Physical Chemistry Flashcards

Question 1

Q

The Bohr model assigns electrons to …

To move between orbits, electrons must …

Answer

A

The Bohr model assigns electrons to fixed orbits of defined energy.

To move between orbits, electrons must emit or absorb electromagnetic radiation of a particular frequency.

Question 2

Q

The Evolution of the Atomic Model

in order

Answer

A

The Evolution of the Atomic Model

Indivisible spheres model

Plum pudding model

Nuclear model

Bohr model

Quantum theory

Question 3

Q

Electrons

An electron has a charge of __.

The mass of an electron is so small it is usually approximated to ____

The mass of an electron is approximately __________.

Answer

A

Electrons

An electron has a charge of -1e.

The mass of an electron is so small it is usually approximated to zero.

The mass of an electron is approximately 0.00055amu.

Question 4

Q

Neutrons

A neutron has a mass of _______.

A neutron has no electric charge.

A neutron is very slightly _______ than a proton, but the difference is so small we often take the masses to be the same, and equal to ____.

Answer

A

Neutrons

A neutron has a mass of 1.0087amu.

A neutron has no electric charge.

A neutron is very slightly heavier than a proton, but the difference is so small we often take the masses to be the same, and equal to 1amu.

Question 5

Q

Protons

A proton has a mass of ________
It has a charge of ___.

Answer

A

Protons

A proton has a mass of 1.0073amu.
It has a charge of +1e.

Question 6

Q

Units

Subatomic particles are so small that conventional SI units aren’t very useful.

We define certain units for use with atoms - the atomic mass unit (amu) is defined as one twelfth the mass of a single carbon-12 atom.

The elementary charge unit (e) is equal to the charge on an _________.

Answer

A

Units

Subatomic particles are so small that conventional SI units aren’t very useful.

We define certain units for use with atoms - the atomic mass unit (amu) is defined as one twelfth the mass of a single carbon-12 atom.

The elementary charge unit (e) is equal to the charge on an electron.

Question 7

Q

Electron shells

Electron shells are further split into _______.

Each sub-shell has a slightly different _____.

Electron shells occupy most of the space of the atom.

Answer

A

Electron shells

Electron shells are further split into sub-shells.

Each sub-shell has a slightly different energy.

Electron shells occupy most of the space of the atom.

Question 8

Q

Contents of atoms

Nucleus
Electron shells

Answer

A

Contents of atoms

Nucleus
Protons
Neutrons

Electron shells
Sub-Shells
Orbitals
Electrons

Question 9

Q

Features of the Bohr model:

3 points

Answer

A

Features of the Bohr model:

has a positive nucleus containing protons and neutrons

is outdated but still useful

fixes electrons in orbits

Question 10

Q

What symbol are these numbers often given?

A

Z

Answer

A

What symbol are these numbers often given?

A
mAss number.

Z
Proton number.

Question 11

Q

Electroneutrality

The principle of electroneutrality states that …

Answer

A

Electroneutrality

The principle of electroneutrality states that cations will always be accompanied by anions.

Question 12

Q

In electrochemistry, cations are attracted to the ______.

Answer

A

In electrochemistry, cations are attracted to the cathode.

Question 13

Q

What letter corresponds to the proton number?

Answer

A

What letter corresponds to the proton number?

Z

Question 14

Q

The charge on an ion is calculated as …

Answer

A

The charge on an ion is calculated as the number of protons minus the number of electrons.

Question 15

Q

What is the correct definition of mass number?

Answer

A

What is the correct definition of mass number?

Mass number = number of neutrons + number of protons

Question 16

Q

Isotopes have a different number of _______.

different isotopes have very similar ________ to each other

Answer

A

Isotopes have a different number of neutrons.

different isotopes have very similar chemistry to each other

Question 17

Q

Isotopes are atoms with the same ______ number, but a different ..

Answer

A

Isotopes are atoms with the same proton number, but a different mass number.

Question 18

Q

Explain why isotopes can have different physical properties

Answer

A

Explain why isotopes can have different physical properties

physical properties depend on the mass of an atom.
isotopes have different masses, so can have different physical properties.

Question 19

Q

Describe how J.J. Thompson’s model of the atom was different from Dalton’s model.

Answer

A

Describe how J.J. Thompson’s model of the atom was different from Dalton’s model.

Dalton describes atoms as sold spheres. J.J. Thompson suggested that atoms were not solid spheres - he thought they contained small negatively charged particles (electrons) in a positively charged “pudding”

Question 20

Q

Explain how Rutherford’s gold foil experiment provided evidence that Thompson’s model was wrong.

Answer

A

Explain how Rutherford’s gold foil experiment provided evidence that Thompson’s model was wrong.

if Thompson’s model was correct the alpha particles fired at the sheet of gold should have been deflected very slightly by the positive “pudding”that made up most of he atom.

Instead, most o the alpha particles passed straight through the gold atoms, and a very small number were deflected backwards. so the plum pudding model couldn’t be correct.

Question 21

Q

Describe Rutherford’s model of the atom

Answer

A

Describe Rutherford’s model of the atom

Rutherford’s model has a tiny positively charged nucleus at the centre surrounded by a “cloud” of negative electrons.

