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Physical chemistry Flashcards

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1
Q

3.1.1
What are the fundamental particles? relative charges? relative masses?

A

Proton:
- relative charge = +1
- relative mass = 1

neutron:
- relative charge = 0
- relative mass = 1

Electron:
- relative charge = -1
- relative mass = 0.0005 (negligible - very small)

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2
Q

3.1.1
define: Atomic number, Mass number, Isotopes, Relative Atomic mass, Relative isotopic mass

A

Atomic number:
The number of protons in the nucleus of an atom. Denoted by Z

Mass number:
The number of protons + neutrons in the nucleus Denoted by A

Isotopes:
atom with a different no. of neutrons but same no. of protons.

Relative Atomic mass:
the mass of a single atom of an element relative to 1/12 mass of an atom of carbon-12.

Relative isotopic mass:
the mass of a single isotope of an element relative to 1/12 mass
of an atom of carbon-12.

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3
Q

3.1.1
What are the uses of time-of-flight (ToF) Mass Spectrometer?

A
  • used to identify elements
  • used to identify elements
  • detecting illegal drugs
  • forensic science
  • space exploration
  • carbon-14 dating
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4
Q

3.1.1
ToF MS stage 1 - Ionisation (Electrospray & Electron Impact) What happens?

A

Electrospray ionisation:
- The sample is dissolved in a volatile solvent and injected through a fine hypodermic needle (Very fine needle) giving a fine mist.
- A high voltage is applied to the tip of the needle, causing each particle to gain a proton as it leaves the needle.
- The solvent evaporates (and is therefore removed) producing gaseous, positive ions
- Equation = x + H+ –> xH+

Electron impact ionisation:
- The sample is vaporised
- an ‘electron gun’ (which is just an exposed filament of wire) is used to fire high energy electrons at the sample.
- This knocks one electron off each particle, so they become +1 ions.
- Has to be in gaseous state
- Equation = x(g) –> x+(g) + e-

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5
Q

3.1.1
ToF MS stage 2 - Acceleration What happens?

A
  • An electric field is applied to give all the ions with the same charge a constant kinetic energy.
  • As KE = 0.5mv2, Kinetic energy of the ions only depends on the mass (m) of the particles and their velocity (V).
  • As all particles are given the same KE, heavier particles (larger Mr with a bigger m) move more slowly than lighter particles.
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6
Q

3.1.1
ToF MS stage 3 - Ion drift what happens?

A
  • The ions enter a region with no electric field called the flight tube.
  • Here the ions are separated based on their different velocities; the heavier ions have less KE (Move less quickly) compared to the smaller ions (which travel quicker) and arrive at the detector first.
  • Equation (used to workout length of ion drift section) = velocity = d/t
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7
Q

3.1.1
ToF MS stage 4 - Ion detection What happens?

A
  • The detector records the different flight times of the ions. The positively charged ions arrive at the negatively charged plate on detector and gain an electron, inducing a current in the detector and also becoming an atom again
  • It’s important to know that the more abundant a given Ion is, the larger the current it induces. This means a bigger peak in the spectrum - he abundance of ions is proportional to size of current
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8
Q

3.1.1
General information about a Mass Spectrum (graph produced from MS)

A

This is what the machine gives you after those four Stages. It’s a series of peaks on a graph. The vertical axis is relative abundance and the horizontal axis is the m/z (Mass / Charge)

The relative abundance is a measure of how many of each ion is present. Often the highest peak (known as the base peak) is given a value of 100 and the other peaks are worked out relative to this value.

The m/z for a single charge ion = mass of the ion, as mass divided by 1 = mass. The m/z values for isotopes of an element are the relative isotopic masses for these isotopes.

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9
Q

3.1.1
What is an Electron Configuration, how does it work?

A

Principle energy levels = shells.
Basic information
* 1st energy level (moving out of nucleus) is called n=1; 2nd n=2 etc. This no. is called the principal quantum number.
* The energy levels get closer together as you move further from the nucleus. Distance between n=1 and n=2 is larger than the distance
between n=2 and n=3. This cont. further from the nucleus.
* Each energy level is divided into sub-shells: contains an orbital or a combination of orbitals.
* Orbital: a 3-D space + each orbital can hold up to 2 electrons.
* 2 electrons in the same orbital spin in opp. directions to minimise repulsions.

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10
Q

3.1.1
What is the Aufbau Principle?

A

It states that the lowest energy sub-levels are occupied first

1s
2s 2p
3s 3p 3d So, the order is: 1s 2s 2p 3s 3p 4s 3d 4p
4s 4p …

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11
Q

3.1.1
How can electron configurations be shortened?

A

Noble gas symbols in square brackets, such as [Ar], are sometimes used as shorthand in electron configurations

E.g. calcium (1s22s22p63s23p64s2) can be written as [Ar]4s2, where [Ar] = 1s22s22p63s23p6

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12
Q

3.1.1
A typical sub-shell diagram

A
  • 4s sub-shell is at a lower energy level than the 3d sub-shell ∴ 4s fills before 3d.
  • The electrons are indicated by ↿ and ⇂. They represent the different directions of spin

When electrons are in their lowest possible energy levels, the atom is said to be in the ground state.

Electrons repel each other, so will only form pairs in an orbital when they have to

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13
Q

3.1.1
What are Ionisation Energies?

A

The electrons in atoms and ions are attracted to the positive nucleus. Energy is required to overcome this attraction and remove electrons. Ionisation is the process of removing electrons

First ionisation energy: energy required to remove 1 mole of electrons from 1 mole of gaseous atoms, forming 1 mole of gaseous ions with a charge of +1.
- Successive ionisations give the 1st, 2nd, 3rd, etc. ionisation energies. Only 1 mole of electrons is removed with each ionisation.

Measured in kJmol-1. All ionisation energy values will be +ve as they are endothermic as energy is required to remove an electron from the attractive power of the nucleus.

The higher the value, the more energy is required to remove 1 mole of electrons.

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14
Q

3.1.1
Using successive ionisation energies

A

The existence of energy levels is proven by the large gaps in the successive ionisation energies as these correspond to the removal of electrons from energy levels closer to the nucleus + so more energy is required to remove the electron.

Successive ionisation energies are an indicator of the group to which an element belongs.

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15
Q

3.1.1
Explain trends in ionisation energies

A

The patterns in ionisation energies can be explained by the electronic structures of the atoms. 3 main factors which can be used to explain these patterns are…
…Atomic radius
* Atomic radius ↑ down a group + ↓ across a period.
* The further an outer electron is from the attractive power of the nucleus the less energy required to ionise it.
…Nuclear charge
* The greater the no. of protons, the greater the nuclear charge.
* A greater nuclear charge leads to a stronger attraction to the outer electron, so more energy is required to ionise it.
…Shielding by inner electrons
* The attractive power of the nucleus can be shielded by inner electrons.
* The more inner electrons there are, the more the nucleus is shielded + the less energy is required to ionise the outer electron.

