F325 Equalibria, Energetic and Elements

Rates, Equalibria and PH

How fast?The rate of reaction is the change in concentration in mol dm-3s-1

This is the rate of change of the concentration of the reactant. This can be worked out from a concentration/time graph.

The rate of reaction is influenced by temperature, concentration and use of catalyst. The collision theory explains all of these. Concentration, temperature and catalysts increase the rate of reaction.

For A + B C + Drate = k [A]m [B]n

k is the rate constant and is affected by temperature. A large k ⟹ fast rate of reaction and small k ⟹ slow rate of reaction.m and n are the orders of the reaction with respect to reactant A and B.

Orders tell you how a reactant’s concentration affects the rate. Found from experiments (i.e. doing the experiment twice with double of one reactant) or a rate of reaction against concentration graph.

Initial Rates Method1. If you double the reactant’s concentration and the rate stays the same, the order with respect to

that reactant is 02. If you double the reactant’s concentration and the rate doubles, the order with respect to that

reactant is 1. Orders of 1 have a constant half life.3. If you double the reactant’s concentration and the rate quadruples, the order with respect to that

reactant is 2

Rate-Concentration Graph

If rate = k [A]m[B]n then obviously the concentration against time graphs will show the following graphs for first, second and third order respectively.

Concentration-Time Graph

For 0 order, concentration falls at a steady rate. The half life decreases with time. For 1st order, the half life is constant. For 2nd order, the half life increases with time.

The rate-determining step is the slowest step in the reaction. The rate equation can tell us about the reaction mechanism. The orders in the rate equation match the number of species involved in the rate-determining step. E.g. in a 2 step mechanism, the rate equation indicates the number of mols of each reactant involved in the slow step. Slow + fast step = balanced equation.e.g. for 2H2 + 2NO 2H2O + N2 the rate equation is rate = k[H2(g)][NO(g)]2 and this tells us the rate-determining step is H2 + 2NO H2O + N2 + O

How far?A dynamic equilibrium is one which is reached when the rate of the forward reaction equals the rate of the reverse reaction such that the composition of the mixture remains constant by the reagents and products constantly changing. Chatelier’s principle states that in a closed system in equilibrium, any change to the equilibrium, forces it to move in such a way as to minimise the change.

Temperature depends on whether the forward reaction is exothermic. Pressure depends on the number of moles on both sides of the equilibrium. Concentration moves the equilibrium to the side of which has had no change in equilibrium. Catalysts speed up both sides by the same rate ⟹ no shift.

The exact position of the equilibrium is calculated; using Kc; the in aA + bB ⇌ cC + dD

K c=¿¿The units depend on each specific case. If Kc is large the equilibrium favours the forward reaction. If small, it favours the backward reaction.

Kc is temperature dependant. In an exothermic reaction, Kc decreases with increasing temperature and in endothermic, Kc increases with increasing temperature.

Kc is unaffected by changes in concentration (if the concentration of one thing changes, the others must change to keep the value of Kc) or pressure (changing the pressure on one side shifts the

equilibrium to the side with more or fewer gas molecules, restoring original pressure) or using a catalyst.

Acids An acid base reaction involves a proton transfer. A molecule of an acid contains a hydrogen that can be released as a positive hydrogen ion or proton

H+. Acids and bases are linked by H+ as conjugate pairs such that the conjugate acid donates H+ and

the conjugate base accepts H+. By mixing an acid with a base, and equilibrium is set up between two acid-base conjugate pairs.

(acid 1) CH3COOH (aq) + (base 1) H2O (l) ⇄ (acid 2) H3O+ + (base 2) CH3COO-(aq)

The acid-base equilibrium of an acid, HA, in water is:HA(aq) + H2O(l) ⇌ H3O+

(aq) + A-(aq)

Or to emphasise the loss of a proton by dissociation;HA(aq) ⇌ H+

(aq) + A-(aq)

A strong acid is one that is highly ionised in aqueous solution.A weak acid is one that is only partially ionised in aqueous solution.

Acids vary in the ease with which they release their hydrogen ions. HNO3(aq) ⇌ H+(aq) + NO3

-(aq)

Therefore strong acids like this undergo complete dissociation as in the above equation.The equilibrium position is far to the right.

