Physical Chem

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General advice for students in physical chemistry:

Develop an inquisitive mind. Seek an intuitive understanding of the knowledge that you have learnt, and you will be able to apply the fundamental natural principles to solve physical chemistry problems pretty easily.

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0. List of Chemistry definitions

List Definition Chemistry

1. ATOMS, MOLECULES AND STOICHIOMETRY – FAQ

Q: How do we calculate empirical formula?
o Empirical formula of a compound may be calculated from experimental data

Composition by mass in a compound

Combustion data

Q: What is the empirical formula used for?
o If the molar mass or Mr of the compound is also known as well as the empirical formula,
we can determine the molecular formula from the empirical formula.

Q: How to determine the limiting reactant and yield of product?
1. Write a balanced equation for the reaction.
2. Calculate the number of moles of each reactant (i.e. A and B) available.
3. Assume reactant A is limiting, calculate the amount of reactant B needed to react completely
with A. Hence, identify the limiting reactant.
4. Use the number of moles of the limiting reactant to calculate the theoretical yield of product.

Q: What  are the steps in back titration calculations?
0. Write balanced equations for the reactions (if it is not given in question).
1. Calculate the initial amount of reactant A added.
2. Calculate the amount of reactant A remaining (from amt reacted with standard reagent C).
3. Calculate the amount of reactant A that reacted with reactant B
(initial known amt of A – amt reacted with C i.e. answer from Step 1 – answer from Step 2)

2. ATOMIC STRUCTURE – FAQ

Q: Why is the angle of deflection directly proportional to charge of the particle?
A: The greater the charge of the particle, the greater would be the attractive force exerted
on it from the oppositely charged plate. Hence greater would be the deviation from its
original direction of motion.

Q: Why is the angle of deflection inversely proportional to the mass of the particle?
A: If two particles are moving at the same speed but one is more massive than the other,
the heavier particle would have a greater kinetic energy. Thus it requires more energy
to be exerted on the heavier particle to cause it to deflect. Since the applied electric
field is exerting the same amount of force on these two different particles with different
masses, the heavier particle would be deflected to a lesser extent.

Q: Why do isotopes react similarly?

A: This is because in chemical reaction, it is the electrons that are transferred between different

atoms; atoms either gain or loose or share electrons. The nucleus remain intact

Q: What is the meaning of this symbol 8O2–?

A: This represents an oxide ion. The subscript on the left is the atomic number whereas
the superscript on the right indicates the extra electrons the oxygen has acquired. For
this 8O2– species, the number of protons (8) is not equal to the number of electrons
(10). The number of electrons would be the same as the number of protons for a
neutral atom only.
Q: What is the meaning of this symbol O2 2–?
A: This represents a peroxide ion. The subscript on the bottom right indicates that there are
2 oxygen atoms in this species covalently bonded to each other. The superscript on the
right indicates the number of extra electrons this species have acquired.

Q: It seems that the relative isotopic mass is a whole number. Is it really so?
A: The word ‘relative’ means that the value obtained is measured with respect to some
other thing, hence this number has no physical unit. In this case here, the relative
isotopic mass of an isotope is the mass measured with respect to 1/12 of the mass of a
12C atom, which has a value of 1 unit (1/12 × 12). As the mass of an atom arises mainly
from the nucleus and since the total number of nucleons in 35Cl is 35, the relative
isotopic mass seems to have a value of 35 too. But in reality, the actual relative isotopic
mass is less than the mass of all nucleons added up. This phenomenon is known as
mass defect. The differences in the two masses (less than 1%) arise because part of the
mass has been converted to binding energy (according to E = mc2) which is necessary
to hold the nucleons together.

Q: Why does the energy of an electron increase as we move away from the nucleus?

A: By convention, when an electron is ‘free’, that is, not subjected to any other
electrostatic interactive forces (attractive or repulsive), it has an assigned zero energy.
This is when the electron is infinitely away from the nucleus. But now if you want to
bring an electron from n=1 Principal Quantum Shell to infinity, you got to do work
against the attractive force of the nucleus, you got to ‘break’ the ‘bond’ between the
electron and the nucleus. Breaking bond needs energy. The energy that you put in while
doing work (energy is conserved from Law of Conservation of Energy) is gained by
this electron, hence its energy has increased. When an electron moves from infinity and
is attracted by the nucleus, a ‘bond’ is formed, and energy will be released. The
following diagram would help in the explanation.

