chemistry 120 stoichiometry: chemical calculations chemistry is concerned with the properties and...

Chemistry 120

Stoichiometry: Chemical Calculations

Chemistry is concerned with the properties and the

interchange of matter by reaction – structure and

change.

In order to do this, we need to be able to talk about

numbers of atoms

Chemistry 120


The mole and atomic mass

The mole is defined as

the number of elementary entities as are present in 12.00 g of 12C.

Numerically, this is equal to Avogadro’s Number

6.022 x 1023

Therefore, in 12.00 g of 12C there are 6.022 x 1023 ‘elementary entities’, in this case atoms.

Chemistry 120


The mole and atomic mass

Atomic masses, in atomic units, u, are defined relative to 12C.

Therefore,

The formula mass of an element or compound contains 1 mole, 6.022 x 1023, of particles

Chemistry 120


Examples

How many particles are there in 5 g of Na?

Chemistry 120


Examples


The particles are atoms – how many atoms are there in 5 g of Na?

Chemistry 120


Examples


The particles are atoms – how many atoms are there in 5 g of Na?

Atomic mass of Na = 22.9898 u

Chemistry 120


Examples


The particles are atoms – how many atoms are there in 5 g of Na

Atomic mass of Na = 22.9898 uAs

1 u = 1/12 x mass (12C) And

1 mole = 6.022 x 1023 particles = number of particles in 12 g 12C

Chemistry 120


Examples


The particles are atoms – how many atoms are there in 5 g of Na

Atomic mass of Na = 22.9898 u

Mass of 1 mole of Na = 22.9898 g

Chemistry 120


Examples

How many particles are there in 5.0000 g of Na?

22.9898 g Na = 1 mole Na

Then 1 g Na = 1 mol Na

22.9898

5 x 1 g Na = 5 x 1 mol Na

22.9898

5 g Na = 0.2175 mol Na

5 g Na = 0.2175 x (6.022 x 1023) particles Na

Chemistry 120


Examples

How many particles are there in 5.0000 g of Na?

1.310 x 1023 atoms

Chemistry 120


Examples

What is the mass of 0.23 mol of butane?

Chemistry 120


Examples


Molecular formula of butane: C4H10

Chemistry 120


Examples



Atomic mass of C = 12.011g

Atomic mass of H = 1.0079g

Chemistry 120


Examples





Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u

= 58.123 u

Chemistry 120


Examples





Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u

= 58.123 u

Relative Molecular Mass of Butane = 58.123 g

Chemistry 120


Examples



1 mole of butane = 58.123 g

0.23 x 1 mole of butane = 0.23 x 58.123 g

Chemistry 120


Examples



1 mole of butane = 58.123 g

0.23 x 1 mole of butane = 0.23 x 58.123 g

0.23 mole of butane = 13.368 g

Chemistry 120


Chemical Equations

These are formulæ which show the chemical change taking place in a reaction.

Sr(s) + Cl2(g) SrCl2(s)

Chemistry 120


Chemical Equations


Physical state


Chemistry 120


Chemical Equations


Physical state


Reactants Product

Chemistry 120


Chemical Equations

As matter cannot be created or destroyed in a chemical reaction, the total number of atoms on one side must be equal to the total number of atoms on the other.

Chemistry 120


Chemical Equations

Example

Cyclohexane burns in oxygen to give carbon dioxide and water

Chemistry 120


Chemical Equations

Example


Reactants: Cyclohexane, C6H12

Oxygen, O2

Chemistry 120


Chemical Equations

Example


Reactants: Cyclohexane, C6H12

Oxygen, O2

Products: Carbon Dioxide, CO2

Water, H2O

Chemistry 120


Chemical Equations

Example

Initially, we can write the reaction as

C6H12 + O2 CO2 + H2O

Chemistry 120


Chemical Equations

Example


C6H12 + O2 CO2 + H2O

This is NOT a correct equation – there are unequal numbers of atoms on both sides

Chemistry 120


Chemical Equations

Example


C6H12 + O2 CO2 + H2O

This is NOT a correct equation – there are unequal numbers of atoms on both sides