Weak acids, such as ethanoic acid, CH3COOH is a poor proton donor. Therefore the equilibrium position is far to the left.

The acid dissociation constant, Ka

The usual way of indicating the strength of an acid is to use the equilibrium constant for its ionisation in water; Ka. This is called the acid dissociation constant.For the reaction HA(aq) ⇌ H+

(aq) + A-(aq);

Ka=¿¿ moldm-3

A large Ka shows the extent of the dissociation is large and the acid is a strong acid. And vice versa.

Calculating pH for strong acids and weak aids

pH is defined by pH=−lo g10¿

For a strong acid, we assume complete dissociation.

E.g. A strong acid, HA, has a concentration of 0.02 moldm-3. What is the pH. Strong acid ⟹ complete dissociation so ¿= 0.02 so pH = -log(0.02) = 1.7

For a weak acid, we need the concentration of the acid and Ka. Only a very small proportion of HA dissociates and hence the amount of undissociated acid is the same as the initial concentration of HA. Also there is a negligible proportion of H+

(aq) from the ionisation of water such that [H+(aq)] = [A-

(aq)].

Using these approximations; Ka=¿¿ ≈ ¿¿

E.g. A weak acid, HA, has a concentration of 0.2 moldm-3, and Ka=1.7 x10-4moldm-3 at 25oC. Find pH. Ka=¿¿ ≈ ¿¿= ¿¿= 1.7x10-4

∴¿

Ionic product of water, Kw

Water ionises very slightly, acting as both an acid and base. H2O(l) ⇌ H+(aq) + OH-

(aq)

Only a very small proportion of molecules dissociates. Thus treating water as a weak acid;Ka = ¿¿ ⟹Ka× [H 2O(l )]=¿

This bit; Ka× [H 2O(l ) ] is a constant and is called Kw. It is temperature dependent.

Kw can be used to calculate the pH of water. At 25oC, Kw=1x10-14.1x10-14= [H+

(aq)][OH-(aq)]. Assume that [H+

(aq)]=[OH-(aq)] and it follows that at 25oC, pH = 7.0

The pH is below 7 with high temperatures and above 7 with low temperatures.Kw can be used to calculate the pH of strong Alkalis. At 25oC, Kw=1x10-14.

A strong Alkali, KOH has a concentration of 0.5 moldm-3. What is the pH at 25oC?KOH(aq) K+

(aq) + OH-(aq) and since the alkali is strong, we assume [OH-] = [KOH(aq)]=0.5 moldm-3.

Kw = [H+(aq)][OH-

(aq)] = 1x10-14 and so [H+(aq)] = 10-14/0.5 and it follows that pH = 12.7

Buffer Solutions

A buffer solution is a system that minimises pH changes on addition of small amounts of an acid or a base. The buffer solution maintains a near constant pH by removing most of any acid or alkali that is added to the solution.

It is a mixture of a weak acid, HA and it’s conjugate base, A-;HA(aq)

⇌ H+(aq) + A-

(aq)

An example for a common buffer solution is a mixture of CH3COOH and CH3COO-Na+ (conjugate base). The pH at which the buffer operates depends on the Ka of the weak acid and the relative concentrations of the weak acid and the conjugate base.

CH3COOH(aq) ⇌ CH3COO-(aq) + H+

(aq) CH3COO-Na+(aq) CH3COO-

(aq) + Na+(aq)

The high [CH3COO-(aq)] forces the equilibrium back to the LHS and results in the buffer solution containing

very low [H+(aq)] and very high [CH3COOH(aq)] and [CH3COO-

(aq)]. The high concentrations of [CH3COOH(aq)] and [CH3COO-

(aq)] resist any changes in pH. Hence a buffer solutions cancels out the effect of any added acid or alkali and minimises the change in pH.

Hence a buffer solution contains: High concentration of weak acid e.g. CH3COOH High concentrations of conjugate base e.g. CH3COO-

Low concentration of H+

The pH of a buffer solution depends (and hence can be adjusted) by the acid dissociation constant Ka, of the buffer system and the ratio of the weak acid to conjugate base. Ka=[SALT ] ¿¿. Then pH= -log([H+]).