Q: Why does the 4s sub-shell have a lower energy than the 3d sub-shell given that n=4
Principal Quantum Shell should have a higher energy than n=3 Principal Quantum
Shell ?
A: n=4 should have a higher energy than n=3. But the s sub-shell has a relatively lower
energy than the d sub-shell for the same Principal Quantum Shell. These 2 different
opposing factors counteract each other, resulting in the 4s sub-shell having a lower
energy than the 3d sub-shell. The same explanation account for the relative energies of
the 5s and 4d sub-shells.

Q: Does that mean that the 4s sub-shell is now closer to the nucleus than the 3d sub-shell?
A: No. On the average, n=4 Principal Quantum Shell is still further away from the nucleus
than n=3. So although the energy of 4s sub-shell is lower than the 3d sub-shell, it does
not imply that the distances have changed. Remember a Principal Quantum Shell is
actually a band, not a single discrete line.

Q: Why do we need to know the electronic configuration of an element?
A: Knowing the correct electronic configuration would enable us to know which electron
is to be removed and which orbital does it reside in. This is important as the removal of
different electrons from different orbitals need different amounts of energy.

Q: What is electron spin?
A: You can imagine an electron like the Earth, rotating at a particular axis.

Q: Why can’t two electrons in the same orbital have the same spin?
A: Well, when an electron spins, this spinning charged-particle creates a magnetic field. If
two electrons spin in the same direction, the magnetic field created would be repulsive
in nature and the energy level of these two electrons would be higher as compare to if
they spin in opposite directions to create an attractive magnetic field. An analogy can
be used here: Picture a spinning electron as moving in one particular direction. Two
such spinning electrons will be moving in opposite directions and the chances of them
‘meeting’ will be lower. This results in lesser inter-electronic repulsion.

Q: Why do we need to first place electrons in empty orbitals of the same sub-shell before
pairing them in an orbital?
A: Well, electrons repel each other. By occupying different orbitals, the electrons remain
as far apart as possible from one another, thus minimizing electron-electron repulsion.
Take note that each orbital represent a particular region of space, hence two different
orbitals would be two different region of space separated from each other. Take for
instance the three p orbitals in a p sub-shell, each is oriented perpendicularly from each
other, occupying different regions in space.

Q: Why is the electronic configuration of 25Mn not 1s2 2s2 2p6 3s2 3p6 3d7?
A: The 4s orbital is filled first before the 3d orbitals. This is because the 4s orbital has a
lower energy level as compared to the 3d orbitals.

Q: If the 4s is filled before the 3d, why the electronic configuration of 25Mn not
1s22s2p63s23p63d54s23d5?
A: The electronic configuration is always written in the order of increasing Principal
Quantum Number. This also indicates the order of increasing energy level of the
various sub-shells.

Q: Why is symmetrical distribution of similar charge preferred?
A: If similar charges are distributed symmetrically, this would mean that all charges are
spread out evenly and as far apart as possible. Such situation would result in a similar
amount of electrostatic repulsion at each point in space. Consequently, such state
would have a lower energy as compared to a state of asymmetrical distribution.

Q: Why must the atom be in the gaseous state?
A: When we carry out ionization, the species must be gaseous atoms. In the gaseous
state, the atoms have very minimal interaction with each other. Thus the energy input
would solely be responsible for removing the electron and not in overcoming other
types of bond. So remember that the gaseous state symbol is very important here.

Q: What is a valence electron?
A: Valence refers to the outermost. Thus a valence electron ‘sits’ in the outermost
Principal Quantum Shell and is furthest from the nucleus. The Principal Quantum
Number for the valence shell corresponds to the Period Number of the element. All
other Principal Quantum Shell of electrons before the valence Principal Quantum
Shell are known as the inner core electrons.

Q: What is the Effective Nuclear Charge ENC on each of the valence electrons of an oxygen atom?
A: The electronic configuration of an O atom is 1s22s22p4. The nuclear charge consists of 8
protons and the number of inner core electrons is 2 (since there is only one Principal
Quantum Shell of electrons before the n=2 valence shell), therefore the ENC is ≈ 6.