Most of the atom is empty space.

Question 22

Q

Describe the main features of Bohr’s model of the atom

Answer

A

Describe the main features of Bohr’s model of the atom

in Bohr’s model the electrons only exist in fixed shells and not anywhere in between.

Each shell has a fixed energy. when an electron moves between shells electromagnetic radiation is emitted or absorbed.

Because the energy of the shell is fixed, the radiation will have a fixed frequency

Question 23

Q

Find the relative atomic mass (Ar) of the following:

Rubidium:
Mercury:
Zinc:

Answer

A

Find the relative atomic mass (Ar) of the following:

Rubidium: 85.5
Mercury: 200.6
Zinc: 65.4

Question 24

Q

find the relative formula mass of the following compounds:

NH3
CO2
C2H4O6N2

Answer

A

find the relative formula mass of the following compounds:

NH3
N = 14.0
H = 1.0

14.0 + ( 1.0 x 3 )
=17.0

CO2
C = 12.0
O = 16.0

12.0 + ( 16.0 x 2 )
=44.0

C2H4O6N2
C = 12.0
H = 1.0
0 = 16.0 
N = 14.0

( 12.0 x 2 ) + ( 1.0 x 4 ) +( 16.0 x 6 ) + ( 14.0 x 2 )
=152.0

Question 25

Q

a sample of tungsten is 0.1% 180W, 26.5% 182W, 14.3% 183W, 30.7% 184W and 28.4%186W. Calculate the Ar of tungsten

Answer

A

a sample of tungsten is 0.1% 180W, 26.5% 182W, 14.3% 183W, 30.7% 184W and 28.4%186W. Calculate the Ar of tungsten

( 0.1 x 180 ) + ( 26.5 x 182 ) + ( 14.3 x 183 ) + ( 30.7 x 184 ) + ( 28.4 x 186 )
=183.891
18389.1/100

=183.9 (to 1 d.p)

Question 26

Q

what is relative atomic mass?

what is relative molecular mass?

what is relative formula mass?

Answer

A

what is relative atomic mass? (Ar)
The average mass of an atom of an element

what is relative molecular mass? (Mr)
the average mass of a molecule

what is relative formula mass?
the average mass of a formula unit

Question 27

Q

Two techniques to ionise the sample:

Answer

A

Two techniques to ionise the sample:

1
Electrospray ionisation
2
Electron impact ionisation

Question 28

Q

Electrospray ionisation is a _________ tec hnique and prevents ____________.
It’s typically used for _________ and …

The sample is … and a …
The … and …
… are now positively charged ions.

Answer

A

Electrospray ionisation is a gentler technique and prevents fragmentation.
It’s typically used for polymers and biological materials like DNA.

The sample is dissolved in a solvent and a high voltage is applied.
The high voltage rips a proton off the solvent and attaches it to the sample molecules.
The sample molecules are now positively charged ions.

Question 29

Q

In electron impact ionisation, the sample is first __________ and then …

The electrons …

The _________ are now positively charged ions.

This method often causes the sample to _______.

Answer

A

In electron impact ionisation, the sample is first vaporised and then hit with electrons from an electron gun.

The electrons knock off electrons from the molecule

The molecules are now positively charged ions.

This method often causes the sample to fragment.

Question 30

Q

2) Acceleration

Molecules are accelerated to all have …

Kinetic energy = 1/2mv^2

All the molecules have the same kinetic energy, so the speed is …
Lighter particles move ______ and are detected before ______ particles.

Answer

A

2) Acceleration

Molecules are accelerated to all have the same kinetic energy.

Kinetic energy = 1/2mv^2

All the molecules have the same kinetic energy, so the speed is dependent on the mass of the molecule.

Lighter particles move faster and are detected before heavier particles.

Question 31

Q

3) Ion drift

The time of flight is given by:
Time of flight =

This leads to an equation for the time travelled that depends on mass:

Lighter ions take less time as the time is dependent on …

Answer

A

3) Ion drift

The time of flight is given by:
Time of flight = distance / velocity

This leads to an equation for the time travelled that depends on mass:
Time = distance × the square root of this –>mass / 2KE

Lighter ions take less time as the time is dependent on the square root of the mass.

Question 32

Q

4) Detection

The ions hit …

This causes a ______ and the size of this current gives a measure of the …

This gives the …

Answer

A

4) Detection

The ions hit a negatively charged plate.

This causes a current and the size of this current gives a measure of the number of molecules hitting the plate.

This gives the abundance of the molecule.

Question 33

Q

Analysis to Identify Molecules

Once a sample has passed through the mass spectrometer, we can analyse the data to identify the _______.

Spectrum produced

When the sample has passed through the mass spectrometer, a _________ is produced by the spectrometer.

On this spectrum:
The x-axis is …
The y-axis is …

Answer

A

Analysis to Identify Molecules

Once a sample has passed through the mass spectrometer, we can analyse the data to identify the molecule.

Spectrum produced

When the sample has passed through the mass spectrometer, a spectrum is produced by the spectrometer.