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16
Q

3.1.1
Why does the first ionisation energy decreases down a group?

A
  • Atomic radius ↑ - outer electrons are further away from the nucleus (at higher energy levels), so held less strongly by nucleus.
  • Shielding from nuclear charge by inner electrons ↑
  • Less energy required to remove the electrons ∴ 1st ionisation energy ↓
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17
Q

3.1.1
Why does the first ionisation energy increases across a period?

A
  • Atomic radius ↓- outer electrons are closer to the nucleus (at lower energy levels), so held more strongly by nucleus.
  • Nuclear charge ↑
  • Shielding by inner electrons is the same (as outer electron is in the same energy level).
  • More energy required to remove the electrons ∴ 1st ionisation energy ↑
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18
Q

3.1.1
Why is the 1st ionisation energy lower than expected for elements in Group 3 and Group 6

A

Atoms of elements in group 3 and 6 show a lower 1st ionisation energy than would be expected. This gives us evidence that 2nd + 3rd energy levels are divided into 2 sub-shells, the 1st taking 2 electrons and the
2nd taking 6, ∴ proving sub-shells exist.
* Group 3 electrons have an s2p1 arrangement - Outer p1 electron is further from the nucleus. - Inner s2 electrons ↑ shielding so less energy is required to ionise the outer p1 electron
- ∴ 1st ionisation energy ↓
* Group 6 atoms have a p4 arrangement - The repulsion of 2 electrons in the same p orbital leads to less energy being required to ionise the outer electron.
- ∴ 1st ionisation energy ↓

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19
Q

3.1.1
Ionisation Energy in Group 1 and Group 0 elements

A
  • Atoms of group 1 elements have the lowest 1st ionisation energy in every period as they have the greatest atomic radius + the lowest nuclear charge in a particular period.
  • Atoms of group 0 elements have the highest 1st ionisation energy in every period as they have the smallest atomic radius + the highest nuclear charge in a period.
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20
Q

3.1.1
What are the Patterns in second ionisation energies?

A

Patterns in 1st ionisation energies are shifted one to the left when the patterns of 2nd ionisation energies are considered. Where a group 1 element would have the lowest 1st ionisation energy, it would have the highest 2nd ionisation energy.
- Group 1 elements have the highest 2nd ionisation energy in a particular period as the 2nd electron is being removed from an energy level closer to the nucleus. Group 2 elements have the lowest 2nd ionisation energy in a particular period

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21
Q

3.1.2
Relative Atomic Mass and Relative Molecular Mass

A

Relative atomic mass (Ar): the average mass of an atom of an element relative to 1/12 the mass of an atom of carbon-12.(Unit: none)
Relative atomic mass (Mr): the average mass of an atom of a molecule relative to 1/12 the mass of an atom of carbon-12.(Unit: none)

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22
Q

3.1.2
The following all have the same numerical value:

A
  • Relative molecular mass (Mr): the average mass of a molecule relative to 1/12 the mass of an atom of carbon-12 (carbon was chosen as it is a solid at room temp., non-toxic, and easy to separate).(Unit: none)
  • Relative formula mass (RFM) may be used for ionic compounds but Mr is accepted for all compounds. (Unit: none)
  • Molar mass: the mass of 1 mole. (Unit: gmol-1)
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23
Q

3.1.2
The Mole and the Avogadro Constant

A

Avogadro constant: the no. of particles in one mole = 6.02 x 1023

So, for solids and pure liquids (not solutions): mass/Ar = moles =

number of particles in Avogadro’s constant

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24
Q

3.1.2
Calculating Masses in Reactions

A

1) Calculate the number of moles of the species with enough data to do this.
2) Use stoichiometry of the equation to deduce the number. of moles of target species (the one we are being asked about).
3) Convert moles of target species to the mass.

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25
Q

3.1.2
The Ideal Gas Equation

A

The ideal gas equation is: pV = nRT, in this equation:
* p = pressure (1atm = 101325Pa = 101325Nm-2) (Unit: Pa)
* V = volume (1m3 = 1,000dm3 = 1,000,000cm3) (Unit: m3)
* n = no. of moles (k = kilo = x1,000; M = mega = x1,000,000) (Unit: mol)
* R = gas constant = 8.31JK-1 mol-1
* T = temperature (K = °C + 273; °C = K - 273) (Unit: K)

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26
Q

3.1.2
define: Empirical and Molecular Formula

A
  • Empirical formula: the simplest whole no. ratio of atoms of each element in a compound.
  • Molecular formula: the actual no. of atoms of each element in a compound.
  • Molecular formulae are simple whole no. multiples of empirical formulae.
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27
Q

3.1.2
Calculating Empirical Formula

A

1) Convert masses to moles: moles = mass/Ar
2) Divide each molar quantity by the smallest.
3) If one of the numbers are not close enough to round (not close enough to whole by 2 d.p.)multiply it by a factor accordingly.

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28
Q

3.1.2
Ionic Equations

A

Some chemical reactions (ones involving ionic compounds) are reactions between only some of the ions involved in a reaction. The balanced equation can be rewritten as an ionic equation leaving out the ions which do not take part in the reaction (Those non-partakers are called spectator ions)

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29
Q

3.1.2
What is Percentage Yield?
- how is it calculated
- which one is smaller and why ( actual or theoretical)

A

Percentage yield: the percentage of the theoretical yield which is achieved in the reaction.
* % yield = actual yield/theoretical yield x 100%
- Theoretical yield: the amount of product that we expect to obtain based on our calculations.
- Actual yield: the true amount of product obtained.
–> Actual yield is less than the theoretical yield because:
* loss by mechanical transfer (transferring from one container to another)
* loss during a separating technique, e.g. filtration
* side reactions occurring
* reaction not being complete

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30
Q

3.1.2
What is Atom Economy?

A

Atom economy: a measure of how efficiently the atoms in the reactants are used in a chemical reaction. It can be calculated as a % using the following equation: % atom economy = Mr of desired product/sum of Mr of all reactants x 100%

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31
Q

3.1.2
Concentrations

A

concentration = moles of solute/volume of solution (dm3)
This means: moles = [ ] x vol (dm3)
(The units of [ ] are moldm-3 - also written as M - and gdm-3)

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32
Q

3.1.2
Calculations Involving Gases

A

✴ Avogadro’s hypothesis: Equal volumes of any gas measured at the same temp and pressure contain the same number of moles of gas.

Molar volume: the vol. of 1 mole of gas under the conditions of the reaction. moles = volume/molar volume (mol. = V/Vm)

–> standard conditions
1) Standard Temperature and Pressure (STP), Vm = 22.4dm3 (273K (0°C), 1atm)
2) Room Temperature and Pressure (RTP), Vm = 24.0dm3
(298K (25°C), 1atm)

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33
Q

3.1.3
What is Melting?

A

The change of state from solid to liquid. Temperature that this occurs = the melting point m.p.)
* Energy is taken in when a substance melts to overcome bonds or intermolecular forces. The stronger the forces or bonds in a substance, the greater the energy required to melt the substance.

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34
Q

3.1.3
What is Freezing?

A

change of state from liquid to solid. Temperature that this occurs = the melting point also
* Energy is released when a substance freezes as forces or bonds are formed. The stronger the forces or bonds formed on freezing, the more energy is released.

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35
Q

3.1.3
What is Boiling?

A

Change of state from liquid to gas. Temperature that this occurs = the boiling point (b.p.)
* Energy is taken in when a substance boils. The stronger the bonds or intermolecular forces in a liquid substance, the greater the energy required to boil the substance.

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36
Q

3.1.3
What is Condensing?

A

change of state from gas to liquid. Temperature that this occurs = The boiling point (b.p.)
* Energy is released when a substance condenses. The stronger the bonds or forces formed on condensing, the more energy is released.

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37
Q

3.1.3
What is Subliming / sublimating?

A

Change of state from solid to gas on heating; change of state from gas to solid on cooling.
* Substances which sublime for the A-Level = solid iodine and solid carbon dioxide (solid carbon dioxide is known as ‘dry ice’)

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38
Q

3.1.3
What is Ionic Bonding?

A

The electrostatic attraction between oppositely charged ions in an ionic lattice. These ions are formed by electrons being
transferred from one atom to another.

