5.2 rates of reactions- enthalpy

Question 1

lattice enthalpy

Answer 1

a measure of the strength of ionic bonding in a giant ionic lattice
the enthalpy change that accompanies the formation of one mole of an ionic compound from its gaseous ions under standard conditions

Question 2

born-haber cycle

Answer 2

lattice enthalpy must be calculated indirectly using known energy changes in an energy cycle
2 routes

Question 3

born-haber cycle route 1

Answer 3

formation of gaseous atoms:
-changing the elements in their standard states into gaseous atoms
-this change is endothermic as it involves bond breaking
formation of gaseous ions:
-changing the gaseous atoms into positive and negative gaseous ions
-overall this change is endothermic
lattice formation:
-changing the gaseous ions into the solid ionic lattice
-this is the lattice enthalpy and is exothermic

Question 4

born-haber cycle route 2

Answer 4

converts the elements in their standard states directly to the ionic lattice
there is just one enthalpy change, the enthalpy change of formation and this is exothermic

Question 5

standard enthalpy change of formation

Answer 5

the enthalpy change that takes place when one mole of a compound is formed from its elements under standard conditions, with all reactants and products in their standard states
Na(g) + 1/2Cl2(g) –> NaCl(s)

Question 6

standard enthalpy change of atomisation

Answer 6

the enthalpy change that takes place for the formation of one mole of gaseous atoms from the element in its standard state under standard conditions
Na(s) –> Na(g)
1/2Cl2(g) –> Cl(g)
always an endothermic process as bonds are being broken to form gaseous atoms

Question 7

first ionisation energy

Answer 7

the enthalpy change required to remove one electron from each atom in one mole of gaseous atoms to form one mole of gaseous 1+ ions
endothermic as energy is required to overcome the attraction between a negative electron and the positive nucleus

Question 8

what is the opposite of ionisation energy

Answer 8

electron affinity

-measures the energy to gain elecs

Question 9

first electron affinity

Answer 9

enthalpy change that takes place when one elec is added to each atom in one mole of gaseous atoms to form one mole of gaseous 1- ions
exothermic as the elec being added is attracted towards the nucleus

Question 10

second electron affinities

Answer 10

endothermic
a second elec is being gained by a neg ion, which repels the elec away
so energy must be put in to force the neg charged elec onto the neg ion

Question 11

standard enthalpy change of solution

Answer 11

the enthalpy change that takes place when one mole of a solute dissolves in a solvent
if the solvent is water, the ions from the ionic lattice finish up surrounded with water molecules as aqueous ions
Na+Cl-(s) + aq –> Na+(aq) + Cl-(aq)

Question 12

dissolving process

Answer 12

when a solid ionic compound dissolves in water, two processes take place:

• the ionic lattice breaks up
• water molecules are attracted to, and surround, the ions

Question 13

what types of energy change are involved in the dissolving process

Answer 13

1- the ionic lattice is broken up forming separate gaseous ions. this is the opposite energy change from lattice energy, which forms the ionic lattice from gaseous ions
2- the separate gaseous ions interact with polar water molecules to form hydrated aqueous ions. the energy change involved is called the enthalpy change of hydration

Question 14

enthalpy change of hydration

Answer 14

enthalpy change that accompanies the dissolving of gaseous ions in water to form one mole of aqueous
ions.

Question 15

factors effecting lattice enthalpy

Answer 15

ionic size

ionic charge

Question 16

effect of ionic size on lattice enthalpy

Answer 16

as ionic size increases:

• ionic radius inc
• attraction between ions dec
• lattice energy less negative
• MP dec

Question 17

effect of ionic charge on lattice enthalpy

Answer 17

as ionic charge inc:

• attraction between ions inc
• lattice energy becomes more neg
• MP inc

Question 18

factors effecting hydration

Answer 18

ionic size

ionic charge

Question 19

effect of ionic size on hydration enthalpy

Answer 19

as ionic size inc:

• ionic radius inc
• attraction between ion and water molecules dec
• hydration energy less neg

Question 20

effect of ionic charge on hydration enthalpy

Answer 20

as ionic charge inc:

• attraction with water molecules dec
• hydration energy becomes more neg

Question 21

entropy

Answer 21

dispersal of energy in a system which is greater,
the more disordered a system
units are JK-1 mol-1

Question 22

entropies of diff states

Answer 22

solid- smallest entropies

gases- greatest

Question 23

predicting entropy changes

Answer 23

• if a system changes to become more random, energy can be spread out more- there will be an entropy change which will be positive
• if a system changes to be less random, energy becomes more concentrated- the entropy change will be negative

Question 24

changes of state - entropy changes

Answer 24

entropy incs during changes in state that give a more random arrangement of particles
solid -> liquid -> gas

Question 25

when any substance changes state from solid to liquid to gas..