On this spectrum:
The x-axis is mass/charge ratio.
The y-axis is % abundance.

Question 34

Q

Main peak

The spectrum produces lots of peaks, but the most important is the …

This is the peak of the greatest mass/charge ratio.

This represents the … value of the molecule we are analysing.

Answer

A

Main peak

The spectrum produces lots of peaks, but the most important is the molecular ion peak.

This is the peak of the greatest mass/charge ratio.

This represents the mass/charge value of the molecule we are analysing.

Question 35

Q

Isotopes

Smaller peaks will cluster around the …

These are from the same molecules but with different …

The isotopic molecules have different ______ and so different … values.

Answer

A

Isotopes

Smaller peaks will cluster around the molecular ion peak.

These are from the same molecules but with different isotopes in them.

The isotopic molecules have different masses and so different mass/charge ratio values.

Question 36

Q

Any smaller and significantly lighter peaks in the spectrum are because of _________.

The ________ can fragment in the spectrometer.

Answer

A

Any smaller and significantly lighter peaks in the spectrum are because of fragmentation.

The molecule can fragment in the spectrometer.

Question 37

Q

Analysis to Calculate

Once we have the mass spectrum, we can calculate the relative atomic masss

Example - boron

This is the mass spectrum of a sample of elemental boron.

Boron has two isotopes, 10B and 11B.

You can see from the spectrum that approximately 20% of the boron is 10B and 80% is 11B.
You can use this to work out the relative atomic mass:
…

Answer

A

Analysis to Calculate

Once we have the mass spectrum, we can calculate the relative atomic masss

Example - boron

This is the mass spectrum of a sample of elemental boron.

Boron has two isotopes, 10B and 11B.

You can see from the spectrum that approximately 20% of the boron is 10B and 80% is 11B.
You can use this to work out the relative atomic mass:
(80% × 11)+ (20% × 10) = 10.8

Question 38

Q

Electron shells

electrons are arranged in _____.

These shells are defined by the symbol ‘_’

The lowest energy shell has ____

Other shells have higher energy and higher _.

The higher the n of an electron, the further from …

Answer

A

Electron shells

electrons are arranged in shells.

These shells are defined by the principal quantum number, given the symbol ‘n’.

The lowest energy shell has n = 1.

Other shells have higher energy and higher n.

The higher the n of an electron, the further from the nucleus it orbits.

Question 39

Q

Sub-shells

For atoms with more than ___ electron, shells are split into sub-shells that have slightly different _______.

The difference in energy between sub-shells is much less than the difference in energy between ______.

A shell with a given n will have n ….

E.g. the n = 3 electron shell has ….

Answer

A

Sub-shells

For atoms with more than one electron, shells are split into sub-shells that have slightly different energies.

The difference in energy between sub-shells is much less than the difference in energy between shells.

A shell with a given n will have n sub-shells.

E.g. the n = 3 electron shell has three sub-shells.

Question 40

Q

Orbitals

________ are composed of orbitals.

Orbitals in the same sub-shell have the same _____.

Each orbital can hold a maximum of ___ electrons.

Answer

A

Orbitals

Sub-shells are composed of orbitals.

Orbitals in the same sub-shell have the same energy.

Each orbital can hold a maximum of two electrons.

Question 41

Q

Bringing it all together

Electrons occupy _______.

An ______ can hold a maximum of two electrons.

A set of … is called a sub-shell.

Sub-shells make up different …

Different sub-shells do not have the same _____.

Answer

A

Bringing it all together

Electrons occupy orbitals.

An orbital can hold a maximum of two electrons.

A set of orbitals of the same energy is called a sub-shell.

Sub-shells make up different shells of electrons.

Different sub-shells do not have the same energy.

Question 42

Q

electronic structure

Answer

A

electronic structure

electron shells

sub-shells

orbitals

electrons

Question 43

Q

Types of orbitals

Orbitals are labelled by letters.

The first three orbitals are called s, p and d:

An s sub-shell is made of …
A p sub-shell is made of …
A d sub-shell is made of …

Orbitals of exactly the same energy are called ‘________’.

Answer

A

Types of orbitals

Orbitals are labelled by letters.

The first three orbitals are called s, p and d:

An a sub-shell is made up of one s orbital.
A p sub-shell is made of three p orbitals.
A d sub-shell is made of five d orbitals.

Orbitals of exactly the same energy are called ‘degenerate’.

Question 44

Q

Shells and sub-shells

The n = 1 shell has ___ sub-shell.
It is an s sub-shell.

The n = 2 shell has ___ sub-shells.
it has an …

The n = 3 shell has ____ sub-shells.
It has an …

Answer

A

Shells and sub-shells

The n = 1 shell has one sub-shell.
It is an s sub-shell.

The n = 2 shell has two sub-shells.
They are an s sub-shell and a p sub-shell.

The n = 3 shell has three sub-shells.
It has an s sub-shell, a p sub-shell and a d sub-shell.

Question 45

Q

Filling orbitals

Orbitals are filled from the ______ energy to the ______ energy.