Q: Does that mean that each of the 6 valence electrons is attracted by 1/6 of the ENC,
which is 1 proton?
A: No. Each of the 6 valence electrons is attracted by 6 protons. This is because the 6
valence electrons are moving round the nucleus within the same distance from the
nucleus and the nucleus is considered a point charge. Therefore, the ENC is the same
on each of the valence electrons.

Q: Does that mean that the ENC on each the two 1s electrons is equivalent to 8 protons?
A: Yes. There are no other inner core electrons before the n=1 Principal Quantum Shell,
therefore there is no shielding effect. The ENC on each of the electron in the n=1 subshell
is the same, i.e. equivalent to 8 protons.

Q: The ENC of an O atom is greater than the N atom. So shouldn’t this cause the O atom
to have a higher 1st I.E.?
A: Yes, the ENC of O atom is greater than the N atom and this should cause it to have a
higher 1st I.E. But experimentally, it has been found that O atom has a lower 1st I.E that
what was expected. There must be some other factors that have yet to be considered.
The only reason we can use to explain the data collected would be to employ the interelectronic
repulsion factor. As shown here, there is the interplay of two opposing
factors but it seems that the inter-electronic repulsion is a more dominant factor than
the ENC effect.

Q: Why is the domineering effect of the inter-electronic repulsion over the ENC factor
being observed in the oxygen versus nitrogen case but not in the beryllium-lithium
scenario?
A: Remember we mention before that a p orbital is 1/3 the size an s orbital from the same
Principal Quantum Shell? Thus because of this different in size, the inter-electronic
repulsion is more prominent in a p orbital than the much bigger s orbital.

Q: The ENC of a B atom is higher than that of a Be atom but there is inter-electronic
repulsion faced by the electrons in the 2s sub-shell of Be. Are these two factors less
dominant than the ‘difference in energy level’ factor?
A: You are right. Based on the ENC factor, B is expected to have a higher 1st 1.E. than Be.
But this not what experimental data makes of it. If we are to consider inter-electronic
repulsion factor, Be should have a lower 1st I.E but observed data proves otherwise.
Hence, the only factor that could be used to account for the observed experimental data
would be because the 2p electron is at a higher energy level than the 2s electron.

Q: Why would attractive force be weaker, the further the electron is from the nucleus?

A: The electrostatic force (F) between the nucleus and electron can be approximated to F ∝1/r2,where r is the distance of separation between the charges. Thus, as r increases, the strength of the electrostatic force decreases drastically.

3. CHEMICAL BONDING – FAQ

Q: What are Chemical bonds?
– Binding forces of attraction between particles (atoms, ions or molecules) resulting in a lower energy arrangement.
– The formation of a bond involves the re-distribution of the outer electrons of the atoms concerned.

Q: What is the Octet Rule?
Atoms tend to lose, gain or share electrons until they are surrounded by eight valence electrons. Atoms try to achieve the same number of electrons as the noble gases closest to them in the Periodic Table.

Q: Explain which of the following ionic compound has the stronger ionic bond.
(i) sodium fluoride or sodium chloride
Sodium fluoride has stronger ionic bond as fluoride is smaller than chloride so that the shorter distance between Na+ and F– resulted in stronger ionic bond.

(ii) sodium fluoride or magnesium fluoride
MgF2 has stronger ionic bond as Mg2+ is more highly charged than Na+, resulting in greater electrostatic attraction in MgF2.

Q: Why is it that both NCl3 and PCl3 exist, but only PCl5 exist and not NCl5?
Such expansion of octet is observed in some compounds formed by elements of Period 3 (and beyond) This is due to the availability of vacant, low-lying orbitals. The energy required to promote an electron from 3s or 3p to 3d is not very large.

However elements in Period 2 (e.g. O and N) do not have low-lying vacant orbitals for expansion of octet. Promotion of electrons to the next quantum shell requires too much energy and hence Period 2 elements can accommodate only a maximum of eight valence electrons.

Point to note
A single covalent bond consists of one s bond.
• A double covalent bond consists of one s bond and one pie bond.
• A triple covalent bond consists of one s bond and two pie bonds.

Q: Why does a stream of chloromethane (polar substance) running from a burette be deflected towards a negatively charged rod?

The polar molecules will align themselves such that the d+ ends of the molecules will face the negatively charged rod. The electrostatic force of
attraction causes the stream to be deflected towards the rod.

