Energy matters

Unit 1

Reaction rates

• From standard grade you should remember that a reaction can be speeded up by;

• Decreasing particle size

• Increasing concentration

• Increasing temperature

Following the course of a reaction

• In general to measure the rate of a reaction we must choose some measurable quantity which changes as the reaction proceeds.

e.g mass of reactants in a flask, volume of gas produced, colour intensity, concentration of reagent

Following the course of a reaction

• If we react marble chips (Calcium carbonate) with hydrochloric acid we can monitor the course of the reaction.

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

Marble chips & acid

• As we are producing a gas, it will escape from the vessel causing the total mass to drop.

• If we measure this change in mass over a fixed period of time we can calculate the rate of the reaction.

Marble chips

Balance

Cotton woolHCl(aq)

Time (s) Total mass of flask (g)

Decrease in mass (g) or Mass of CO2

produced

0

30

60

90

120

150

180

210

240

270

300

149.00

147.75

147.08

146.60

146.26

145.94

145.68

145.48

145.32

145.19

145.08

-

1.25

1.92

2.40

2.76

3.06

3.32

3.52

3.68

3.81

3.92

Decrease in mass (g) -

00.5

11.5

22.5

33.5

44.5

0 50 100 150 200 250 300 350

Time (seconds)

Decre

ase in

mass

(g)

Average rate of reaction

• It is difficult to measure the actual rate at any one instant since the rate is always changing.

• We can calculate average rate over a certain period of time.

Average reaction rate = Change in mass of product Time taken for change

Example

• Calculate average rate of reaction between 30 and 60 seconds.

Average reaction rate = Change in mass of product

Time taken for change

Average reaction rate = 1.92 - 1.25

30

Average reaction rate = 0.022gs-1

Collision theory

• For a chemical reaction to occur, the reactants must collide.

• Any factor that increases the number of collisions per second is likely to increase reaction rate.

Particle size

• More collisions occur if the particle size of a solid reactant is decreased, since its overall surface area is increased.

• Powdered marble (calcium carbonate) reacts much faster than marble chips.

Concentration

• If concentration is increased, there are more reactant particles.

• The more particles there are in one space, the more collisions.

Raising temperature

• Raising the temperature at which a reaction takes place does more than merely raise the number of collisions.

• Temperature is a measure of the average kinetic energy of particles in a substance.

• Therefore at higher temperatures, particles have greater kinetic energy and they collide with more force.

Collisions

• Not all collisions cause a reaction to occur e.g. nitrogen & oxygen particles in the air.

• The colliding particles must have a minimum amount of kinetic energy for a reaction to occur.

• This minimum kinetic energy is called the Activation energyActivation energy (EA)

Activation energy

• Activation energy required varies from one reaction to another.

• If the activation energy of a reaction is high, only few particles will have enough energy to successfully collide.

• Conversely, a reaction with low activation energy will be very fast.

Kinetic energy• At a given temperature (T1) individual molecules of

a gas have widely different kinetic energies.

• Most molecules will have energy near to the average energy but some will be well below average, and some well above.

Activation energy

• The shaded area represents the all of the molecules which have kinetic energy greater than the activation energy.

• The shaded area represents the portion of molecules that will react

EA

Temperature• Distribution of energy changes when the temperature

changes.• A small rise from T1 to T2 considerably increases the

number of particles capable of reacting.• Hence increasing the reaction rate.

T2

 

Kinetic energy

Catalysts

• Substance that alters rate of reaction without being used up.

• Homogeneous catalyst: Same state as the reactants.

• Heterogeneous catalyst: Different state as the reactants.

Heterogeneous catalyst

• The catalyst has a large surface area.• Catalysis occurs at certain points on the

catalyst called ‘active sites’. • At these sites reactant molecules are

adsorbed onto the surface of the catalyst.• At least 1 reactant is held in place on active

site, making collision more likely.

Catalyst poisoning

• Occurs when reactants or impurities become preferentially adsorbed or even permanently attached to the catalyst surface.

• Hence reducing number of active sites and therefore rendering the catalyst as useless.

Catalytic converters• Petrol engine cars must now be fitted

with a catalytic converter.

• The contains a honeycomb network of platinum, converting harmful gases into less harmful ones.

