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chemistry > topic five/fifteen > Flashcards

topic five/fifteen Flashcards

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1
Q

heat definition

A

the transfer of energy between objects of different temperature.

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2
Q

second law of thermodynamics

A

heat will spontaneously flow from an object of higher temperature to an object of lower temperature (Figure 1). Once the two objects reach the same temperature, which is known as thermal equilibrium, no more energy will be transferred.

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3
Q

The total enthalpy of a system cannot be measured, but

A

changes in enthalpy can.

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4
Q

standard conditions (Ɵ)

A

A pressure of 100 kPa.

A temperature of 25°C (298 K).

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5
Q

temperature definition

A

the average kinetic energy of the particles in a substance.

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6
Q

closed system

A

only energy is able to move between the system and the surroundings.

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7
Q

the law of conservation of energy

A

energy cannot be created or destroyed; it is converted from one form to another. Therefore, the total amount of energy in the universe is constant.

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8
Q

first law of thermodynamics

A

is embodied in the law of conservation of energy. This is highly relevant when studying energy changes during chemical reactions.

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9
Q

The enthalpy level diagram for an exothermic reaction

A
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10
Q

waht is enthalpy change of a reaction also known as

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔHƟc = −890 kJ mol−1

A

the standard enthalpy change of combustion, or the molar enthalpy of combustion (ΔHƟc).

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11
Q

The enthalpy level diagram for an endothermic reaction.

A
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12
Q

why use polystyrene in practicals

A

is a good heat insulator, therefore, it reduces heat loss to the surroundings

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13
Q

q = mc∆T

A

q is the heat absorbed or released in J

c is the specific heat capacity of the solution in J g−1 °C−1 or J g−1 K−1

m is the mass of solution in g

∆T is the change in temperature in oC or K

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14
Q

what is specific heat capacity

A

the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius or one kelvin. The units of specific heat capacity are either joules per gram per degree Celsius (J g−1 °C−1) or joules per gram per kelvin (J g−1 K−1).

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15
Q

Metals tend to have lower specific heat capacity values which mean that

A

they heat up quickly but also lose heat quickly.

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16
Q

how to graphically compensate for heat loss in a reaction

A

This can be compensated for by plotting a graph of the change in temperature against time and extrapolating the line back to the point at which the reactants were mixed. In Figure 2, you can see that the temperature of the reaction mixture was read every minute, and the zinc was added after three minutes. The temperature then increased to a maximum, after which it decreased in a linear fashion. By extrapolating the linear part of the cooling phase back to the time when the reaction started, and assuming the heat loss has been linear throughout, the maximum temperature without any heat loss can be determined.

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17
Q

what is enthalpy change of neutralisation

A

the enthalpy change when an acid and base react together to form one mole of water

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18
Q

A simple method for determining the enthalpy change of neutralisation (ΔHn) involves

A

mixing known volumes and concentrations of a strong acid and a strong base and measuring the temperature increase.

A measured volume of a strong alkali is placed into the polystyrene cup, and an equal volume of a strong acid is added. The temperature of the reaction mixture increases until neutralisation is complete. Continued addition of acid produces a cooling effect, as no further reaction takes place. The maximum temperature can be determined from the graph in Figure 3 by extrapolating both lines towards each other and finding the point where they intersect. This point gives both the maximum temperature reached and the volume of acid that neutralises the alkali. This method, known as a thermometric titration, can be used to find the end-point of a titration (section 8.2.2).

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19
Q

waht is the molar enthalpy of combustion, or the standard enthalpy of combustion (ΔHƟc),

A

the enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions.

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20
Q

A simple experiment to calculate and compare the molar enthalpy of combustion of a range of alcohols can be carried out using the apparatus shown in Figure 4.

A

A known mass of an alcohol is measured into a pre-weighed spirit burner. The alcohol is burned and the heat released increases the temperature of a known volume of water in the calorimeter. The temperature increase is measured for a certain time period and the experiment is then stopped. The spirit burner and its contents are then re-weighed. The mass of the alcohol burned to produce the temperature increase is recorded and the molar enthalpy of combustion of the alcohol can be calculated.

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21
Q

Percentage error =

A

((experimental value - theoretical value) ÷ theoretical value) × 100

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22
Q

You should be aware of the limitations of calculating enthalpy changes in a school laboratory. These include but are not limited to:

A

Heat loss to the surroundings and heat absorbed by the calorimeter

Incomplete combustion of the fuel

Assumptions made about the specific heat capacity and density of aqueous solutions.

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23
Q

what does hess law state

A

the total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place, as long as the initial and final conditions are the same.

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24
Q

An enthalpy cycle for the conversion of reactants to products.