Reactants: 6 C, 12 H, 2 O

Products: 1 C, 2 H, 3 O

Chemistry 120


Balancing the equation

C6H12 + O2 CO2 + H2O

Chemistry 120

Stoichiometry: Chemical CalculationsBalancing the equation

6 C, 12 H, 2 O 1 C, 2 H, 3 O6 C on LHS means there must be 6 C on the RHS

C6H12 + O2 CO2 + H2O

C6H12 + O2 6CO2 + H2O

6 C, 12 H, 2 O 6 C, 2 H, 13 O

13 O on RHS means there must be 13 O on LHSC6H12 + 13/2 O2 6CO2 + H2O

6 C, 12 H, 13 O 6 C, 2 H, 13 O

Chemistry 120

Stoichiometry: Chemical CalculationsBalancing the equation

C6H12 + 13/2 O2 6CO2 + H2O

6 C, 12 H, 13 O 6 C, 2 H, 13 O

12 H on RHS means there must be 12 H on LHS

C6H12 + 13/2 O2 6CO2 + 6H2O

6 C, 12 H, 13 O 6 C, 12 H, 18 O

18 O on RHS means there must be 18 H on LHS

C6H12 +9O2 6CO2 + 6H2O

6 C, 12 H, 18 O 6 C, 12 H, 18 O

Chemistry 120

Stoichiometry: Chemical CalculationsThe final balanced equation is

and the coefficients are known as the

stoichiometric coefficients.

These coefficients give the molar ratios for reactants and products

This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction

C6H12 +9O2 6CO2 + 6H2O

Chemistry 120


If cyclohexane were burnt in an excess of oxygen,

the quantity of oxygen used would be the same

although O2 would be left over.

Chemistry 120

Stoichiometry: Chemical CalculationsThe final balanced equation is

and the coefficients are known as the

stoichiometric coefficients.

These coefficients give the molar ratios for reactants and products

This is a stoichiometric reaction – one where exactly the correct number of atoms is present in the reaction

Chemistry 120

The Exam

Chemistry 120

Solutions

A solution is a homogenous mixture which is composed of two or more components

the solvent

- the majority component

and

one or more solutes

- the minority components

Chemistry 120

Solutions

Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.

Some are solids where both the solvent and the solute are solids. Brass is an example

Chemistry 120

Solutions

Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent.

Some are solids where both the solvent and the solute are solids. Brass is an example

Cu

ZnHere copper is the solvent, zinc the solute.

Chemistry 120

Solutions

Gas-Solid solution: Hydrogen in palladium

Steel

Chemistry 120

Solutions

Common laboratory solvents are usually organic liquids such as acetone, hexane, benzene or ether or water.

Solutions in water are termed aqueous solutions and species are written as E(aq).

Water is the most important solvent. The oceans cover ~ ¾ of the surface of the planet and every cell is mainly composed of water.

Chemistry 120

Solutions

Aqueous Solutions

Water is one of the best solvents as it can dissolve many molecular and ionic substances.

The properties of solutions which contain molecular and ionic solutes are very different and give insight into the nature of these substances and solutions.

Chemistry 120

Solutions

Ionic Solutions

An ionic substance, such as NaClO4, contain ions – in this case Na+ and ClO4

-.

The solid is held together through electrostatic forces between the ions.

In water, the solid dissolves and the particles move away from each other and diffuse through the solvent. This process is termed

Ionic Dissociation

Chemistry 120

Solutions

Ionic Solutions

In an ionic solution, there are therefore charged particles – the ions – and as the compound is electrically neutral, then the solution is neutral.

When a voltage is applied to the solution, the ions can move and a current flows through the solution.

The ions are called charge carriers and whenever electricity is conducted, charge carriers are present.

Chemistry 120

Solutions

Molecular Solutions

A molecular solution does not conduct electricity as there are no charge carriers present.

The bonding in a molecule is covalent and involves the sharing of atoms and there is no charge separation.

Chemistry 120

Solutions

Electrolytes

A solute that, when dissolved, produces a solution that conducts is termed an electrolyte, which may be strong or weak.

A strong electrolyte is one which is fully dissociated in solution into ions

A weak electrolyte is one which is only partially dissociated.

Chemistry 120

Solutions

Moles and solutions

When a substance is dissolved in a solvent, we relate the quantity of material dissolved to the volume of the solution through the concentration of the solution.

The concentration is simply the number of moles of the material per unit volume:

C = n V

n = number of moles; V = volume of solvent

Chemistry 120

Solutions

Moles and solutions

The units of concentration are:

C = n = moles V L3

and we define a molar solution as one which has 1 mole per liter.

Alternatively,

Concentration = Molarity = number of molesvolume of solution

Chemistry 120

Solutions

Example

4 g of Na2SO4(s) is dissolved in 500 ml of water.

What is the concentration of the solution?