Q: Explain the trend in the boiling points of the halogens.

The halogens from F2 to I2 are non-polar so that the intermolecular attraction between their molecules is due to instantaneous dipole–induced dipole (id−id)interactions.

The number of electrons of the halogens increases down the group from F2 to I2.\ Ease of polarisability of electron clouds and strength of id–id interactions also increase down the group, with I2 having electron clouds which are most easily polarised resulting in greatest id–id interactions.
Since boiling involves overcoming intermolecular forces, the boiling points of halogens increases down the group.

Ease of donation of lone pair of electrons on Y
Q: Why must Y be N, O or F?

To interact strongly with the d+ H on H-X, Y needs to be highly electronegative and the lone pair of electrons on Y must not be too diffuse in space. Hence, Y also needs to be small
must be N, O or F.

 

Do you know?

The fact that ice is less dense than water causes water to freeze downwards. This helps living organisms in a body of water to survive freezing conditions!

As the temperature of water near the surface drops, the density of water increases. Cold water sinks while warmer water, which is less dense, rises. This convection motion of water continues until the temperature of the water reaches 4 °C. Below this temperature, the density of water
decreases with decreasing temperature so that water colder than 4 °C no longer sinks. On further cooling, the water begins to freeze at the surface. The ice layer formed does not sink as it is less dense than water.

This layer of ice helps to insulate the water underneath from further heat loss, thus keeping the water below it from freezing solid. Hence, living things can survive in ponds and rivers even when the temperature falls below freezing.

Q: Explain why the boiling point of propane, dimethyl ether and ethanol deviates so greatly even though their electron cloud sizes are similar.

Since the electron cloud sizes of the three compounds are similar, the magnitudes of instantaneous dipole–induced dipole (id–id) interactions in the three compounds are similar.

Propane is non–polar so that the intermolecular forces between propane molecules are due only to id–id interactions.

Dimethyl ether is polar so that besides id–id interactions, permanent dipole–permanent dipole (pd–pd) interactions also exist between the molecules. Since boiling involves breaking intermolecular forces, more energy is required to overcome the stronger pd–pd interactions between dimethyl ether molecules compared to the weaker id–id interactions between propane molecules. Hence the boiling point of dimethyl ether is significantly higher than propane.

Ethanol contains a hydrogen atom covalently bonded to the small and highly

electronegative O atom so that hydrogen bonding exists between ethanol molecules. Since ethanol contains hydrogen bonding besides pd–pd interactions and id–id interactions, intermolecular forces between ethanol molecules are the strongest,hence the energy required to overcome intermolecular interactions in ethanol is greater than in propane and dimethylether. Hence ethanol has a much greater boiling point than both propane and dimethylether.

Q:  Explain why iodine has a higher boiling point than water even though iodine has only instantaneous dipole–induced dipole interactions and water can form hydrogen bonds.

The strength of instantaneous dipole–induced dipole interactions depends on the size of the electron cloud. Iodine has a much larger electron cloud than water so that the instantaneous dipole–induced dipole interactions between iodine molecules are significantly stronger than the hydrogen bonding between water molecules which have much smaller number of electrons.

Since boiling involves overcoming the intermolecular attractions between molecules, a greater amount of energy is required to overcome the id−id forces between iodine molecules compared to the id−id forces and the hydrogen bonds between water molecules.

Q: Explain why ionic compounds do not dissolve in non-polar solvents. 

Answer: The ion−solvent interaction is much too weak to overcome the
strong electrostatic attraction between the ions in the crystal lattice.

 

4. THE GASEOUS STATE

Content
 Ideal gas behaviour and deviations from it
 pV = nRT and its use in determining a value for Mr

Learning Outcomes
Candidates should be able to:
(a) state the basic assumptions of the kinetic theory as applied to an ideal gas
(b) explain qualitatively in terms of intermolecular forces and molecular size:
(i) the conditions necessary for a gas to approach ideal behaviour
(ii) the limitations of ideality at very high pressures and very low temperatures
(c) state and use the general gas equation pV = nRT in calculations, including the determination of Mr

5. CHEMICAL ENERGETICS

Definitions

1. The enthalpy change of formation, Hf , of a compound is defined as the enthalpy change when 1 mole of the compound is formed from its elements under standard conditions of 298 K and 1 atm.