Application of Buffers

In blood, the pH is about 7.4. If the pH becomes >7.8 or <7.0 the results are fatal. Blood contains a mixture of buffering solutions, the major one being carbonic acid-hydrogen carbonate (bicarbonate) system.

H2O(l) + CO2(g) ⇌ H2CO3 ⇌ HCO3- + H+

Carbonic acid Hydrogen Carbonate

Adding an acid to the system will increase the concentration of H+(aq) driving the equilibrium to the

LHS. This increases the concentration of the carbonic acid, H2CO3, which is in turn decreased by an increased rate of breathing such that more CO2(g) is exhaled resulting in H2CO3 moving further to the left to replace it. The 2 equilibria together resist increase in acidity.

Adding an alkali to the system will decrease the concentration of H+(aq), driving the equilibrium to

the RHS. This decreases the concentration of the carbonic acid, H2CO3, which is in turn increased by a decreased rate of breathing such that less CO2(g) is exhaled resulting in H2CO3 being replaced as the equilibrium shifts to the H2CO3 side. The 2 equilibria together resist increase in basicity.

PH changes and indicators

Indicators are substances that change colour with a change in pH. Many indicators are weak acids (some are weak bases though) and can be represented by HIn. They have many different colours.

HIn (aq) ⇌ H+(aq) + In-

(aq)

e.g. for methyl orange RED

YELLOW

At the end point of a titration HIn and In- are present in equal concentrations. Using a methyl orange as indicator:

At the end point [HIn] (red) = [In-] (yellow) Hence the colour at the end point is orange FYI, the pH of the end point is called the pKIn

of the indicator.

When the acid and base have completely reacted this is known as the equivalence point. Here there is a sharp change in pH for a very small addition of acid or base.

Experiments to distinguish the strengths of acids and bases

A pretty cool way is to test electrical conductivity. In solution, ions conduct electricity. ;)The standard enthalpy of neutralisation is the enthalpy change that occurs when 1 mole of water is produced in the reaction of an acid with an alkali under std. conditions.The reaction between all acids and alkalis is H+

(aq) + OH-(aq) H2O(l)

Since all strong acids and alkalis are completely dissociated, the enthalpy change of neutralisation is the same. If the acids and alkalis are weak, then the standard enthalpy of neutralisation will be less.

Energy

Lattice EnthalpyStandard enthalpy changes are the enthalpy changes when 1 mole of a substance is formed/burnt in their natural state/in excess oxygen under standard conditions of 298K and 100kPa. The average bond enthalpy is the breaking of 1 mole of a covalent bond in a gaseous molecule under standard conditions. The enthalpy change of formation of ethane and the combustion of ethane are:

2C(s)+3H 2(g)⟶1C2H 6(g) and 1C2H 6( g)+312O 2

(g )⟶2C O2(g )+3H 2O(l)

Bond enthalpies are averages and don’t take into account the chemical environment.

Activation energy is the minimum energy required in a collision between particles to react. Breaking bonds is an endothermic process requiring energy. Hess’s law states that the enthalpy change for a reaction is the same irrespective of the route taken provided that the initial and final conditions are the same.

Lattice enthalpy indicates the strength of the ionic bonds in an ionic lattice.The lattice enthalpy (∆ H ¿

Θ) of an ionic compound is the enthalpy change that accompanies the formation of 1 mole of an ionic compound from its constituent gaseous ions. (∆ H ¿

Θ is exothermic)Na+

(g) + Cl-(g) → Na+Cl-

(s)

It is almost impossible to measure lattice enthalpy experimentally such that lattice enthalpy is calculated using a Born-Haber cycle. A Born-Haber cycle is similar to a Hess’s cycle and enables the calculation of changes that cannot be measured directly by experiment. The lattice enthalpy of NaCl can be calculated by considering the standard enthalpy of formation of NaCl(s). In order to form an ionic solid, both sodium and chlorine have to undergo a number of changes;

All the changes in the cycle can be measured experimentally except the formation of the Na+Cl-(s) lattice

from the gaseous ions i.e. the lattice enthalpy. However all the other steps in the cycle the lattice enthalpy can be calculated. The above cycle is adapted with a Hess’s cycle to make the Born-Haber cycle.