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39
Q

3.1.3
What are the formulars of the 5 compound ions?

A

Formulae of Compound Ions:
Ammonium: NH4+
Carbonate: CO3 2-
Hydroxide: OH-
Nitrate: NO3-
Sulfate: SO4 2-

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40
Q

3.1.3
What is a Giant Ionic Structure?

  • Properties?
A

1) Particles of the 3-D lattice are cations & anions.
2) Strong electrostatic attractions hold cations & anions together.

Ionic compounds are hard, because the anions & cations are strongly attracted to each other therefore difficult to separate - but if force is applied and lattice shifts, same charges will repel

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41
Q

3.1.3
What are the Melting + Boiling Points of giant ionic lattices?

A

The giant ionic lattices are held together by a large number of strong electrostatic attractions between oppositely charged ions. Therefore, large amount of energy required to overcome the forces. That means ionic compounds tend to have high b.p.’s and m.p.s
* Importantly, the smaller the ions (in terms of actual size) and the higher the charge on the ions, the stronger the ionic bond.

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42
Q

3.1.3
What is the rules of general Electrical Conductivity?

A

To conduct electricity, they must:
1. The substance must contain charged particles.
2. The charged particles must be free to move throughout the substance.

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43
Q

3.1.3
Do ionic lattices conduct electricity?

A

Ions are charged particles. BUT, they are fixed in positions in the lattice by strong ionic bond. This means ionic compounds do not conduct in solid state.
* In molten state, or in an aqueous solution, the ions are free to move. So the solution containing the ions, e.g. salt water (NaCl in H2O), will conduct.

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44
Q

3.1.3
Explain ionic lattice solubility

A
  • Ionic compounds are polar molecules, this means they tend to dissolve in polar solvents, such as water
  • When moving water molecules hit an ionic lattice, they can knock ions off the outer layers
  • The water molecules then surround the ions, causing the substance to break up and dissolve
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45
Q

3.1.3
What is Covalent Bonding?

A

The electrostatic attraction between a bonding pair of electrons and the nuclei of the 2 atoms involved in the bond.
* Atoms make 1 covalent bond for each electron that it needs to get a ‘full’ shell.

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46
Q

3.1.3
What is a molecular covalent structure?

Give example…

A

1) Particles of the 3-D lattice are covalent
molecules.
2) The forces between the molecules are weak van der Waals forces.
- for example I2 (Iodine)

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47
Q

3.1.3
- Hardness
- Melting / Boiling Points of a molecular covalent structure

A
  • Relatively soft as they do not have strong intermolecular forces.
  • Relatively low because the van der Waals forces between the molecules are weak. This means it only takes a little energy required to overcome them.
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48
Q

3.1.3
- Solubility
- Electrical Conductivity
of a molecular covalent structure

A
  • They’ll dissolve in non polar solvents, because they themselves are not polar. This means they will not dissolve in water. Your typical solvent will be some non-polar organic substances.
  • Covalent things do not conduct electricity; covalent things don’t contain any charged particles which are free to move.
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49
Q

3.1.3
What is a Macromolecular (Giant Covalent) Structure?

Give examples…

A

1) The atoms are held together by strong covalent bonds in a giant network.
- for example diamond and graphite

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50
Q

3.1.3
- Hardness
- Melting / Boiling Points of a Macromolecular (Giant Covalent) Structure

A
  • Very hard due to many strong covalent bonds + rigid 3-D structure holding surface atoms in place.
  • You need very high temperatures to melt and boil giant covalent things. This is because ALL the strong covalent bonds holding the giant lattice have to be broken which needs a lot of energy. Compare this to simple covalent structures, where only VDW’s or Dipole Dipole IMF’s have to be broken and you quickly see why the M.P./B.P’s are so different
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51
Q

3.1.3
- Solubility
- Electrical Conductivity
of a Macromolecular (Giant Covalent) Structure

A
  • These lot are insoluble. All the strong covalent bonds of the giant covalent structure need to be broken to free the atoms and let them move through the solvent.
  • There are no charged particles, since the lattice particles are atoms, so giant covalent crystals are insulators. Graphite is an exception to this rule because it has a ‘spare electron’. It doesn’t really break the rule as an electron IS a charged particle and in graphite’s case, it can move!
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52
Q

3.1.3
What is Metallic bonding?

A

It is the electrostatic attraction between metal cations (metal positive ions, cat means +) delocalised electrons in a lattice

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53
Q

3.1.3
What is a metallic structure?

A

1) Particles of the 3-D lattice are positive ions surrounded by a ‘sea’ of delocalised electrons.
2) The positive ions are held together by metallic bonding; metallic bonding literally is the electrostatic attraction between metal cations and the associated delocalised electrons.

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54
Q

3.1.3
Metallic structure
- Hardness
- Solubility

A
  • Metals are typically hard, due to strong electrostatic attraction between metal cations and delocalised electrons.
  • Metals are Insoluble, except in other liquid metals. This is, again, because of the strength of metallic bonds
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55
Q

3.1.3
Metallic structure
- Melting + Boiling Points

A

Metals have relatively high M.P./B.P. because there are strong forces of attraction between cations and delocalised electrons
* The greater the number of delocalised electrons per atom, the stronger the metallic bond so therefore the higher the m.p./ b.p.
* The smaller the ions in the metallic lattice, the stronger the attraction between the delocalised electrons + the cations therefore the higher the m.p./b.p.
* Transition metals have much higher m.p./b.p. than the main group metals (group 1 and 2) due to the large number of d sub-shell electrons. These electrons may become delocalised, creating an even stronger metallic bond.
* This principle : smaller ion + more charge = stronger bond is the same as the ionic bonding principle

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56
Q

3.1.3
Metallic structure
- Conductivity

A
  • Metals have delocalised electrons in their structures, these can move. An electric current = the movement of charged particles.
  • Heat is conducted when particles can move and are close enough together to pass the heat energy from one to another. The delocalised electrons enable heat energy to be passed through the metal because they’re nice and close. So metals are good conductor of heat.
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57
Q

3.1.3
Metallic structure
- Malleable + Ductile

A
  • Because of the layered structure of the lattice, metallic layers can slide over each other without disrupting the bonding. This makes metals malleable
  • They’re also ductile (this means you can draw them out into a wire)
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58
Q

3.1.3
What is an alloy?

A
  • An Alloy is simply a mixture of metals.
  • Alloy’s aren’t malleable because different metals are different in size
  • This makes the structure of an Alloy irregular, so it is much more difficult for layers to slide.
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59
Q

3.1.3
What do shapes of Simple Molecules and Ions depend on?

A
  • The shapes of a covalent molecule or ion depends on the repulsion of the electrons around a central atom.
  • The electron pairs are charge clouds, areas where you have a big chance of finding an electron around an atom that repel each other.
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60
Q

3.1.3
What are the 2 types of electron pairs?

A

There are 2 types of electron pairs:
* Bonding pairs: pairs of electrons which are shared in covalent bonds.
* Lone (non-bonding) pairs: pairs of electrons which are unshared.

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61
Q

3.1.3
Explain the Valence Shell Electron Repulsion Theory

A

✴ Pairs of electrons in the outer shell of atoms arrange themselves as far apart as possible to minimise repulsion.
✴ Lone pairs are held closer to the central atom so they have a greater repulsive effect on the other pairs of electrons.
- This means lp–lp repulsion is greater than lp–bp repulsion, which is greater than bp–bp repulsion.

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62
Q

3.1.3
Shapes of Molecules
- What occurs?
- What determines the molecules’ shape?

A

The molecule or ion will take up a shape which minimises electron pair repulsions. The shape of the molecule is determined from:
* The total amount of electron pairs around a central atom.
* The amount of (sets of) bonding pairs of electrons.
* The amount of lone pairs of electrons.