The usual energy order is from …

The ________ metals are an anomaly. The __ sub-shell is of a lower energy than the __ sub-shell, and so here the filling order is: …

Answer

A

Filling orbitals

Orbitals are filled from the lowest energy to the highest energy.
The usual energy order is from s ⇒ p ⇒ d.
The transition metals are an anomaly. The 4s sub-shell is of a lower energy than the 3d sub-shell, and so here the filling order is: 3s ⇒ 3p ⇒ 4s ⇒ 3d.

Question 46

Q

Electron shells

n=1
_ sub-shell
__ orbital

n=2
_ sub-shell
_ sub-shell
\_\_ orbital
\_\_\_\_\_ orbitals

n=3
_ sub-shell
_ sub-shell
_ sub-shell
\_\_ orbital
\_\_\_\_\_\_ orbitals
\_\_\_\_\_\_\_ orbitals

Answer

A

Electron shells

n=1
s sub-shell
1s orbital

n=2
s sub-shell
p sub-shell
2s orbital
3 × 2p orbitals

n=3
s sub-shell
p sub-shell
d sub-shell
3s orbital
3 × 3p orbitals
5 × 3d orbitals

Question 47

Q

An atom has 16 electrons to hold.
How are they arranged within the sub-shells?

1
\_\_ orbital contains _ electrons
2
\_\_ orbital contains _ electrons
3
\_\_ orbital contains _ electrons
4
\_\_ orbital contains _ electrons
5
\_\_ orbital contains _ electrons

Answer

A

An atom has 16 electrons to hold.
How are they arranged within the sub-shells?

1
1s orbital contains 2 electrons
2
2s orbital contains 2 electrons
3
2p orbital contains 6 electrons
4
3s orbital contains 2 electrons
5
3p orbital contains 4 electrons

Question 48

Q

the ionisation energy equation.

This is best shown by a few examples.

The first ionisation energy equation of Na:
__() → __ + __ +()

The second ionisation energy equation of Mg:

Answer

A

the ionisation energy equation.

This is best shown by a few examples.

The first ionisation energy equation of Na:
Na(g) → e− + Na +(g)

The second ionisation energy equation of Mg:
Mg(g) → e− + Mg 2+(g)

Question 49

Q

The electrons

Ionisation is the reaction where …

There will be ____ electron in the product of the equation.

So we will always have an e− term on the _____-hand side of the equation.

Answer

A

The electrons

Ionisation is the reaction where one electron is removed from an atom.

There will be one electron in the product of the equation.

So we will always have an e− term on the right-hand side of the equation.

Question 50

Q

The charges

We must understand the charges on the atom during the reaction.

During ionisation, we remove one electron from an atom.

As the negative electron is removed, the atom’s charge will …

E.g. the second ionisation of Na converts Na+ → ______

Answer

A

The charges

We must understand the charges on the atom during the reaction.

During ionisation, we remove one electron from an atom.

As the negative electron is removed, the atom’s charge will increase by 1.

E.g. the second ionisation of Na converts Na+ → Na2+.

Question 51

Q

The state symbols

Ionisation energies are always defined as being in the ____ phase.

So atomic species of any charge, on either side of the equation, will have the state symbol (_).

We do not define the state of the _________ - it has no state symbol.

Answer

A

The state symbols

Ionisation energies are always defined as being in the gas phase.

So atomic species of any charge, on either side of the equation, will have the state symbol (g).

We do not define the state of the electron - it has no state symbol.

Question 52

Q

Ionisation Energies

A specific amount of energy is needed to …

Answer

A

Ionisation Energies

A specific amount of energy is needed to remove an electron from an atom or ion.

Question 53

Q

First ionisation energy

The first ionisation energy is the energy required to … to produce …

Answer

A

First ionisation energy

The first ionisation energy is the energy required to remove an electron from every atom in a mole of atomic gas, to produce a mole of unipositive gaseous ions.

Question 54

Q

Second ionisation energy

The second ionisation energy is the energy required to …, to produce …

Answer

A

Second ionisation energy

The second ionisation energy is the energy required to remove an electron from every ion in a mole of unipositive gaseous ions, to produce a mole of dipositive gaseous ions.

Question 55

Q

What do clustered small peaks around the molecular ion peak represent

Answer

A

What do clustered small peaks around the molecular ion peak represent

Isotopic molecules

Question 56

Q

Shielding

The greater the number of …, the lower …

This is because the …

This means that the … the lower …

Answer

A

Shielding

The greater the number of electrons between the nucleus and the outer electrons, the lower the effective nuclear charge.
This is because the positive charge felt by the electron is reduced by the electrons in between.
This means that the greater the number of electrons, the lower the ionisation energy.

Question 57

Q

Nuclear charge

The greater the …

A greater attraction of the electron means …

This means …

Answer

A

Nuclear charge

The greater the number of protons in the nucleus, the greater the attraction of the electron to the nucleus.

A greater attraction of the electron means more energy is needed to remove the electron.

This means that ionisation energy is greater.

Question 58

Q

Distance from the nucleus

Electrostatic attraction …

This means that …

This means that as …

In practice, this means that the higher … the lower …

Answer

A

Distance from the nucleus

Electrostatic attraction decreases sharply with distance.