Using Hess’s law; A+B+C+D+E=F

∆ H atΘ N a(g )+∆ H at

ΘC l(g)+∆H IEΘ N a(g)+∆ HEA

Θ C l(g)+E=∆ H fΘN a

+¿C l( s)−¿¿ ¿

⇒+107+122+496+(−349 )+E=−411

⟹ Lattice enthalpyof N a+¿C l( s)

−¿E=−787KJmol−1¿ ¿

F; the std. enthalpy change of formation; ∆ H fΘ;

Formation of an ionic compoundthe enthalpy change when 1 mole of the compound in its standard state is formed from its constituent elements in their standard states under standard conditions. Exothermic for an ionic compound.

A&B; the std. enthalpy change of atomisation; ∆ H atΘ ; Formation of gaseous atoms

the enthalpy change that accompanies the formation of one mole of gaseous atoms from the element in its standard state. The standard enthalpy change of atomisation is always endothermic.For gaseous molecules, this enthalpy change can be determined from the bond dissociation enthalpy- the enthalpy required to break and separate one mole of bonds so the resulting gaseous atoms exert no force on each other.

C; the first ionisation energy; ∆ H IEΘ ; Formation of positive ions

the enthalpy change that accompanies the removal of one electron from each atom in 1 mole of gaseous atoms to form 1 mole of gaseous 1+ ions. Always endothermic.

D; the first electron affinity; ∆ H EAΘ ; Formation of negative ions

the enthalpy change that accompanies the addition of one electron to each atom in 1 mole of gaseous atoms to form 1 mole of gaseous 1- ions. Always exothermic.

C; the lattice enthalpy; ∆ H ¿Θ; Formation of ionic compounds

the enthalpy change that accompanies the formation of 1 mole of an ionic compound from its constituent gaseous ions. Always exothermic.

For each different ionic lattice it may have to be adapted. E.g. for MgO(s) there has to be a second electron affinity to form the O2- ion.

The strength of an ionic lattice and the value of it lattice enthalpy depends upon ionic radius and ionic charge.

The size of the halide ion effects the ionic radius. A smaller halide ion decreases the (already negative) lattice enthalpy.

The small, highly charged ions (greatest charge density) are the strongest ionic lattices.

Enthalpy change of hydrationThe concept of a Born-Haber cycle can be extended to provide a partial explanation of the solubility of substances in water. The enthalpy of hydration is the enthalpy change when 1 mol of gaseous ions is completely hydrated by water under standard conditions. i.e. X (g )

n+¿⟶ X (aq)n+¿¿¿

In the case of hydration, the attraction is either between a positive ion and the oxygen atom of the water molecule or a negative ion and the hydrogen. This is because of the dipoles due to hydrogen bonding.

The enthalpy depends on the ionic radius and the size of the charge on the ion. The greater charge density, the greater attraction.

Values of lattice enthalpies and enthalpies of hydration relate to the enthalpy of solution, which is the enthalpy change that occurs when an ionic solid dissolves in water.

Applying Hess’s law, ∆ H 2+∆ H 1=∆ H 3 where ∆ H 1 is the enthalpy

of solution, ∆ H 2 is the lattice enthalpy of NaCl and ∆ H 3 is the enthalpy of hydration of the Na+ ion + the enthalpy of hydration of the Cl- ion. ∆ H 1 is endothermic. It may seem odd that dissolving sodium chloride is endothermic. Another factor is encouraging dissolving to take place. It is an energy term called entropy.

EntropyKnowledge of enthalpy changes alone is insufficient to provide a certain answer as to whether a reaction will take place or not. Another energy change, entropy, takes place during a reaction.

When a reaction occurs, energy is absorbed (endo) or released (exo). In addition, some is absorbed or released as a result of the re-distribution of the particles when the products are formed. The extent of this energy relies on the physical state of the substances and the temperature. Entropy (S) is used to measure this. It is a measure of the disorder of a system.

Solids are more ordered than liquids and gases and so the most entropy/energy is required to hold a solid in its ordered state. H 2O(s)⟶H 2O(l)∆ H=+6.02kJ mol−1 at 0oc.This endothermic reaction happens because the change in entropy, as melting occurs, releases sufficient energy to counteract the positive enthalpy. The energy required to hold the rigid structure of the ice in place is released as the less restrained molecules of water are produced.