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63
Q

3.1.3
Describe all of the 5 shapes without any lone pairs

A
  • Linear ; BP=2, LP=0, BA= 180
  • Trigonal planar ; BP=3, LP=0, BA=120
  • Tetrahedral ; BP=4, LP=0, BA= 109.5
  • Trigonal Bipyramidal ; BP=5, LP=0, BA= 90 + 120
  • Octahedral ; BP=6, LP=0, BA= 90
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64
Q

3.1.3
Describe all of the 4 shapes with lone pairs

A
  • Pyramidal ; BP=3, LP=1, BA= 107
  • Bent ; BP=2, LP=2, BA= 104.5
  • Trigonal planar ; BP=3, LP=2, BA= 120
  • Square planar ; BP=4, LP=2, BA= 90
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65
Q

3.1.3
What are Intermolecular forces?
Where do they exist?

A

These are forces that DON’T require the exchange of an electron. They exist because of bonds, they aren’t bonds themselves. They exist between molecule.

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66
Q

3.1.3
What is Electronegativity?

How is it measured?

A

The measure of attraction by an atom to a pair of electrons in a covalent bond.
* Electronegativity is measured on the Pauling Electronegativity Scale. A higher number on the scale means an element is better able to attract the bonding electrons in a covalent bonds; a higher number means it likes electrons more.

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67
Q

3.1.3
What 3 factors does electronegativity depend on?

A

1) The distance of the bonding electrons from the attractive power of the nucleus (atomic radius).
2) The size of the nuclear charge (i.e. atomic no.).
3) The attractive power of the nucleus being shielded by inner electrons - shielding

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68
Q

3.1.3
Electronegativity across a period…

A

INCREASES. This is because…
* atomic radius goes down across a period, giving a progressively stronger attraction between the positive nucleus and the 2 electrons in the covalent bond.
* nuclear charge goes up across a period. This will cause a greater attraction for the electrons in the covalent bond.

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69
Q

3.1.3
Electronegativity down a group…

A

DECREASES. This is because:
* atomic radius goes up down a group, giving a progressively weaker attraction between the positive nucleus and the electrons in the covalent bond.
* shielding of the nuclear charge increases down a group as there are more electrons in inner energy levels.

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70
Q

3.1.3
What is a non-polar covalent bond?

A

The electronegativities of the 2 atoms in the covalent bond are the same. This means that the bonding pair of electrons are shared equally, i.e. electron distribution is symmetrical. E.g. H-H, O=O etc.

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71
Q

3.1.3
What is a polar covalent bond?

A

One atom is more electronegative than the other. The electrons in the covalent bond will therefore be closer to one atom than the other, i.e. electron distribution is unsymmetrical. This means one atom will be partially positive and the
other partially negative. E.g. H-Cl, C-Cl, Be-Cl, C=O, etc.

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72
Q

3.1.3
What is a dipole?

A

It is a difference in electronegativity between 2 atoms caused by a shift in electron density in the bond.

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73
Q

3.1.3
What occurs in polar bonds?

A
  • In a polar bond, the difference in electronegativity between the 2 atoms causes a dipole.
  • If charge is distributed unevenly over a whole molecule, then the molecule will have a permanent dipole. Molecules that have permanent dipoles = polar molecules. Whether or not a molecule is polar depends on whether it has any polar bonds + its overall shape.
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74
Q

3.1.3
When do the bonds cancel each other out - preventing a charge?

A
  • two polar C=O bonds exactly cancel each other out, so the molecule has no permanent dipole moment (O=C=O)
  • If the polar bonds are arranged so that they all point roughly in the same direction + they don’t cancel each other out, then charge will be arranged unevenly across the whole molecule = molecule has a permanent dipole + is polar.
75
Q

3.1.3
What are Induced dipole–dipole (van der Waals /dispersion/London) forces?

A

These are attractive forces between all molecules and atoms caused by induced dipoles, which are temporary and constantly shifting.

76
Q

3.1.3
How do the Van der Waal forces work?

A
  • Electrons in charge clouds are always in rapid motion + at any one time they may be distributed more on one side of the molecule than the other. Another molecule approaching this side of the molecule will have
    its electrons repelled - this creates a temporary or induced dipole.
  • The 2 dipoles are then attracted to each other, and the 2nd dipole can cause yet another dipole in a 3rd atom.
  • larger the molecule ➜ higher the Mr ➜ greater the no. of electrons ➜ greater the induced dipoles ➜ greater the van der Waals forces between atoms
77
Q

3.1.3
What are Permanent dipole–dipole forces?

A

These are weak electrostatic forces of attraction between the δ+ end of the dipole on one molecule and the δ- end of the dipole on another neighbouring molecule.

78
Q

3.1.3
What are Permanent dipole–dipole forces responsible for?

A
  • The permanent dipole-dipole forces are responsible for holding together the molecules in polar substance, but there are also VDW’s forces between these polar
    molecules as well.
79
Q

3.1.3
Explain results of a charged rod and liquids experiment

A

If you put an electrostatically charged rod next to a jet of a polar
liquid (e.g. water), the liquid will move towards the rod - this is because polar liquids contain molecules w/ permanent dipoles.
* The more polar the liquid, the stronger the electrostatic attraction between rod + jet, so the greater the deflection will be.
* By contrast, liquids made up of non-polar molecules (e.g. hexane) will not be affected at all when placed near a charged rod.

80
Q

3.1.3
What is Hydrogen bonding?

A

Hydrogen bonds occur between a δ+ H atom which is covalently bonded to O, N or F of one molecule + the lone pair of electrons of an O, N or F atom of another molecule.
- Stronger than induced + permanent dipole-dipole forces.

81
Q

3.1.3
How does Hydrogen bonding work?
- What does Hydrogen have to be bonded with to be called H-bonding?

A
  • O, N + F are very electronegative, so they draw the bonding electrons away from the hydrogen atom.
  • This causes the bond to be very polarised, + because hydrogen has such a high charge density due to its small size, the hydrogen atoms form weak bonds w/ lps on F, N or O atoms of other molecules.
  • Only O-H and N-H groups are found in covalent molecules as H-F is only found in hydrogen fluoride.
82
Q

3.1.3
What are the properties of water?

A

The hydrogen bonds between water molecules explain many of its physical properties.
- Higher boiling point + melting point than would be expected for group 6 hydrides

83
Q

3.1.3
Explain why there is a higher boiling point + melting point than would be expected for group 6
hydrides?

  • Why would we think that, based on other hydroxides, water should have a lower b.p.?
A
  • The ↑ seen from H2S ➜ H2Se ➜ H2Te is caused by an ↑ in the van der Waals forces of attraction. As you descend group 6, each atom has more electrons so the induced dipoles increases. Hence van der Waals forces between molecules increase.
  • It would be expected based on the other hydrides that water should have a very low b.p. - this would be the case if the only intermolecular
    forces between water molecules were van der Waals forces of attraction
    + permanent dipole-dipole forces.
  • However, the hydrogen bonds between water molecules mean that it
    requires much more energy to break the bonds, hence giving it a higher b.p. + m.p.
84
Q

3.1.3
Why does ice have a lower density than water?

A

As liquid water cools to form ice, the molecules make more hydrogen bonds and arrange themselves into a more open, regular 3-D lattice structure.
- Since hydrogen bonds are relatively long, the average distance between water molecules is greater in ice than in liquid water - so ice is less dense than liquid water.

85
Q

3.1.4
What is Enthalpy change (∆H)?

A

The heat energy change measured under constant pressure. (Unit: kJ mol-1)

86
Q

3.1.4
What is an exothermic reaction?

A

These give out energy to their surroundings, so the temperature in the reaction goes up. The products of the reaction end up with less energy than the reactants. This means that the enthalpy change for the reaction, ∆H, will be negative.

