General Chemistry 2 - Intermolecular Forces (PDF)

THIRD QUARTER — GENERAL CHEMISTRY 2 : INTERMOLECULAR FORCES LESSON 01 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ I. KINETIC MOLECULAR MODEL OF LIQUIDS AND SOLIDS Intermolecular forces — are attractive forces that act...

THIRD QUARTER — GENERAL CHEMISTRY 2 : INTERMOLECULAR FORCES LESSON 01 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ I. KINETIC MOLECULAR MODEL OF LIQUIDS AND SOLIDS Intermolecular forces — are attractive forces that act between molecules or particles in the solid or liquid offers a description of the microscopic properties of states; are much weaker than bonding forces. atoms (or molecules) and their interactions, leading to when a solid melts, or a liquid boils, the particles move observable macroscopic properties (such as pressure, away from each other volume, temperature). ○ intermolecular forces of attraction are broken application: helps to explain why matter exists in the stronger the intermolecular forces to be broken, the different phases (solid, liquid, and gas) and how matter larger the amount of energy needed to break them, can change from one phase to the next. ○ hence, the higher the melting point for solid to liquid transformation, and boiling point for A. KINETIC MOLECULAR THEORY liquid to gas transformation. states that “matter is made up of particles that are Intermolecular force — bond that connects molecules constantly moving” ○ ex: NaCl all particles have energy, but energy varies depending Intramolecular force — bond that connects atoms; on the temperature the sample of matter is in. greater than intermolecular determines whether the substance exists in the solid, ○ ex: H2O liquid, or gaseous state. molecules in solid phase — least amount of energy III. PHASE CHANGES gas particles — greatest amount of energy temperature of a substance = a measure of the 1. freezing - liquid to solid average kinetic energy of the particles. 2. melting - solid to liquid energy of the particles is changed = change in phase 3. evaporation - liquid to gas there are spaces between particles of matter. 4. condensation - gas to liquid ○ the average amount of empty space between 5. sublimation - solid to gas molecules gets larger as a sample of matter 6. deposition - gas to solid moves from solid to liquid and gas phases. ──── TYPES OF INTERMOLECULAR FORCES ──── II. INTERMOLECULAR FORCES I. London Dispersion : forces of attraction result from electrostatic in nature; that is, they arise from the temporary dipole moments induced in ordinarily interaction between positively and negatively charged nonpolar molecules; weakest bond. species. → present between all types of molecules due to the are attractive forces between atoms/molecules, and movement of electrons. these become stronger as the particles move closer → as electrons move around the nucleus, an uneven together. distribution causes momentary charge separations. gases, liquids and solids are all made up of atoms, → slightly positive sides of molecule are attracted to molecules, and / or ions, but the behaviour of these slightly negative sides of adjacent molecule. particles differ in the three phases. polarisability — extent to which a dipole moment can → particles are well separated with no regular be induced in a molecule; atom or molecule refers to Solid arrangement the ease with which the electron distribution can be → vibrate and move freely at high speeds. distorted. → particles are close together with no regular Liquid arrangement. HOW TO IDENTIFY & EXAMPLES: → vibrate, move about, and slide past each other covalent compounds (covalent bonds form between → particles are tightly packed, in a regular pattern nonmetals, ionic bonds form between metals and Gas → vibrate (jiggle) but generally do not move from nonmetals) place to place non-polar (no leftover) halogens and noble gases → Liquids and solids are often referred to as condensed phases ○ F2, Cl2, Br2, I2, He, Ne, Ar, Kr because the particles are very