ion
AgBr(s) + 2NH3(aq) -> [Ag(NH3)2]+(aq) + Br - (aq)
Colourless solution

Silver iodide does not react with ammonia – it is too insoluble

35
Q

Testing for the Presence of a carbonate:

A

Add any dilute acid and observe effervescence. Bubble gas through limewater to test for CO2 – will turn limewater cloudy 2HCl + Na2CO3 -> 2NaCl + H2O + CO2

Fizzing due to CO2 would be observed if a carbonate was present

36
Q

Testing for the presence of a sulfate:

A

Acidified BaCl2 solution is used as a reagent to test for sulfate ions. If barium chloride is added to a solution that contains sulfate ions a white precipitate forms.
Ba2+ (aq) + SO42-(aq) -> BaSO4(s).
Other anions should give a negative result which is no precipitate forming.

The acid is needed to react with carbonate impurities that are often found in salts which would form a white barium carbonate precipitate and so give a false result.

Sulfuric acid cannot be used to acidify the mixture because it contains sulfate ions which would form a precipitate.

37
Q

Testing for halide ions with silver nitrate:

A

This reaction is used as a test to identify which halide ion is present. The test solution is made acidic with nitric
acid, and then silver nitrate solution is added dropwise.

The role of nitric acid is to react with any carbonates present to prevent the formation of the precipitate Ag2CO3. This would mask the desired observations.
2 HNO3 + Na2CO3 -> 2 NaNO3 + H2O + CO2

Fluorides produce no precipitate

Chlorides produce a white precipitate
Ag+(aq) + Cl- (aq) -> AgCl(s)

Bromides produce a cream precipitate
Ag+(aq) + Br- (aq) -> AgBr(s)

Iodides produce a pale yellow precipitate
Ag+(aq) + I- (aq) -> AgI(s)

The silver halide precipitates can be treated with ammonia solution to help differentiate between them if the colours look similar:

Silver chloride dissolves in dilute ammonia to form a complex ion
AgCl(s) + 2NH3(aq) -> [Ag(NH3)2]+(aq) + Cl- (aq)
Colourless solution

Silver bromide dissolves in concentrated ammonia to form a complex ion
AgBr(s) + 2NH3(aq) -> [Ag(NH3)2]+(aq) + Br - (aq)
Colourless solution

Silver iodide does not react with ammonia – it is too insoluble.

Hydrochloric acid cannot be used to acidify the mixture because it contains chloride ions which would form a precipitate

The sequence of tests required is carbonate, sulfate then halide. (This will prevent false results of as both BaCO3 and Ag2SO4 are insoluble.)

38
Q

Testing for positive ions (cations):

A

Test for ammonium ion NH4+, by the reaction with warm NaOH(aq), forming NH3 gas Ammonia gas can be identified by its pungent smell or by turning red litmus paper blue

39
Q

What is an enthalpy change?

A

If an enthalpy change occurs then energy is transferred between the system and surroundings. The system is the chemicals and the surroundings is everything outside the chemicals.

40
Q

What is exothermic and endothermic?

A

In an exothermic change energy is transferred from the system (chemicals) to the surroundings. The products have less energy than the reactants. In an exothermic reaction, the ∆H is negative
Common oxidation exothermic processes are the combustion of fuels and the oxidation of carbohydrates such as glucose in respiration

In an endothermic change, energy is transferred from the surroundings to the system (chemicals). They require an input of heat energy e.g. thermal decomposition of calcium carbonate. The products have more energy than the
reactants. In an endothermic reaction, the ∆H is positive

41
Q

What is activation energy (EA)

A

The Activation Energy is defined as the minimum energy which particles need to collide to start a reaction

42
Q

What are standard conditions?