2. The standard enthalpy change of combustion, Hco , is defined as the enthalpy change when bone mole of a compound is completely burnt in oxygen under standard conditions of 298K and 1 atm.

3. The standard enthalpy change of hydration, Hhydo, of an ion is defined as the enthalpy change when 1 mole of the gaseous ions is dissolved in a large amount of water under standard conditions of 298 K and 1 atm

4. The standard enthalpy change of solution, Hsolo , is defined as the enthalpy change when 1 mole of a substance dissolves in such a large volume of solvent that addition of more solvent produces no further heat change under standard conditions of 298 K and 1 atm.

5. The standard enthalpy change of neutralisation, Hno, is defined as the enthalpy change when 1 mole of water is formed in the neutralisation between an acid and an alkali, the reaction being carried out in aqueous solution under standard conditions of 298 K and 1 atm.
Always negative (exothermic reaction)

6. The standard enthalpy change of atomisation, Hato, is defined as the enthalpy change when 1 mole of separate gaseous atoms of the element is formed from the element under standard conditions of 298 K and 1 atm.

7. Bond energy is the energy required to break the covalent bond between 2 atoms in the gaseous state (for dissociation).It is usually measured in kJ mol-1 and is an average value.

8. Ionisation energy is the energy required to remove one electron from each atom in a mole of gaseous atoms producing one mole of gaseous cations

9. The electron affinity is the energy change associated with the formation of an anion from the gaseous atom, measured in kJ mol-1.

 

6. ELECTROCHEMISTRY

Content
 Redox processes: electron transfer and changes in oxidation number (oxidation state)
 Electrode potentials
(i) Standard electrode (redox) potentials, E ; the redox series
(ii) Standard cell potentials, cell E , and their uses
(iii) Batteries and fuel cells
 Electrolysis
(i) Factors affecting the amount of substance liberated during electrolysis
(ii) The Faraday constant; the Avogadro constant; their relationship
(iii) Industrial uses of electrolysis

Learning Outcomes
Candidates should be able to:
(a) describe and explain redox processes in terms of electron transfer and/or of changes in
oxidation number (oxidation state)
(b) define the terms:
(i) standard electrode (redox) potential
(ii) standard cell potential
(c) describe the standard hydrogen electrode
(d) describe methods used to measure the standard electrode potentials of:
(i) metals or non-metals in contact with their ions in aqueous solution
(ii) ions of the same element in different oxidation states
(e) calculate a standard cell potential by combining two standard electrode potentials
(f) use standard cell potentials to:
(i) explain/deduce the direction of electron flow from a simple cell
(ii) predict the feasibility of a reaction
(g) understand the limitations in the use of standard cell potentials to predict the feasibility of a reaction
(h) construct redox equations using the relevant half-equations (see also Section 9.4)
(i) predict qualitatively how the value of an electrode potential varies with the concentration of the aqueous ion
(j) state the possible advantages of developing other types of cell, e.g. the H2/O2 fuel cell and improved batteries (as in electric vehicles) in terms of smaller size, lower mass and higher voltage
(k) state the relationship, F = Le, between the Faraday constant, the Avogadro constant and the charge on the electron
(l) predict the identity of the substance liberated during electrolysis from the state of electrolyte (molten or aqueous), position in the redox series (electrode potential) and concentration
(m) calculate:
(i) the quantity of charge passed during electrolysis
(ii) the mass and/or volume of substance liberated during electrolysis, including those in the electrolysis of H2SO4(aq); Na2SO4(aq)
(n) explain, in terms of the electrode reactions, the industrial processes of:
(i) the anodising of aluminium
(ii) the electrolytic purification of copper
[technical details are not required]

7. EQUILIBRIA

Content
 Chemical equilibria: reversible reactions; dynamic equilibrium
(i) Factors affecting chemical equilibria
(ii) Equilibrium constants
(iii) The Haber process
 Ionic equilibria
(i) Brønsted-Lowry theory of acids and bases
(ii) Acid dissociation constants, Ka and the use of pKa
(iii) Base dissociation constants, Kb and the use of pKb
(iv) The ionic product of water, Kw
(v) pH: choice of pH indicators
(vi) Buffer solutions
(vii) Solubility product; the common ion effect