close together. ○ CH4, CO2, C4H10 rai — 01  II. Ion — Dipole: stronger than london dispersion → the intermolecular force of attraction between a charge ion (cation or anion) and a molecule. → involved in solutions where an ionic compound is dissolved into a polar solvent → often the result of ionic compounds being dissolved into polar substances HOW TO IDENTIFY & EXAMPLES: ionic compounds (ionic bonds form between metals and nonmetals) examples: ○ NaCl ○ between metal cation like Mg+2 or Zn+2 III. Dipole — Dipole: attractive forces between polar molecules (molecules that possess dipole moments) → partial negative side of one molecule is attracted to the partial positive side of another molecule. → nature of attraction is electrostatic and can be understood in terms of Coulomb’s law → larger dipole moment = stronger attraction HOW TO IDENTIFY & EXAMPLES: covalent bond occurs when there is an overall uneven distribution of electrons across a molecule. examples: ○ HCl, HF, H2O, HI IV. Hydrogen Bonds : a special type of dipole-dipole interaction between the hydrogen atom in a polar bond, such as ○ N‒H, O‒H ○ F‒H ○ electronegative O, N, or F atom. → hydrogen bonds between water molecules are particularly strong. → those which occur within one single molecule HOW TO IDENTIFY & EXAMPLES: exists in a water molecule examples: ○ Water (H₂O) ○ Ethanol (C₂H₅OH) ○ Ammonia (NH₃) ○ Hydrogen fluoride (HF) — lesson 01 — rai — 02  THIRD QUARTER — GENERAL CHEMISTRY 2 : PROPERTIES OF LIQUIDS & SOLIDS LESSON 02 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ ABSTRACTION AND GENERALIZATION IV. VAPOR PRESSURE unique properties of liquids that depend on the nature molecules in the gaseous state over its liquid create a of intermolecular interactions. pressure. greater the number of gaseous particles = greater the I. SURFACE TENSION pressure exerted by the gas. pressure exerted by the gas in equilibrium with a liquid measure of the elastic force in the surface of a liquid in a closed container at a given temperature. is the amount of energy required to stretch or increase equilibrium vapor pressure — the maximum vapor the surface of a liquid by a unit area. pressure of a liquid at a given temperature and that it is manifested as some sort of skin on the surface of a constant at a constant temperature. liquid or in a drop of liquid. ○ increases with temperature ○ floating paper clip on water independent of the amount of liquid as well as the ○ drops of liquid on a coin surface area of the liquid in contact with the gas. ○ water beads on apple examples: ○ dew drops ○ boiling tea in a kettle ○ soap bubbles in the air ○ humid air & smell of brandy or whisky molecules within a liquid — pulled in all directions by ○ drying clothes intermolecular forces. ○ water present in lakes molecules at the surface — pulled downward and sideways by other molecules, not upward away from V. MOLAR HEAT OF VAPORIZATION the surface. (ΔHvap) is the energy required to vaporize 1 mole of a II. CAPILLARY ACTION liquid at a given temperature examples: the tendency of a liquid to rise in narrow tubes or be ○ opening a sealed bottle of water and leaving it drawn into small openings such as those between out for days & heat you feel when in contact grains of a rock. with steam from cooking also known as capillarity, is a result of intermolecular attraction between the liquid and solid materials. VI. BOILING POINT examples: ○ water rising spontaneously in capillary tubes temperature at which the liquid converts into a gas ○ placing a straw into a glass of water liquid boils when its vapor pressure equals the pressure ○ dipping paper towels into water acting on the surface of the liquid. ○ transferring water from one jar to another boiling points of substances often reflect the strength of using yarn. the intermolecular forces operating among the types of forces involved in capillary action: molecules ○ Cohesion is the intermolecular attraction boiling point of a liquid depends on the external between