87
Q

3.1.4
What is an endothermic reaction?

A

These take in energy from their surroundings, so the temperature in the reaction falls. The products of the reaction end up with more energy than the reactants. This means that the enthalpy change for the reaction, ∆H, will be positive.

88
Q

3.1.4
What are the standard conditions of enthalpy change?
- pressure
- temperature

A
  • Standard conditions are represented by the symbol Ɵ, (read ‘feye’) - Standard conditions are 100 kPa (about 1 atm) and 298K.
  • State symbols should always be included in chemical equations to represent enthalpy changes.
89
Q

3.1.4
Standard enthalpy change of reaction (∆rHƟ)?

A

The enthalpy change when a reaction occurs in the molar quantities shown in the chemical equation, under standard conditions with all reactants and products in their standard states
* E.g. CaO(s) + H2O(l) → Ca(OH)2(s)

90
Q

3.1.4
What is Standard enthalpy change of formation (∆fHƟ)?

A

The enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions.
* E.g. 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l)
- The enthalpy change of formation of an element is zero if the element is in its standard state. E.g. the enthalpy of formation of oxygen, O2(g) , is zero.

91
Q

3.1.4
What is Standard enthalpy change of combustion (∆cHƟ)?

A

The enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions w/ all reactants + products in their standard states.
* E.g. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

92
Q

3.1.4
What is Standard enthalpy change of neutralisation (∆neutHƟ)?

A
93
Q

3.1.4
What are calorimetry experiments?

A

They are experiments that measure the heat given out by reactions

94
Q

3.1.4
In a calorimetry experiment, for reactions that happen in solution….

A

Put the reactants (with a known conc or mass) in a polystyrene cup (to reduce heat loss or gain through the sides), that is in a beaker (for stability). Use a thermometer to measure the temperature of the mixture at regular intervals. Stir the solution to make sure it’s evenly heated.

95
Q

3.1.4
In a calorimetry experiment, for combustion reactions….

A

Energy given out by a combustion reaction can be calculated by measuring the temperature change it causes as it burns. To find the enthalpy of combustion of a flammable liquid, you burn it in a calorimeter

96
Q

3.1.4
Why is less energy always used up to heat the water

A
  • Heat is always lost to the surroundings, however well your calorimeter is insulated.
  • When you burn a fuel, some of the combustion that takes place may be incomplete
  • Flammable liquids are often quite volatile, so some of the fuel may be lost by evaporation.
97
Q

3.1.4
What are the 5 steps of graphical extrapolation, to work out the temperature change in calorimetry experiments?

A

1) Measure the temperature at regular intervals, beginning a couple of minutes before the start of the reaction so that you have recorded down temperatures before and after the reaction.
2) Plot the graph of your results.
3) Draw two lines of best fit: one going through the points before the reaction started, one going through the points after the reaction started.
4) Extrapolate both lines so that they both pass the time when the reaction started.
5) Distance between the 2 lines at the time the reaction started (before any heat was lost) is the accurate temp. change for the reaction.

98
Q

3.1.4
How do you calculate the Standard Enthalpy Change of Combustion?

A

1) Calculate the heat change, q, in a reaction. This is given by the equation q = mc∆T
* Where m (unit: g) is the mass of the substance that has a temp. change ∆T (unit: K or °C) + a specific heat capacity c (unit: JK-1g-1).
2) Change the units of q from J to kJ, because standard enthalpies of combustion are always given in units of kJmol-1.
3) Calculate the no. of moles of fuel that caused this enthalpy change, from the mass that reacted. Use the equation moles = mass/Mr
4) Calculate the standard enthalpy change of combustion, ∆cHƟ (in kJ mol-1), using q (in kJ), + the no. of moles of fuel that burned. Use the equation: ∆cHƟ = q / no. of moles reacting

99
Q

3.1.4
What is Hess’ Law?

A

✴ The enthalpy change is independent of the reaction route taken

100
Q

3.1.4
Hess’ Law explained

A

Hess’s Law says that the total enthalpy change for route 1 is the same as for route 2 - this is true so long as the starting and ending conditions are the same. Calculations using enthalpies of formation ∆H = [sum of ∆fHƟ products] - [sum of ∆fHƟ reactants]
* ∆fHƟ for elements is zero, as the element is being formed from the element.

101
Q

3.1.4
How do you calculate enthalpies of combustion?

A

∆H = [sum of ∆cHƟ reactants] - [sum of ∆cHƟ products]
* Remember that you can ignore ∆cHƟ for oxygen, because you can’t burn 1 mole of oxygen in oxygen.
E.g. calculate ∆fHƟ of ethanol using e.g. enthalpy values:
2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l)
∴ ∆H = ∆cHƟ [C] x 2 + ∆cHƟ [H2] x 3 - ∆cHƟ [C2H5OH]
= (-394 x 2) + (-286 x 3) - (-1367)
= -279 kJ mol-1

102
Q

3.1.4
What is a bond enthalpy?

A

It is the energy required to break a bond.

103
Q

3.1.4
What is the mean bond enthalpy?

A

It is the average energy required to break one mole of a covalent bond (in the gaseous state) averaged over a range of compounds containing the bond.

104
Q

3.1.4
Explain the bond enthalpies in exothermic and endothermic reactions?

A
  • Energy is required to break bonds ∴ bond breakage = endothermic ∴ ∆H = +ve.
  • Stronger bonds take more energy to break.
  • Energy is released to form bonds ∴ bond formation = exothermic ∴ ∆H = -ve.
  • Stronger bonds release more energy when they form.
  • The enthalpy change for a reaction is the overall effect of these two changes. If you need more energy to break bonds than is released when bonds are made, ∆H = +ve. If it’s less, ∆H = -ve.
105
Q

3.1.4
What are the general points for bond enthalpies

A
  • Generally the shorter the covalent bond, the stronger the bond.
  • Triple covalent bonds are generally shorter (+ hence stronger) than double covalent bonds which are shorter (+ hence stronger) than single covalent bonds.
106
Q

3.1.4
What is the equation to calculate Enthalpy Changes using Mean Bond Enthalpies?

A

∆H = sum of mean bond enthalpies of bonds broken - sum of mean bond enthalpies of bonds made

107
Q

3.1.4
When calculating enthalpy changes using the mean bond enthalpies, why are the reactions considered to be in gas phase?

A
  • The reactions are considered in the gas phase so that only the covalent bonds are involved in the calculations. In liquid or solid phase, intermolecular forces would also be involved.
108
Q

3.1.4
Why are the enthalpy change values using mean bond enthalpy values less accurate?

A
  • ∆H values calculated using this method will be less accurate than using enthalpy of formation or combustion data because the mean bond enthalpies are not exact.
    ∴ Any deviation of the answer to a calculation from the quoted value may be due to the mean bond enthalpy values used or the reactants not being in their gaseous states in the calculation.
109
Q

3.1.5
What does Collision theory state?

A

Collision theory states that reactions only occur when collisions take place between particles with sufficient energy - this minimum energy is called the activation energy.