Chemistry 120

Solutions

Example



Formula mass of Na2SO4(s):

Molar Atomic Mass of Na: 22.9898 gmol-1

Molar Atomic Mass of S: 32.064 gmol-1

Molar Atomic Mass of O: 15.9994 gmol-1

Chemistry 120

Solutions

Example



Formula mass of Na2SO4(s):

(2 x 22.9898)+ 32.064+(4x15.9994)=142.041gmol-1

1 mole of Na2SO4(s) = 142.041g

1/142.041 mole of Na2SO4(s) = 1 g

Chemistry 120

Solutions

Example



1/142.041 mole of Na2SO4(s) = 1 g

Therefore 4 g of Na2SO4(s) = 4/142.041 mole

= 2.82 x 10-2 mole

Chemistry 120

Solutions

Example



2.82 x 10-2 mole is therefore dissolved in 500 ml of

water;

So in 1 L, there are 2 x 2.82 x 10-2 mole

Molarity of solution = 5.64 x 10-2 molL-1

Chemistry 120

Solutions

Example

The equation for the dissolution of Na2SO4(s) is

So if we have 5.64 x 10-2 molL-1 Na2SO4(s), we must

have 1.13 x 10-1 moles Na+(aq)

and 5.64 x 10-2 mol SO42-

(aq) as there are 2 Na

cations for every sulfate ion

Na2SO4(s)H2O

2Na+(aq) + SO4

2-(aq)

Chemistry 120

Solutions

If we change the volume of the solution then we change the concentration:

If the Na2SO4 solution is diluted with 500ml of water, the concentration or molarity would be halved:

2.82 x 10-2 mole is therefore dissolved in 1000 ml of water

Molarity = 2.82 x 10-2 molL-1

Chemistry 120

Solutions

Dissolution on an atomic level.

Solids are held together by very strong forces.

NaCl(s) melts at 801oC and

boils at 1465 oC but it

dissolves in water at room

temperature.

Chemistry 120

Solutions


When we dissolve NaCl(s) in water we break the bonds between ions but make bonds between the ions and the water

Chemistry 120

Solutions


When we dissolve NaCl(s) in water we break the bonds between ions but make bonds between the ions and the water

The ions are hydrated or solvated in solution and these bonds between solvent and solute make the dissolution energetically possible

If something does not dissolve then the energetics are wrong for it do do so.

Chemistry 120

Solutions

Solubility rules

All ammonium and Group I salts are soluble.

All Halides are soluble except those of silver, lead and mercury (I)

All Sulfates are soluble except those of barium and lead.

All nitrates are soluble.

Everything else is insoluble

Chemistry 120

Solutions

• Solutions are homogenous mixtures in which the

majority component is the solvent

and the

minority component is the solute

• Solutions are normally liquid but solutions of gases in solids and solids in solids are known.

• Ionic compounds dissolve in water to give conducting solutions – they are electrolytes

Chemistry 120

Solutions

• Electrolytes are either strong or weak depending on the degree of dissociation in solution

• Molecular solutions do not conduct as molecules do not dissociate in solution

• The concentration or molarity of a solution is defined by

C = n = moles V L3

and the units are molL-1 or moldm-3

Chemistry 120

Solutions

• When ionic substances dissolve,

bonds between particles in the solid break

and

bonds between the solvent and the ions are made

• There are general rules for the solubilities of ionic compounds

Chemistry 120

Reactions in Solution

Reactions in solution include

• Acid – base reactions

• Precipitation reactions

• Oxidation- reduction reactions

Chemistry 120

Reactions in Solution

Reactions and equilibria

Reactions are often written as proceeding in one

direction only – with an arrow to show the direction

of the chemical change, reactants to products.

Not all reactions behave in this manner and not all

reactions proceed to completion.

Even those that do are dynamic.

Chemistry 120

Reactions in Solution: Acid - Base

NaI*(s)

NaI(aq)

A saturated solution of NaI is placed in contact

with Na131I(s), which is radioactive.

Chemistry 120


NaI*(s)

NaI(aq)

A saturated solution of NaI is placed in contact

with Na131I(s), which is radioactive.

After time, the activity

in the solution is

measured and ..........

Chemistry 120

I-

I-

I-

I-

I-

I-

I-

I-I-

I-

I-

I-I-I-

I-

I- I-

Na+

Na+

Na+

Na+

Na+

Na+Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+Na+

I-

I-

I-

I-

I-

I-

I-

I-I-

I-

I-

I-I-I-

I-

I- I-

Na+

Na+

Na+

Na+

Na+

Na+Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+

Na+Na+


Radioactivity is found in the solution, even though

the concentration of I-(aq) has not changed.