CO, NOx, O2

CO2

H2OO2

Industrial catalystCatalystCatalyst ProcessProcess ImportanceImportance

Vanadium(v) oxide

Contact Manufacture of H2SO4

Iron Haber Manufacture of ammonia

Platinum Oxidation of ammonia

Manufacture of nitric acid

Nickel Hydrogenation Manufacture of margarine

Enzymes

• Biological catalyst.

• Examples of enzymes:– Amylase, catalyses the hydrolysis of starch.– Catalase, catalyses the decomposition of

hydrogen peroxide. Catalyase is found in the blood, preventing build up of hydrogen peroxide in the body.

Enzymes continued

• Enzymes are highly specific. • Enzymes work best at their

optimum temperature & pH.• Optimum temperature for

human enzymes will be 37°C.

• Greatly exceeding either of these will result in the protein being denatured.

Industrial enzymesEnzymeEnzyme ProcessProcess

Lipase Enhance flavour of cheese, ice-cream &

chocolate

Rennin Cheese production

Protease Tenderising meat

Amylase Desizing (removing starch from fabric)

Potential energy• Potential energy is the energy

possessed by the reactants.

• In an exothermic reaction, the products have less potential energy than reactants.

Potential energy

• In an endothermic reaction, the opposite is true. Reactants must absorb energy from their surroundings.

• Products have more energy than the reactants.

Enthalpy

• The difference in potential energy between reactant and product is called the enthalpy enthalpy changechange (ΔH)

• Enthalpy changes are normally quoted in kJ mol-1

Activation energy

• The rate of reaction depends on the height of the Ea barrier.

• Rate of reaction does not depend on the enthalpy change ( )

HH

Catalyst• Catalysts provide alternative reaction

pathways.

• Thus lowering the activation energy.

Energy

Reaction pathway

Activated complex

• When reactants change into products, they pass through a very unstable state known as the activated activated complexcomplex. (Situated at the maximum potential energy).

• The activated complex is a highly energetic arrangement of atoms that exists for a short time.

• The activated complex loses this energy by either forming products or reforming as reactant particles.

Activated complex

Density

• The amount of material packed into a given volume.

• Density values are much larger for Solid & liquid elements.

• Density increases down each group.• Across the period from L to R,

density increases towards the centre of the period, then decreases again towards the noble gases.

Atomic size: Groups• Atomic size is measured in covalent

radius. This is the distance from the nucleus to the outer electrons.

• As you move down a group the atomic radius increases.

• This is due to the increased number of occupied electron shells.

Atomic radius: Periods• Across a period atomic number and electron number

increase by one.

• Although the number of outer electrons is increasing across the period, the atomic radius decreases.

• This is due to the increasing attraction between the nucleus and the outermost electrons.

Ionisation energies

• The attraction between the nucleus and the outer electrons means that energy is required to remove electrons from the atom.

• Ionisation energy is a measure of the nuclear attraction for outer electrons.

First ionisation energy

• Energy required to remove an electron from one mole of free atoms in a gaseous state.

• K(g) K+(g) + e-

Second ionisation energy

• Energy required to remove an electron from one mole of ions with a charge of 1+ in the gaseous state.

• K+(g) K2+(g) + e

Third ionisation energy

• Energy required to remove an electron from 1 mole of ions with 2+ charge in the gaseous state.

• K2+ K3+(g) + e

Ionisation energies• The first ionisation energy decreases as

you go down a group.• This is due to the increasing atomic radius.• As the radius increases, the attraction

between the nucleus and the outermost electrons decreases.

• Screening / Shielding effect.• Therefore the energy required to remove

that electron decreases.

Li

Na

K

e-

e-

e-

Bonding, structure and properties of

compounds

Metallic Bonding

Covalent bonding

Polar covalent bonding

Ionic bonding

Metallic Lattice

Covalent molecular

Carbon atoms

Covalent bonds

Covalent network

Electronegativity

• The greater the difference in electronegativity between two elements, the less likely they are to share electrons and form covalent bonds.

• Caesium fluoride is the compound with the greatest degree of ionic bonding.

• Formed when atoms of different electronegativities bond to form a covalent compound.

• Bonding electrons are not shared equally.

• The atom with the greater share of electrons becomes slightly negative (δ-)

• The other atom becomes slightly positive (δ+)

• These molecules have a permanent dipole.