A

Enthalpy changes are normally quoted at standard conditions.
Standard conditions are :
* 100 kPa pressure
* 298 K (room temperature or 25oC)
* Solutions at 1mol dm-3
* all substances should have their normal state at 298K

43
Q

Define enthalpy change of reaction (^rH)

A

the enthalpy change when the number of moles of reactants as specified in the balanced equation reacts together

44
Q

Define Standard enthalpy change of formation

A

The standard enthalpy change of formation of a compound is the enthalpy change when 1 mole of the compound is formed from its elements under standard conditions (298K and 100kpa), all reactants and products being in their standard states

Mg (s) + Cl2(g) -> MgCl2(s)
2Fe (s) + 1.5 O2(g) -> Fe2O3(s)

The enthalpy of formation
of an element = 0 kJ mol-1

45
Q

Define standard enthalpy change of combustion

A

The standard enthalpy of combustion of a substance is defined as the enthalpy change that occurs when one mole of a substance is combusted completely in oxygen under standard conditions. (298K and 100kPa), all reactants and products being in their standard states
Symbol ^cH

CH4 (g) + 2O2(g)-> CO2(g) + 2 H2O (l)
Incomplete combustion will lead to soot (carbon), carbon monoxide and water. It will be less exothermic than complete combustion.

46
Q

Define enthalpy change of neutralisation

A

The standard enthalpy change of neutralisation is the enthalpy change when solutions of anacid and an alkali react together under standard conditions to produce 1 mole of water. Symbol ^neutH

47
Q

What is the equation for energy change?

A

energy change = mass of solution x heat capacity x temperature change
Q (J) = m (g) x cp(J g-1K-1) x ^T ( K)

This equation will only give the energy for the actual quantities used. Normally this value is converted into the energy change per mole of one of the reactants. (The enthalpy change of reaction, ^rH)

48
Q

What is the calorimetric method?

A

One type of experiment is one in which substances are mixed in an insulated container and the temperature rise is measured.
This could be a solid dissolving or reacting in a solution or it could be two solutions reacting together

General method:
 washes the equipment (cups and pipettes etc) with the solutions to be used
 dry the cup after washing
 put a polystyrene cup in a beaker for insulation and support  Measure out desired volumes of solutions with
volumetric pipettes and transfer them to an insulated cup
 clamp the thermometer into place making sure the thermometer bulb is immersed in the solution
 measure the initial temperatures of the solution or both solutions if 2 are used. Do this every minute for 2-3
minutes
 At minute 3 transfer the second reagent to the cup. If a solid reagent is used then add the solution to the cup
first and then add the solid weighed out on balance.
 If using a solid reagent then use ‘before and after’ weighing
method
 stirs mixture (ensures that all of the solution is at the same temperature)  Record temperature every minute
after addition for several minutes

If the reaction is slow, the exact temperature rise can be difficult to obtain as cooling occurs simultaneously. To counteract this we take readings at regular time intervals
and extrapolate the temperature curve/line back to the time the reactants were added together.
We also take the temperature of the reactants for a few minutes before they are added together to get a better average temperature. If the two reactants are solutions then the temperature of both solutions needs to be measured before addition and an average temperature is used.

Errors in this method
* energy transfer from surroundings (usually loss)
* approximation in specific heat capacity of solution. The method assumes all solutions have the heat capacity of
water.
* neglecting the specific heat capacity of the calorimeter- we ignore any energy absorbed by the apparatus.
* reaction or dissolving may be incomplete or slow.
* The density of the solution is taken to be the same as water.

Calorimetric method Practical Read the question carefully. It may be necessary to describe:
* Method
* Drawing of graph
with extrapolation
* Description of the
calculation

49
Q

Calculating the enthalpy change of reaction, ^rH from experimental data

A

General method
1. Using q= m x cp x T calculate energy change for quantities used
2. Work out the moles of the reactants used
3. Divide q by the number of moles of the reactant, not in excess to give H
4. Add a sign and unit (divide by a thousand to convert Jmol-1 to kJmol-1

The heat capacity of water is 4.18 J g-1K-1. In any reaction where the reactants are dissolved in water we assume that the heat capacity is the same as pure water. Also, assume that the solutions have the density of water, which is 1g cm-3. Eg 25cm3 will weigh 25 g

Example 1. Calculate the enthalpy change of reaction for the reaction where 25.0cm3 of 0.200 mol dm-3
copper sulfate was reacted with 0.01mol (excess of zinc). The temperature increased to 7oC.