110
Q

3.1.5
What is Activation energy (Ea)

A

It is the minimum amount of energy reacting particles must have for a successful reaction to take place.
- This means that most collisions do not lead to a reaction; only a small proportion of particles collide with energy that’s greater than the activation energy

111
Q

3.1.5
What is meant by rate of reaction?

A

It’s the change in concentration of a reactant or product in a given amount of time (concentration ÷ time)

112
Q

3.1.5
What are the 5 main factors that affect the rate of a reaction?

A
  • Concentration
  • Pressure
  • Temperature
  • Surface area of solid reactants
  • Presence of a catalyst
113
Q

3.1.5
How does concentration affect Rate of reaction?

A
  • If concentration of a reactant is increased,
  • More particles of that reactant are present in a given volume, leading to more successful collisions between reactant particles in a given period of time
  • This causes the rate of reaction to increase
  • Concentration UP = Rate UP
114
Q

3.1.5
How does pressure affect Rate of reaction?

A
  • If pressure on a gaseous reaction is increased,
  • The particles are forced closer together,
  • Leading to more successful collisions between reactant particles in a given period of time,
  • This causes the rate of reaction to increase
  • Pressure UP = Rate UP
115
Q

3.1.5
How does temperature affect Rate of reaction?

A
  • If temperature goes up, the particles gain more KE and move faster,
  • Leading to more frequent successful collisions between reactant particles in a given period of time, this causes the rate of reaction to increase
  • Temperature UP = Rate UP
116
Q

3.1.5
How does the surface area of solid reactants affect Rate of reaction?

A
  • If the surface area of solid reactants increases, the exposed surface of the reactants increases too.
  • This increases the number of successful collisions in a given period of time
  • Which increases the rate of the reaction.
  • Surface area UP = Rate UP
117
Q

3.1.5
How does the presence of a catalyst affect Rate of reaction?

A
  • A catalyst increases the rate of reaction without being used up.
  • A catalyst works by providing an alternative reaction pathway of lower activation energy.
  • By lowering the activation energy, more collisions are successful in a given period of time, so the rate of reaction increases.
  • Catalysts are important in many industrial processes by speeding up the process + so reducing costs. - Catalyst added = Rate UP
118
Q

3.1.5
What is a catalyst?

A

It is a substance that increases the rate of a chemical reaction without being changed in chemical composition or amount.

119
Q

3.1.5
What is the Maxwell-Boltzmann distribution Graph?

A

a plot of the number of gaseous molecules
against the energy they have at a fixed temperature.
* A single plot on the graph shows the distribution of molecular energies
at a constant temp.

120
Q

3.1.5
On the Maxwell-Boltzmann distribution Graph what does the peak show?

A

The peak of the curve represents the most likely energy of any single molecule
- most of the molecules are moving at a moderate speed so their energies are in the middle

121
Q

3.1.5
On the Maxwell-Boltzmann distribution Graph what does the dotted line tot he right of the peak show?

A

It shows the mean (average) energy of all the molecules is a bit to the right of the peak

122
Q

3.1.5
On the Maxwell-Boltzmann distribution Graph what is different about the molecules that are after the activation energy line?

A

Some molecules have more than the activation energy.
- These are the only particles/molecules that can react

123
Q

3.1.5
On the Maxwell-Boltzmann distribution Graph why does the curve start at (0,0)?

A

The curve starts at (0,0) because no molecules have zero energy

124
Q

3.1.5
What is the effect of different temps. on the Maxwell-Boltzmann Distribution graph?

A
  • When temp. of the reactant molecules is increased, this will ↑ the energy of the gaseous reactant molecules, changing the shape of the Maxwell Boltzmann distribution.
  • A small ↑ in temp. can cause a large ↑ in the rate of reaction as there is a significant ↑ in the no. of molecules w/ enough energy to undergo a successful collision.
  • At higher temps., more molecules have at least the activation energy
125
Q

3.1.5
What should sketching a distribution for the same sample of gas at a lower or higher temp. on the Maxwell-Boltzmann Distribution graph?

A
  • Lower temp. distributions are moved to the left + the peak is higher.
  • Higher temp. distributions are moved to the right + the peak is lower.
  • The graphs should only cross once.
  • The curves should always start at the origin + should end up being asymptotic to the energy axis at higher energy levels.
126
Q

3.1.5
What is the effect of different temps. on the Maxwell-Boltzmann Distribution graph?

A
  • Increasing the conc. means there are more molecules available to collide in a given volume . There will also be more molecules with energies above the Ea
  • Rate of reaction increases with an increase in concentration, but the effect is much less significant than is achieved when increasing the temperature. This is because only a few more reactant molecules with enough energy to undergo a successful collision.
127
Q

3.1.5
What should sketching a distribution for the same sample of gas at diff. conc. on the Maxwell-Boltzmann Distribution graph?

A
  • The curve retains the same basic shape. This means that the most probable energy of the molecules remains the same so the peak should be higher but at the same energy value on the horizontal axis.
  • As there are more reactant molecules at the same temp., the overall area under the curve increases.
  • This increases the number of reactant molecules which have enough energy to undergo a successful reaction, leading to an increases in the rate of reaction.
128
Q

3.1.5
What is the effect of a catalyst on the Maxwell-Boltzmann Distribution graph?

A

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy.
* The area above the activation energy with a catalyst is greater, indicating that there are more molecules with sufficient energy to undergo a successful collision. This increases the rate of reaction.
- more molecules have at least the activation energy in the catalysed reaction

129
Q

3.1.5
What should sketching a distribution for the same sample of a catalyst on the Maxwell-Boltzmann Distribution graph?

A
  • Adding a catalyst to a reaction mixture lowers the activation energy
    *A catalyst does not affect the shape of the distribution as long as the temperature and total number of molecules are not changed.
  • the line of the activation energy point will be moved more to the left
130
Q

3.1.6
What is dynamic equilibrium?
- what occurs

A
  • The forward and reverse (backwards) reactions proceed at equal rates
  • the concentrations of reactants and products remain constant.
    A dynamic equilibrium can only happen in a closed system which is at constant temperature.
131
Q

3.1.6
What is Le Chatelier’s Principle?

A

✴ If a factor is changed which affects a system in equilibrium, the position of equilibrium will shift in the direction opposing that change.

132
Q

3.1.6
According to Le Chatelier’s Principle, what happens to the position of equilibrium when pressure changes?

A
  • An increase in pressure at constant temperature shifts the position of equilibrium to the side with fewer moles of gas. This is opposing the change by decreasing the pressure of the system.
  • Logically, a decrease in pressure at constant temperature shifts the position of equilibrium to the side with more moles of gas. This opposes the change by increasing the pressure.
133
Q

3.1.6
According to Le Chatelier’s Principle, what happens to the position of equilibrium when temp. changes?
- For a reaction in equilibrium at constant pressure where the forward
reaction is exothermic

A
  • an increase in temperature would shift the position of equilibrium in the direction of the reverse endothermic reaction. This moves the position of equilibrium from right to left in order to oppose change, because the reverse reaction absorbs heat. (it’s endothermic!)
  • a decrease in temperature would shift the position of equilibrium in the direction of the forward exothermic reaction. The position of equilibrium would move from left to right, opposing the change and increasing temperature.
134
Q

3.1.6
According to Le Chatelier’s Principle, what happens to the position of equilibrium when temp. changes?
- For a reaction in equilibrium at constant pressure where the forward
reaction is endothermic

A
  • an increase in temperature would shift the position of equilibrium in the direction of the forward endothermic reaction and the position of equilibrium would move from left to right in order to oppose change. Decreasing the temperature by absorbing heat.
  • a decrease in temperature would shift the position of equilibrium in the direction of the reverse exothermic reaction and the position of equilibrium would move from right to left. This opposes the change and increases the temperature by giving out heat.
135
Q

3.1.6
According to Le Chatelier’s Principle, what happens to the position of equilibrium when concentration changes?

A
  • An increase in concentration of the reactants would shift the position of equilibrium from left to right in order to oppose change by increasing the yield of products and, therefore, decreasing the concentration of reactants.
  • An increase in concentration of the products would shift the position of equilibrium from right to left in order to oppose change by increasing the yield of reactants and therefore decreasing concentration of products.
136
Q