Polar covalent bonding

Polar covalent bonding

Ionic bonding

• Different elements have different attraction for bonding electrons, (electronegativity values).

• One atom may attract electrons very strongly and another atom may attract them very weekly and lets them go.

Ionic bonding

Summary

Intermolecular forces of attraction

Covalent molecular

Intermolecular interactions

• Van der Waal forces are a result of electrostatic attraction between temporary dipoles and induced dipoles caused by movement of electrons in atoms and molecules.

• All covalent molecules interact by van der Waals bonding, as all molecules possess temporary dipoles.

Halogens

• All halogen have 1 unpaired electron in the outer shell. Therefore form 1 pure covalent bond. E.g. F2, Cl2, Br2, I2

• These molecules interact only weakly by van der Waals’ mechanism, this makes them very volatile. (Fluorine & chlorine are gaseous).

                                          

Permanent dipole

A molecule can be described as polar if it has a permanent dipole. A permanent dipole is due to a difference in electronegativity between the atoms involved in a covalent bond.

Symmetry

• Some molecules have a symmetrical arrangement of polar bonds.

• This cancels out the polarity over the molecule as a whole.

Polar or Non-polar?

Boiling point• Polar molecules have higher boiling

points than non-polar molecules with similar molecular mass.

Hydrogen bonds• Bonds consisting of a hydrogen atom bonded to an

atom of a strongly electronegative element such as fluorine, oxygen or nitrogen.

Water molecules

Ice

Glycerol Sulphuric Acid

Phosphoric Acid

Covalent molecular

Carbon atoms

Covalent bonds

Covalent network

Diamond

Fullerenes• Discrete covalently

bonded molecules

• Consisting of pentagonal & hexagonal panels.

Graphite

Bonding, structure & properties of elements

Groups 1,2 & 3• Not enough electrons to achieve full outer shell.

• Elements contribute electrons to a common ‘pool’ of delocalised electrons.

• This binds the resultant positive ions.

• Bonding is less directional, therefore metals are more ductile & malleable.

• Delocalised electrons, therefore conduct electricity.

Metallic Bonding

1 exception: Boron• Structure made up of

B12 groups, interbonded with other groups.

• This results in an element almost as hard as diamond.

Group 4

• Standard structure: Infinite 3D network or lattice, e.g. diamond, silicon.

• Therefore exceptionally hard & rigid.

• No discrete molecules, each atom joined to another.

Diamond

Graphite

Fullerenes• Discrete covalently

bonded molecules

• Consisting of pentagonal & hexagonal panels.

Phosphorus (group 5)

• Phosphorous bonds to 3 other phosphorous atoms to form tetrahedral P4 molecules.

• Fewer electrons in P4 than S8 make van der Waals forces weaker in phosphorous, therefore lower m.p.

Group 6

• Oxygen: 2 unpaired electrons, therefore forms 2 pure covalent bonds.

• Intermolecular interactions are weak van der Waals, therefore volatile & gaseous.

Sulphur

• Sulphur atoms can bond to more than one other sulphur, forming an 8 member ring.

• Van der Waals forces strong enough to make sulphur a solid at room temperature.

Groups 5, 6 & 7

• Intra molecular forces (bonds within molecules) are covalent.

• Intermolecular forces are very weak van der Waals forces.

• Therefore most elements are volatile even if solid at room temperature.

• This is due to the little energy required to break intermolecular forces in order to melt/boil.

Bonding in elements: Noble gases

• There are no covalent or ionic bonds between atoms in group 8.

• Uneven distribution of electrons within the atom produce temporary (or transient) dipoles on the atom.

Solvent action

• In general polar solvents dissolve polar substances and ionic substances.

Non polar solvents…(e.g hexane)

• Dissolve non polar solvents

The Avagadro constant

• 1 mole of any element contains the same number of atoms.

• This number is known as the Avagadro constant.

• This constant is given the symbol (L) after the first person to calculate a numerical value.

Avagadro constant (L)

• One mole of any substance contain L, 6.02x1023 formula units.

Formula units• For metals & monatomic species e.g. Noble

gases, a formula unit is an atom.

• Thus 4g helium

40g of calcium

197g of gold

Contain L Contain L (6.02x10(6.02x102323) ) atomsatoms

Covalent substances• A formula unit is a molecule• The total number of atoms can be found by

multiplying L by the number of atoms in the molecule.

