Chemistry form 6 semester 1

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 1 962 /1 - CHEMISTRY SEMESTER 1 CHAPTER 1 - 6 : REVISION NOTES BY : MR CHEW CHIN KUEN CLASS : LOWER 6 TEACHERS LESSON 1 : 7 TH APR 2017 (FRIDAY) TIME : 6.30 - 9.30 PM NAME : CHAPTER 1 : MATTER 1. The elementary particles (also known as sub-articles) in a carbon-12 atom Diagram 1.a b) The table 1.dbelow shows the description of elementary particles and their properties. Elementary particles Symbol Charge Mass (kg) Relative Coulumb unit mass proton p +1.6022 x 10-19 +1 1.673 x10-24 1 neutron n 0 0 1.675 x 10-24 1 electron e- –1.6022 x 10-19 –1 9.11 x 10-28 1/1840 2. Deflection of sub-particles in electrical field and (electro)magnetic field. Deflection in electrical field Deflection in (electro)magnetic field a) The factors that affect the deflection of proton / electron are mass and charge ch arg e (e ) mass (m) distance of deflection Therefore, distance of deflection become further, when the mass is greater or the charge of the species is lower. 3 a) Isotopes : Atoms with the same atomic numbers (number of protons) but different mass numbers (nucleon numbers) Element Isotopes No of proton No of neutron % of abundance Hydrogen Protium, 1 1H 1 0 99.0 Deuterium, D : 2 1H 1 1 0.99 Tritium, T : 3 1H 1 2 0.01 Oxygen Oxygen-16 : 16 8O 8 8 98.9 Oxygen-17 : 17 8O 8 9 1.00 Oxygen-18 : 18 8O 8 10 0.01 4. Mass spectrometer : instrument used to analyse components in a chemical products. It can be used to determine relative isotopic mass relative abundance of isotopes relative atomic mass relative molecular / formula mass fragments of a large molecule a) Mass spectrum : a graph that shows the analyses of the species involved i. From the graph : ratio of 35Cl+ to 37Cl+ is 100 : 33.33 or 3 : 1. From the ratio, we can determine the relative atomic mass of chlorine : RAM of Cl = (35 x 100) + (37+33.33) / 100 + 33.33 ; RAM of Cl = 35.5 ii. Also from the mass spectrum, we can see that there are 3 peaks at m/e 70, 72 and 74. The abundance ratio of 3 peaks are measured at m/e ratio Species involved Probability Percentage of abundance 70 ( 35Cl – 35Cl)+ 16 9 4 3 4 3 P 100% 56.25% 16 9 72 ( 35Cl – 37Cl)+ and ( 37Cl – 35Cl)+ 16 6 2 4 3 4 1 P 100% 37.5% 16 6 74 ( 37Cl – 37Cl)+ 16 1 4 1 4 1 P 100% 6.25% 16 1

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 2 5. Mole concept and stoichiometry a) In a chemical reaction, the amount of reactants required to form certain amount of products can be expressed by a balanced equation. For example : 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l) From the equation, when 4 mol of ammonia reacted with 5 mol of oxygen, 4 mol of nitrogen monoxide will formed, together with 6 mol of waters. b) Using mole concepts and calculations, the moles between reactants were compared, in order to determine which is the limitant or excess. For example, if 0.4 mol of NH3 reacts with 0.6 mol of O2, Reactants NH3 (g) O2 (g) Stoichiometry Ratio 4 5 Reacting mole 0.4 0.6 Theory reactants 0.4 0.5 Role Limitant Excess (by 0.1) c) Say if the reaction above were carried out in laboratory and produced 0.3 mol of NO, therefore, 0.3 mol of NO produced is called as actual yield, the amount of product formed in laboratory. Theoretically, the amount of NO produced is : Reactants NH3 (g) (lim) NO (g) Stoichiometry Ratio 4 4 Reacting mole 0.4 0.3 Theory reactants / products 0.4 0.4 Therefore, the theoretical yield, amount of product that would result if all the limiting reagent reacted, produced should be 0.4 mol. Therefore, there is a difference between actual yield and theoretical yield. This can be then used to calculate the percentage of yield 100% 75% 0.4 0.3 100% @ % yield theoretica l yield actual yield % yield CHAPTER 2 : ELECTRONIC STRUCTURE OF ATOMS 1. Important equations for calculation of hydrogen specturm. a) Rydberg’s Equation 2 2 2 1 1 1 1 n n RH b) ∆E = h f ; ∆E = h c / h = Planck constant = 6.63 x 10-34 J s ; c = speed of light = 3.0 x 108 m s-1 . f = frequency ; f = c / . c) ∆H = ∆E x Na Na = Avogadro's constant = 6.02 x 1023 particles per mole 2. Emission line spectra is formed when electrons falls from higher energy level to ground state level, where the spectrum formed consist of discreet lines. Eventually, higher the energy level, the lines in the series converge. a) Different light source caused electrons fall at different ground energy level. Ultraviolet falls at ground level n = 1 ; visible light falls at n = 2 ; while infrared falls at n = 3. Table below shows the difference of the line spectra produced by ultraviolet and visible light. Ultraviolet rays Visible rays Produced Lyman series Produced Balmer series Usually use to calculate ionisation energy of hydrogen gaseous atom Can be used to determine wavelength produced by each spectra given by dispersion of light In emission spectrum, electrons from higher energy level settled at n = 1 In emission spectrum, electrons from higher energy level settled at n = 2 Series of convergence lines produced have higher frequency Series of convergence lines produced have lower frequency λ = wavelength (in m) n1 = ground state energy level n2 = energy level where electron fall from compared to ground state RH = Rydberg constant = 1.097 x 107 m -1