This means that less energy is needed to remove electrons which are further away.

This means that as distance increases, ionisation energy decreases

In practice, this means that the higher the principal quantum number of an electron, the lower its ionisation energy.

Question 59

Q

The three main factors affecting ionisation energies:

Answer

A

The three main factors affecting ionisation energies:

1
The nuclear charge
2
The distance of the electron from the nucleus
3
The electron shielding

Question 60

Q

Breaking down the trend

Consider the sodium atom of electron configuration 1s2 2s2 2p6 3s1

Its first ionisation energy corresponds to …

Its second ionisation energy corresponds to …

Between the third and ninth ionisation energies, …

Its tenth ionisation energy corresponds to …

Answer

A

Breaking down the trend

Consider the sodium atom of electron configuration 1s2 2s2 2p6 3s1

Its first ionisation energy corresponds to the removal of an electron from the n = 3 shell.

Its second ionisation energy corresponds to the removal of an electron from the n = 2 shell.

Between the third and ninth ionisation energies, an electron is being removed from the n = 2 shell of an increasingly positive ion.

Its tenth ionisation energy corresponds to the removal of an electron from the n = 1 shell.

Question 61

Q

Explaining the patterns

Ionisation energy increases with …

The n = 1 shell will have a higher ionisation energy than …

The n = 2 shell will have a higher ionisation energy than …

Within the n = 2 shell, successive ionisation energies will rise as …

Answer

A

Explaining the patterns

Ionisation energy increases with proximity to the nucleus.

The n = 1 shell will have a higher ionisation energy than n = 2 shell.

The n = 2 shell will have a higher ionisation energy than n = 3 shell.

Within the n = 2 shell, successive ionisation energies will rise as the ion becomes more positively charged.

Question 62

Q

Trend down a group

First ionisation energy decreases down a group.

This is because …

The electron is further from the nucleus, and so …

There are also …

This means the outer electrons experience …

This also provides evidence for …

Answer

A

Trend down a group

First ionisation energy decreases down a group.

This is because the atomic radius increases.

The electron is further from the nucleus, and so experiences lower attraction.

There are also more electrons between the nucleus and the outer electrons.

This means the outer electrons experience greater shielding from the nucleus.

This also provides evidence for the existence of shells.

Question 63

Q

Trend across a period

First ionisation energy increases across a period.

This is because the …

Proton number increases across a period, so …

The electron is closer to the nucleus, and so experiences …

… increases across a period

Answer

A

Trend across a period

First ionisation energy increases across a period.

This is because the atomic radius decreases.

Proton number increases across a period, so electrons are more attracted to the nucleus.

The electron is closer to the nucleus, and so experiences a greater attraction.

Nuclear charge increases across a period.

Question 64

Q

Why does ionisation energy increase across a period?

Answer

A

Why does ionisation energy increase across a period?

Atomic radius is smaller and so there is a greater attraction

Answer 65

A

How does ionisation energy vary within the periodic table?

Decreases down a group

Increases across a period

Answer 66

A

Example of sub-shell impact

An example of the impact of the sub-shell structure is the difference in ionisation energies between magnesium and aluminium within Period 3.

They are in the same period and so we would expect aluminium’s ionisation energy to be greater due to a greater nuclear charge.

Answer 67

A

They are in the same period and so we would expect aluminium’s ionisation energy to be greater due to a greater nuclear charge

But the ionisation energies are:
Mg = 738 kJ/mol.
Al = 578 kJ/mol.

These values do not align with the general trend we predicted.

We need to look at the sub shell structures of these atoms to understand what is going on.

Answer 68

A

Mg vs Al sub-shell structure

Mg = 1s2 2s2 2p6 3s2
So the first ionisation removes an electron from the 3s orbital.

Al = 1s2 2s2 2p6 3s2 3p1
So the first ionisation removes an electron from the 3p orbital.

Answer 69

A

Sub-shell energies

The 3p orbital is further from the nucleus on average than the 3s orbital.

So this orbital has a higher energy and its electrons have a lower electrostatic attraction to the positive nucleus.

It is also slightly shielded by the 3s2 electrons.

So it’s easier to ionise an electron in the 3p orbital than in the 3s orbital.

This explains why it is easier to ionise Al than Mg.

This effect outweighs the general trend we looked at first.

Answer 70

A

Why is it easier to ionise a 3p electron than a 3s electron?

1
3p orbital is higher in energy
2
3p electron has a lower electrostatic attraction to the positive nucleus
3
3p electrons are shielded by 3s electrons

Answer 71

A

The impact of electron repulsion can be seen in the ionisation energies of Group 5 and Group 6.

Phosphorus and sulfur

Both P and S have their outer electron in a 3p orbital.

P = 1s2 2s2 2p6 3s2 3p3
S = 1s2 2s2 2p6 3s2 3p4

The shielding for each atom’s outer electron is identical.

S has a higher nuclear charge and so you might expect S to have a higher ionisation energy. This is NOT the case.