Chemistry 120


The equilibrium here is composed of two reactions:

So we write

Na131I(s)H2O

Na+(aq) + 131I-

(aq)

H2ONa+

(aq) + I-(aq) NaI(s)

H2ONa+

(aq) + I-(aq)NaI(s)

Chemistry 120


Such reactions are termed equilibria and all chemical reactions are equilibria.

The symbol for an equilibrium is a double-headed arrow

Chemistry 120

+ =

Forward reaction

Reverse reaction


Such reactions are termed equilibria and all chemical reactions are equilibria.

The symbol for an equilibrium is a double-headed arrow

Chemistry 120


Equilibria are important in the chemistry of acids and bases

Strong acids and bases are completely ionized

But.....

Weak acids and bases are not.

Chemistry 120


The Arrhenius definition of acid and bases are:

an acid is a compound which dissolves

in water or reacts with water to give

hydronium ions, H3O+(aq)

a base is a compound which dissolves

in water or reacts with water to give

hydroxide ions, OH- (aq)

Svante Arrhenius

(1859 – 1927)

Chemistry 120


A strong acid is a compound which dissolves and dissociates completely in water or reacts with water to give hydronium ions, H3O+

(aq)

- the double arrow implies that the reaction can go both ways – it is an equilibrium.

As a strong acid, the reaction is completely on the RHS:

HCl(g)H2O

H3O+(aq) + Cl-(aq)

HCl(g)H2O

H3O+(aq) + Cl-(aq)

Chemistry 120

Reactions in Solution: Acid - BaseA strong base is a compound which dissolves and dissociates completely in water or reacts with water to give hydroxide ions, OH- (aq)

Again, we could write this reaction as an equilibrium with a double headed arrow, but the base is strong and the reaction is completely over to the right hand side.

NaOH(s)H2O

Na+(aq) + OH-

(aq)

Chemistry 120

Reactions in Solution: Acid - BaseIn a reaction such as

we write the reaction as going from LHS to RHS.

Chemical reactions run both ways, so in this reaction, there are two reactions present:

Ionization

Recombination

H2OMeCO2H H3O+

(aq) + MeCO2-(aq)

H2OMeCO2H H3O+

(aq) + MeCO2-(aq)

H2OMeCO2HH3O+

(aq) + MeCO2-(aq)

Chemistry 120

Reactions in Solution: Acid - BaseWe write the reaction for acetic acid, MeCO2H, as

an equilibrium to include the ionization and recombination. Ionization

Recombination

As the amount of ionization and recombination are the same, the concentrations of the ions and the molecular form are constant

H2OMeCO2H H3O+

(aq) + MeCO2-(aq)

H2OMeCO2HH3O+

(aq) + MeCO2-(aq)

H2OMeCO2H H3O+

(aq) + MeCO2-(aq)

Chemistry 120


In solution, weak acids establish an equilibrium between the un-ionized or molecular form and the ionized form:

un-ionizedmolecular form

ionized

H2OMeCO2H H3O+

(aq) + MeCO2-(aq)

Chemistry 120


In solution, strong acids are completely ionized and even though there is an equilibrium, it lies entirely on the RHS and recombination is negligible:

un-ionizedmolecular form

ionized

H2OHBr(g) H3O+

(aq) + Br-(aq)

Chemistry 120


Acids with more than one ionizable hydrogen are termed

Polyprotic

The common polyprotic acids are

H3PO4 Phosphoric acid

H2SO4 Sulfuric acid

Chemistry 120


Polyprotic acids can ionize more than once

H3PO4

Each proton is ionizable and the anions, dihydrogen phosphate (H2PO4

-(aq))

and hydrogen phosphate (HPO42-

(aq)) both act as acids, though H3PO4 is a weak acid.

H2SO4(aq) H3O+(aq)

+ HSO4-(aq)H2O

HSO4-(aq) H3O+

(aq) + PO4

2-(aq)H2O

HPO42-

(aq) H3O+(aq)

+ PO42-

(aq)H2O

Chemistry 120


Polyprotic acids can ionize more than once

H3PO4

H2SO4

H3PO4(aq) H3O+(aq)

+ H2PO4-(aq)H2O

H2PO4-(aq) H3O+

(aq) + HPO4

2-(aq)H2O

HPO42-

(aq) H3O+(aq)

+ PO42-

(aq)H2O

H2SO4(aq) H3O+(aq)

+ HSO4-(aq)H2O

HSO4-(aq) H3O+

(aq) + PO4

2-(aq)H2O

Chemistry 120


In contrast, H2SO4 is a strong acid and hydrogen

sulfate (HSO4-(aq)) is also a strong acid.