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 3 3. Orbitals are spaces where there is a high probability to find an electron. Energy level, n Sublevel, l orbital, ml Number of orbital Atomic Orbital Designation 1 0 0 1 1s 2 0 1 0 –1, 0 +1 1 3 2s 2px , 2py , 2pz , 3 0 1 2 0 –1, 0 +1 –2, –1, 0 , +1 , +2 1 3 5 3s 3px , 3py , 3pz , 3dxy , 3dyz , 3dxz , 3dx2-y2 , 3dz2 a) Shape of orbitals for s and p orbitals 4. Electronic configuration of an elements are written based on three rules and Principles a) Aufbau's principle ~ electrons are filled up in orbitals from the lowest energy orbital available ( 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < …..) b) Pauli Exclusion's Principle ~ an atomic orbital can hold a maximum of two electrons with opposing quantum spins c) Hund's Rule ~ when orbitals of equal energy are available, the most stable configuration has the maximum number of unpaired electrons with parallel spins. Examples : Particles Electronic configuration 7N 1s2 2s2 2p3 14Si 1s2 2s2 2p6 3s2 3p2 19K 1s2 2s2 2p6 3s2 3p6 4s1 22Ti 1s2 2s2 2p6 3s2 3p6 3d2 4s2 24Cr 1s2 2s2 2p6 3s2 3p6 3d5 4s1 29Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1 7N 3- 1s2 2s2 2p6 14Si2+ 1s2 2s2 2p6 3s2 24Cr3+ 1s2 2s2 2p6 3s2 3p6 3d3 29Cu2O 1s2 2s2 2p6 3s2 3p6 3d10 CHAPTER 3 ~ CHEMICAL BONDING Interaction between elements Metal-non metal Metal-metal Non-metal - non-metal Form ionic bond Form metallic bond Form covalent bond Electrostatic attraction forces formed between metal cation and non-metal anion. Formed when metal ions interact with delocalised electron sea Formed when electrons are shared when orbitals of 2 atoms are overlapped. Main factors affecting the intermolecular forces Lattice energy r r Q Q L.E. n n - LE is high when ions have high charge and small ionic radius. Band theory - greater the number of electrons delocalised , greater the conductivity. d-block elements will always have higher conductivity than sblock elements. Weak Van Der Waals' forces measure the interaction between 2 molecules. Greater molecular mass, greater weak VDW forces, higher boiling point. Properties - high melting points - hard yet brittle - insulator at solid state, when molten, it can conduct electricity - relatively high melting point - soft, ductile and malleable. - conductor in solid, or in molten state - Generally have low melting and boiling point - an insulator regardless of what state of matter. (except for C)

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 4 1. Hybridisation in covalent molecules sp3 hybridisation sp2 hybridisation sp hybridisation Hybridisation sp3 sp 2 sp Meaning 1 s and 3 p merge to overlap with orbitals of 4 surrounding atoms 1 s and 2 p merge to overlap with orbitals of 3 surrounding atoms 1 s and 1 p merge to overlap with orbitals of 2 surrounding atoms Example CH4 BCl3 BeF2 Ground state of element 2p 2s 2p 2s 2p 2s Excited state 2s 2p 2s 2p 2s 2p Hybridised state sp3 2pz sp2 2py 2pz sp Hybrid diagram Angle 109.5o 120o 180o Shape tetrahedral trigonal planar linear Formation of multiple bonds in ethene ethyne Hybridisation sp 2 sp Ground and excited state 2p 2s 2s 2p Hybridised state 2pz sp2 2py 2pz sp Orbitals diagram 2. Valence Shell Electron Pair Repulsion (VSEPR) Theories are based on 3 postulates a) Surrounding atoms repel themselves as far as possible to occupy their own space b) Double bond and triple bond are considered as one bond c) bond-pair electrons - bond-pair electrons repulsion < bond-pair electrons - lone-pair electrons repulsion < lone pair electrons - lone pair electrons repulsion Arrangement of e- pairs linear trigonal planar trigonal planar No of lone pair and bond pair electrons 2 bond pair electrons 0 lone pair electrons 3 bond pair electrons 0 lone pair electrons 2 bond pair electrons 1 lone pair electrons Lewis diagram Geometrical shape linear trigonal planar bent Angle 180o 1200 < 1200 Arrangement of e- pairs tetrahedral tetrahedral tetrahedral No of lone pair and bond pair electrons 4 bond pair electrons 0 lone pair electrons 3 bond pair electrons 1 lone pair electrons 2 bond pair electrons 2 lone pair electrons Lewis diagram Geometrical shape tetrahedral trigonal pyramidal bent Angle 109.5o 1070 104.50 Arrangement of e- pairs trigonal bipyramidal trigonal bipyramidal trigonal bipyramidal No of lone pair and bond pair electrons 5 bond pair electrons 0 lone pair electrons 4 bond pair electrons 1 lone pair electrons 3 bond pair electrons 2 lone pair electrons Lewis diagram Geometrical shape trigonal bipyramidal see saw T-shape Angle 120 o & 90o Less than 1200 Less than 1200

