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 Stoichiometry: Chemical Calculations The mole and atomic mass The mole is defined as the number of elementary entities as are present in 12.00 g of 12 C. Numerically, this is equal to Avogadro s Number 6.022 x 10 23 Therefore, in 12.00 g of 12 C there are 6.022 x 10 23 elementary entities, in this case atoms. Stoichiometry: Chemical Calculations The mole and atomic mass Atomic masses, in atomic units, u, are defined relative to 12 C. Therefore, The formula mass of an element or compound contains 1 mole, 6.022 x 10 23, of particles
Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? The particles are atoms how many atoms are there in 5 g of Na? Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? The particles are atoms how many atoms are there in 5 g of Na? Atomic mass of Na = 22.9898 u
Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? The particles are atoms how many atoms are there in 5 g of Na Atomic mass of Na = 22.9898 u As 1 u = 1 / 12 x mass ( 12 C) And 1 mole = 6.022 x 10 23 particles = number of particles in 12 g 12 C Stoichiometry: Chemical Calculations Examples How many particles are there in 5 g of Na? 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 Stoichiometry: Chemical Calculations 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 10 23 ) particles Na
Stoichiometry: Chemical Calculations Examples How many particles are there in 5.0000 g of Na? 1.310 x 10 23 atoms Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10
Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C 4 H 10 = (4x12.011)+(10x1.0079)u = 58.123 u Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C 4 H 10 = (4x12.011)+(10x1.0079)u = 58.123 u Relative Molecular Mass of Butane = 58.123 g
Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 1 mole of butane = 58.123 g 0.23 x 1 mole of butane = 0.23 x 58.123 g Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 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 Stoichiometry: Chemical Calculations Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr (s) + Cl 2(g) SrCl 2(s)
Stoichiometry: Chemical Calculations Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr (s) + Cl 2(g) SrCl 2(s) Physical state Stoichiometry: Chemical Calculations Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr (s) + Cl 2(g) SrCl 2(s) Reactants Physical state Product Stoichiometry: Chemical Calculations 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.
Stoichiometry: Chemical Calculations Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Stoichiometry: Chemical Calculations Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C 6 H 12 Oxygen, O 2 Stoichiometry: Chemical Calculations Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C 6 H 12 Oxygen, O 2 Products: Carbon Dioxide, CO 2 Water, H 2 O
Stoichiometry: Chemical Calculations Chemical Equations Example Initially, we can write the reaction as C 6 H 12 + O 2 CO 2 + H 2 O Stoichiometry: Chemical Calculations Chemical Equations Example Initially, we can write the reaction as C 6 H 12 + O 2 CO 2 + H 2 O This is NOT a correct equation there are unequal numbers of atoms on both sides Stoichiometry: Chemical Calculations Chemical Equations Example Initially, we can write the reaction as C 6 H 12 + O 2 CO 2 + H 2 O 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
Stoichiometry: Chemical Calculations Balancing the equation C 6 H 12 + O 2 CO 2 + H 2 O Stoichiometry: Chemical Calculations Balancing the equation C 6 H 12 + O 2 CO 2 + H 2 O 6 C, 12 H, 2 O 1 C, 2 H, 3 O 6 C on LHS means there must be 6 C on the RHS C 6 H 12 + O 2 6CO 2 + H 2 O 6 C, 12 H, 2 O 6 C, 2 H, 13 O 13 O on RHS means there must be 13 O on LHS C 6 H 12 + 13 / 2 O 2 6CO 2 + H 2 O 6 C, 12 H, 13 O 6 C, 2 H, 13 O Stoichiometry: Chemical Calculations Balancing the equation C 6 H 12 + 13 / 2 O 2 6CO 2 + H 2 O 6 C, 12 H, 13 O 6 C, 2 H, 13 O 12 H on RHS means there must be 12 H on LHS C 6 H 12 + 13 / 2 O 2 6CO 2 + 6H 2 O 6 C, 12 H, 13 O 6 C, 12 H, 18 O 18 O on RHS means there must be 18 H on LHS C 6 H 12 +9O 2 6CO 2 +6H 2 O 6 C, 12 H, 18 O 6 C, 12 H, 18 O
Stoichiometry: Chemical Calculations The final balanced equation is C 6 H 12 +9O 2 6CO 2 +6H 2 O 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 Stoichiometry: Chemical Calculations If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O 2 would be left over. Stoichiometry: Chemical Calculations The 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
The Exam 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 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
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 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 Here copper is the solvent, zinc the solute. Cu Zn Solutions Gas-Solid solution: Hydrogen in palladium Steel
Solutions Common laboratory solvents are usually organic liquids such as acetone, hexane, benzene or ether or water. Water is the most important solvent. The oceans cover ~ ¾ of the surface of the planet and every cell is mainly composed of water. Solutions in water are termed aqueous solutions and species are written as E (aq). 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. Solutions Ionic Solutions An ionic substance, such as NaClO 4, contain ions in this case Na + and ClO 4-. 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
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. 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. 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.