like molecules (the liquid molecules). pressure. ○ Adhesion is an attraction between unlike molecules (such as those in water and in the ──── UNIQUE PROPERTIES OF WATER ──── particles that make up the glass tube) cohesive forces greater than the adhesive forces I. Water is a good solvent : its ability to dissolve a large = surface of the liquid is convex variety of chemical substances. II. Water has a high specific heat III. VISCOSITY → specific heat — the amount of heat or energy needed to raise the temperature of one gram of a measure of a fluid’s resistance to flow. substance by 1oC the greater the viscosity, the slower the liquid flow III. Boiling point of water unusually high :strong expressed in units of centipoise intermolecular forces in water allow it to be a liquid at a examples: large range of temperatures. ○ water, milk, fruit juice (flow very easily) IV. Solid water is less dense, and floats on liquid ○ vegetable oil, maple syrup, , peanut butter, water: molecules in solid water are actually farther and dish soap (more viscous, resisting flow apart than they are in liquid water. and pouring out more slowly) rai — 03 ABSTRACTION AND GENERALIZATION III. Molecular Crystals : made of atoms, such as in noble gases, or molecules; atoms or molecules are held solids can be categorized into two groups: the together by a mix of hydrogen bonding / dipole dipole crystalline solids and the amorphous solids. and dispersion forces. ○ C12H22O11 CRYSTALLINE AMORPHOUS ○ iodine and I2 ○ naphthalene & C10H8 arranged in fixed geometric have a random orientation of patterns or lattices (NaCl, particles (glass, plastic, coal, CuSO4) rubber) have structures formed by do not have long range order, repeating three dimensional but may have a limited, patterns of atoms, ions, or localized order in their IV. Covalent Network Crystals : are made of atoms in molecules. structures. which each atom is covalently bonded to its nearest physical properties of the soften gradually when they neighbors; atoms can be made of one type of atom or crystalline solids change are heated; melt over a wide can be made of different atoms sharply. range of temperature ○ SiO2 & BN ○ diamond ──── FOUR TYPES OF CRYSTALS ──── I. Metallic Crystals : made of atoms that readily lose electrons to form positive ions (cations), but no atoms in the crystal would readily gain electrons. ○ the strongest ○ gold, aluminum,copper, etc. ○ high melting point and dense — lesson 02 — II. Ionic Crystals : are made of ions (cations and anions); these ions form strong electrostatic interactions that hold the crystal lattice together. ○ KF, KCl, KBr, KI, NaF rai — 04  THIRD QUARTER — GENERAL CHEMISTRY 2 : PHASE DIAGRAM & CURVES LESSON 03 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ ABSTRACTION AND GENERALIZATION Phase diagram — a graphical representation of the physical states of a substance under different conditions of temperature and pressure. between A & B = material is a solid between B & C = melting (solid to liquid) point C = fully change to liquid between C & D = temp of liquid starts to rise between D & E = boiling (liquid to gas) point E = fully change to gas between E & F = temperature of the gas increases II. COOLING CURVE pressure in atm and temperature in celsius exothermic process (heat is released) I. HEATING CURVE graph of the variation of the temperature with time as it is allowed to cool endothermic process when the system is heated, energy is transferred into it. In response to the energy it receives, the system changes, for example by increasing its temperature. if the temperature of a material is monitored during heating, it varies with time. — lesson 03 — going up = change in temperature ○ q = mSΔt ○ m = mass ○ s = specific heat ○ Δt = final temp — initial temp going straight (straight line) = change in phase ○ q = nHf (melting) ○ q = nHv (boiling / evaporation) rai — 05  THIRD QUARTER — GENERAL CHEMISTRY 2 : PHYSICAL PROPERTIES LESSON 04 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ I. CONCENTRATION OF A SOLUTION 3. Mole Fraction — is a way of describing solution composition and is the ratio of the number of moles of measure of the amount of solute in a given amount of one component of a mixture to the total number of solvent or solution. moles of all components. sometimes solutions are referred to as “dilute” or formula : n = mole nS = nA + nB “concentrated” dilute — there is a relatively small amount of solute in a Xa = nA Xb = nB solvent nS nS concentrated — there is a relatively large amount of EXAMPLE: What is the mole fraction of the solute in a 40% by solute in a solvent mass ethanol (C2H6O) solution in water (H2O)? the substance with the lesser amount is always the solute. → Given: (Assume 100g sample) → Asked: mole fraction of concentration expressions — often based on the mA = 40g C2H6O solute. number of moles of one or more components of a mB = 60g H2O solution. → Solution: Xa = nA → nA = 40g C2H6O _= 0.87 mol ○ three most commonly used: molarity, mole nS 46 g/mol C2H6O fraction, molality → nS = nA + nB = 0.87 mol + 3.3 mol = 4.17 mol → Xa = 0.87 mol = 0.2080 = 0.2 mole fraction II. WAYS TO EXPRESS CONCENTRATION 4.17 mol symbols : 4. Molarity — a useful concentration unit for many ○ A = solute applications in chemistry and is defined as the number ○ B = solvent of moles of solute and exactly 1 liter of the solution. ○ S = solution unit : M or mol / L formula : n = mass _ 1. Mass Percent – concentration of a solution that molar mass describes the mass of the solute every 100 g of M = nA = moles of solute solution. The mass in grams of the solution is the sum MmA VL liter of solution of the mass of the solute and the mass of the solvent EXAMPLE: Determine the molarity of the solution with 0.124 mol unit : percent (%) CoCl2 in 0.654 L of solution mass formula : → Given: nCoCl2 = 0.124 mol → Asked: molarity %MA = MA (100) %MB = MB (100) VL = 0.654 L MS MS → Solution: M = nA → 0.124 mol CoCl2 = 0.190 M EXAMPLE : MmA VL 0.654 L sol → Given: KCl = 0.894 g → Asked: Mass percent of KCl H20 = 54.8 g 5. Molality — is the ratio of the number of moles of solute per kilogram of solvent it is not the same as molarity → Solution: %MA = MA (100) = 0.894 g (100) even if their names are very similar; molality does not MS 54.8 g + 0.894 g change with the solutions temperature. → %MA = 0.0160 (100) = 1.61 % formula : m = nA _ = moles of solute _ mB in kg kilogram of solvent 2. Volume Percent – is the volume of solute in every 100 ml of solution. EXAMPLE: The density of a 2.45 M aqueous solution of methanol (CH3OH) is 0.976 g/mL. What is the molality of the volume formula : solution? %VA = VA (100) %VB = VB (100) VS VS → Given: Msol = 2.45 M → Asked: molality of the Psol = 0.976 g/mL solution. Assume 1 L solution → same process with mass percent Mm CH3OH = 32.04 g/mol rai — 06 → Solution: a. m = nA _ ; to find nA = mass _ mB in kg molar mass → since we don’t have mass: mass = Molarity (Molar mass) (1 Liter) mass = 2.45 M (32.04 g/mol) (1L) mass = 78.498 g = 78.50 g b. nA = 78.50 g _ = 2.450062422 = 2.4501 mol 32.04 g/mol c. since mB in kg is not given, we will use the density of the solution and we will assume 1 Liter (1000 mL) solution. density (P) = mass → 0.976 g/mL = mass _ volume 1000 mL mass (m) = 0.976g/mL (1000 mL) mass (m) = 976 g mass of solution d. to find the mB in kg: mB = 976 g - 78.50 g = 897.5 g or 0.8975 kg e. therefore, molality is: m= nA _ = 2.4501 mol = 2.73 mol/kg = 2.73 m mB in kg 0.8975 kg 6. PPM (parts per million) — is The measurement of the concentration of solution. 