A
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25
Q

hess law ΔH1 is equal to waht

A

Hess’s law tells us that the enthalpy change of the reaction for the direct route (ΔH1) is equal to the enthalpy change for the indirect route (ΔH2 + ΔH3). Hess’s law still applies, regardless of how many steps there are in the indirect route. This can be shown in equation form for this enthalpy cycle as:

ΔH1 = ΔH2 + ΔH3

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26
Q

waht is the standard enthalpy of formation

A

the enthalpy change when one mole of a compound is formed from the elements in their standard states under standard conditions.

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27
Q

The enthalpy cycle used to determine the enthalpy change of formation

A
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28
Q

Enthalpy of formation values are useful in that they indicate

A

he stability of compounds in relation to their elements

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29
Q

does hgaving a double bond increase or decrease reactivity and why

A

incease

This region of high electron density is the site of chemical reactivity within the molecule. This means that alkenes undergo addition reactions that take place across the carbon-carbon double bond.

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30
Q

enthalpy change of a reaction using combustion

A

ΔH⦵ = ΣΔH⦵c (reactants) − ΣΔH⦵c (products)

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31
Q

bond breaking is

A

endothermic (releases energy)

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32
Q

bond making is

A

exothermic (absorbs energy)

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33
Q

bond enthalpy aka

A

bond dissociation energy

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34
Q

bond enthalpy defintiion

A

It is defined as the energy required to break one mole of chemical bonds in the gaseous state

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35
Q

average bond enthalpy defintiion

A

when one mole of bonds are broken in the gaseous state averaged for the same bond in similar compounds.

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36
Q

The enthalpy change for a reaction can be calculated using average bond enthalpy values.

A

ΔH = ΣE(bonds broken) − ΣE(bonds formed)

37
Q

the atmosphere contains two forms of oxygen:

A

di oxygen and tri oxygen

38
Q

whcih is stronger oxygen or ozone

A

Bond energy/kJ mol-1 498 364
oxygen

39
Q

which has a higher wavelength oxygen or ozone

A

ozone

oxygen
λ < 242 nm
higher energy radiation of shorter wavelength
ozone
λ < 330 nm
lower energy radiation of longer wavelength

40
Q

benefit of the decomposition of oxygen

A

The two reactions in this cycle depend on different wavelengths of UV radiation, and the effect is to remove the higher energy radiation (λ < 242 nm), so that only the longer wavelength, which is less damaging radiation, reaches the Earth’s surface.

41
Q

thes strong double covalent bond in oxygen is disrupted by

A

the suns high energy UV-C raditation to form atoms which are free radicals since they have an unpaired electron. Such oxygen radicals can then react with an oxygen molecule to form ozone. The bonds in ozone, being weaker, can then be broken by the less energetic UV-B radiation (of longer wavelength) to reform oxygen and an oxygen free-radical.

42
Q

The key feature here is that the surface of the Earth is protected by these breakdown of ozone reactions from the very damaging effects of

A

UV-B and UV-C radiation.

43
Q

The lattice enthalpy (ΔH⦵lat) is

A

the enthalpy change when one mole of a solid ionic compound breaks down to form gaseous ions under standard conditions.

44
Q

The magnitude of the lattice enthalpy of an ionic compound can be thought of as

A

a measure of the strength of the ionic bonds between the ions.

45
Q

The magnitude of the lattice enthalpy depends on two factors:

A

the charge of the ions (ionic charge)

the size of the ions (ionic radii).

46
Q

the lattice enthalpy is directly proportional to

A

the product of the ionic charges and inversely proportional to the distance between the nuclei of the ions.

47
Q

lattice enthalpy increases alongside

A

ionic charge increasing or ionic radii decreasing

48
Q

Which of the following is the correct definition of the term lattice enthalpy?

A

The enthalpy change that occurs when one mole of an ionic solid is broken down into its gaseous ions.

49
Q

urpose of born harber

A

to calculate lattice enthalpy

50
Q

Enthalpy of formation, ΔH⦵f:

A

the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

51
Q

Enthalpy of atomisation, ΔH⦵at:

A

the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.

52
Q

Bond dissociation energy, E

A

the energy required to break one mole of bonds in the gaseous state.

53
Q

Ionisation energy, ΔH⦵IE:

A

the energy required to remove one mole of electrons from one mole of gaseous atoms.