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 5 Arrangement of e- pairs trigonal bipyramidal octahedral octahedral No of lone pair and bond pair electrons 2 bond pair electrons 3 lone pair electrons 6 bond pair electrons 0 lone pair electrons 5 bond pair electrons 1 lone pair electrons Lewis diagram Geometrical shape linear octahedral Square pyramidal Angle 180o 90 0 < 90 0 Arrangement of e- pairs octahedral No of lone pair and bond pair electrons 4 bond pair electrons 2 lone pair electrons Lewis diagram Geometrical shape square planar Angle 900 3. There are generally 4 steps that can be used to draw a Lewis structure of a covalently bonded molecules. Step 1 : Calculate the total number of valence electrons from all atoms in a molecule. If it is a molecular ion, depend on the type of ions, the number of electrons are added / subtracted. If the molecular ion is positively charged, the total number of electrons are subtracted, however, if its negatively charged, the total number of electrons are added. Step 2 : Arrange all the atoms surrounding the central atom by using a pair of electron per bond (- 2 electrons per bond formed) Step 3 : Assign the remaining electrons to the terminal atoms so that each terminal atom has 8 electrons [except for hydrogen] (- 6 electrons per surround atom) Step 4(a) : Place any left-over electron on the central atom. (or) Step 4(b) : Form multiple bonds if there are not enough electrons to give the central atom an octet of electrons c) iodine tetrachloride ion, ICl4 - S2. Electrons bond = 4 x 2eElectrons left = 36 - 8 = 28 eS4a ) The 4e- remain is placed at the center atom P S1. Total valence electrons I = 7 e- ; 4 Cl = 4 x 7e- ; + 1e- accept ; Total electrons = 36 S3. e- at surroatom = 4 x 6eElectrons left = 28 - 24 = 4 e4 surround atoms 2 lone pair electrons Arrangement : octahedral Shape : square planar 4. Flow chart below shows how to determine whether a molecule is polar or non-polar 5. Fajan Rule ~ ionic or covalent character can be deduced by its charge and ionic radius Ionic compound Low positive charge Large cation Small anion Covalent compound High positive charge Small cation Large anion Aluminium compound - Since Al3+ has high charge and small ionic radius, therefore, it has high charge density, hence high polarisation power. This will results aluminium compound to have high covalent characteristic. - AlF3 is ionic compound due to the small ionic radius of F- , hence a low polarisability for the anion. Therefore Al3+ ion is not able to polarise the small Fhence an ionic compound. AlCl3 is a covalent compound as Clhas larger ionic radius, hence higher polarisability. This will allow Al3+ to polarise the electron cloud of Clhence forming covalent bond. This will explain why AlF3 has much higher melting point than AlCl3 - Al2O3 is an amphoteric oxide, an oxide that can act as acid as well as a base. This is due to Al2O3 is an ionic compound with high covalent characteristic, as Al3+ has high charge and small ionic radius, therefore, it has high charge density, hence a high polarisation power. This will explain why Al2O3 has high melting point, yet it is insoluble in water. Equation of Al2O3 as base : Al2O3 + 6 HCl → 2 AlCl3 + 3 H2O Equation of Al2O3 as acid : Al2O3 + 2 NaOH + 3 H2O → 2 Na[Al(OH)4] Al 3+ F - Cl-

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 6 6. Metallic bonding ~ valence electrons of the metal atoms are delocalised to form sea of electrons and attract all the metal cations. - Greater the valence electrons, greater the number of electrons delocalised, stronger the metallic bond. a) Conductivity of a substance depend on the distance between conduction band and valence band. b) Difference between conductor and semi conductor are compared below Conductor Semi conductor Made of metals Made of mostly metalloid Valence band and conduction band are overlapping Valence band and conduction band has a small gaps between the band As T increase, electrons vibrate faster, hence impede its movement, causing the conductivity activity decreased As T increased, more electrons are able to excite from valence band to conduction band, increasing the conductivity. Conductivity activity can also increased by adding a dopant such as germanium or phosphorous to semiconductor. 7. Intermolecular forces between covalent molecules. Covalent molecules Hydrogen bond Weak Van Der Waals' forces Molecular mass (or size) Polar molecule (higher dipole moment) Non-polar molecule (lower dipole moment) Straight chain (higher total surface area) Branched chain (lower total surface area) CHAPTER 4 : STATE OF MATTER 1. Gas Law ~ Gas Law Factors Formula Graph Boyle's Law V 1 P P1V1 = P2V2 Charles' Law V T 2 2 1 1 T V T V Avogadro's Law V n 2 2 1 1 n V n V Usually no graph is sketched Based on the 3 gas laws, the relationship between P, V and T can be described as 2 2 2 1 1 1 T P V T P V Avogadro's Law Boyle's Law Charles' Law V n P 1 V V T As conclusion : P nT V ; P nT V R ; where R is gas constant, 8.31 J mol-1 K -1 Ideal Gas Equation : PV = nRT Given density, d = m/ V ; therefore P dRT MR

