stoichiometry: chemical calculations · 1 stoichiometry: chemical calculations chemistry 120...

1

Chemistry 120Stoichiometry: 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


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.


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

2


Examples

How many particles are there in 5 g of Na?


Examples


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


Examples


The particles are atoms – how many atomsare there in 5 g of Na?Atomic mass of Na = 22.9898 u

3


Examples


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


Examples


The particles are atoms – how many atomsare there in 5 g of NaAtomic mass of Na = 22.9898 uMass of 1 mole of Na = 22.9898 g


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 Na22.9898

5 x 1 g Na = 5 x 1 mol Na22.9898

5 g Na = 0.2175 mol Na5 g Na = 0.2175 x (6.022 x 1023) particles Na

4


Examples

How many particles are there in 5.0000 g of Na?

1.310 x 1023 atoms


Examples

What is the mass of 0.23 mol of butane?


Examples


Molecular formula of butane: C4H10

5


Examples



Atomic mass of C = 12.011g

Atomic mass of H = 1.0079g


Examples





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

= 58.123 u


Examples





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

= 58.123 u

Relative Molecular Mass of Butane = 58.123 g

6


Examples



1 mole of butane = 58.123 g

0.23 x 1 mole of butane = 0.23 x 58.123 g


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


Chemical Equations

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

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

7


Chemical Equations


Physical state



Chemical Equations


Physical state


Reactants Product


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.

8


Chemical Equations

Example

Cyclohexane burns in oxygen to give carbon dioxide and water


Chemical Equations

Example


Reactants: Cyclohexane, C6H12

Oxygen, O2


Chemical Equations

Example


Reactants: Cyclohexane, C6H12

Oxygen, O2

Products: Carbon Dioxide, CO2

Water, H2O

9


Chemical Equations

Example

Initially, we can write the reaction as

C6H12 + O2 CO2 + H2O


Chemical Equations

Example


C6H12 + O2 CO2 + H2O

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


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

10


Balancing the equation

C6H12 + O2 CO2 + H2O

Chemistry 120Stoichiometry: 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 + H2O6 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 120Stoichiometry: 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 + 6H2O6 C, 12 H, 13 O 6 C, 12 H, 18 O

18 O on RHS means there must be 18 H on LHSC6H12 +9O2 6CO2 + 6H2O

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

11

Chemistry 120Stoichiometry: 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


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 120Stoichiometry: 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

12

Chemistry 120The Exam

Chemistry 120Solutions

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


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

13







Cu

ZnHere copper is the solvent, zinc the solute.


Gas-Solid solution: Hydrogen in palladium

Steel

14


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.


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.


Ionic Solutions

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

-.

The solid is held together through electrostaticforces 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

15


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.


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.


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.

16


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 = nV

n = number of moles; V = volume of solvent


Moles and solutions

The units of concentration are:

C = n = molesV L3

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

Concentration = Molarity = number of molesvolume of solution


Example

4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution?

17


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


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


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

18


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


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) 2Na+(aq) + SO4

2-(aq)

H2O


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

19


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.



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



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.

20


Solubility rules

All ammonium and Group I salts are soluble.

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

All Sulfates are soluble except those of barium and lead.

All nitrates are soluble.

Everything else is insoluble


• 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


• Electrolytes are either strong or weakdepending 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 = molesV L3

and the units are molL-1 or moldm-3

21


• 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 120Reactions in Solution

Reactions in solution include

• Acid – base reactions

• Precipitation reactions

• Oxidation- reduction reactions

Chemistry 120Reactions 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.

22

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+

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

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+


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+

Reactions in Solution: Acid - BaseRadioactivity is found in the solution, even though

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

23

Chemistry 120Reactions in Solution: Acid - Base

The equilibrium here is composed of two reactions:

So we write

Na131I(s) Na+(aq) + 131I-

(aq)

Na+(aq) + I-

(aq) NaI(s)

H2O

H2O

Na+(aq) + I-

(aq)NaI(s)H2O


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

The symbol for an equilibrium is a double-headed arrow

Forward reaction




+

Forward reaction

24




+

Forward reaction

Reverse reaction

Chemistry 120

+ =

Forward reaction

Reverse reaction





Equilibria are important in the chemistry of acids and bases

Strong acids and bases are completely ionized

But.....