Answer 72

A

Singly vs doubly occupied orbitals
The P electron is removed from a singly occupied orbital but the S electron is removed from a doubly occupied orbital.
S has a lower ionisation energy because electrons in the same orbital repel each other.
This means that an electron is easier to remove from a doubly occupied orbital.
This gives more evidence for the electronic structure model we are using.

Answer 73

A

It’s easier to remove an electron from a doubly occupied orbital because the electrons repel each other

Answer 74

A

Concentration

Concentration is defined as moles / unit volume.

The usual units of concentration are moles per litre.

Litre is often written as dm3.

So moles per litre is mol ÷ dm3.

This is often written as moldm-3.

Answer 75

A

Three moles of NaCl are dissolved in half a litre of water.
Workout the concentration

Concentration = number of moles ÷ volume

Concentration = 3 mol ÷ 0.5 dm3

Concentration = 6 moldm-3

Answer 76

A

Mole calculations

We can calculate the number of moles present in a sample if we know its mass, and its Mr:

Moles = mass ÷ Mr

Remember mr mole!

Answer 77

A

Benzene has an Mr of 78. How many moles of benzene are in 7.8 g of pure benzene?

Moles = mass ÷ Mr
Moles = 7.8 g ÷ 78
Moles = 0.1 mol

Answer 78

A

Calculating masses

1) . Balance the equation
2) . Work out how many miles of the reactant you have
3) . Use the molar ratio from the balanced equation to work out the number of miles of product that will be formed
4) . Calculate mass of that many miles of product

Answer 79

A

Find the mass of iron oxide produced when 28.0 g of iron is burnt in air

1). Balance equation: 2Fe + 1 1/2 O2 —-> Fe2O3

2). Work out number of moles of iron you have - Ar of Fe = 55.8
Moles = mass / Ar
= 28.0/55.8
=0.502

3). The molar ratio of Fe: Fe2O3 is 2:1
For every 2 moles of Fe you will produce 1 mole of Fe2O3
But you only have 0.502 / 2 = 0.251

4). Now find the mass of 0.251 moles of Fe2O3
Mr of Fe2O3 = (2x55.8)+(3x16) = 159.6
Mass = moles x Mr
0.25x159.6 = 40.0 g of iron oxide

Answer 80

A

Definition of enthalpy change

An enthalpy change is a measure of the heat given out or taken in during a process.

When objects are heated, they use energy to expand.

Enthalpy takes into account the energy used in the expansion.

Answer 81

A

Enthalpy vs energy changes

We use enthalpy instead of energy because we cannot easily measure energy changes.

This is because objects expand when heated.

To measure an energy change, we would have to fix the volume of the object.

Enthalpy is much easier to use as it allows for expansion.

Enthalpy changes are instead measured under constant pressure.

The atmosphere is at a constant pressure, so we require no extra equipment.

Answer 82

A

Endothermic vs exothermic reactions

An exothermic reaction is one which gives out heat energy.
An exothermic reaction has a negative enthalpy change.
An endothermic reaction is one which takes in heat energy.
An endothermic reaction has a positive enthalpy change.

Answer 83

A

Combustion is exothermic

Combustion is an exothermic process as it gives out heat!
E.g. Burning methane:
CH4 + 2O2 → CO2 + 2H2O
ΔH = −882.00kJmol-1

Answer 84

A

Bond enthalpies

During a reaction, some chemical bonds must be broken and made.
The energy to break or make a bond is known as the bond enthalpy.
Energy is needed to break a bond so it is an endothermic process.
Energy is given off when a bond is made so it is an exothermic process.
The enthalpy change of a reaction is a sum of the individual bond enthalpies being broken and made during the reaction.

Answer 85

A

The dominating term

During a chemical reaction, we must determine whether the reaction requires more energy to break bonds or to make bonds.
This will give an overall reaction enthalpy that is either positive or negative.

Answer 86

A

Endothermic vs exothermic

Since bond breaking is endothermic and bond making is exothermic:
More energy required to break bonds in a reaction will lead to an overall endothermic reaction.
More energy released making bonds in a reaction will lead to an overall exothermic reaction.

Answer 87

A

When calculating reaction enthalpies, we must add the bond enthalpies of all the bonds that are being broken and then minus the bond enthalpies of all the bonds that are being made.

Answer 88

A

Bond breaking and making:

1
Bond breaking is an endothermic process
2
Bond making is an exothermic process

Answer 89

A

Enthalpy change
ΔH = +
Endothermic
E.g. Thermal decomposition of calcium carbonate
ΔH = −
Exothermic
E.g. Combustion of methane

Answer 90

A

Standard conditions:
For many measurements, we can record them under standard conditions. This is true for enthalpy changes - we call it the standard enthalpy change.

1
Pressure of 1bar
2
Temperature of 298K
3
Substances in their most stable state under the above conditions

Answer 91

A

The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under the standard conditions of 298K and 100 kPa (1bar).

Answer 92

A

Bomb calorimetry

Bomb calorimetry uses a machine called a bomb calorimeter to measure enthalpy changes of combustion.
This process involves burning a sample of a compound in a sealed vessel and measuring the temperature change.
Often the calorimeter will just determine the temperature change in the vessel and you will have to calculate the enthalpy change of combustion.