If enthalpy is released in a reaction, ∆ H is negative. If entropy is released in a reaction, ∆ S is positive. The energy unit of enthalpy is kJ and entropy J.

Entropy always increases (∆ S is positive) when there is a greater opportunity for energy to be spread out as a result of change (e.g. ‘disorderness’). Entropy increases when:

A solid becomes a liquid. Entropy is released and ∆ S is positive; more disordered. A liquid becomes a gas. Entropy is released and ∆ S is positive; more disordered A solid dissolves in a liquid to form a solution. Entropy is released and ∆ S is positive; more

disordered. A reaction resulting in products with more movement so ∆ S is positive; more disordered. A reaction producing more particles of the same state; ∆ S is positive and more disordered. As temperature rises, entropy is released and ∆ Sis positive, more disordered.

∆ S=Σ (entropy of∏ ucts )−Σ(entropy of reactants)

The change in the entropy of a reaction can be combined with the change in enthalpy to provide an answer as to how a chemical reaction is feasible. Free energy (G) relates to the enthalpy and entropy changes;

ΔG=ΔH−T Δ S (T is temperature)And ΔG provides a certain answer to whether a reaction is feasible. If ΔG<0⇒ Reactionis feasible

If ∆G>0⇒Reaction is not feasible(at least at this temp erature)

At ∆G=0 the system will be at equilibrium and ∆ H=T ∆ S. The enthalpy change for the melting of ice to water at 0oc is

e.g. N H 3(g )+HC l(g)⇌N H 4C l(s)

∆ H∅r=−176000 J mo l−1 and ∆ S∅=−284.5J mol−1 and so at 298K,

∆G=−176000− (298 ) (−284.5 )=−91.2kJ mol−1 is the reaction would happen at 298K. This is not true at 500K.

Ionic Equations Ionic substances that are solid has no free-moving ions so their ions cannot react independently of

each other. Their complete formula must be given. Compounds of metals and strong acids in aquous solutions will be split into ions. Covalent compounds exist as complete molecules and are always shown as complete entities in the

ionic equation.Acid + Base Salt + Water MgO(s) + 2H+ Mg2+

(aq) + H2O(l) (Ionic Equation)Acid + base Salt + carbon dioxide + water CO3

2-(aq) + 2H+

(aq) CO2(g) + H2O(l) (Ionic Eqation)

Oxidation number identifies whether a substance has undergone oxidation or reduction. All elements in their natural state have oxidation number 0. The oxidation numbers of any molecules add up to 0 and ions add to their charge. Group 1, 2, 3, 4, 5, 6 are always +1, +2, +3, +4, +5, +6. Fluorine is always -1, Hydrogen is usually +1, Oxygen is usually -2, Chlorine is usually -1.e.g. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) . Cu goes from +2 to 0, Zn from 0 to +2 hence ;Zn(s) Zn(aq)

2+ + 2e- and Cu2+(aq) + 2e- Cu(s)

Consider Cu(s) + AgNO3(aq) Cu2+(aq) + Ag(s). Half equations are Cu(s) Cu2+

(aq) + 2e- and Ag+(aq) + e- Ag(s).

Eliminating the electrons gives Cu(s) + 2Ag+(aq) Cu2+

(aq) + 2Ag(s)

Electrode Potentials The standard electrode potential E∅ is the potential difference between one half-cell (e.g. metal in

contact with its metal ions) and the standard hydrogen electrode (when measured under standard conditions).

The standard cell potential is the voltage formed when two half-cells are connected and the voltage measured using a voltmeter (of very high resistance under standard conditions).

The standard hydrogen electrode looks like:To provide a better surface for the hydrogen, the platinum electrode is usually coated with very finely divided platinum black. This cell is connected via an external circuit and through a salt bridge to the other cell. The voltage measured then gives the electrode potential of the cell on a scale with the half-reaction 2H+

(aq) + 2e- H2(g) being given the arbitrary value of 0.

For metals the system to the left applies. For non-metals/ions of the same element in different oxidation states the system to the right applies.