Step 1: Calculate the energy change for the amount of reactants in the calorimeter.
Q = m x cp x T
(Note the mass is the mass of the copper sulfate solution only. Do not include the mass of zinc powder.)
Q = 25 x 4.18 x 7
Q = 731.5 J
Step 2: calculate the number of moles of the reactant not in excess.
moles of CuSO4 = conc x vol
= 0.2 x 25/1000
= 0.005 mol
If you are not told what is in excess, then you need to work out the moles of both reactants and work out using the balanced equation which one is in excess.
Step 3: Calculate the enthalpy change per mole
^H = Q/ no of moles
= 731.5/0.005
= 146300 J mol-1 = 146 kJ mol-1 to 3 sf
Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign e.g. –146 kJ mol-1
Remember in these questions: sign, unit, 3 sig figs.

Example 2. 25.0cm3 of 2.00 mol dm-3 HCl was neutralised by 25.0cm3 of 2.00 mol dm-3 NaOH. The temperature
increased by 13.5oC. Calculate the enthalpy change per mole of HCl.

Step 1: Calculate the energy change for the amount of reactants in the calorimeter.
Q = m x cp x T
(Note the mass equals the mass of acid + the mass of alkali, as they are both solutions.)
Q = 50 x 4.18 x13.5
Q = 2821.5 J
Step 2: calculate the number of moles of the HCl.
moles of HCl = conc x vol
= 2 x 25/1000
= 0. 05 mol
Step 3: calculate ^H the enthalpy change per mole which can be called the enthalpy change of neutralisation
^H = Q/ no of moles
= 2821.5/0.05 = 56430 J mol-1 = -56.4 kJ mol-1 to 3 sf , Exothermic and so is given a minus sign
Remember in these questions: sign, unit, 3 sig figs.

50
Q

Measuring Enthalpies of Combustion using Calorimetry:

A

Enthalpies of combustion can be calculated by using calorimetry. Generally, the fuel is burnt and the flame is used to heat water in a metal cup.

Example 3. Calculate the enthalpy change of combustion for the reaction where 0.650g of propan-1-ol was completely combusted and used to heat 150g of water from 20.1 to 45.5oC
Step 1: Calculate the energy change used to heat the water.
Q = m x cp x T
(Note the mass is the mass of water in the calorimeter and not the alcohol)
Q = 150 x 4.18 x 25.4
Q = 15925.8 J
Step 2: calculate the number of moles of alcohol combusted.
moles of propan-1-ol = mass/ Mr = 0.65 / 60 = 0.01083 mol
Step 3: Calculate the enthalpy change per mole which is called cH (the enthalpy change of combustion)
^H = Q/ no of moles
= 15925.8/0.01083
= 1470073 J mol-1 = 1470 kJ mol-1 to 3 sf
Finally add in the sign to represent the energy change: if temp increases the reaction is exothermic and is given a minus sign eg –1470 kJ mol-1
Remember in these questions: sign, unit, 3 sig figs.

Errors with this method:
* Energy losses from calorimeter * Incomplete combustion of fuel * Incomplete transfer of energy
* Evaporation of fuel after weighing
* Heat capacity of calorimeter not included
* Measurements not carried out under standard conditions as H2O is gas, not liquid, in this experiment

51
Q

Mean Bond energies:

A

Definition: The mean bond enthalpy is the enthalpy change when one mole of bonds of (gaseous covalent) bonds is broken (averaged over different molecules) These values are positive because energy is required to break a bond.

The definition only applies when the substances start and end in the gaseous state.

We use values of mean bond energies because every single bond in a compound has a slightly different bond energy. E.g. In CH4 there are 4 C- H bonds. Breaking each one will require a different amount of energy. However, we use an average value for the C-H bond for all hydrocarbons.

52
Q

Calculating enthalpy changes:

A

In general (if all substances are gases) ^H = Σ bond energies broken - Σ bond energies made

H values calculated using this method will be less accurate than using formation or combustion data because the mean bond energies are not exact

53
Q

What is Hess’ Law?

A

Hess’s law states that total enthalpy change for a reaction is independent of the route by which the chemical change takes place
Hess’s law is a version of the first law of thermodynamics, which is that energy is always conserved.

On an energy level diagram, the directions of the arrows can show the different routes a reaction can proceed by.