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 7 2. Partial pressure : pressure exerted by each individual gas a) Dalton's Law partial pressure : In a gaseous mixture that is at equilibrium, total pressure is the summation of the partial pressure exerted by each individual gas Ptot = PA + PB + …. b) For the partial pressure of each gases, it can be calculated from the formula A A B A x n n n 3. In the Maxwell distribution of speed (or energy), two factors are used to explain. - temperature - molecular mass Area at 3 temperature are the same, indicating number of particles are the same at 3 diff temp, where each distribution graph shows its peak at middle, indicate - most molecules travel at average speed (energy) but little molecule travel at high speed (energy) As T increased, peak become lowered while base become broader, indicate lesser particles travel at average peak, but more molecule travel at higher speed. Peak shift to the right meaning at higher T, molecules has higher average speed, and molecules with higher speed also increased gradually. 4. Deviation of gases from ideal behaviour a) All gases obey ideal gas law at high temperature and low pressure b) Gases are ideal due to these characteristic - no intermolecular forces (attraction or repulsion) between particles - gases has negligible volumes c) Graph below shows the relationship between the factors of PV/RT against P for ideal gases and all other gases. - When P initially increased, attraction forces take place between particles. As a result, attracted particles collide less frequently with wall of container and caused the pressure to be lesser than expected. Therefore, PV/RT < 1, hence a negative deviation. - Greater the molecular mass, greater the weak Van Der Waals forces, stronger the attraction force - As pressure gradually increased, repulsion forces started to take place between particles. As a result, particles will repulse each other greatly, and caused the collision of particles to collide more rapidly, causing PV / RT > 1, hence a positive deviation. 5. Vapour Pressure ~ pressure exerted when vapour particles collide with the wall of a closed container. Saturated vapour pressure is achieved when the rate of vaporation is the same with the rate of condensation. a) When a liquid has high vapour pressure, it is easier to vapourise. For such liquid, it is called as a volatile liquid, a liquid which vaporised easily at room temperature. In another words, for a liquid with high vapour pressure, it will have a low boiling point. b) Boiling is achieved when the pressure of the system is the same with external pressure. Therefore, if the external pressure is lowered, the pressure of system will also be lowered, hence a lower boiling point. 6. Primitive unit cell Rhombohedral Tetragonal Triclinic Monoclinic Orthorhombic Hexagonal No of particles / unit cell = 8 x corner 8 1 8 = 1 particle / unit cell No of particles / unit cell = (8 x corner) + 1 body center 1(1) 8 1 8 = 2 particles / unit cell No of particles / unit cell = (8 x corner) + 6 face center 2 1 6 8 1 8 = 4 particles / unit cell PA = xA . Ptot

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 8 7. Allotropy is elements that can exist in more than one structure under the same state of matter. For examples : a) oxygen can exist as oxygen gas (O2) gas and ozone (O3) gas b) carbon (C) can exist as three allotropes : diamond, graphite and fullerene Diamond (sp3 hybridised) Graphite (sp3 hybridised) - no delocalised electron (insulator) - contain one unhybridised orbital with an electron delocalised (conductor) - use as diamond cutter, jewelery - use as electrode, lubricant 8. Phase diagram - a graph of pressure against temperature which explain how solid, liquid and gas are formed between each other. a) Phase diagram of water, H2O b) phase diagram of carbon dioxide, CO2 b) Note in the phase diagram of water, the melting curve slopes to the left with increasing pressure showing that as the pressure is increased, the melting point of ice decreases slightly. This is unusual because an increase in pressure usually favours the formation of solid. However, this behaviour is due to ice having an open structure. As the pressure exerted is increased, the hydrogen bonds between the H2O molecules in ice are broken down, changing the ice into a denser liquid phase which occupies a smaller volume. Same things applied when temperature increased. Hydrogen bond between ice broken, and the open structure of ice broken. As a result, distance between water molecules become closer, hence volume decreased, which increase the density of water. c) For carbon dioxide, its solid is called as dry ice, which is used as refrigerant. Fumy effect is due to as dry ice sublimed when surrounding heat is absorbed, surrounding temperature decreased. As a result, water vapour from the surrounding condensed hence form the fume. CHAPTER 5 : KINETIC CHEMISTRY 1. Kinetic chemistry ~ study of changes of concentration in a chemical reaction at a given time. Says a chemical reaction take place below x A (aq) → y B (aq) a) rate of reaction of A and B can be expressed below dt d[A] rate dt d[B] rate 2. 3 postulates for particles to react from the angle of kinetic theory : a) Particles must collide in order to react - This is generally influenced by 2 factors, concentration and temperature. Higher the concentration, greater the number of particles collide, higher the rate of reaction. - Higher the temperature, greater the kinetic energy of particles, faster the particles move and collide with greater energy, hence higher the rate of reaction. b) Particles collide at the right orientation :- when 2 particles of reactant collide, it must be collide at the right position, in so that the reaction can take place. Example : c) Particles collides with minimum amount of energy called as activation energy. Activation energy is the minimum amount of energy required to initiate a chemical reaction. 3. During a chemical reaction take place, a substance will be formed during chemical reaction. For example, in the hydrolysis of haloalkane : a) From the reaction, is called as intermediate, a substance formed during chemical reaction, but will not formed as a product 4. - Rate law is a way to expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers - Rate constant, k, is the proportionality constant of the reaction, in which the value remain constant under constant temperature - order of reaction, x and y, is the power to which the concentration of the reactants (in this case, [A] and [B] respectively) is raised to in the rate equation