The salt bridge is made of porous material soaked in a saturated solution of KNO3. It completes the circuit.

Half-Cell E∅ /V F2(g) + 2e- ⇄ 2F-

(aq) 2.87Cl2(g) + 2e- ⇄ 2Cl-

(aq) 1.36Cu2+

(aq) + 2e- ⇄ Cu(s) 0.342H+

(aq) + 2e- ⇄ H2(g) 0.00K+

(aq) + e- ⇄ K(s) -2.92

Half cells with a positive E∅ favour the forward reaction and gain electrons. Half cells with a negative E∅ favour the reverse reaction and lose electrons.

F2(g) has a high E∅ value and so readily gains electrons to form F-(aq) ions. F2 is a powerful oxidising agent. K(s)

has a low E∅ value and so readily loses electrons to form K+(aq) and is a powerful reducing agent.

The standard electrode cell can be calculated by using any two half-cells. Zn2+

(aq) + 2e- ⇄ Zn(s) -0.76V Cu2+(aq) + 2e-

⇄ Cu(s) +0.34VWhich indicated the Cu2+

(aq) moves in the forward reaction whilst the Zn(s) favours the reverse reaction.So hence we can write each half cell as:

Cu2+(aq) + 2e- Cu(s) +0.34V Zn(s) Zn2+

(aq) + 2e- +0.76 Cell potential +1.10 V

e.g. Acidified H+(aq)/MnO4

-(aq) is a good oxidising agent and can be used to preare Cl2(g) by the oxidation of

Cl-(aq) ions. Find the cell potential and deduce the balanced equation.

MnO4-(aq) + 8H+

(aq) + 5e- ⇄ Mn2+ + 4H2O(l) +1.52V Cl2(g) + 2e- ⇄ 2Cl-(aq) +1.36V

Both are positive but MnO4-(aq) is more, so moves to the right such that:

MnO4-(aq) + 8H+

(aq) + 5e- Mn2+ + 4H2O(l) +1.52V 2Cl-(aq) ⇄ Cl2(g) + 2e-

(aq) -1.36V Cell potential +0.16VWe can eliminate the electrons by multiplying by 2 and 5 respectively to give;2MnO4

-(aq) + 16H+

(aq) + 10Cl- 2Mn2+ + 8H2O(l) + 5Cl2(g)

A positive cell potential indicates the reaction is feasible but gives no indication to its rate. On the example above, despite a low cell potential, this reaction happens relatively quickly. Cell potentials are assumed under standard conditions. If they are not, this may alter the reaction.For Fe3+

(aq) + e- ⇄ Fe2+(aq) (Ee=+0.77V) a lower concentration of Fe2+

(aq) will cause the equilibrium to shift to the RHS and cause the electrode potential to increase. And vice versa. Any increase/decrease in concentration actually needs to be quite large to have any effect on Ee.

Storage cells and fuel cellsStorage cells are batteries. Electrode potential can be used to predict the possible voltage of a battery. Commercial batteries tend to avoid solutions that can leak and use pastes that surround the electrode.

An alkaline battery has a cathode (half cell that receives electrons) made of graphite and manganese(IV) oxide and an anode (provides electrons) made of zince or nickel-plated steel. Electrolyte is potassium hydroxide.

m

At the anode;Zn + 2OH- ZnO + H2O + 2e-

At the cathode;2MnO2 + H2O + 2e- Mn2O3 + 2OH-

By eliminating the electrons we have Zn + 2MnO2 ZnO + Mn2O3. The overall voltage is about 1.5V. Both OH- and H2O are eliminated so their concentrations should remain constant.

Another battery is rechargeable nickel cadmium (Ni-Cd) cell. While it is supplying electricity the reactions taking place are:

At the anode;Cd + 2OH- Cd(OH)2 + 2e-

At the cathode;2NiO(OH) + 2H2O + 2e- 2Ni(OH)2 + 2OH-

The overall reaction is Cd + 2NiO(OH) + 2H2O Cd(OH)2 + 2Ni(OH)2. The batteries can be recharged by supplying an additional voltage which reverses the reactions above. A disadvantage is that Caladium is toxic. Voltage of 1.2 V is produced.