Often Hess’s law cycles are used to measure the enthalpy change for a reaction that cannot be measured
directly by experiments. Instead, alternative reactions are carried out that can be measured experimentally.

54
Q

Using Hess’s law to determine enthalpy changes from enthalpy changes of formation:

A

^H reaction = Σ ^fH products - Σ ^fH reactants

55
Q

Using Hess’s law to determine enthalpy changes from enthalpy changes of combustion:

A

^H reaction = Σ ^cH reactants - Σ ^cH products

56
Q

What is Collision Theory?

A

Reactions can only occur when collisions take place between particles having sufficient energy. The energy is
usually needed to break the relevant bonds in one or either of the reactant molecules. This minimum energy is called the Activation Energy

57
Q

Define Activation Energy:

A

The Activation Energy is defined as the minimum energy which particles need to collide to start a reaction

58
Q

Effect of Increasing Concentration and Increasing Pressure:

A

At higher concentrations(and pressures) there are more particles per unit volume and so the particles collide
with a greater frequency and there will be a higher frequency of effective collisions.

59
Q

Measuring Reaction Rates:

A

The rate of reaction is defined as the change in concentration of a substance in unit time. Its usual unit is mol dm-3s-1
When a graph of the concentration of reactant is plotted vs time, the gradient of the curve is the rate of reaction. The initial rate is the rate at the start of the reaction where it is fastest.
Reaction rates can be calculated from graphs of concentration of reactants or products, by drawing a tangent to the curve (at different times) and calculating the gradient of the tangent.

In the experiment between sodium thiosulfate and hydrochloric acid, we usually measure the reaction rate as
1/time where the time is the time taken for a cross placed underneath the reaction mixture to disappear due
to the cloudiness of the sulfur.
Na2S2O3 + 2HCl -> 2NaCl + SO2 + S + H2O
This is an approximation for the rate of reaction as it does not include concentration. We can use this because
we can assume the amount of sulfur produced is fixed and constant.

60
Q

Effect of Catalyst:

A

Definition: Catalysts increase reaction rates without getting used up.

Explanation: They do this by providing an alternative route or mechanism with lower activation energy so more molecules have energy above activation energy.

Comparison of the activation energies for an uncatalysed reaction and the same reaction with a catalyst present.

A heterogeneous catalyst is in a different phase from the reactants
A homogeneous catalyst is in the same phase as the reactants

Heterogeneous catalysis
Heterogeneous catalysts are usually solids whereas the reactants are gaseous or in solution. The reaction occurs at the surface of the catalyst.
Homogeneous catalysis
When catalysts and reactants are in the same phase, the reaction proceeds through an intermediate species.

Benefits of Catalysts:
-Catalysts speed up the rate of reaction. This means that the use of a catalyst may mean lower temperatures and pressures can be used. This can save energy costs as there is reduced energy demand for providing high temperatures and less electrical pumping costs for producing pressure. This can mean fewer CO2 emissions from the burning of fossil fuels.
-Catalysts can enable different reactions to be used, with better atom economy and with reduced waste, fewer undesired products or less use of hazardous solvents and reactants.
-Catalysts are often enzymes, generating very specific products, and operating effectively close to room temperatures and pressures.

61
Q

Effect of Increasing Temperature

A

At higher temperatures, the energy of the particles increases. They collide more frequently and often with energy greater than the activation energy.
More collisions result in a reaction As the temperature increases, the graph shows that a significantly bigger proportion of particles have energy greater than the activation energy, so the frequency of successful collisions
increases

62
Q

Effect of Increasing Surface Area

A

Increasing surface area will cause collisions to occur more frequently between the reactant particles and this increases the rate of the reaction.

63
Q

What is equilibrium?

A

All reversible reactions reach a dynamic equilibrium state.
Many reactions are reversible
N2 + 3H2 <=>2NH3
Dynamic equilibrium occurs when forward and backward reactions occur at equal rates. The concentrations of reactants and products stay constant and the reaction is continuous.
We use the expression ‘position of equilibrium’ to describe the composition of the equilibrium mixture.
If the position of equilibrium favours the reactants (also described as “towards the left”) then the equilibrium mixture will contain mostly reactants.