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 Solutions Moles and solutions The units of concentration are: C = n = moles V L 3 and we define a molar solution as one which has 1 mole per liter. Alternatively, Concentration = Molarity = number of moles volume of solution Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution?
Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na 2 SO 4(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 Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na 2 SO 4(s) : (2 x 22.9898)+ 32.064+(4x15.9994)=142.041gmol -1 1 mole of Na 2 SO 4(s) = 142.041g 1 / 142.041 mole of Na 2 SO 4(s) = 1 g Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? 1 / 142.041 mole of Na 2 SO 4(s) = 1 g Therefore 4 g of Na 2 SO 4(s) = 4 / 142.041 mole = 2.82 x 10-2 mole
Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? 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 Solutions Example The equation for the dissolution of Na 2 SO 4(s) is Na 2 SO 4(s) 2Na + 2- (aq) + SO 4 (aq) H 2 O So if we have 5.64 x 10-2 moll -1 Na 2 SO 4(s), we must have 1.13 x 10-1 moles Na + (aq) and 5.64 x 10-2 mol SO 2-4 (aq) as there are 2 Na cations for every sulfate ion Solutions If we change the volume of the solution then we change the concentration: If the Na 2 SO 4 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
Solutions Dissolution on an atomic level. Solids are held together by very strong forces. NaCl (s) melts at 801 o C and boils at 1465 o C but it dissolves in water at room temperature. Solutions Dissolution on an atomic level. When we dissolve NaCl (s) in water we break the bonds between ions but make bonds between the ions and the water Solutions Dissolution on an atomic level. 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.
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 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 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 L 3 and the units are moll -1 or moldm -3
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 Reactions in Solution Reactions in solution include Acid base reactions Precipitation reactions Oxidation- reduction reactions 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.
Reactions in Solution: Acid - Base A saturated solution of NaI is placed in contact with Na 131 I (s), which is radioactive. Na + I - I - Na + I - I - Na + Na + I - I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na + NaI (aq) NaI * (s) Reactions in Solution: Acid - Base I - Na + I - I - Na + I - I - Na + Na + I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na + NaI (aq) NaI * (s) A saturated solution of NaI is placed in contact with Na 131 I (s), which is radioactive. After time, the activity in the solution is measured and... Reactions in Solution: Acid - Base Radioactivity is found in the solution, even though the concentration of I - (aq) has not changed. I - Na + I - I - Na + I - I - Na + Na + I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na + I - Na + I - I - Na + I - I - Na + Na + I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na +
Reactions in Solution: Acid - Base The equilibrium here is composed of two reactions: Na 131 I (s) H 2 O Na + (aq) + 131 I - (aq) Na + (aq) + I - (aq) H 2 O NaI (s) So we write NaI (s) H 2 O Na + (aq) + I - (aq) Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction +
Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction + Reverse reaction Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction + = Reverse reaction Reactions in Solution: Acid - Base Equilibria are important in the chemistry of acids and bases Strong acids and bases are completely ionized But... Weak acids and bases are not.