Answer 25

MP and BP incs the randomness of particles | energy is spread out more and change in entropy is positive

Question 26

change in number of gaseous molecules - entropy changes

Answer 26

reactions that produce gases result in an inc in entropy - production of a gas incs the disorder of particles - energy is spread out more and change in entropy is positive

Question 27

standard entropies

Answer 27

the entropy of one mole of a substance, under standard conditions (100kPa and 298K) - units of JK-1 mol-1 - always positive

Question 28

feasibility

Answer 28

used to describe whether a reaction is able to happen and is energetically feasibly

Question 29

free energy

Answer 29

the overall change in energy during a chemical reaction is called the free energy change is made up of two types of energy: -enthalpy change ΔH. this is the heat transfer between the chemical system and the surroundings -entropy change at the temp of the reaction TΔS. this is the dispersal of energy within the chemical system itself

Question 30

gibb's equation

Answer 30

relationship between the two types of energy, ΔH and TΔS. ΔG = ΔH - TΔS free energy change= enthalpy change with surroundings - temperature x entropy change of system

Question 31

condition for feasibility

Answer 31

feasibility of a reaction depends on the balance between ΔH and TΔS in the gibb's equation for a reaction to be feasible, there must be a decrease in free energy: ΔG < 0

Question 32

limitations of predictions made for feasibility

Answer 32

many reactions have a negative ΔG and do not seem to take place although it indicates the thermodynamic feasibility, it takes no account of the kinetics or rate of a reaction

Question 33

oxidising agent

Answer 33

takes elecs from the species being oxidised | contains the species that is reduced

Question 34

reducing agent

Answer 34

adds elecs to the species being reduced | contains the species that is oxidised

Question 35

construction of redox equations using half equations

Answer 35

write out two half equations reduction and oxidation balance electrons in each equation add and cancel electrons cancel any species that are on both sides of the equation

Question 36

construction of redox equations using oxidation numbers

Answer 36

assign oxidation numbers to identify the atoms that change the ox number balance only the species that contain the elements that have changed ox number e.g. to match the total increase of +6 for sulfur, you need a total decrease of -6 from nitrogen balance any remaining atoms

Question 37

predicting products of redox reactions

Answer 37

in equations you might not know all the species involved in the reaction and you might need to predict any missing reactants or products in aqueous redox reactions, H2O is often formed other likely products are H+ and OH- ions, depending on the conditions used ensure both sides of the equations are balanced by charge

Question 38

manganate (VII) titration procedure

Answer 38

1. a standard solution of potassium manganate (VII), KMnO4, is added to the burette 2. using a pipette, add a measured volume of the solution being analysed to the conical flask. an excess of dilute sulfuric acid is also added to provide the H+ ions required for the reduction of MnO4- ions 3. during the titration the manganate solution reacts and is decolourised as it is being added. the end point of the titration is judged by the first permanent pink colour, indicating when there is an excess of MnO4- ions present. 4. repeat the titration until you obtain concordant titres

Question 39

reading the meniscus - manganate (VII) titration

Answer 39

KMnO4 is a deep purple colour and is very difficult to see the bottom of the meniscus through it in manganate (VII) titrations, burette readings are read from the top rather than the bottom of the meniscus titre is the same

Question 40

examples of manganate (VII) titrations

Answer 40

used for the analysis of many diff reducing agents, e.g: - iron (II) ions, Fe2+ - ethanedioic acid, (COOH)2

Question 41

iodine/thiosulfate titration procedure

Answer 41

1. add a standard solution of Na2S2O3 to the burette. 2. prepare a solution of the oxidising agent to be analysed. using a pipette, add this solution to a conical flask. then add an excess of potassium iodide. the oxidising agent reacts with iodine ions to produce iodine, which turns the solution a yellow-brown colour 3. titrate this solution with the Na2S2O3. during te titration, the iodine is reduced back to I- ions and the brown colour fades gradually, making it difficult to decide on a end point. this problem is solved by using starch indicator. when the end point is being approached, the iodine colour has faded enough to become a pale straw colour.

Question 42

using starch for the end point

Answer 42

iodine/thiosulfate titration when the end point is being approached and the iodine has faded enough to become a pale straw colour, a small amount of starch indicator is added. a deep blue-black colour forms to assist with the identification of the end point. as more sodium thiosulfate is added, the blue-black colour fades at the end point, all the iodine will have just reacted and the blue-black colour disappears

Question 43

examples of iodine-thiosulfate titrations

Answer 43

for the analysis of different oxidising agents e.g: - chlorate (I) ions, ClO- - copper (II) ions, Cu2+

Question 44

Analysis of copper

Answer 44

Iodine/thiosulfate titrations can be used to determine the copper content of copper (II) salts or alloys. For copper (II) salts, Cu2+ ions are produced simply by dissolving the compound in water Insoluble copper (II) compounds can be reacted with acids to form Cu2+ ions Cu2+ ions react with I- to form a solution of iodine I2 and a white precipitate of copper (I) iodide CuI. The mixture appears a brown colour The iodine in the brown mixture is then titrated with a standard solution of sodium thiosulfate

Question 45

Thiosulfate

Answer 45

S2O3^2-

Question 46

Half cell

Answer 46

Contains the chemical species present in a redox half-equation A voltaic cell can be made by connecting together two different half cells, which then allows electrons to flow In the cell, the chemicals in the two half cells must be kept apart

Question 47

What happens if chemicals in two half cells mixed?