H2SO4(aq) H3O+(aq)

+ HSO4-(aq)H2O

HSO4-(aq) H3O+

(aq) + PO4

2-(aq)H2O

Chemistry 120

Reactions in Solution: Acid - BaseStrong or weak?

All acids can be assumed to be weak except the following:

HCl(aq) hydrochloric acid

HBr(aq) hydrobromic acid

HI(aq) hydriodic acid

HClO4(aq) perchloric acid

HNO3(aq) nitric acid

H2SO4(aq) sulfuric acid

Chemistry 120

Reactions in Solution: Acid - BaseHydrogens attached to carbon are not ionizable in water

Acetic acid, MeCO2H (or CH3CO2H) has the

structure H

H

H

O

O H

Chemistry 120


Only the hydrogen attached to oxygen is ionized in aqueous solution

The methyl hydrogens are NOT ionizable in aqueous solution.

H

H

H

O

O HH2O

H

H

H

O

O

O

HHH+

Chemistry 120


Strong bases are those which ionize in solution of react to generate hydroxide ion. The common strong bases are those which already contain the OH- ion in the solid. 2 Li

3

3 Na11 Mg12

4 K19

Ca20

5 Rb37

Sr38

6 Cs55

Ba56

Strong bases are therefore the hydroxides of the group I and II metals

Chemistry 120


Weak bases are the majority and are usually amines and ammonia. These react with water and deprotonate it, forming hydroxide ion and an ammonium ion:

Trimethylamine

Trimethylammonium

N

H3C CH3

CH3H2O

N

H3C CH3

CH3

H

+ OH-

Chemistry 120

Reactions in Solution: Acid - BaseReactions in Solution: Acid - Base

Neutralization reactions and titrations Hydroxide and hydronium ions will react to form water.

From the stoichiometry of the balanced equation, the hydroxide and hydronium react in a 1:1 ratio.

We can therefore neutralize a known concentration of base or acid with the same quantity of acid or base. This is an example of a titration.

2H2O(l)H3O+(aq) + OH-

(aq)

Chemistry 120

Reactions in Solution: Acid - BaseReactions in Solution: Acid - Base

Neutralization reactions and titrations We use an indicator to determine the acidity or basicity of a solution:

An indicator is a compound which changes color strongly at a certain level of acidity.

Chemistry 120

Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseNeutralization reactions and titrations

We add acid or base – the titrant - to a solution of unknown concentration containing a few drops of the indicator solution.

When the solution is still acid, no color change occurs; when the indicator changes color, we know the equivalence point – the point where the acidity or basicity has been neutralized.

By knowing the concentration and the volume of the titrant, we can calculate the concentration of the of the unknown solution.

Chemistry 120

Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acid

A solid base, such as Ca(OH)2(s), will dissolve with

reaction in an acid. The anion, hydroxide, reacts

with the acid to form the calcium salt of the acid:

Chemistry 120





Is this balanced?

Chemistry 120





Is this balanced? No

Chemistry 120





Is this balanced? No

2Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + 2H2O(l)

Chemistry 120


Some anions also decompose in acid. These are usually anions which are derived from gases which are not soluble in water:

CO32-

(aq) carbonate CO2(g)

HCO3-(aq) hydrogen carbonate CO2(g)

S2-(aq) sulfide H2S(g)

HS-(aq) hydrogen sulfide H2S(g)

SO32-

(aq) sulfite SO2(g)

HSO3-(aq) hydrogen sulfite SO2(g)

Chemistry 120

Gases

Properties of Gases

Kinetic Molecular Theory of Gases

Pressure

Boyle’s and Charles’ Law

The Ideal Gas Law

Gas reactions

Partial pressures

Chemistry 120

Gases

Properties of Gases

All elements will form a gas at some temperature

Most small molecular compounds and elements are either gases or have a significant vapor pressure.