Reactions in Solution: Acid - Base 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, H 3 O + (aq) a base is a compound which dissolves in water or reacts with water to give hydroxide ions, OH - (aq) Svante Arrhenius (1859 1927) Reactions in Solution: Acid - Base A strong acid is a compound which dissolves and dissociates completely in water or reacts with water to give hydronium ions, H 3 O + (aq) HCl (g) H 3 O + (aq) + Cl - (aq) H 2 O - 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) H 2 O H 3 O + (aq) + Cl - (aq) Reactions in Solution: Acid - Base A strong base is a compound which dissolves and dissociates completely in water or reacts with water to give hydroxide ions, OH - (aq) NaOH (s) H 2 O Na + (aq) + 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.
Reactions in Solution: Acid - Base In a reaction such as H 3 O + - MeCO 2 H (aq) + MeCO 2 (aq) H 2 O 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 H 3 O + - MeCO 2 H (aq) + MeCO 2 (aq) H 2 O Recombination H 3 O + (aq) + MeCO 2 - (aq) H 2 O MeCO 2 H Reactions in Solution: Acid - Base We write the reaction for acetic acid, MeCO 2 H, as an equilibrium to include the ionization and recombination. Ionization MeCO 2 H H 2 O H 3 O + (aq) + MeCO 2 - (aq) Recombination H 3 O + (aq) + MeCO 2 - (aq) H 2 O MeCO 2 H MeCO 2 H H 2 O H 3 O + (aq) + MeCO 2 - (aq) As the amount of ionization and recombination are the same, the concentrations of the ions and the molecular form are constant Reactions in Solution: Acid - Base In solution, weak acids establish an equilibrium between the un-ionized or molecular form and the ionized form: MeCO 2 H un-ionized molecular form H 2 O H 3 O + (aq) + MeCO 2 - (aq) ionized
Reactions in Solution: Acid - Base In solution, strong acids are completely ionized and even though there is an equilibrium, it lies entirely on the RHS and recombination is negligible: HBr (g) un-ionized molecular form H 2 O H 3 O + (aq) + Br - (aq) ionized Reactions in Solution: Acid - Base Acids with more than one ionizable hydrogen are termed Polyprotic The common polyprotic acids are H 3 PO 4 H 2 SO 4 Phosphoric acid Sulfuric acid Reactions in Solution: Acid - Base Polyprotic acids can ionize more than once H 3 PO 4 H 2 SO 4(aq) H 2 O H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) H 2 O H 3 O + (aq) + SO 4 2- (aq) HPO 4 2- (aq) H 2 O H 3 O + (aq) + PO 4 2- (aq) Each proton is ionizable and the anions, dihydrogen phosphate (H 2 PO - 4 (aq) ) and hydrogen phosphate (HPO 2-4 (aq) ) both act as acids, though H 3 PO 4 is a weak acid.
Reactions in Solution: Acid - Base Polyprotic acids can ionize more than once H 3 PO 4 H 2 SO 4(aq) H 2 O H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) H 2 O H 3 O + (aq) + PO 4 2- (aq) HPO 4 2- (aq) H 2 O H 3 O + (aq) + PO 4 2- (aq) H 2 SO 4 H 2 SO 4(aq) H 2 O H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) H 2 O H 3 O + (aq) + PO 4 2- (aq) Reactions in Solution: Acid - Base In contrast, H 2 SO 4 is a strong acid and hydrogen sulfate (HSO 4 - (aq) ) is also a strong acid. H 2 SO 4(aq) H 2 O H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) H 2 O H 3 O + (aq) + PO 4 2- (aq) Reactions in Solution: Acid - Base Strong or weak? All acids can be assumed to be weak except the following: HCl (aq) HBr (aq) HI (aq) HClO 4(aq) HNO 3(aq) H 2 SO 4(aq) hydrochloric acid hydrobromic acid hydriodic acid perchloric acid nitric acid sulfuric acid
Reactions in Solution: Acid - Base Hydrogens attached to carbon are not ionizable in water Acetic acid, MeCO 2 H (or CH 3 CO 2 H) has the structure H O H H O H Reactions in Solution: Acid - Base Only the hydrogen attached to oxygen is ionized in aqueous solution H H H O O H H 2 O H H H O O + H O H H The methyl hydrogens are NOT ionizable in aqueous solution. Reactions in Solution: Acid - Base 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 Strong bases are therefore the hydroxides of the group I and II metals 3 4 5 6 Li 3 Na11 K 19 Rb 37 Cs 55 Mg1 2 Ca 20 Sr 38 Ba 56
Reactions in Solution: Acid - Base 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: H H 3 C N CH 3 CH 3 H 2 O H 3 C N CH 3 CH 3 + OH - Trimethylamine Trimethylammonium Reactions in Solution: Acid - Base Neutralization reactions and titrations Hydroxide and hydronium ions will react to form water. H 3 O + (aq) + OH - (aq) 2H 2 O (l) 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. Reactions 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.