3.1.6
What is the effect of a catalyst being used in a reversible reaction?
- in terms of dynamic equilibrium

A
  • A catalyst increase the rate of the forward and reverse reactions equally therefore allowing equilibrium to be reached faster.
  • However, it has no effect on the position of equilibrium.
137
Q

3.1.7

A

Ethanol can be produced by the hydration of ethene with steam, which is a
reversible reaction:
C2H4(g) + H2O(g) ⇌ C2H5OH(g) ∆H = -46kJmol-1

138
Q

3.1.7
What 5 conditions are used to increase + maximise yield of ethanol?

A

1) Concentrated phosphoric acid catalyst.
2) High pressure - (60-70atm.)
3) Temperature of 300°C
4) Excess ethene/steam can be used
5) Ethanol can be condensed out of the reaction mixture

139
Q

3.1.6
How does concentrated phosphoric acid catalyst increase + maximise yield of ethanol?

A
  • Allows equilibrium to be attained more rapidly as catalyst increasing the rate of reaction (hence also allowing lower temperature to be used, meaning lower energy costs).
  • No effect on position of equilibrium, therefore no effect on yield of product
140
Q

3.1.6
How does High pressure - (60-70atm.) increase + maximise yield of ethanol?

A
  • Increases equilibrium yield of ethanol (as 2 moles on the left and 1 mole on the right, therefore position of equilibrium shifts to the right) and increased rate of reaction (think collision theory)
  • However, too high a pressure is expensive (increased electrical pumping costs and requires expensive equipment to withstand the pressure.)
141
Q

3.1.6
How does a temperature of 300°C increase + maximise yield of ethanol?

A
  • Forward reaction = exothermic, so increased temperature will shift position of equilibrium in the direction of the reverse endothermic reaction leading to a lower yield. SO lower temperature means higher yield.
  • However, a lower temperature gives a slower rate of reaction, again, think collision theory
  • So in conclusion: 300°C is a compromise temperature; it’s a compromise between high rate and high yield.
142
Q

3.1.6
How does excess ethene/steam can being used increase + maximise yield of ethanol?

A
  • Excess ethene/steam will shift the position of equilibrium to the right to remove the excess reactant so increase yield of ethanol.
  • Any unreacted ethene is recycled back into the reaction mixture to increases overall yield.
143
Q

3.1.6
How does Ethanol being condensed out of the reaction mixture increase + maximise yield of ethanol?

A
  • If a product is removed from an equilibrium reaction, the position of equilibrium will shift to the right to replace it, therefore increasing the overall yield.
144
Q

3.1.6
What occurs in Homogeneous and Heterogeneous Reactions?
- what are they

A
  • A homogeneous reaction is one in which all the reactants and products are in the same physical states.
  • A heterogeneous reaction (hetro meaning different, geneous meaning state) is one in which all the reactants and products are NOT in the same physical states.
145
Q

3.1.6
Explain the equilibrium constant, Kc

A
  • A SQUARE BRACKET [] AROUND A CHEMICAL MEANS CONCENTRATION OF THAT CHEMICAL.
  • With that, at dynamic equilibrium, the ratio: [reactants]:[products] is constant. The ratio’s value is known as Kc. (Constant of concentration) It’s always the same for a particular reaction under the same conditions.
146
Q

3.1.6
What is the Kc equation?

A

Kc = [products]
[reactants]

147
Q

3.1.6
What does it mean if Kc is bigger than 1 and if is smaller than 1?
- would equilibrium lie to RH or LH

A
  • If Kc>1, [products]>[reactants] and we say that the equilibrium lies to the RH side.
  • If Kc<1, [products]<[reactants] and we say that the equilibrium lies to the LH side.
148
Q

3.1.6
What type of reactions can Kc be calculated for?

A

It can be calculated for reactions in solution OR homogeneous gaseous reactions, as the concentration of a solution or gas can be calculated as the number of moles in a certain volume (in dm3).

149
Q

3.1.6
How would Kc be written out for the general reaction: aA + bB ⇌ cC + dD?

A

For the general reaction:
aA + bB ⇌ cC + dD
Kc = [C]c[D]d
[A]a[B]b

150
Q

3.1.6
How are the units written in the Kc equation?

A

The units are in terms of conc. in mol dm3 but the overall power depends on the balancing no.s in the eq. for the reaction.
SO, units of Kc = (mol dm-3)(c+d)
(mol dm-3)(a+b)
Therefore, if Kc has no units: there are equal numbers of moles on both
sides of the reaction equation.

151
Q

3.1.6
What factors effect Kc?

A

Kc will only change when temperature changes - other factors have no effect on Kc.

152
Q

3.1.7
What are oxidation States?

A

Oxidation state: the hypothetical charge an atom has, assuming that the bonding is completely ionic.
* It is a numerical value of the ‘degree of oxidation or reduction’ of an atom.

153
Q

3.1.7
What are the rules for assigning Oxidation States?

A
  • Uncombined elements have an oxidation state of 0. (An atom of lithium, for example, has an oxidation state of 0)
  • Elements bonded to identical atoms also have an oxidation state of 0. (for example both oxygen atoms in the Oxygen-Oxygen / O2 molecule)
  • The oxidation state of a simple monatomic ion is the same as its charge. (For example, an Li+ ion has an oxidation state of 1)
154
Q

3.1.7
What must the oxidation states of all elements in a compound and all elements in a molecular ion, add up to?

A
  • The total sum of the oxidation states of all the elements in a compound must add up to zero if it’s not an ion. (for example: NaCl, Sodium is Na+ and chlorine (chloride) is Cl- the oxidation state of sodium is +1 and chlorine (chloride) is -1. +1 + -1 = 0)
  • The total sum of the oxidation states of all the elements in a molecular ion must add up to the charge on the ion. (for example: SO42-. Oxygen has a default 2- oxidation state, so 4 x -2 = -8 and, therefore, sulphur has an oxidation state of +6)
155
Q

3.1.7
What are the oxidation states of the groups when in compounds?

A
  • Following on from that example, Oxygen has an oxidation state of -2 in all compounds except in peroxides, where it has an oxidation state of -1 and in OF2 where it has an oxidation state of +2.
  • Hydrogen has an oxidation state of +1 in all compounds except in hydrides where it has an oxidation state of -1.
  • Group 1 elements have an oxidation state of +1 in all compounds.
  • Group 2 elements have an oxidation state of +2 in all compounds.
  • The oxidation states of transition elements and p block elements vary.
  • In a simple binary compound (2 elements), the more electronegative element has the negative oxidation state.
156
Q

3.1.7
How do you find Oxidation States from Formulae?

A

E.g. find the oxidation state of Zn in Zn(OH)2.
* Zn(OH)2 has an overall charge of 0 therefore overall oxidation state = 0
* Oxidation state of O = -2 and oxidation state of H = +1
* Hence oxidation state of Zn in Zn(OH)2 = 0 - 2 x (-2 + 1) = +2

157
Q

3.1.7
How do you find oxidation States from Systematic Names?

A
  • The charge of an ion of a d block element is given using Roman numerals. They decided to be annoying and can have multiple oxidation states (e.g. copper(II) = +2, in the same breath copper(III) = +3).
158
Q

3.1.7
What is Oxidation and Reduction?
- OILRIG

A

An increase in oxidation state is oxidation; a decrease in oxidation state is reduction.
* Oxidation: the process of electron loss.
* Reduction: the process of electron gain.
* OIL RIG: Oxidation is loss, Reduction is gain (of electrons)
A redox reaction is one in which an oxidation + reduction occur simultaneously in the same reaction.