1 H1 Room Temperature Gases He

2

2 Li3

Be4

B5

C6

N7

O8

F9

Ne10

3 Na11

Mg12

Al13

Si14

P15

S16

Cl17

Ar18

4 K19

Ca20

Sc21

Ti22

V23

Cr24

Mn25

Fe26

Co27

Ni28

Cu29

Zn30

Ga31

Ge32

As33

Se34

Br35

Kr36

5 Rb37

Sr38

Y39

Zr40

Nb41

Mo42

Tc43

Ru44

Rh45

Pd

46

Ag47

Cd48

In49

Sn50

Sb51

Te52

I53

Xe54

6 Cs55

Ba56

Lu71

Hf72

Ta73

W74

Re75

Os76

Ir77

Pt78

Au79

Hg80

Tl81

Pb82

Bi83

Po

84

At85

Rn86

Chemistry 120

Gases

Properties of Gases

As the temperature rises, all elements form a gas at some point.

In the following diagram,

Blue represents solids

Green represents liquids

Red represents gases

At O K, all elements are solids

At 6000 K, all are gases

Chemistry 120

Gases

Chemistry 120

Gases

Properties of Gases

Gases have no shape and no volume.

They take the volume and shape of the container

Their densities are low – usually measured in gL-1

The atoms or molecules of the gas are far further apart than in a solid or a liquid.

Chemistry 120

Gases

Gases as an ensemble of particles

The attractive forces between liquids and solids are very strong

LiF: M.p.: 848°C B. p.: 1676°C

In a gas, the forces between particles are negligible and as there are no attractive forces, a gas will occupy the volume of the container.

Solid Liquid Liquid Gas

Chemistry 120

GasesGases as an ensemble of particles

The structures of liquids and solids are well ordered on a microscopic level

CaCl2Ethanol, C2H5OH

Chemistry 120


In a gas, there is no order and all the properties of the gas are isotropic – all the properties of the gas are the same in all directions.

Gas particles are distributed uniformly throughout the container.

They can move throughout the container in straight line trajectories.

Chemistry 120


The directions of the motions of the gas particles are random

and

The velocities form a distribution – there is a range of possible velocities around an average value.

The trajectories of the gas particles are straight lines and there are two possible fates for a gas molecule......

Chemistry 120


A gas particle can collide with

– the walls of the container

Or

– another gas molecule

When this happens, the gas particle changes direction.

Chemistry 120


Kinetic energy can be transferred between the two colliding particles

– one can slow down and the other speed up –

but the net change in kinetic energy is zero.

These collisions are termed elastic, meaning that there is no overall change in kinetic energy.

Chemistry 120

Gases

The average kinetic energy for a given gas is determined by the temperature alone and the width and peak maximum is also determined by the temperature.

The Maxwell-Boltzmann distribution for He

Chemistry 120


The force exerted by the gas particles on the walls of the container gives rise to the pressure of the gas.

We define pressure as the force exerted per unit area:

P = Force = F Area A

The unit of pressure is the Pascal (Pa)

1 Pa = 1 Nm-2

In practice, the Pascal is too small - kPa or GPa

Chemistry 120

GasesPressure measurement

Pressure is also measured in several other non – SI units:

In industry: Pounds per square in (p.s.i.)

In research: Pascal, atmosphere, bar, Torr

Chemistry 120

GasesPressure conversion factors

Atmospheric pressure = 101,325 Pa

1 Atmosphere = 101,325 Pa = 1 bar

1 Atmosphere = 101,325 Pa

= 1 bar

= 760 Torr

= 760 mmHg

= 14.7 p.s.i.

Chemistry 120

GasesPressure Measurement

Pressure is measured using a manometer or barometer

– either one containing Hg or an electronic gauge

A mercury manometer is a U–tube connected to the gas vessel, with the other end either evacuated or open to the atmosphere.

The measurement of the height difference between the mercury levels on both sides of the ‘U’ gives the pressure........

Chemistry 120


Let the height difference between the two Hg levels be h

Then the gas pressure is given by

Pgas = P0 + h

As

P = Force = F = mg where g = 9.81 ms-2

Area A A

Chemistry 120


How is the height difference related to the pressure?

As density, = m V

Then m = V

The volume of the column of mercury is

V = A.h

And so m = V = A.h

Chemistry 120


The pressure above the baseline pressure P0 is therefore

Pgas = mg =gA.h = gh A A

Chemistry 120


Kinetic energy can be transferred between the two colliding particles

– one can slow down and the other speed up –

but the net change in kinetic energy is zero.

These collisions are termed elastic, meaning that there is no overall change in kinetic energy.

Chemistry 120

The Gas Laws

The factors that control the behavior of a gas are

• The nature of the gas

Chemistry 120

The Gas Laws



• The quantity of the gas - n

Chemistry 120

The Gas Laws




• The pressure - P

Chemistry 120

The Gas Laws





• The temperature - T

Chemistry 120

The Gas Laws





• The temperature - T

• The volume of a gas -V

Chemistry 120

The Gas Laws

These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures.