Reactions in Solution: Acid - Base Neutralization 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. Reactions in Solution: Acid - Base Decompostion 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: Ca(OH) 2(s) + H 2 SO 4(aq) Ca 2 SO 4(s) + H 2 O (l) Reactions in Solution: Acid - Base Decompostion 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: Ca(OH) 2(s) + H 2 SO 4(aq) Is this balanced? Ca 2 SO 4(s) + H 2 O (l)
Reactions in Solution: Acid - Base Decompostion 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: Ca(OH) 2(s) + H 2 SO 4(aq) Is this balanced? No Ca 2 SO 4(s) + H 2 O (l) Reactions in Solution: Acid - Base Decompostion 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: Ca(OH) 2(s) + H 2 SO 4(aq) Is this balanced? No 2Ca(OH) 2(s) + H 2 SO 4(aq) Ca 2 SO 4(s) + H 2 O (l) Ca 2 SO 4(s) + 2H 2 O (l) Reactions in Solution: Acid - Base Decompostion in acid Some anions also decompose in acid. These are usually anions which are derived from gases which are not soluble in water: CO 2-3 (aq) carbonate CO 2(g) HCO 3 - (aq) hydrogen carbonate CO 2(g) S 2- (aq) sulfide H 2 S (g) HS - (aq) hydrogen sulfide H 2 S (g) SO 3 2- (aq) sulfite SO 2(g) HSO 3 - (aq) hydrogen sulfite SO 2(g)
Gases Properties of Gases Kinetic Molecular Theory of Gases Pressure Boyle s and Charles Law The Ideal Gas Law Gas reactions Partial pressures 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 H He 1 Room Temperature Gases 2 2 Li 3 3 Na 11 4 K 19 5 Rb 37 6 Cs 55 Be 4 Mg 12 Ca 20 Sr 38 Ba 56 Sc 21 Y 39 Lu 71 Ti 22 Zr 40 Hf 72 V 23 Nb 41 Ta 73 Cr 24 Mo 42 W 74 Mn 25 Tc 43 Re 75 Fe 26 Ru 44 Os 76 Co 27 Rh 45 Ir 77 Ni 28 Pd 46 Pt 78 Cu 29 Ag 47 Au 79 Zn 30 Cd 48 Hg 80 B 5 Al 13 Ga 31 In 49 Tl 81 C 6 Si 14 Ge 32 Sn 50 Pb 82 N 7 P 15 As 33 Sb 51 Bi 83 O 8 S 16 Se 34 Te 52 Po 84 F 9 Cl 17 Br 35 I 53 At 85 Ne 10 Ar 18 Kr 36 Xe 54 Rn 86 Gases Properties of Gases As the temperature rises, all elements form a gas at some point. In the following diagram, Blue represents solids Greenrepresents liquids Red represents gases At O K, all elements are solids At 6000 K, all are gases
Gases 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. 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 Solid Liquid Liquid Gas 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.
Gases Gases as an ensemble of particles The structures of liquids and solids are well ordered on a microscopic level CaCl 2 Ethanol, C 2 H 5 OH Gases Gases as an ensemble of particles 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. Gases Gases as an ensemble of particles 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...
Gases Gases as an ensemble of particles A gas particle can collide with the walls of the container Or another gas molecule When this happens, the gas particle changes direction. Gases Gases as an ensemble of particles 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. 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
Gases Gases as an ensemble of particles 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 Gases Pressure 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 Gases Pressure 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.