1 ppm is part by weight or volume of solute in 1 million parts by weight or volume of solution In weight / volume terms: ○ 1 ppm = 1g In weight / weight terms: ○ 1 ppm = 1mg EXAMPLE: — lesson 04 — rai — 07  THIRD QUARTER — GENERAL CHEMISTRY 2 : COLLIGATIVE PROPERTIES LESSON 05 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ I. COLLIGATIVE PROPERTIES If volatile: → Psolution = (XA PA) + (XB PB) are properties of a solution that depend only on the → XA = mole fraction of solute number and not on the identity of the solute particles. → PA = vapor pressure of solute depend on the collective effect of the concentration of solute particles present in an ideal solution. EXAMPLE: Heptane and octane form ideal solutions. are very useful for characterizing the nature of a solute What is the vapor pressure at 40 degree Celsius of a after it is dissolved in a solvent and for determining the solution that contains 3.00 mol of heptane and 5.00 mol of molar masses of substance because of their direct octane? At 40 degree Celsius, the vapor pressure of relationship with the number of solute particles heptane is 0.121atm and the vapor pressure of octane is 0.041 atm. COLLIGATIVE PROPERTIES 1. vapor pressure lowering → Given: mass heptane = 3.00 mol 2. boiling point elevation vapor pressure of heptane = 0.121 atm at 40oC 3. freezing point depression mass octane = 5.000 mol vapor pressure of octane = 0.041 atm at 40oC COLLIGATIVE PROPERTIES FOR NONVOLATILE SOLUTES 1. vapor pressure is always lower → Solution: since we’re dealing with volatiles 2. boiling point is always higher Psolution = (XA PA) + (XB PB) 3. freezing point is always lower 4. osmotic pressure drives solvent from lower a. mole fraction: concentration to higher concentration → XA = nA = 3.00 mol = 0.375 nS 5.00 + 3.00 EFFECT OF SOLUTE CONCENTRATION ON THE → XB = nB = 5.00 mol = 0.625 COLLIGATIVE PROPERTIES OF SOLUTIONS: nS 5.00 + 3.00 the concentration or amount of non-volatile solute in the b. Psolution = (XA PA) + (XB PB) solution has an effect on the colligative properties of → Psolution = (0.375 0.121) + (0.625 0.041) solution. → 0.709 = 0.71 atm the effect would depend on the ratio of the number of particles of solute and solvent in the solution and not on B. BOILING POINT ELEVATION the identity of the solute. Is the difference in temperature between the boiling A. VAPOR PRESSURE LOWERING point of the pure solvent and that of the solution Molal boiling point constant (Kb) — is the boiling point vapor pressure — a direct measure of escaping elevation of the solvent in a 1 molal solution of a non tendency of molecules. volatile nonelectrolyte solute. a pure liquid or solvent in a closed container will Boiling point elevation (∆tb) — is the difference establish equilibrium with its vapor. between the boiling points of the pure solvent and vapor pressure — pressure exerted by the vapor when non-electrolyte solution of the solvent and is directly equilibrium is reached. proportional to the molal concentration of the solution RAOULT’S LAW FORMULA states that a solvent's partial vapor pressure in a ∆tb = Kbm solution (or mixture) is equal or identical to the vapor ∆tb = boiling point elevation pressure of the pure solvent multiplied by its mole Kb = molal boiling point constant fraction in the solution m = molality FORMULA: BOILING POINT ELEVATION OF SOLUTION: If nonvolatile: tbsolution = Normal tb + ∆tb → Psolution = XB (PB) → XB = mole fraction of solvent → PB = vapor pressure of solvent rai — 08  C. FREEZING POINT DEPRESSION freezing point of a 1 molar solution of any non B. freezing point depression electrolyte solute in water is found by experiment to be tfsolution = Normal tf - ∆tf 1.86oC lower than the freezing point of water a. ∆tb = Kbm Molal freezing point constant (Kf) — is the freezing → 5.12oC / molal (0.2077 m) point depression of the solvent in a 1 molal solution of a → 1.063oC nonvolatile nonelectrolyte solute. freezing point depression (∆tf) — is the difference between the freezing points of the pure solvent and b. tfsolution = Normal tf - ∆tf non-electrolyte solution of the solvent and is directly → 5.5oC – 1.063oC proportional to the molal concentration of the solution → 4.377 = 4.38oC FORMULA D. OSMOTIC PRESSURE ∆tf = Kfm ∆tb = freezing point depression osmosis — movement of solvent through a Kf = molal freezing point constant semipermeable membrane from the side of lower solid m = molality concentration to the side