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 9 Zero order First order Second order Rate equation rate = k [A]0 = k rate = k [A]1 rate = k [A]2 or rate = k [A] [B] Unit of k mol dm-3 s -1 s -1 mol-1 dm3 s -1 Graph of [reactant] vs time [reactant] k = - gradient time [reactant] [A]0 [A]0 /2 [A]0 /4 t½ t½ time [reactant] [A]0 [A]0 /2 [A]0 /4 t½ 2t½ time Graph of rate against [reactant] rate [reactant] rate gradient = k [reactant] rate [reactant] Equation [A]0 – [A]t = k t ln [A]0 – ln [A]t = kt kt [A] 1 [A] 1 t 0 Half-life equation t½ = [A]0 / 2k t½ = 0.693 / k t½ = 1 / k [A]0 Linear plot [A]t against time [A]t k = gradient time ln [A]t against time ln [A]t k = gradient time 1 / [A]t against time 1 / [A]t k = gradient time Example of calculation to determine order of reaction O2 (g) + 2 NO (g) → 2 NO2 (g) The rate equation, expressed in general form, is rate = k [O2] x [NO]y Note that the order of reaction cannot be determined directly from the stoichiometry of the reaction. To find out the orders of reactant with respect to each O2 and NO, we run series of experiments, starting each one with a different set of reactant concentrations and obtaining an initial rate in each case. Exp Initial concentration of reactant Initial rate (mol dm-3 s -1 O ) 2 / mol dm-3 NO / mol dm-3 1 1.10 x 10-2 2.50 x 10-2 2.40 x 10-3 2 2.20 x 10-2 2.50 x 10-2 4.80 x 10-3 3 1.10 x 10-2 5.00 x 10-2 9.60 x 10-3 From each experiment, the rate equations are expressed individually, where Experiment 1 : 2.40 x 10-3 = k (1.10 x 10-2 ) x (2.50 x 10-2 ) y Experiment 2 : 4.80 x 10-3 = k (2.20 x 10-2 ) x (2.50 x 10-2 ) y Experiment 3 : 9.60 x 10-3 = k (1.10 x 10-2 ) x (5.00 x 10-2 ) y Comparing Exp 2 to Exp 1 : 4.80 x 10-3 = k (2.20 x 10-2 ) x (2.50 x 10-2 ) y 2.40 x 10-3 = k (1.10 x 10-2 ) x (2.50 x 10-2 ) y 2 = (2)x Order of reaction with respect to O2 ; x = 1 Comparing Expe 3 to Exp 1 : 9.60 x 10-3 = k (1.10 x 10-2 ) x (5.00 x 10-2 ) y 2.40 x 10-3 = k (1.10 x 10-2 ) x (2.50 x 10-2 ) y 4 = (2)x Order of reaction with respect to NO ; x = 2 From the order of reaction deduced, the rate equation is rate = k [O2][NO]2 . The overall order of reaction = 1 + 2 = 3 Using any experiment, rate constant can be calculated. For example, in experiment 1 2.40 x 10-3 = k (1.10 x 10-2 ) (2.50 x 10-2 ) 2 k = 349 mol-2 dm6 s -1 . 5 Chemical reactions may occur in one way reaction or a reversible reaction. In a one way reaction, process may be taken in multiple steps, and therefore irreversible back to the reactants, as they might involve in steps that required higher activation energies. Example 2 NO (g) + O2 (g) 2 NO2 (g) Mechanism Reaction Equation Rate equation Step 1 2 2 slow 2 NO N O rate = k [NO]2 Step 2 2 fast 2 2 2 2 NO N O O rate = k [N2O2] [O2] In a mechanism, slow step is rate determining step. Therefore, the rate equation for the overall reaction can be written as rate = k [NO]2 .