Weak acids and bases are not.

25


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)


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) H3O+(aq) + Cl-(aq)

H2O

HCl(g) H3O+(aq) + Cl-(aq)

H2O

Chemistry 120Reactions 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) Na+(aq) + OH-

(aq)H2O

26

Chemistry 120Reactions 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

H3O+

(aq) + MeCO2-(aq)

H2OMeCO2H

H3O+

(aq) + MeCO2-(aq)

H2OMeCO2H

H3O+(aq) + MeCO2

-(aq)

H2OMeCO2H

Chemistry 120Reactions 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

H3O+(aq) + MeCO2

-(aq)

H2OMeCO2H

H3O+(aq) + MeCO2

-(aq)

H2OMeCO2H

H3O+

(aq) + MeCO2-(aq)

H2OMeCO2H

Chemistry 120Reactions in Solution: Acid - BaseIn solution, weak acids establish an equilibrium between the un-ionized or molecular form and the ionized form:

un-ionizedmolecular form

ionized

H3O+

(aq) + MeCO2-(aq)

H2OMeCO2H

27

Chemistry 120Reactions in Solution: Acid - BaseIn 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

HBr(g) H3O+

(aq) + Br-(aq)

H2O

Chemistry 120Reactions in Solution: Acid - BaseAcids with more than one ionizable hydrogen are termed

PolyproticThe common polyprotic acids are

H3PO4 Phosphoric acid

H2SO4 Sulfuric acid

Chemistry 120Reactions in Solution: Acid - BasePolyprotic 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)

HSO4-(aq) H3O+

(aq) + SO4

2-(aq)

HPO42-

(aq) H3O+(aq)

+ PO42-

(aq)

H2O

H2O

H2O

28

Chemistry 120Reactions in Solution: Acid - BasePolyprotic acids can ionize more than once

H3PO4

H2SO4

H2SO4(aq) H3O+(aq)

+ HSO4-(aq)

HSO4-(aq) H3O+

(aq) + PO4

2-(aq)

HPO42-

(aq) H3O+(aq)

+ PO42-

(aq)

H2O

H2O

H2O

H2SO4(aq) H3O+(aq)

+ HSO4-(aq)

HSO4-(aq) H3O+

(aq) + PO4

2-(aq)

H2O

H2O

Chemistry 120Reactions in Solution: Acid - BaseIn contrast, H2SO4 is a strong acid and hydrogen sulfate (HSO4

-(aq)) is also a strong acid.

H2SO4(aq) H3O+(aq)

+ HSO4-(aq)

HSO4-(aq) H3O+

(aq) + PO4

2-(aq)

H2O

H2O

Chemistry 120Reactions 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

29

Chemistry 120Reactions 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


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 H

H

H

H

O

O

OH H

H+H2O


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.

Sr38

Rb37

5

Ba56

Cs55

6

Ca20

K19

4

Mg12

Na113

Li3

2

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

30


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

CH3N

H3C CH3

CH3

H

+ OH-

H2O

Chemistry 120Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseNeutralization 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 120Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseNeutralization 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.

31

Chemistry 120Reactions 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 120Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acidA 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:

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



Is this balanced?

32



Is this balanced? No



Is this balanced? No

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

Chemistry 120Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion in acidSome 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)

33

Chemistry 120Gases

Properties of Gases

Kinetic Molecular Theory of Gases

Pressure

Boyle’s and Charles’ Law

The Ideal Gas Law

Gas reactions

Partial pressures

Chemistry 120Gases

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.