Answer 47

Electrons would flow in an uncontrolled way and heat energy would be released rather than electrical energy

Question 48

Voltaic cell

Answer 48

Electrochemical cell | Converts chemical energy into electrical energy

Question 49

Metal/metal ion half-cells

Answer 49

A half cell that consists of a metal rod dipped into a solution of its aqueous metal ion At the phase boundary where the metal is in contact with its ions, an equilibrium will be set up In an isolated half cell, there is no net transfer of electrons either into or out of the metal When two half cells are connected, the direction of electron flow depends on the relative tendency of each electrode to release electrons

Question 50

Ion/ion half cells

Answer 50

Contains ions of the same element in different oxidation states There is no metal to transport electrons either into or out of the half cell, so an inert metal electrode made out of platinum is used

Question 51

Equilibrium in a half cell

Answer 51

Written so that the forward reaction shows reduction and the reverse shows oxidation

Question 52

Negative electrode

Answer 52

The electrode with the more negative metal loses electrons and is oxidised

Question 53

Positive electrode

Answer 53

The electrode with the less reactive metal gains electrons and is reduced

Question 54

Standard electrode potential

Answer 54

Measures the tendency to be reduced and gain electrons The standard chosen is a half cell containing hydrogen gas and a solution containing H+ ions An inert platinum electrode is used to allow electrons into and out of the half cell The Standard electrode potential is the emf of a half cell connected to a standard hydrogen half cell under standard conditions The Standard electrode potential of a standard hydrogen electrode is exactly 0V

Question 55

Standard electrode potential standard conditions

Answer 55

Solutions have a conc of 1 mol dm-3 298K 100kPa

Question 56

Measuring a Standard electrode potential

Answer 56

The half cell is connected to a standard hydrogen electrode - the two electrodes are connected by a wire to allow a controller flow of electrons - the two solutions are connected with a salt bridge which allow ions to flow. The salt bridge typically contains a concentrated solution of an electrolyte that does not react with either solution. E.g. strip of filter paper soaked in aqueous potassium nitrate KNO3 The more negative the E^o value the greater the reactivity of a metal in losing electrons The more positive the E^o value the greater the reactivity of a non metal in gaining electrons

Question 57

Cell potentials

Answer 57

Cells can easily be assembled using any half cells | The emf measured is then a cell potential, Ecell

Question 58

Measuring standard cell potentials

Answer 58

1. Prepare two standard half cells 2. Connect the metal electrodes of the half cells to a voltmeter using wires 3. Prepare a salt bridge by soaking a strip of filter paper in a saturated aqueous solution of potassium nitrate 4. Connect the two solutions of the half cells with a salt bridge 5. Record the standard cell potential from the voltmeter

Question 59

Calculation of a standard cell potential from standard electrode potentials

Answer 59

E^o cell= E^o (positive electrode) - E^o (negative electrode)

Question 60

Predicting redox reactions

Answer 60

Predictions can be made about feasibility of any potential redox reactions using standard electrode potentials - the most negative system has the greatest tendency to be oxidised and lose electrons - the most positive system has the greatest tendency to be reduced and gain electrons Oxidising agents take elecs away from the species being oxidised, so are reduced and are on the left Reducing agents adds elecs to the species being reduced, so are oxidised and are on the right You can predict the feasibility of redox reactions from E^o values

Question 61

Strong reducing agent

Answer 61

More negative E^o value | As is oxidised so loses elecs more easily

Question 62

Strong oxidising agent

Answer 62

More positive E^o value | As is reduced so gains elecs more easily

Question 63

Limitations of using E^o values

Answer 63

Reaction rate- electrode potentials May indicate the thermodynamic feasibility of a reaction but they give no indication of the rate of a reaction Conc- standard electrode potentials are measured using concs of 1moldm-3. Many reactions take place using concentrated or dilute solutions. If conc different then value of electrode potential will be diff from the standard value

Question 64

Standard electrode potentials and ΔG

Answer 64

ΔG^o = -nFE^ocell n is moles of elecs transferred in the balanced equation F is faradays constant

Question 65

Primary cells

Answer 65

Non rechargeable When in use, electrical energy is produced by oxidation and reduction at the electrodes However, the reactions cannot be reversed Eventually the chemicals will be used up, voltage will fall, the battery will go flat, and the cell will be discarded or recycled Are alkaline based on zinc and manganese dioxide, Zn/MnO2, and a potassium hydroxide alkaline electrolyte

Question 66

Secondary cells

Answer 66

Rechargeable Cell reactions producing electrical charge can be reversed during recharging The chemicals in the cell are then regenerated and the cell can be used again Examples: - lead-acid batteries used in car batteries - lithium ion cells used in laptops, phones etc

Question 67

Fuel cells

Answer 67

Uses energy from the reaction of a fuel with oxygen to create a voltage - the fuel and oxygen flow into the fuel cell and the products flow out. The electrolyte remains in the cell - fuel cells can operate continuously provided that the fuel and oxygen are supplied into the cell - don’t have to be recharged

Question 68

Hydrogen fuel cells

Answer 68

Can have either an alkali or acid electrolyte. | Produce no carbon dioxide during combustion, with water being the only combustion product.