Gases Pressure 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... Gases Pressure Measurement Let the height difference between the two Hg levels be h Then the gas pressure is given by As P gas = P 0 + h P = Force = F = mg where g = 9.81 ms -2 Area A A Gases Pressure Measurement How is the height difference related to the pressure? As density, ρ = m V Then m = ρ V The volume of the column of mercury is V = A. h And so m = ρ V = ρ A. h
Gases Pressure Measurement The pressure above the baseline pressure P 0 is therefore P gas = mg =ρga. h = ρg h A A Gases Gases as an ensemble of particles 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. The Gas Laws The factors that control the behavior of a gas are The nature of the gas
The Gas Laws The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The Gas Laws The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The pressure - P The Gas Laws The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The pressure - P The temperature - T
The Gas Laws The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The pressure - P The temperature - T The volume of a gas -V 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 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.
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. The Gas Laws Boyle s Law Mathematically, PV = a constant as long as n and T are constant The Gas Laws Boyle s Law Mathematically, PV = a constant, k or P = k V as long as n and T are constant.
The Gas Laws Boyle s Law A graph of Boyle s data shows this relationship: PV = k 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 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.
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. The Gas Laws Charles Law Mathematically, we express this as V = k T And a graph of Charles Law is a straight line: 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 that V nt P
The Gas Laws The Combined Gas Law for a Perfect Gas V nt P Or V =K nt P Rearranging we find PV = a constant nt The Gas Laws The Combined Gas Law for a Perfect Gas The constant is termed the Universal Gas Constant, R, and takes the value R = 8.314 Jmol -1 K -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 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 P 1 V 1 = n = P 2 V 2 RT 1 RT 2 Or P 1 V 1 = P 2 V 2 T 1 T 2
The Gas Laws Using the Combined Gas Law The advantage of this expression is that the units do not matter; the units used for P 1,V 1, and T 1 will be returned in the calculation for P 2,V 2, and T 2. 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 m 3 ; the value for R is then 8.314 Jmol -1 K -1 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 o C through T/K = T/ o C + 273.16 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
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 0 o C and 1 atm pressure occupies 22.4 L 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. The Gas Laws Partial pressures In a mixture of gases, we can measure the total pressure of the mixture P Total 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
The Gas Laws Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: The Gas Laws Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: P Total = P 1 + P 2 + P 3 + P 4 +... The Gas Laws Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: P Total = P 1 + P 2 + P 3 + P 4 +... As PV = nrt then
The Gas Laws Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: P Total = P 1 + P 2 + P 3 + P 4 +... As PV = nrt then n Total RT = n 1 RT + n 2 RT + n 3 RT + n 4 RT +... The Gas Laws Partial pressures So P Total = P 1 + P 2 + P 3 + P 4 +... n Total RT = n 1 RT + n 2 RT + n 3 RT + n 4 RT +... The Gas Laws Partial pressures So P Total = P 1 + P 2 + P 3 + P 4 +... n Total RT = n 1 RT + n 2 RT + n 3 RT + n 4 RT +... n Total = n 1 + n 2 + n 3 + n 4 +...
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. The Gas Laws Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P 4 +... The Gas Laws Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P 4 +... n Total = n 1 + n 2 + n 3 + n 4 +...
The Gas Laws Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P 4 +... n Total = n 1 + n 2 + n 3 + n 4 +... P 1 = n 1 RT The Gas Laws Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P 4 +... n Total = n 1 + n 2 + n 3 + n 4 +... P 1 = n 1 RT So, P 1 = n 1 P Total n 1 + n 2 + n 3 + n 4 +... The Gas Laws Partial pressures P 1 = n 1 P Total n 1 + n 2 + n 3 + n 4 +... The fraction on the RHS is called the mole fraction and is written as x 1 so we can write Or P 1 = n 1 P Total n 1 + n 2 + n 3 + n 4 +... P 1 = x 1 P Total
Thermochemistry Energy Energy is defined as the ability to do work. There are several forms of energy Kinetic energy energy due to motion Thermochemistry Energy Energy is defined as the ability to do work. There are several forms of energy Kinetic energy energy due to motion E K = 1 / 2 mv 2 Potential energy the energy due to the position of a particle in a field e.g. Gravitational, electrical, magnetic etc. Thermochemistry Energy The unit of energy is the Joule (J) and 1 J = 1 kgm 2 s -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.
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 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 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.
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 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