159
Q

3.1.7
Explain REDOX example:
- 6HI + H2SO4 → 3I2 + S + 4H2O

A

In the reaction: 6HI + H2SO4 → 3I2 + S + 4H2O
* Iodine (I) is oxidised, it shows an increase in oxidation state: from -1 (in HI) to 0 (in I2).
* S is reduced as it shows a decrease in oxidation state: from +6 (in H2SO4) to 0 (in S).
* This means (UNBELIEVABLY!) that this is a redox reaction; both oxidation and reduction occur together in the same reaction.

160
Q

3.1.7
What is a disproportionation reaction?
- give an example, Cl

A

In the reaction: Cl2 + H2O → HCl + HOCl
* Cl is oxidised as it shows an increase in oxidation state from 0 (in Cl2) to +1 (in HOCl).
* Cl is also reduced as it shows a decrease in oxidation state from 0 (in Cl2) to -1 (in HCl).
* This is a redox reaction, but a special type called a DISPROPORTIONATION REACTION, as chlorine is oxidised and reduced in the same reaction; for it to be a disproportionation reaction, the same species must be oxidised and reduced.

161
Q

3.1.7
What are Oxidising Agents and Reducing Agents?
- also in terms of electrons

A
  • Oxidising agent: an electron acceptor. They cause another substance to lose electrons. They oxidises another thing, but are reduced themselves
  • Reducing agent: an electron donor. - They cause another substance to gain electrons. They reduce another thing, but are oxidised themselves
162
Q

3.1.7
What do half-equations show?
- give examples

A

Half-equations show oxidation or reduction. The electrons are shown in a half-equation so that the charges balance.
E.G.
* Oxidation of iron: Fe → Fe3+ + 3e-
* Reduction of chlorine: Cl2 + 2e- → 2Cl

163
Q

3.1.7
How do you combine half-equations for oxidising and reducing agents?

A

Half-equations for oxidising and reducing agents can be combined to make full equations.
- However, make sure before combining the two that both half-equations have the same number of electrons in them (when they do, combine them and cancel out on each side).

164
Q

3.1.7
How can half-equations be written to identify oxidation and reduction processes?

A

Half-equations can be written to identify the oxidation and reduction processes in redox reactions.
- When doing so, make sure the atoms and charges balance.

165
Q

3.1.7
In what situations must H2O and H+ ions need to be added when balancing?

A

Sometimes half-equations need to be written for a more complicated reaction where the oxidising or reducing agent contains oxygen or hydrogen. If so, H2O and H+ ions may need to be added to balance the equation.

166
Q

3.1.9
What are rate equations?
- what is the equation

A

The rate (concentration change ÷ time change) of a chemical reaction is related to the concentration of reactants by a rate equation of the form:
Rate = k [A]m [B]n

167
Q

3.1.9
What is order of reaction?

A

The power to which the concentration of a reactant is raised in the rate equation.
➜ The orders m and n are restricted to the values 0, 1, and 2 for the exam.

168
Q

3.1.9
What do the letters stand for in the rate equation?

A
  • Where m and n are the orders of reaction with respect to reactants A and B. K is known as the rate constant, it’s the same for a given temperature.
169
Q

3.1.9
What is rate constant?

A

The proportionality constant which links the rate of reaction to the concentrations in the rate equation

170
Q

3.1.9
Use B + C + D → E + F as an example and put into a rate equation.
- what does it mean

A

rate = k [B]2 [D]
This means:
* the order of reaction with respect to B is 2.
* the order of reaction with respect to D is 1.
* the order of reaction with respect to C is 0.

171
Q

3.1.9
Why does [C] not appear in the rate equation?

A
  • This is because [C] does not appear in the rate equation; if you do ALevel maths you’ll know anything ‘to the power of 0’ is — having [C]0 is the same as multiplying everything by 1, so we leave it out.
172
Q

3.1.9
What is the overall order?

A

The overall order is all the ‘orders’ added up: 2 + 1 + 0 = 3

173
Q

3.1.9
What does changing the concentration of each letter in the rate equation do to the rate?

A
  • Doubling the conc. of B would quadruple the rate as (2)2 = 4. - Doubling the conc. of C would have no effect on the rate as (2)0 = 1. - Doubling the conc. of D would double the rate as (2)1 = 2. - Doubling the conc. of B, C, and D would increase the rate by a factor of (2)3 = 8.
174
Q

3.1.9
What are the units of the rate constant?

A
  • The units of rate are mol dm-3 s-1 | I.E. concentration ÷ time
  • the units of conc. are mol dm-3
  • The units of the rate constant K depend on equation
175
Q

3.1.9
Using this equation, Rate = k [A]2 [B], what are the units?

A

➜ the overall order = 3
➜ If we replace everything with units:
➜ mol dm-3 s-1 = k (mol dm-3)3
➜ expand those brackets:
➜ mol dm-3 s-1 = k * (mol dm-3 * mol dm-3 * mol dm-3)
➜ Simplify
➜ mol dm-3 s-1 = k * mol3 dm-9
➜ Then rearrange that equation so we have “k = something”
➜ k = mol dm-3 s-1 ÷ mol3 dm-9
➜ When you divide by an index (power) you take them away So,
➜ K = mol dm-3 s-1 ÷ mol3 dm-9 BECOMES: K = = mol-2 dm6 s-1
➜ Final units for K
➜ mol-2 dm6 s-1

176
Q

3.1.9
Using the equation, Rate = k [A]2 [B], what is the rate constant when the overall order is 1 and when it is 2?

A
  • Therefore: when overall order = 1, units of rate constant = s-1
  • And when overall order = 2, units of rate constant = mol-1 dm3 s-1
177
Q

3.1.9
What is the effect of temperature on the rate constant?

A

The rate constant is dependent on temperature; if a constant temperature is not been maintained, the rate will vary based on changes in concentrations of reactants as well as temperature
* As temperature increase, the rate constant increases too.

178
Q

3.1.9
How does the Arrhenius equation link the rate constant?

A

The Arrhenius equation links the rate constant with activation energy and temperature. It’s written below:
k = Ae–Ea/RT

179
Q

3.1.10
How do you calculate Mole Fractions?

A
  • The mole fraction of one of the gasses, A, in a mixture of gasses is written as: xA.
  • It’s calculated by dividing the amount, in moles, of A by the total amount of moles of gas in the mixture.
180
Q

3.1.10
What is partial pressure?

A

The partial pressure of one of the gasses in a mixture of gases is the contribution which that gas makes to the total pressure of the gas mixture.
* The partial pressure of A is denoted p(A) and its units are Pa (or kPa
= 103 Pa)

181
Q

3.1.10
What is the Equilibrium Constant (Kp)?
- how do you calculate it

A

The equilibrium constant Kp is deduced from the equation for a reversible reaction occurring in the gas phase. Kp is calculated from partial pressures for a system at constant temperature (in a very similar way to Kc)

182
Q

3.1.10
Turn this equation, aA(g) + bB(g) ⇌ cC(g) + dD(g), into the Kp formula
- show the units

A

Kp = p(C)^c p(D)^d/p(A)^a p(B)^b
Therefore, units of Kc = (Pa)^(c+d)/(Pa)^(a+b)

183
Q

3.1.10
What are the qualitative effects of Temperature changes on Kp?

A
  • Kp is affected by changes in temp. only.
184
Q

3.1.10
What are the qualitative effects of Temperature changes on Kp?

A
  • Whilst a catalyst can affect the rate of attainment of an equilibrium, it does not affect the value of the equilibrium constant.

AQA Chemistry A level (3 decks)

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