Fuel cells can produce electrical power from the chemical reaction of a fuel with oxygen. They operate as storage cells except fuels are supplied as gases externally. The hydrogen/oxygen fuel cell is widely used.

- Anode; the platinum catalyst splits the H2 into protons and electrons.- Polymer electrolyte membrane only allows the H+ across and this

forces the e- to travel around the circuit.- An electric current is created in the circuit which is used to power

something.- Cathode; O2 combines with the H+ from the anode and the e- from the

circuit to make H2O. This is the only waste product.

Scientists in the car industry are making cars fuelled by hydrogen gas and hydrogen rich fuels. They have less pollution and less CO2 and greater efficiency.The hydrogen could be stored;

- As a liquid under pressure.- Absorbed on the surface of a solid material.- Absorbed within a solid material.

Limitations include- Storing and transporting hydrogen. Safety, feasibility of a pressurised liquid and the

limited life cycle of a sold ‘absorber’ or ‘adsorber’ are disadvantages.- The fuel cell has a limited lifetime.- Toxic chemicals are used in production.

- Hydrogen economy may contribute to future energy but this requires public acceptance of hydrogen as a fuel, handling and maintenance of hydrogen systems and the initial manufacture of hydrogen which requires energy.

Transition Elements

PropertiesA d-block element that forms one or more stable ions with an incomplete d-sub-shell.The 4s sub shell is filled before the 3d sub shell as the 4s sub shell is at a lower energy. The orbitals in the 3d-sub shell are first occupied singly to prevent any repulsion caused by pairing.

Scandium and Zinc are not transition metals because Zinc forms one ion, Zn2+ with [Ar] 3d10 and Scabdauan forms one ion, Sc3+ with [Ar] 3d0.

Variable Oxidation States; transition elements have compounds with two or more oxidation states. This is due to the fact that successive ionisation energies of transition metals only increase gradually. All the transition metals can form an ion of oxidation state +2, representing the loss of two 4s-electrons. The maximum oxidation state possible cannot exceed the total number of 4s- and the 3d-electrons in their electron configuration.

Colour of compounds; transition elements have at least one oxidation state in which their compounds and ions are coloured. The colours are often distinctive and can be used as a means of identification. Cu2+ ions are blue; Cr3+ ions are green; Cr2O7

2- are orange; MnO4- are purple.

Catalysis; Transition metals are frequently used as heterogeneous catalysts. This is a result of the use of their d-orbital’s to bind other molecules or ions to their surface. E.g. Fe in Haber process

Complex ions and ligands; Ions of transition elements form complex ions with ligands. Other properties; Denser (due to smaller atoms); Higher melting and boiling points due to stronger

metallic bond due to smaller ions meaning greater free electron density.

Simple precipitation reactions from transition metal hydroxidesTakes place between aqueous alkali e.g. NaOH and an aqueous solution of a metal. Results in formation of precipitate of the metal hydroxide with a colour.

Cu2+(aq

+ 2OH- (aq) Cu(OH)2(S) (Pale Blue)Fe2+

(aq) + 2OH-(aq) Fe(OH)2(S) (Pale green gelatinous precipitate)

Fe3+(aq) + 3OH-

(aq) Fe(OH)3(S) (Orange-Brown gelatinous precipitate)Co2+

(aq) + 2OH-(aq) Co(OH)2(s) (Blue precipitate)

Transition element complexes

Transition metal ions are small and densely charged and can strongly attract electron-rich species called ligands forming complex ions.

A ligand is a molecule or ion that bonds to a metal ion by forming a dative covalent bond by donating a lone pair of electrons into a vacant d-orbital. Common ligands are H2O: , :Cl- , :NH3 and :CN- all of which have at least one lone pair of electrons.

A complex ion is a central metal ion surrounded by ligands

In [Cu(H2O)6]2+, the six electron pairs surrounding the Cu2+ ion repels one another as far apart to create an octahedral shape around the complex ion.

In [CuCl4]2- the four electron pairs surrounding the Cu2+ ion repel one another as far as possible to create a tetrahedral shape such that the bond angles are 109.5o.