Answer 93

A

Inaccuracy

Bomb calorimetry can be inaccurate due to:
Heat lost to the surroundings.
Any incomplete combustion that may take place.
Loss of some reactant that evaporates before it combusts.

Answer 94

A

Calculating enthalpy changes

The equation to calculate enthalpy changes from temperature changes is:
q = m × c × ΔT
q is the heat energy.
m is the mass of the sample.
c is the specific heat capacity.
ΔT is the temperature change.
If the pressure is constant, q = ΔcH

Answer 95

A

Calculating enthalpy changes

We have calculated q, the energy given off to the surroundings (the enthalpy change).
The units of q are Joules.
To determine the enthalpy change of combustion, we must convert Joules into the unit of enthalpy change, Joules per mole.
Calculate the number of moles:
Moles = mass ÷ Mr
ΔH = \frac{q}{moles} 
moles
q

Answer 96

A

Calorimetry is used to physically measure changes in enthalpy.

Answer 97

A

A student carried out a simple laboratory experiment to measure the enthalpy change
for Process 3. The student showed that the temperature of 200 g of water increased
by 8.0 °C when 0.46 g of pure ethanol was burned in air and the heat produced was
used to warm the water.
Use these results to calculate the value, in kJ mol–1, obtained by the student for this
enthalpy change. (The specific heat capacity of water is 4.18 J K–1 g–1)
Give one reason, other than heat loss, why the value obtained from the student’s
results is less exothermic than a data book value.

q = m x c x delta temp

= 200 x 4.18 x 8.0

= 6688

moles = mass / Mr

0.46 / 46

= 0.01

6.688 / 0.01
kj/mol

=668.8 kjmol^-1

(to convert an number of moles to one is to divide it b themselves 0.01 / 0.01 = 1)

the reaction is exothermic so we add a negative

Answer 98

A

For best accuracy, we must plot the temperature loss after the reaction is complete. Then we can extrapolate the line to find the most accurate temperature change value.

Answer 99

A

Why are the initial and final temperatures in a calorimeter inaccurate?

heat is always being lost to surroundings

Answer 100

A

Combustion of cyclohexane

Cyclohexane fuel is burned completely in a calorimeter.
There are 200 g of water in the calorimeter.
There are 0.5 moles of cyclohexane burnt.
The temperature of the water was raised from 298 K to 368 K.

q = 200 x 4.18 x 70
= 58,520 j

enthalpy change of combustion
ΔH = q / moles
ΔH = −58520 J ÷ 0.5 moles

Note the minus sign added. This is because we know the reaction is exothermic since the water’s temperature was increased.
ΔH = −117040 Jmol-1
ΔH = −117.04 kJmol-1
Note the final units of kJmol-1 as this is more standard.

Answer 101

A

Neutralisation reaction

Calculate the heat lost/gained during the reaction between H2SO4(aq) and NaOH(aq):
20 cm3 of the acid is added to an insulate container.
30 cm3 of the base is then added.
The temperature change is recorded to be 40 K.
Assume the density of the solutions to be the same as water, 1 gcm-3.
Assume the specific heat capacity is the same as water’s, 4.18 Jg-1K-1.

The calculation

Because we have assumed that the density is the same as water, we can calculate the mass of the solution as:
1 cm3 = 1 g
(20 + 30) cm3 = 50 g
The heat change:
q = mcΔT
q = (50) g x 4.18 Jg-1K-1 x 40 K
q = 8360 Joules
q = 8.36 kJ

Answer 102

A

Two things that change if calculating temperature changes in reaction mixtures:

1
The density is no longer equal to the density of water (1 cmg-1)
2
The specific heat capacity may not be equal to 4.18 Jg-1K-1

Answer 103

A

What is the only case where we can state that a reaction mixture has the same density as water or the same specific heat capacity as water?

if the q lets us assume so

Answer 104

A

The units of q are Joules.

The units of enthalpy change are Joules per mole.

Answer 105

A

IIt is NOT accurate to just measure the initial and final temperatures on the calorimeter.

Answer 106

A

The enthalpy change of X to Y is 10 Jmol-1. The enthalpy change of Y to Z is 30 Jmol-1. What is the enthalpy change of X to Z?
1
By Hess' Law:
ΔH(X → Z) = ΔH(X → Y) + ΔH(Y → Z)
2
Substitute the values in:
ΔH(X → Z) = 10 Jmol-1 + 30 Jmol-1
ΔH(X → Z) = 40 Jmol-1

Answer 107

A

The enthalpy of formation for any element is zero.

Answer 108

A

hess’ law allows us to indirectly calculate enthalpy change of a complex reaction by considering simpler reactions

Answer 109

A

Hess’ Law states that the enthalpy change of a reaction is independent of the pathway.

Answer 110

A

Bond enthalpy is the enthalpy change when one mole of bonds is broken in the gas phase.
This varies from molecule to molecule and from bond to bond.

Answer 111

A

Bond enthalpy

Bond enthalpy is the enthalpy change when one mole of bonds is broken in the gas phase.
This varies from molecule to molecule and from bond to bond.