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 10 6. Generally, the rate of reaction increased with temperature as it affect the rate by increased the rate constant of the reaction. The dependence of the rate constant of a reaction on temperature can be expressed by using Arrhenius equation RT EA k A e k = rate constant A = Arrhenius constant T = temperature EA = activation energy R = gas constant (8.31 J mol-1 K -1 ) a) When Arrhenius equation is derived, the equation can be re-expressed as : ln A T 1 R E ; rearrange the equation ln k RT E ln k ln A A A b) When a graph of ln k against 1/T (in Kelvin) is plotted, the gradient of the graph can be used to calculate the activation energy of the reaction. c) Activation energy can also be calculated when the rate constant at two different temperature. 2 1 A 2 1 2 A 1 A 1 2 T 1 T 1 R E k k ln A ln RT E ln A RT E ln k ln k 7. Catalyst is a substance added to a chemical reaction to alter the rate of reaction. - catalyst is added to lower the activation energy by providing an alternative route for the reaction to take place. However, amount of product will remain the same. - the route provided by catalyst is specific. However, even without the addition of catalyst, the reaction will still take place, its just that catalyst will make reaction occur faster. a) Energy profile of endothermic and exothermic reaction with and without the addition of catalyst is described in the diagram below b) Autocatalyst is formed when product itself act as catalyst. An example of reaction for autocatalyst is the reaction between sodium ethanedioate and potassium manganate (VII) 2 MnO4 - (aq) + 5 C2O4 2- (aq) + 16 H+ (aq) → 2 Mn2+ (aq) + 10 CO2 (g) + 8 H2O (l) CHAPTER 6A : CHEMICAL EQUILIBRIUM 1. Dynamic equilibrium ~ In a reversible reaction, a system is to say reaches dynamic equilibrium when the concentration of reactants and products remain constant, while the reaction is ongoing at the rate of forward reaction is the same as the rate of backward reaction. a) For example : w A (aq) + x B (aq) ↔ y C (aq) + z D (aq) equilibrium constant of concentration, w x y z C [A] [B] [C] [D] K b) If a gaseous system is involved, not only it can be expressed as Kc, but it can also be expressed as equilibrium constant of partial pressure, Kp N2 (g) + 3 H2 (g) 2 NH3 (g) Equilibrium constant of concentration, KC Equilibrium constant of partial pressure, KP 3 2 2 2 3 C [N ][H ] [NH ] K 3 N2 H2 2 NH3 P (P ) (P ) (P ) K 2. Method of calculating Kc and Kp for examples of reaction are stated below a) Dissociation of phosphorous (V) pentachloride PCl5 (g) ↔ PCl3 (g) + Cl2 (g) Initial mol 1 mol 0 0 degree of dissociation, - + + At equilibrium 1 - If calculating Kc of reaction ( ) ( ) ( ) If calculating KP of reaction 1. First calculate their mol fraction total mol = 1 - + + = 1 + XPCl5 = ( ) XPCl3 = ( ) XCl2 = ( ) 2. The find its partial pressure (total pressure = Ptot) PPCl5 = ( )Ptot PPCl3 = ( ) Ptot PCl2 = ( ) Ptot 3. The substitute all partial pressure into Kp KP = b) Another example is the dissociation of dinitrogen tetraoxide, N2O4 (g) to form nitrogen dioxide gas, NO2 (g) N2O4 (g) ↔ 2 NO2 (g) Initial mol 1 mol 0 degree of dissociation, - + 2 At equilibrium 1 - 2

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 11 If calculating Kc of reaction ( ) ( ) If calculating KP of reaction 1. First calculate their mol fraction total mol = 1 - + 2 = 1 + XN2O4 = ( ) XNO2 = ( ) 2. The find its partial pressure (total pressure = Ptot) PN2O4 = ( )Ptot PNO2 = ( ) Ptot 3. The substitute all partial pressure into Kp KP = 3. Le Châtelier’s principle states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. a) Concentration :- changes of concentration will shift position to counter the changes toward the concentration of the substance involved. However, Kc and Kp will remain unchanged. System is disturbed - added at right side. Equilibrium will shift to the left to balance back the equilibrium. b) Pressure : If the pressure of the system is increased, equilibrium will adjust to decrease the pressure by shifting to the position with less total mol of gas. - If the pressure of the system is decreased, equilibrium will adjust to increase the pressure by shifting to the position with more total mol of gas. i. When an inert gas is added to an equilibrium system under constant volume, equilibrium will not be disturbed, hence position of equilibrium will not shift. ii. When an inert gas is added to an equilibrium system under constant pressure, the total pressure will not change , and adding inert gas will decrease the partial pressure of the gaseous mixture, hence position of equilibrium will shift its position to increase the pressure, therefore position with more total mol of gas. However, regardless how the pressure changes, KC and KP will remain constant. c) Temperature :- when temperature increased, system will respond to decrease the temperature - by shifting to the direction of decreasing temperature, hence an endothermic path. Conversely, when temperature decreased, system will respond to increase the temperature - by shifting to the direction of increasing temperature, hence an exothermic reaction 4. Temperature is the only factor that can change the equilibrium constant of a chemical equilibrium. This can be deduced using Van't Hoff equation, which is expressed below C RT H ln K a) Based on the expression above, if a graph of ln K against 1/T is plotted, a liner graph will be obtained. Using this graph, the enthalpy change of a chemical reaction can be determined. Table below summarised the graph of endothermic and exothermic process Exothermic process Endothermic process K Temperature / K K Temperature / K Given A + B ↔ C + D ∆H = - ve Increasing temperature decreased K Given A + B ↔ C + D ∆H = + ve Increasing temperature increased K ln K + gradient = - ∆H / R ∆H = negative 1/T / K-1 ln K - gradient = - ∆H / R ∆H = positive 1/T / K-1 b) At two different temperature, ∆H of the reaction can also be calculated using the formula 1 1 2 2 T 1 T 1 R H K K ln ∆H = enthalpy change of reaction R = gas constant (8.31 J mol-1 K -1 ) T = temperature in Kelvin (x 0C + 273)