of higher solute concentration. osmosis occurs whenever two solutions of different FREEZING POINT DEPRESSION OF SOLUTION: concentrations are separated by a semipermeable tfsolution = Normal tf - ∆tf membrane. Osmotic pressure — is dependent on the concentration EXAMPLE: What are the boiling point and freezing point of a of solute particles and not on the type of solute solution prepared by dissolving 2.40g of biphenyl ( molar mass particles 154) in 75.0g of benzene? FORMULA → Given: mass of biphenyl = 2.40 g π = MRT ; where: molar mass of biphenyl = 154 g/mol ○ M = Molality moles of biphenyl = 2.40 / 154 = 0.01558 ○ R = gas constant = 0.0821 L atm / mol K mass of benzene = 75.0 g ○ T = temperature in Kelvin KB benzene = 2.53oC / molal TB benzene = 80.1oC EXAMPLE: An aqueous solution contains 30.0g of a protein in KF benzene = 5.12oC / molal 1.00L. What is the osmotic pressure of the solution at 25 degree TF benzene = 5.50C Celsius if the approximate molecular weight of the protein is 43900g. → Solution: A. boiling point elevation → Solution: tbsolution = Normal tb + ∆tb a. M= mA _ = 30.0 g = 6.834 x 10-4 M a. ∆tb = Kbm MmA VL 43,900 L → since molality is not given: → m = nA _ = 0.01558 = 0.2077 m b. R = 0.0821 L atm / mol K mB in kg 0.075 kg c. T = oC + 273.15 = 25oC + 273.15 = 298.15 oK b. ∆tb = Kbm → 2.53oC / molal (0.2077 m) d. π = MRT → 0.5421oC → π = (6.834 x 10-4 M) (0.0821 L atm / mol K) (298.15oK) → 0.015626828 c. tbsolution = Normal tb + ∆tb → 0.0156 atm → 80.1oC + 0.5421oC → 80.6421 = 80.6OC — lesson 05 — rai — 09  THIRD QUARTER — GENERAL CHEMISTRY 2 : THERMODYNAMICS LESSON 06 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ ABSTRACTIONS AND GENERALIZATIONS SAMPLE PROBLEM: The Law of Conservation of Energy or First Law of Thermodynamics — states that energy can neither be A. Combustion of Propane: C3H8 + O2 → CO2 + H2O created nor destroyed. It can only be transformed from one form to another, meaning, the energy of the a. balance the equation: universe is constant. C3H8 + 5O2 → 3CO2 + 4H2O Exothermic change — a change when energy is released to the surroundings (-- ∆H) b. given enthalpy: Endothermic change— a change that involves ○ C3H8 = -104 KJ / mol absorption of energy from the surroundings (+∆H). ○ O2 = 0 KJ / mol calorimeter — measure of energy absorbed or ○ CO2 = -393.5 KJ / mol released as heat in a chemical or physical change ○ H2O = -241.8 KJ / mol Temperature — a measure of the average kinetic energy of the particles in a sample of matter. c. substitute: the greater the kinetic energy of the particles in a ○ ∆H = 3 (-393.5 KJ/mol) + 2 (-241.8 KJ/mol) sample, the higher the temperature and the hotter it — 1(104 KJ/mol) feels. = – 2043.7 KJ I. ENTHALPY CHANGE B. 2C + O2 → CO2 the heat transferred by a process that occurs at a. Given that: constant pressure. i. C + O2 → CO2 : ∆H = -393.5 KJ heat can be thought of as the energy transferred ii. 2CO + O2 → 2CO2 : ∆H = -566.0 KJ between samples of matter because of a difference in their temperatures b. C + O2 → CO2 enthalpy of elemental diatomics is always 0 KJ / mol i. multiply by 2 since the final equation has 2 Carbons. II. HEAT CAPACITY AND SPECIFIC HEAT ii. 2C + 2O2 → 2CO2 quantity of energy transferred as heat during a iii. ∆H : -393.5 (2) = -787 KJ temperature change depends on the nature of the material changing temperature, the mass of the c. 2CO + O2 → 2CO2 material changing temperature, and the size of the i. reverse the equation since the CO is on the temperature change. reactant, it should be on the product side. ii. 2CO2 → 2CO + O2 III. SPECIFIC HEAT iii. ∆H : -566 will be 566 KJ (since we reversed the equation, we must also reverse the sign) the amount of energy required to raise the temperature of one gram of substance by one Celsius degree (1°C) d. add both the equations: or one kelvin (1 K) (because the sizes of the degree divisions on both scales are equal. 