Learning Outcomes
Candidates should be able to:
(a) explain, in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium
(b) state Le Chatelier’s Principle and apply it to deduce qualitatively (from appropriate information) the effects of changes in concentration, pressure or temperature, on a system at equilibrium
(c) deduce whether changes in concentration, pressure or temperature or the presence of a catalyst affect the value of the equilibrium constant for a reaction
(d) deduce expressions for equilibrium constants in terms of concentrations, Kc, and partial pressures, Kp
[treatment of the relationship between Kp and Kc is not required]
(e) calculate the values of equilibrium constants in terms of concentrations or partial pressures from appropriate data
(f) calculate the quantities present at equilibrium, given appropriate data (such calculations will not require the solving of quadratic equations)
(g) describe and explain the conditions used in the Haber process, as an example of the importance of an understanding of chemical equilibrium in the chemical industry
(h) show understanding of, and apply the Brønsted-Lowry theory of acids and bases, including the concept of conjugate acids and conjugate bases
(i) explain qualitatively the differences in behaviour between strong and weak acids and bases in terms of the extent of dissociation
(j) explain the terms pH; Ka; pKa; Kb; pKb; Kw and apply them in calculations, including the relationship Kw = KaKb
(k) calculate [H+(aq)] and pH values for strong acids, weak monobasic (monoprotic) acids, strong bases, and weak monoacidic bases
(l) explain the choice of suitable indicators for acid-base titrations, given appropriate data
(m) describe the changes in pH during acid-base titrations and explain these changes in terms of the strengths of the acids and bases
(n) (i) explain how buffer solutions control pH
(ii) describe and explain their uses, including the role of H2CO3/HCO3– in controlling pH in blood
(o) calculate the pH of buffer solutions, given appropriate data
(p) show understanding of, and apply, the concept of solubility product, Ksp
(q) calculate Ksp from concentrations and vice versa
(r) show understanding of the common ion effect

8. REACTION KINETICS

Content
 Simple rate equations; orders of reaction; rate constants
 Concept of activation energy
 Effect of concentration, temperature, and catalysts on reaction rate
 Homogeneous and heterogeneous catalysis
 Enzymes as biological catalysts

Learning Outcomes
Candidates should be able to:
(a) explain and use the terms: rate of reaction; rate equation; order of reaction; rate constant; half-life of a reaction; rate-determining step; activation energy; catalysis
(b) construct and use rate equations of the form rate = k[A]m[B]n (limited to simple cases of single-step reactions and of multi-step processes with a rate-determining step, for which m and n are 0, 1 or 2), including:
(i) deducing the order of a reaction by the initial rates method
(ii) justifying, for zero- and first-order reactions, the order of reaction from concentration-time graphs
(iii) verifying that a suggested reaction mechanism is consistent with the observed kinetics
(iv) predicting the order that would result from a given reaction mechanism
(v) calculating an initial rate using concentration data
[integrated forms of rate equations are not required]
(c) (i) show understanding that the half-life of a first-order reaction is independent of concentration
(ii) use the half-life of a first-order reaction in calculations
(d) calculate a rate constant using the initial rates method
(e) devise a suitable experimental technique for studying the rate of a reaction, from given information
(f) explain qualitatively, in terms of collisions, the effect of concentration changes on the rate of a
reaction
(g) show understanding, including reference to the Boltzmann distribution, of what is meant by the
term activation energy
(h) explain qualitatively, in terms both of the Boltzmann distribution and of collision frequency, the
effect of temperature change on a rate constant (and hence, on the rate) of a reaction
(i) (i) explain that, in the presence of a catalyst, a reaction has a different mechanism, i.e. one
of lower activation energy, giving a larger rate constant
(ii) interpret this catalytic effect on a rate constant in terms of the Boltzmann distribution
(j) outline the different modes of action of homogeneous and heterogeneous catalysis, including:
(i) the Haber process
(ii) the catalytic removal of oxides of nitrogen in the exhaust gases from car engines
(see also Section 10.2)
(iii) the catalytic role of atmospheric oxides of nitrogen in the oxidation of atmospheric sulfur
dioxide
(iv) catalytic role of Fe2+ in the –/S2O8
2– reaction
(k) describe enzymes as biological catalysts which may have specific activity
(l) explain the relationship between substrate concentration and the rate of an enzyme-catalysed
reaction in biochemical systems