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 12 CHAPTER 6B : ACID-BASE AND SOLUBILITY EQUILIBRIA 1. Bronsted-Lowry acid : substance which donate proton to base Bronsted-Lowry base : substance which accept proton from acid. Lewis acid : substance which accept lone pair electron from a base Lewis base : substance which donate lone pair electron to an acid. From the angle of Bronsted-Lowry acid-base theory : CH3COOH (aq) + H2O (l) ↔ CH3COO- (aq) + H3O + (aq) B.L acid B.L base conjugate base conjugate acid From the angle of Lewis acid-base theory : 2. For a dissociation of a weak acid/base, the strength of an acid /base can be measured using dissociation constant of acid, Ka or base, Kb. Equation HA (aq) + H2O (l) H3O + (aq) + A- (aq) Initially concentration c (unknown) 1 (constant) 0 0 Degree = - c + c + c At equilibrium c ( 1 - ) c c [HA] [H O ] [A ] K + 3 a Substitute the concentration of [H3O + ], [A- ] and [HA] into the equation c(1 ) (c )(c ) Ka Since is very small (less than 0.05 is consider small), therefore we assume that 1 ≈ 1. c (c )(c ) Ka Since [H3O + ] = c Therefore, [H3O + ] = K c a Ka = c 2 pKa = lg Ka and pH = lg [H3O + ] Equation M (aq) + H2O (l) MH+ (aq) + OH- (aq) At equilibrium c ( 1 - ) c c [MH] [MH ] [OH ] K + 2 b Substitute the concentration of [MH2 +], [OH-] and [MH] into the equation c(1 ) (c )(c ) Kb Since is very small (less than 0.05 is consider small), therefore we assume that 1 ≈ 1. c (c )(c ) Kb Since [OH ] = c c (c )(c ) Kb Kb = c 2 pKb = lg Kb Kb = c 2 3. Ionic product of water is derived from the equation H2O (l) ↔ H + (aq) + OH- (aq). From the reaction, ionic product of water, Kw = [H+ ] [OH- ] a) At 25oC, Kw = 1.0 x 10-14 mol2 dm-6 . Using this figure, Kw is expressed as pKw. Therefore, the equation Kw = [H+ ] [OH- ] is applied with negative logarithm. -lg 1.0 x 10-14 mol2 dm-6 = -lg [H+ ] + (-lg [OH- ] ) @ 14 = pH + pOH b) Under certain derivation, Kw = Ka x Kb 4. Indicator is a weak organic acid added to a solution to show acidity of a solution. A few example of indicator are shown below Indicator Acid Basic End point colour Methyl orange Reddish orange Yellow Orange Bromothymol blue Yellow Blue Green Phenolphthalein Colourless Pink purplish Light pink In the titration curve section shall discuss further of how its applied 5. Buffer solution ~ solution where pH does not change much when a little acid or base is added to the solution a) Acidic buffer [Eg : CH3COOH (weak acid) and CH3COONa (salt of conjugate base) ] Basic buffer [Eg : CH3NH2 (weak base) and CH3NH3Cl (salt of conjugate acid) ] b) Role of buffer solution :- In aqueous solution, both ethanoic acid and sodium ethanoate dissociate accordingly to the following equations : CH3COOH (aq) + H2O (l) ↔ CH3COO- (aq) + H3O + (aq) CH3COONa (aq) → CH 3COO- (aq) + Na+ (aq) When a little acid is added to the solution, H+ added will react with CH3COOto form CH3COOH, therefore does not affect the concentrated of H+ in the buffer system. When a little base is added to the buffer solution, OH- will react with H+ in system, where H + + OH- H2O. Therefore decreased the concentration of H+ . Based on Le Chatelier Principle, equilibrium in (1) will shift to the right, and form H+ to compensate the H+ lost in the system. c) The equation used for determining pH of buffer solution with the appropriate ratio of weak acid and its conjugate base salt used is : [conjugate base](salt) [weak acid] pH pKa lg [conjugate acid] (salt) [weak base] pOH pKb lg 5. Dissociation of polyprotic acid take place in stages of neutralisation. For example in the neutralisation of sulphite acid, H2SO3, with NaOH, it undergoes 2 stages of reaction Stage 1 : NaOH (aq) + H2SO3 (aq) → NaHSO3 (aq) + H2O (l) Stage 2 : NaOH (aq) + NaHSO3 (aq) → Na2SO3 (aq) + H2O (l) Each stage of neutralisation at equal volume. Another example is the reaction of HCl with sodium carbonate, Na2CO3. Stage 1 : HCl (aq) + Na2CO3 (aq) → NaHCO3 (aq) + NaCl (aq) Stage 2 : HCl (aq) + NaHCO3 (aq) → NaCl(aq) + H2O (l) + CO2 (g)