2C + 2O2 → 2CO2 2CO2 → 2CO + O2 IV. HEAT OF REACTION = 2C + O2 → CO2 is the quantity of energy released or absorbed as heat during a chemical reaction. ∆H = -787 + 566 = -221 KJ V. HESS’S LAW basis for calculating heats of reaction overall enthalpy change in a reaction is equal to the — lesson 06 — sum of enthalpy changes for the individual steps in the process. H∑product — H∑reactant rai — 010  THIRD QUARTER — GENERAL CHEMISTRY 2 : CHEMICAL KINETICS LESSON 07 ❛ notes ni rai ‧₊˚✧ STEM 12 - A ❬ subject reviewer ❭ ABSTRACTIONS AND GENERALIZATIONS catalyst — a substance that changes the rate of a chemical reaction without itself being permanently Reactions — take place when particles collide with a consumed certain amount of energy. catalysis — action of catalyst activation energy — minimum amount of energy catalyst provides an alternative energy pathway or needed for the particles to react reaction mechanism in which the potential-energy rate of reaction depends on two things: barrier between reactants and products is lowered. ○ the frequency of collisions between particles ○ the energy with which particles collide. if particles collide with less energy than the activation energy, they will not react. the particles will just bounce off each other. collision theory — way of looking at chemical reactions FACTORS AFFECTING RATE OF REACTION: I. NATURE OF REACTING SUBSTANCES TYPES OF CATALYST rate of a reaction depends on the nature of the participating substances 1. homogeneous catalyst — catalyst that is in the same phase as all the reactants and products in a reaction II. TEMPERATURE system. chemical reactions typically occur faster at higher 2. heterogeneous catalyst — phase is different from that temperatures. of the reactants. ex. the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for metals are often used as heterogeneous catalyst days. catalysis of many reactions is promoted by adsorption an increase in temperature of only 10 °C will of reactants on the metal surfaces, which has the effect approximately double the rate of a reaction in a of increasing the concentration of the reactants. homogeneous system. 3. enzymes or biological catalyst — proteins that III. CONCENTRATION OF REACTANTS accelerate biochemical transformations by lowering the activation energy of reactions. the rates of many reactions depend on the concentrations of the reactants rates — usually increase when the concentration of one VI. RATE OF REACTION AND RATE LAW or more of the reactants increases tells how fast a given amount of a reactant or product IV. SURFACE AREA changes with time. can be expressed either as the disappearance of a gaseous mixtures and dissolved particles can mix and reactant or the appearance of the products collide freely is the number of moles of reactant consumed per unit reactions involving them can occur rapidly. time, or the decrease in concentration of reactant with heterogeneous reactions, the reaction rate depends on time. the area of contact of the reaction substances. Rate of reaction = change in concentration the surface area of a solid reactant is an important time factor in determining rate. increase in surface area increases the rate of heterogeneous reactions. V. PRESENCE OF CATALYST reaction rates can be increased dramatically by the presence of a catalyst. rai — 011 ORDER OF REACTIONS: A. Zero - Order Reactions reactions (where order = 0) have a constant rate. is constant and independent of the concentration of reactants. reaction that depends on the concentration of only one reactant other reactants can be present, but each will be zero B. First - Order Reactions reaction (where order = 1) has a rate proportional to the concentration of one of the reactants. a common example of a first-order reaction is radioactive decay, the spontaneous process through which an unstable atomic nucleus breaks into smaller, more stable fragments. C. Second - Order Reaction reaction (where order = 2) has a rate proportional to the concentration of the square of a single reactant or the product of the concentration of two reactants. —- third quarter — rai — 012