He2Room Temperature GasesH

11

Xe54

I53

Te52

Sb51

Sn50

In49

Cd48

Ag47

Pd46

Rh45

Ru44

Tc43

Mo42

Nb41

Zr40

Y39

Sr38

Rb37

5

Rn86

At85

Po84

Bi83

Pb82

Tl81

Hg80

Au79

Pt78

Ir77

Os76

Re75

W74

Ta73

Hf72

Lu71

Ba56

Cs55

6

Kr36

Br35

Se34

As33

Ge32

Ga31

Zn30

Cu29

Ni28

Co27

Fe26

Mn25

Cr24

V23

Ti22

Sc21

Ca20

K19

4

Ar18

Cl17

S16

P15

Si14

Al13

Mg12

Na11

3

Ne10

F9

O8

N7

C6

B5

Be4

Li3

2

Chemistry 120Gases

Properties of Gases

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

Blue represents solidsGreen represents liquidsRed represents gases

At O K, all elements are solids

At 6000 K, all are gases

34

Chemistry 120Gases

Chemistry 120Gases

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 120Gases

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

35

Chemistry 120GasesGases as an ensemble of particles

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

CaCl2 Ethanol, C2H5OH


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.


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......

36


A gas particle can collide with

– the walls of the container

Or

– another gas molecule

When this happens, the gas particle changes direction.


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 120GasesThe 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

37


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 = FArea 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 120GasesPressure 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 120GasesPressure 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.

38

Chemistry 120GasesPressure 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........


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


How is the height difference related to the pressure?

As density, ρ = mV

Then m = ρ V

The volume of the column of mercury is

V = A.∆h

And so m = ρ V = ρ A.∆h

39


The pressure above the baseline pressure P0 is therefore

Pgas = mg =ρgA.∆h = ρg∆hA A


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 120The Gas Laws

The factors that control the behavior of a gas are

• The nature of the gas

40




• The quantity of the gas - n





• The pressure - P






• The temperature - T

41






• The temperature - T

• The volume of a gas -V


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


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.

42


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.


Boyle’s Law

Mathematically,

PV = a constant

as long as n and T are constant


Boyle’s Law

Mathematically,

PV = a constant, k

or

P = kV

as long as n and T are constant.

43


Boyle’s Law

A graph of Boyle’s data shows this relationship:

PV = k


Boyle’s Law

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

V = kP

The graph shows a straight line of slope k


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.

44


Charles’ Law

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


Charles’ Law

Mathematically, we express this as

V = k’T

And a graph of Charles’ Law

is a straight line:

Chemistry 120The Gas LawsThe 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”nP

we can say thatV ∝ nT

P

45


V ∝ nTP

OrV =K nT

PRearranging we find

PV = a constantnT


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 120The Gas LawsUsing 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 = P2V2RT1 RT2

OrP1V1 = P2V2T1 T2

46

Chemistry 120The Gas LawsUsing 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 120The Gas LawsThe 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 120The Gas LawsExample: 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 = mRTPV

47

Chemistry 120The Gas LawsExample: 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 120The Gas LawsExample: 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 120The Gas LawsPartial 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

48


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



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



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

As PV = nRT then

49



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

As PV = nRT then

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


So

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



So

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


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

50


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.


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

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



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

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

51



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

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

P1 = n1RT



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

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

P1 = n1RT

So, P1 = n1

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


P1 = n1

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

The fraction on the RHS is called the mole fractionand is written as x1 so we can write

P1 = n1 PTotal

n1 + n2 + n3 + n4 + .......... Or P1 = x1 PTotal

52

Chemistry 120ThermochemistryEnergy

Energy is defined as the ability to do work.

There are several forms of energy

Kinetic energy – energy due to motion


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.


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.

53

Chemistry 120ThermochemistryThermochemically, 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 120ThermochemistryMatter 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 120ThermochemistryMatter 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.

54

Chemistry 120ThermochemistryHeat 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 120ThermochemistryHeat 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

stoichiometry: chemical calculations · 1 stoichiometry: chemical calculations chemistry 120...

Documents