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 13

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 14

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 15 6. - Solubility is the quantity of solute that dissolved in unit volume of saturated solution at a given temperature - Solubility product, Ksp, of a sparingly soluble electrolyte is the product of the concentrations of the ions in a saturated solution at a given temperature a) The common ion effect is the suppression of the ionization of a weak acid or weak base by the presence of a common ion from a strong electrolyte. For example : Mg(OH)2 (aq) ↔ Mg2+ (aq) + 2 OH- (aq) When strong base, NaOH, is added to Mg(OH)2 aqueous solution, concentration of OHis high, therefore equilibrium shift to left, therefore decrease solubility. b) At a particular temperature, Ksp value indicates the maximum product of ion concentrations in solution at equilibrium. MX (s) M + (aq) + X- (aq) Ionic product, Q = [M+ ] [X- ] a) If ionic product, Q = Ksp → The system is at equilibrium and the solution is saturated, formation of precipitate is observed. b) If ionic product, Q > Ksp → The system is not at equilibrium and the solution is supersaturated. The concentrations of the ions in solution are too high and thus the salt will precipitate. c) If ionic product [M+ ] [X- ] < Ksp → The system is not at equilibrium and the solution is not saturated (ie. more solid can dissolve until equilibrium is achieved), so no precipitate formed Example 1: Will a precipitate form when 50 cm3 of 0.050 mol dm-3 AgNO3 is added to 50 cm3 of 0.10 mol dm-3 KBrO3? [Ksp AgBrO3 = 6.0 x 10-5 mol2 dm-6 ] For [Ag+ ] in mixture, M1V1 = M2V2 (total) (50 50) (0.050)(50) M2 M2 = 0.025 mol dm-3 For [BrO3 - ] in mixture, M1V1 = M2V2 (total) (50 50) (0.10)(50) M2 M2 = 0.050 mol dm-3 Since AgBrO3 (s) ↔ Ag+ (aq) + BrO3 - (aq) ; Therefore, Q = [Ag+ ][BrO3 - ] Q = (0.025)(0.050) = 1.25 x 10-3 mol2 dm-6 Since Q > Ksp ; therefore mixing both solution will form precipitate of AgBrO3. CHAPTER 6C : PHASE EQUILIBRIA 1. Ideal solution ~ mixture of two solutions that obey Raoult's Law. Two characteristics of ideal solutions are : a) Intermolecular forces between two solutions and mixing solutions are the same b) No heat absorb or liberated when mixed. Therefore, total volume = summation of volume of two solution mixture. In vapour (Dalton's Law) PA = xA PT PT = PA + PB. In liquid (Raoult's Law) PA = xA PA 0 2. Diagram of vapour pressure composition graph and boiling point composition graph. Vapour Pressure / kPa liquid gas 0 mol fraction composition 1 Boiling point / oC gas liquid 0 mol fraction composition 1 3. Fractional distillation ~ technique use to separate 2 miscible liquids. Boiling point / oC gas T1 T2 T3 liquid ' a b c 0 mol fraction composition A 1 When a mixture at composition a is brought to fractional distillation, it will first boil at temperature T1, and vapour mixture is at composition b, which then condensed at temperature T2. Fractional distillation continue at the second plate, where liquid mixture boiled at T2 and gives vapour composition at c, then condensed at temperature T3. Note that concentration of A keep increasing as mixture distillate where vaporation - condensation keep on occurring until pure A is distillate, leaving pure B as residue.

PRE–U STPM CHEMISTRY I SEMESTER 1 I Chapter 1 - 6 - Short Note Revision Kursus Kecemerlangan STPM Chemistry I Chew Chin Kuen | 16 Deviation from Ideal solution behaviour. Negative deviation Positive deviation Example HCl – H2O ; HNO3 – H2O ; H2SO4 – HCOOH ; phenol – aniline ; acetone (CH3COCH3) – CHCl3 Propanol – H2O ; ethanol – H2O ; ethanol – toluene ; ethanol and benzene Happen when Intermolecular forces between A---B molecules are stronger than in between pure A---A and pure B---B Intermolecular forces between A---B molecules are weaker than in between pure A---A and pure B---B Partial pressure of mixture Since the intermolecular forces are stronger in A---B ; the partial vapour pressure are lower than expected in Raoult’s Law Since the intermolecular forces are weaker in A---B ; the partial vapour pressure are higher than expected in Raoult’s Law Heat changes Since the intermolecular forces are stronger, A --- B are more stable than their individual forces, thus the process is exothermic Since the intermolecular forces are weaker, A --- B are less stable than their individual force thus the process is endothermic Reasons HCl and H2O The forces hold between HCl – HCl are pure Van Der Waals forces whereas the forces between H2O – H2O are pure hydrogen bonding When HCl and H2O are mixed together, it will ionised to form HCl + H2O H3O + + ClThe ionic interactions between hydrated ions are stronger than both Van Der Waals forces and hydrogen bond. Ethanol, C2H5OH and H2O The intermolecular forces between ethanol – ethanol are hydrogen bonding and same goes to between water – water. However, when this 2 solutions are mixed, the hydrogen between water are broken by ethanol and reduce the intermolecular forces formed between ethanol – water. Volume of mixture Since mixing the solution causes a stronger attraction forces formed between A---B ; so when 50 cm3 A + 50 cm3 B < 100 cm3 . Since mixing the solution causes a weaker attraction forces formed between A---B ; so when 50 cm3 A + 50 cm3 B > 100 cm3 . Vapour pressure composition curves kPa 0 1 mol fraction of A kPa 0 1 mol fraction of A Boiling pointcomposition curve boiling point / oC 0 1 Composition by % mass A boiling point / oC 0 1 Composition by % mass A Fractional distillation of negative deviation 1 st Distillate : Pure B 2 nd distillate : Azeotropic mixture Residue : Pure A + impurities boiling point / oC 0 1 Composition by % mass A 1 st Distillate : Pure A 2 nd distillate : Azeotropic mixture Residue : Pure B + impurities Fractional distillation of positive deviation 1 st Distillate : Azeotropic mixture 2 nd distillate : Pure B Residue : Pure A + impurities boiling point / oC 0 1 Composition by % mass A 1 st Distillate : Azeotropic mixture 2 nd distillate : Pure A Residue : Pure B + impurities Azeotropic mixture is a mixture of two solution that has a constant boiling point, which the liquid state is at equilibrium with gaseous state