Quantity of substance

Number of molecules

No. of atoms per molecule

Total No. of atoms

2g of Hydrogen, H2

18g of Water, H2O

30g of ethane, C2H6

L

L

L

2

3

8

2L

3L

8L

Ionic compounds• Formula unit consists of a ratio of ions

expressed by ionic formula.

Quantity of substance No. of formula

units

No. of +ve and –ve ions

Total No. of ions

58.5g of Na+Cl-

74g of Ca2+ (OH-)2

342g of (Al3+)2(SO42-)3

L

L

L

LNa+ and LCl-

LCa2+ and 2LOH-

2L Al3+ and 3L SO4

2-

2L

3L

5L

Example 1

• How many molecules are there in 8.8g of CO2?

1 mole of CO2 contains L molecules

44g of CO2 contains L molecules

1g of CO2 contains L/44 molecules

8.8g of CO2 contains L/44 x 8.8 molecules

= 1.204 X 10= 1.204 X 102323 molecules molecules

Example 2

• What mass of Nitrogen gas contains 18.06x1022 atoms of Nitrogen?

6.02x1023 molecules of N2 1 mole

6.02x1023 molecules of N2 28g

1 molecule of N2 28/L

18.06x1022 molecules of N2 28/L x 18.06x1022

= 8.4g

Therefore 8.4g of N2 gas contains 18.06x1022 molecules

4.2g of N2 gas contains 18.06x1022 atoms

Molar volume

• The volume occupied by one mole of gas at specific temperature and pressure.

• At room temperature and pressure, i.e. 20°C and 1 atmosphere pressure, the molar volume of any gas is approximately 24 litres mol-1

Molar volume

• Volume of gas changes if the temperature and/or pressure changes.

• Therefore you must specify at what temperature and pressure the volume is being measured.

Mass

Volume Density

GFM

DensityVmol

Density in g l-1Molar volumeIn litres mol-1

Example 1

(a) D = GFM / Vmol

= 20.2 / 24

= 0.84 g l= 0.84 g l-1-1

(b) V = GFM / Vmol

= 17 / 24

= 0.71 g l = 0.71 g l -1-1

• Calculate the density in g l-1 of the following gases at room temperature. The molar volume under these conditions is 24 litres mol-1.

(a) Neon (b) Ammonia

Gas Formula Density (gl-1) Molar volume (litres mol-1)

Methane CH4 0.65

Oxygen O2 1.33

Nitrogen N2 1.15

Carbon dioxide

CO2 1.81

Argon Ar 1.63

24.6

24.0

24.3

24.3

24.5

V

n Vmol

Volume of gasIn litres

No. of moles

Molar volumeIn litres mol-1

Example 2

• The molar volume at 0°C and 1 atmosphere pressure is 22.4 litres mol-1, Calculate

(a) The volume of 0.025 mol of oxygen

(b) The no. of moles of nitrogen in 4.48 litres under these conditions.

(a) Volume of oxygen V = n X Vmol= 0.025 X 22.4= 0.56 litres

= 560cm= 560cm33

(b) No. of moles of Nitrogen in 4.48 litres under these conditions?

No. of moles of nitrogen, n = V/Vmol

= 4.48/22.4

= 0.2 moles= 0.2 moles

Example 1• Calculate (i) the volume of oxygen required for the

complete combustion of 100cm3 of ethane and (ii) the volume of each product.

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)

2 mol 7 mol 4 mol 6 mol2 vol 7 vol 4 vol 6 vol

Simplified ratio1 vol 3.5 vol 2 vol 3 vol

(i) 100cm3 of ethane requires 350cm350cm33 O O22 and (ii) produces 200cm200cm33 CO CO22 and 300cm300cm33 H H22O.O.

Example 2• A mixture of 20cm3 propane and 130cm3 Oxygen was

ignited and allowed to cool. Calculate the volume and composition of the resulting gaseous mixture.

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

1 mol 5 mol 3 mol 4 mol1 vol 5 vol 3 vol -

20cm3 propane requires 5 x 20 = 100cm3 Oxygen. Oxygen in excess by 30cm3 (130cm3 – 100cm3)CO2 formed = 3 x 20 = 60cm3

Resulting gas mixture = 30cm30cm33 O O22 and 60cm and 60cm33 COCO22..