The qualities of an ideal gas are:

• Zero size to the gas particles

We assume that the volume of the container is very much larger than the total volume of the gas molecules

• No attractive forces between atoms

Chemistry 120

The Gas Laws

These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures.

At low temperatures and high pressures gases deviate from ideality.

The ideal gas laws are based on three interdependent laws – Boyle’s Law, Charles’ Law and Avogadro’s Law.

Chemistry 120

The Gas Laws

Boyle’s Law

Robert Boyle experimented with gases in Oxford in 1660.

He discovered that the product of the volume and the pressure of a gas is a constant, so long as the quantity of gas and the temperature are constant.

Chemistry 120

The Gas Laws

Boyle’s Law

Mathematically,

PV = a constant

as long as n and T are constant

Chemistry 120

The Gas Laws

Boyle’s Law

Mathematically,

PV = a constant, k

or

P = kV

as long as n and T are constant.

Chemistry 120

The Gas Laws

Boyle’s Law

A graph of Boyle’s data shows this relationship:

PV = k

Chemistry 120

The Gas Laws

Boyle’s Law

A graph of 1/P as the abscissa and V as the ordinate.

V = k P

The graph shows a straight line of slope k

Chemistry 120

The Gas Laws

Boyle’s Law

As the pressure rises, 1/P becomes smaller and the graph passes through the origin.

This implies that at infinitely large pressure, the volume of a gas is zero.We know that molecules and and atoms have a definite volume, so Boyle’s law must fail at very high pressures.

Chemistry 120

The Gas Laws

Charles’ Law

Jacques Charles was a Feench scientist and aeronaut who discovered (1787) that all gases expand by the same amount when the temperature of the gas rises by the same amount.

Chemistry 120

The Gas Laws

Charles’ Law

Mathematically, we express this as

V = k’T

And a graph of Charles’ Law

is a straight line:

Chemistry 120

The Gas Laws

The Combined Gas Law for a Perfect Gas

Combining Boyle’s Law, Charles’ Law and Avogadro’s Law,

V = k and V = k’T and V = k”n P

we can say thatV nT P

Chemistry 120

The Gas Laws


V nT P

OrV =K nT P

Rearranging we find

PV = a constantnT

Chemistry 120

The Gas Laws


The constant is termed the Universal Gas Constant, R, and takes the value

R = 8.314 Jmol-1K-1

So the Universal Gas Law is written as

PV = nRT

This Law applies to all gases as long as they fulfill the conditions for near ideal behavior – not at high pressure and not at low temperature

Chemistry 120

The Gas Laws

Using the Combined Gas Law

If the quantity of gas is the same, then changes in pressure, temperature or volume can be calculated easily as

P1V1 = n = P2V2 RT1 RT2

Or

P1V1 = P2V2 T1 T2

Chemistry 120

The Gas Laws

Using the Combined Gas Law

The advantage of this expression is that the units do not matter; the units used for P1 ,V1, and T1 will be returned in the calculation for P2 ,V2, and T2.

However, if you have to use PV = nRT, you must use the correct units which are consistent with R.

The easiest way is to convert all temperatures to K, all pressures to Pa and all volumes to m3; the value for R is then 8.314 Jmol-1K-1

Chemistry 120

The Gas Laws

The absolute temperature scale

From Charles’ Law, the decrease in volume per unit temperature is always the same and therefore there must be a minimum temperature that can be reached. This is absolute zero O K, and is the zero point for the absolute temperature scale.

The temperature in K is related to the temperature in oC through

T/K = T/oC + 273.16

Chemistry 120

The Gas Laws

Example: Molecular Mass determinations

If we know the mass of gas in a sample of known volume, pressure and temperature, then we can calculate the relative molecular mass as we can calculate n.

As n = m then, PV = mRT , so RMM = mRT RMM RMM PV

RMM = mRT PV

Chemistry 120

The Gas Laws

Example: Molar volumes

From Avogadro’s Law, equal quantities of gas occupy equal volumes.

The volume of one mole of gas is therefore independent of the nature of the gas, as long as the gas behaves as ideal.

One mole of a perfect gas at 0oC and 1 atm pressure occupies

22.4 L

Chemistry 120

The Gas Laws

Example: Volumes and moles

When we react solids or liquids, the easiest way is to measure the mass of the sample and then convert to moles by dividing by the relative molecular mass.

For gases, the easiest way is to measure the pressure or the volume, as the densities of gases are so low.