64
Q

Le Chatelliers Principle

A

We use Le Chatelier’s principle to work out how changing external conditions such as temperature and pressure affect the position of the equilibrium
Le Chatelier’s principle states that if an external condition is changed the equilibrium will shift to oppose the change (and try to reverse it).

65
Q

Effect of temperature on equilibrium

A

If the temperature is increased the equilibrium will shift to oppose this and move in the endothermic direction to try to reduce the temperature by absorbing heat.
And it’s reverse

Typical exam question: What effect would increasing temperature have on the yield of ammonia?
N2 + 3H2 <=> 2NH3 ^H = -ve exo
If the temperature is increased the equilibrium will shift to oppose this and move in the endothermic, backwards direction to try to decrease temperature. The position of equilibrium will shift towards the left, giving a lower yield of ammonia.

If the temperature is decreased the equilibrium will shift to oppose this and move to the exothermic direction to try to increase the temperature by giving out heat.

Low temperatures may give a higher yield of product but will also result in slow rates of reaction. Often a compromise temperature is used that gives a reasonable yield and rate

66
Q

Effect of pressure on equilibrium

A

Increasing pressure will cause the equilibrium to shift
towards the side with fewer moles of gas to oppose
the change and thereby reduce the pressure.
And it’s reverse

Typical exam question: What effect would increasing pressure have on the yield of methanol?
CO (g) + 2H2(g) <=> CH3OH (g)
If pressure is increased the equilibrium will shift to oppose this and move towards the side with fewer moles of gas to try to reduce the pressure. The position of equilibrium will shift towards the right because there are 3 moles of gas on the left but only 1 mole of gas on the right, giving a higher yield of methanol.

Decreasing pressure will cause the equilibrium to shift towards the side with more moles of gas to oppose the change and thereby increase the pressure.

If the number of moles of gas is the same on both sides of the equation then changing pressure will have no effect on the position of the equilibrium
H2 + Cl2 <=> 2HCl

Increasing pressure may give a higher yield of product and will produce a faster rate. Industrially high pressures are expensive to produce (high electrical energy costs for pumping the gases to make a high pressure) and the equipment is expensive (to contain the high pressures)

67
Q

Effect of catalysts on equilibrium

A

A catalyst does not affect the position of equilibrium, but it will speed up the rate at which the equilibrium is achieved.
It does not affect the position of equilibrium because it speeds up the rates of the forward and backward reactions by the same amount.

68
Q

Importance of equilibrium to industrial processes

A

Common examples:
Haber process:
N2 + 3H2 <=> 2NH3 ^H = -ve exo
T= 450oC, P= 200 – 1000 atm, catalyst = iron
Low temp gives good yield but slow rate:
compromise temp used
High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure

Contact process
Stage 1 S (s) + O2 (g) -> SO2 (g)
Stage 2 SO2 (g) + ½ O2 (g) <=> SO3 (g) ^H = -98 kJ mol-1
T= 450oC, P= 1 to 2 atm, catalyst = V2O5
Low temp gives good yield but slow rate: compromise moderate temp used
High pressure gives a slightly better yield and high rate: too high pressure would lead to too high energy costs for pumps to produce the pressure

Hydration of ethene to produce ethanol
CH2=CH2 (g) + H2O (g) <=> CH3CH2OH(l) ^H = -ve
T= 300oC, P= 70 atm, catalyst = conc H3PO4
Low temp gives good yield but slow rate: compromise temp used.
High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure. High pressure also leads to unwanted polymerisation of ethene to poly(ethene).

Production of methanol from CO
CO (g) + 2H2(g) <=> CH3OH (g) ^H = -ve exo
T= 400oC, P= 50 atm, catalyst = chromium and zinc oxides
Low temp gives good yield but slow rate: compromise temp used
High pressure gives a good yield and high rate: too high pressure would lead to too much energy costs for pumps to produce the pressure.

In all cases, high pressure leads to too high energy costs for pumps to produce the pressure and too high equipment costs to have equipment that can withstand high pressures.
In all cases catalysts speed up the rate, allowing lower temperatures to be used (and hence lower energy costs) but have no effect on equilibrium.
Recycling unreacted reactants back into the reactor can improve the overall yields of all these processes.