The coordination number is defined as the total number of coordinate bonds from the ligands to the central transition metal ion in a complex ion. Complex ions with ligands such as H2O and NH3 are usually 6 coordinate and octahedral in shape. Complex ions with Cl- ligands are usually 4 coordinate and tetrahedral in shape.

Ligands that form two dative coordinate bonds with the central transition metal ion are called bidendate ligands.1,2-Diaminoethane is a common bidendate ligand. Each N has a lone pair of electrons that can form a dative bond. It is often drawn as:

And hence a nickel complex could be drawn as:

Stereoisomerism can occur in complex ions. The square planar structure of tetrahedral shapes allows cis-trans isomerism. Optical isomerism is also possible in the case of complexes coordinated by

polydentate ligands. The superimposable mirror image creates this isomerism.

One particularly interesting use of a complex ion is of the cis form of PtCl2(NH3)2. (The platinum is present as a 2+ ion so no overall charge)It is known as cis-platin and is used during chemotherapy as an anti-cancer drug. A colourless liquid that is put into the blood via a drip and it binds to the DNA of cancerous cells preventing their division. The trans molecule is ineffective.

Ligands SubstitutionA ligand substitution reaction takes place when a ligand in a complex ion exchanges for another ligand. The colour of the solution changes as a different complex is formed.

Exchange between H2O and NH3 ligands; Water and ammonia ligands have similar sizes such that the coordination number does not change. Blue to Deep Blue

Exchange between H2O and Cl- ligands; Water molecules and chloride ions have different sizes and the coordination number changes. Blue to Yellow

A similar reaction takes place when Co2+ replaces Cu2+ such that [Co(H2O)6]2+ (pink) forms [CoCl4]2- (blue)

The strength of binding of a ligand to a cation can be represented quantitatively. When a complex ion such as [Cu(NH3)4(H2O)2]2+ is made it exists in equilibrium with the hydrated copper ion, [Cu(H2O)6]2+ from which it was made. The equation is:

[Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 2H2OThe equilibrium constant is KSTAB. The H2O are NOT included as the whole reaction is done in water. The value of KSTAB is usually very high, indicating the reaction lies well to the right so hence [Cu(NH3)4(H2O)2]2+ is much more stable than [Cu(H2O)6]2+. When KSTAB is applied to a substitution reaction, a large KSTAB results in the formation of a stable complex ion.

Haemoglobin contains Fe2+ ions. The globin bonds to a protein. The sixth lone pair has a water molecule attached to it.In the lungs, the high oxygen concentration replaces the water ligand for an oxygen ligand. This is carried around the body until the oxygen molecule is exchanged for water again. If CO is inhaled, the strong CO ligand bonds and is not substituted. Therefore it cannot carry oxygen any more.

Redox ReactionsThe oxidation number can determine whether a substance has lost electrons (oxidised) or gained electrons (reduced).

A common reaction is the iron(II)-manganate(VII) reaction. Fe2+(aq) and MnO4

-(aq). Here is the derivation.

Write a half-ionic equation for each transition metalFe2+ is oxidised to Fe3+ so Fe2+ Fe2+ + e-

Therefore MnO4- must be reduced. It is reduced to Mn2+. Therefore MnO4

- + 5e- Mn2+ and so MnO4

- + 8H- + 5e- Mn2+ + 4H2O

Rewrite the half-equations so that the number of electrons in both is the sameIn this case, we need to multiply equation (1) by 5 to get:5Fe2+ 5Fe3+ + 5e- and MnO4

- + 8H- + 5e- Mn2+ + 4H2O as before

Add the last half-equations together to cancel out the electrons.5Fe2+ + MnO4

- + 8H+ 5Fe3+ + Mn2+ + 4H2O

Redox titrations involves the transfer of electrons from one species to another.The colour change provides the end point without the need for an indicator. Potassium manganate (VII) contains MnO4

-(aq)

and is often used to oxidise solutions containing iron(II) ions. MnO4

-(aq) (purple) Mn2+

(aq) (colourless)

This reaction is carried out with potassium manganate (VII) solution in the burette and the iron(II) solution in the conical flask. As the manganate(VII) solution is added to the iron(II) solution, it decolourises. The end point indicates there are excess MnO4

- ions present.