For these calculations, you must use the same temperatures and pressures for each gas.

Chemistry 120

The Gas Laws

Partial pressures

In a mixture of gases, we can measure the total pressure of the mixture – PTotal and therefore we can use PV = nRT to determine the total number of moles of gas present.

As the mixture contains more than one gas, we can write the contribution of the pressure of each gas to the total pressure

Chemistry 120

The Gas Laws

Partial pressures

So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture:

Chemistry 120

The Gas Laws

Partial pressures


PTotal = P1 + P2 + P3 + P4 + ...........

Chemistry 120

The Gas Laws

Partial pressures


PTotal = P1 + P2 + P3 + P4 + ...........

As PV = nRT then

Chemistry 120

The Gas Laws

Partial pressures


PTotal = P1 + P2 + P3 + P4 + ...........

As PV = nRT then

nTotalRT = n1RT + n2RT + n3RT + n4RT + ...........

Chemistry 120

The Gas Laws

Partial pressures

So

PTotal = P1 + P2 + P3 + P4 + ...........


Chemistry 120

The Gas Laws

Partial pressures

So

PTotal = P1 + P2 + P3 + P4 + ...........


nTotal = n1 + n2 + n3 + n4 + ...........

Chemistry 120

The Gas Laws

Partial pressures

So the pressures of each component of the gas mixture correlate with the number of moles of the gas component of the mixture – a simple extension of Avogadro’s Law.

Chemistry 120

The Gas Laws

Partial pressures

We can also write the fraction of the total pressure that is due to one of the component:

PTotal = P1 + P2 + P3 + P4 + ...........

Chemistry 120

The Gas Laws

Partial pressures


PTotal = P1 + P2 + P3 + P4 + ...........

nTotal = n1 + n2 + n3 + n4 + ..........

Chemistry 120

The Gas Laws

Partial pressures


PTotal = P1 + P2 + P3 + P4 + ...........

nTotal = n1 + n2 + n3 + n4 + ..........

P1 = n1RT

Chemistry 120

The Gas Laws

Partial pressures


PTotal = P1 + P2 + P3 + P4 + ...........

nTotal = n1 + n2 + n3 + n4 + ..........

P1 = n1RT

So, P1 = n1

PTotal n1 + n2 + n3 + n4 + ..........

Chemistry 120

The Gas Laws

Partial pressures

P1 = n1

PTotal n1 + n2 + n3 + n4 + ..........

The fraction on the RHS is called the mole fraction

and is written as x1 so we can write

P1 = n1 PTotal

n1 + n2 + n3 + n4 + ..........

Or P1 = x1 PTotal

Chemistry 120

Thermochemistry

Energy

Energy is defined as the ability to do work.

There are several forms of energy

Kinetic energy – energy due to motion

Chemistry 120

Thermochemistry

Energy

Energy is defined as the ability to do work.

There are several forms of energy

Kinetic energy – energy due to motion EK = 1/2mv2

Potential energy – the energy due to the position of a particle in a field

e.g. Gravitational, electrical, magnetic etc.

Chemistry 120

Thermochemistry

Energy

The unit of energy is the Joule (J) and

1 J = 1 kgm2s-2

Thermochemistry is the study of chemical energy and of the conversion of chemical energy into other forms of energy.

It is part of thermodynamics – the study of the flow of heat.

Chemistry 120

Thermochemistry

Thermochemically, we define the system as the part of the universe under study and the surroundings as everything else.

Systems come in three forms:

Open The system can exchange matter and energy with the surroundings

Closed The system can exchange energy only with the surroundings

Isolated There is no exchange of matter or of energy with the surroundings

Chemistry 120

Thermochemistry

Matter is continually in motion and has an internal energy that is composed of several different types

There is

Translation Rotation Vibration Potential

between molecules and inside molecules.

The internal energy is written as U

Chemistry 120

Thermochemistry

Matter is continually in motion and has an internal energy that is composed of several different types

There is

Translation Rotation Vibration Potential

between molecules and inside molecules.

The internal energy is written as U

The internal energy is directly connected to heat and the transfer of heat.

Chemistry 120

Thermochemistry

Heat is the transfer of energy between the surroundings and the system or between systems.

The direction of the heat flow is indicated by the temperature

– heat flows along a Temperature gradient

from high temperature to low temperature.

When the temperature of the system and that of the surroundings are equal, the system is said to be

in thermal equilibrium

chemistry 120 stoichiometry: chemical calculations chemistry is concerned with the properties and...

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