54
Q

Electron affinity, ΔH⦵EA:

A

the energy released when one mole of electrons are added to one mole of gaseous atoms

55
Q

steps in born harber

A

Enthalpy of formation, ΔH⦵f

Enthalpy of atomisation, ΔH⦵at

Bond dissociation energy, E

Ionisation energy, ΔH⦵IE

Electron affinity, ΔH⦵EA

56
Q

why do stable ionic compounds typically have large negative values for enthalpy of formation

A

because as more energy is released, the compound becomes more stable

57
Q

ionic solids dissolving

A

break up of the lattice structure, by which the ionic solid is converted to gaseous ions. The enthalpy change for this process is the lattice enthalpy, ΔHlat

Once the lattice structure has been broken down, the separated gaseous ions are hydrated by the surrounding water molecules (Figure 1a). During hydration, ion-dipole force​s are formed between the gaseous ions and the partial charges of the water molecules. The partially negative sides of the water molecules are attracted to the positively-charged ions and partially positive sides of the water molecules are attracted to the negatively-charged ions

58
Q

waht is enthalpy of hydration

A

the enthalpy change when one mole of gaseous ions dissolves in water to form a solution of infinite dilution. It can be represented by the general equation:

X+ (g) → X+ (aq)

A solution of infinite dilution is a solution that has a large excess of water and the addition of more water would not cause any more heat to be released or absorbed.

59
Q

The enthalpy change of solution, ΔH⦵sol,

A

the enthalpy change when one mole of solute dissolves to form a solution of infinite dilution.

60
Q

what isthe enthalpy change of sol thesum of

A

lattice and hydration enthalpies

61
Q

if a substnace has a high positive value fo the enthalpy of solution

A

generaally insoluble

62
Q

factors affectign enthalpy of hydration

A

kionic radiusu and charge on ion

63
Q

the magnitude of enthalpy of hydration decreases whilst

A

ionic radius increases.
This is due to the weaker ion-dipole forces produced as the size of the ion increases.
charge decreases

64
Q

the charge density of an ion increases with

A

greater ionic charge and a decrease in ionic radius

A higher charge density results in a stronger ion-dipole force between the ion and the water molecule, and a greater, more negative value for the enthalpy of hydration.

65
Q

substances with a high positive (endothermic) value of ΔH⦵sol, are

A

less soluble

66
Q

entropy

A

refers to the distribution of energy among the particles in a system.

67
Q

secpnd law of thermodynamics

A

the total entropy of the universe tends to increase.

68
Q

entropy and states of matter

A
69
Q

is melting a positive or negative entropty change?

A

increase, positive tnropy

70
Q

is sublimation a positive or negative entropty change?

A

increase, positive entropy

71
Q

is deposition a positive or negative entropty change?

A

decrease, negative entropy

72
Q

dissolving a solute to form a solution positive or negative entropty change?

A

increase, positive entropy

73
Q

spontaneous meaning

A

A spontaneous process occurs without the addition of energy, other than that required to overcome the initial energy barrier (also known as the activation energy).

74
Q

For a spontaneous process, the total entropy of the system and surroundings (ΔStotal) must increase. This can be represented in equation form as:

A

ΔStotal = ΔSsystem + ΔSsurroundings ≥ 0

75
Q

how can negative entropy reactions be spontaneous

A

the entropy of the surroundings increases to a much greater extent, which gives an overall increase in entropy for the process.

76
Q

For the surroundings, the entropy change is given by the following relationship:

A

ΔSsurroundings=−ΔH/T

77
Q
A

ΔSuniverse=ΔSsystem−ΔH/T

78
Q

any decrease in the enthalpy of the system equates to an addition of

A

heat to the surroundings

79
Q

gibbs free energy change formula

A

ΔG=ΔH−TΔS

80
Q

gibbs free energy

A

the energy associated with a chemical reaction that can be used to do work.

81
Q

This example shows us that there are three factors to be considered when determining the spontaneity of a reaction:

A

the sign of the ΔH

the sign of the ΔS

the temperature at which the reaction takes place.

82
Q

If the Gibbs free energy change, ΔG, is negative, the reaction is

A

spontaneous

83
Q

If the Gibbs free energy change, ΔG, is positive, the reaction is

A

non spontaneous

84
Q

If the Gibbs free energy change, ΔG, is zero, the reaction is

A

at equilibrium

85
Q

For a reaction to be spontaneous, the sign of the Gibbs free energy change (ΔG) must be

A

negative.

86
Q

Summary of the possible combinations of entropy and enthalpy and the spontaneity of a reaction

A
87
Q

The Gibbs free energy of formation, ΔG⦵f, is defined as

A

the change in free energy when one mole of a compound is formed from its elements in their standard states under standard conditions.

88
Q

gibbs free energy change formula

A

ΔG⦵ = ΣΔG⦵f (products) − ΣΔG⦵f (reactants)

89
Q
A

chemistry (13 decks)

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