Answer 112

A

Mean bond enthalpies

Some molecules (e.g. methane - CH4) have multiple identical bonds.
The energy to break the second CH bond in methane will be greater than for the first bond because the molecule will rearrange after the first bond is broken.
The CH bond enthalpy value for methane is the average of all four bond enthalpies.

Answer 113

A

Example

Looking at the previous example of methane:
First C-H broken: 401 kJmol-1
Second C-H broken: 413 kJmol-1
Third C-H broken: 428 kJmol-1
Fourth C-H broken: 439 kJmol-1
So overall, the enthalpy change will be an average:
(439 + 428 + 413 + 401) ÷ 4 = 420.25 kJmol-1

Answer 114

A

Inaccuracies

There are two main reasons mean bond enthalpies are inaccurate:
Mean bond enthalpies are only valid in the gas phase. You might not be doing your reaction in the gas phase.
Bond enthalpies depend on the particular molecule, whereas mean bond enthalpies are quoted generally.
So we can’t expect a perfect answer - they are only good as a guide.

Answer 115

A

Which of the following best describes bond enthalpies?

often mean values

Answer 116

A

In an endothermic reaction, is the energy released by the formation of bonds GREATER or LESS THAN the energy needed to break bonds?

Less than

Answer 117

A

Which of the following is the correct definition for ΔfHθ?

the enthalpy

Answer 118

A

Conditions for a collision to result in a reaction
Activation energy
The collision must have enough energy for a reaction to happen
Steric factor
A collision must take place at the correct location within the molecule for a reaction to happen

Answer 119

A

Enthalpy profiles show the progress of reaction on the x-axis and energy level on the y-axis. They contain the following information:
The relative enthalpies of reactants and products.
The activation energy of a reaction.
The overall enthalpy change of a reaction.
This diagram shows an exothermic reaction.

Answer 120

A

Types of enthalpy profiles:

Exothermic enthalpy profile

Endothermic enthalpy profile

Answer 121

A

The axes in enthalpy profile diagrams:

x
-axis
Progress of reaction.

y
-axis
Enthalpy.

Answer 122

A

“Steric factor” is just a complex phrase to describe how the location of a collision is important.

Answer 123

A

how would u form carbon dioxide (give equation)

         ΔH C(s) + O2(g) ---> CO2(g)

Answer 124

A

how would u form methane (give equation)

   ΔH C(s) + 2H2(g) --->CH4(g)

Answer 125

A

calculate the enthalpy chnage for this reaction

     ΔH=? C2H2(g) + 2H2(g) ---> C2H6(g)

enthalpies of formation:
ΔHf (C2H2) = +228 Kj mol ^-1
ΔHf (C2H6) = -85 kj mol^-1

                     ΔHf = -85 2C(s) + 3H2(g) --------------> C2H6(g)

                   ΔHf = +228 2C(s) + H2(g) -------------------> C2H2(g)

ΔH = -228 + (-85) = 313 kj mol^-1

(see elliot rintoul energetics pt 3 10.46 minns)

Answer 126

A

ΔHc = enthalpy chnage of combustion

Answer 127

A

define ΔHc

when one mole of a compound is completely burned in oxygen with all reactants and products in their standard states and standard conditions

Answer 128

A

complete combustion products

CO2 + H2O

Answer 129

A

incomplete combustion products

CO + H20

Answer 130

A

complete combustion of methane

CH4 + 2O2 —> CO2 + 2H20

Answer 131

A

ENTHALPY CHANGE OF COMBUSTION

                            ΔHc = ?       C2H2(g) + 2H2(g) ----------------> C2H6(g)

ΔHc (C2H2(g)) = - 1301

ΔHc (2H2(g)) = - 285

ΔHc (C2H6(g)) = - 1560

Answer 132

A

when ΔHc arrows go down

when ΔHf arrows go up

Answer 133

A

for a reaction to occur particles must have energy greater than the activation energy

Answer 134

A

The Maxwell-Boltzmann distribution of energies is a handy little graph showing the energy distribution of all the molecules in a gas.

The graph

The number of molecules is on the y-axis.
The kinetic energy of the molecules is on the x-axis.

Answer 135

A

The origin

No molecules have zero energy, so the curve passes through the origin at (0,0).

Answer 136

A

The peak

There is a peak in the middle.
This represents the most likely energy of any molecule.
There are more molecules with this energy than with any other energy.

Answer 137

A

The area under the graph

The area under the graph gives the total number of molecules.
You can think of this as like adding up the number of molecules with every particular energy.

Answer 138

A

Molecule energies

The average energy is to the right of the maximum (peak).
The tail of the graph is asymptotic to the x-axis.
This means it tends to zero, but never touches the x-axis.
This reflects that there is a small possibility of having a very high energy molecule.

Answer 139

A

Activation energies

We can mark the activation energy on the Maxwell-Boltzmann distribution. This allows us to view the number of molecules with the energy to react.
On the left-hand side of the activation energy mark, molecules have less energy than the activation energy and so they can’t react.
On the right-hand side, molecules have more energy than the activation energy and so they can react.

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