CHEM-REVIEWER-1-3rd-Shift PDF

“LEC PPT HEAT CHAPTER 6 - It is a form of energy that is transferred from a body of high temperature to a body of low tem...

“LEC PPT HEAT CHAPTER 6 - It is a form of energy that is transferred from a body of high temperature to a body of low temperature, ENERGY CHANGES IN when brought into contact with each other. CHEMICAL REACTIONS - - heat flow: hot → cold measured by temperature 6.1 Energy and energy changes - SI unit of heat: Joule 1 J = 1 kgm2/s2 6.2 Introduction to thermodynamics - non-SI unit: calorie 1 cal = 4.18 J - Heat is an extensive property. INTENDED LEARNING OUTCOMES - Temperature is an intensive property. 1. perform calculations involving enthalpy - Heat flows spontaneously from the 2. perform calculations involving calorimetry surroundings into the ice, causing it to melt. ENERGY - Heat flows spontaneously from the hot coffee to the cup that contains it and the air around it. TWO WAYS HEAT FLOWS BETWEEN SYSTEM AND SURROUNDINGS Endothermic - heat flows out of the system Exothermic - heat is absorbed by the system - the capacity to do work or to produce heat - Energy is neither created nor destroyed. - It can be transferred from one object to another or THERMODYNAMICS transformed from one form to another - The branch of science that deals with the relations - Law of conservation of energy (The First Law o f between heat and other forms of energy and by Thermodynamics) extension, of the relationship between all forms of - energy SYSTEM VS SURROUNDINGS - study of energy and its interconversion System - part of the universe where attention FIRST LAW OF THERMODYNAMICS is focused, being studied or measured. - The energy of the universe is constant Surroundings - everything else in the universe in Law of Conservation of Energy contact with the system. - Heat is exchanged (flows) between the system and - A “universe” consists of the “system” and the the surroundings. “surrounding”, separated by a “barrier”. THERE ARE 3 TYPES OF SYSTEMS: 1. open system - allow transfer of heat and mass 2. closed system - allow transfer of heat but not mass 3. isolated system - no transfer of heat and mass Internal Energy, E - sum of the kinetic and potential energies of all particles in a system - can be changed by flow of work or heat Take note of the following: Let’s have an example: - ΔHf – heat of formation - Positive H is endothermic (required heat for reaction to take place) Where do we get the enthalpy values of the reactants and products? For a Compound - The standard state of a gaseous substance is a pressure of exactly 1 atmosphere. - For a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid. - For a substance present in a solution, the standard state is a concentration of exactly 1 mole/L (equal to 1 Molar) For an Element - The standard state of an element is the form in which the element exists under conditions of 1 atmosphere 6.3 Enthalpy and 25°C. 6.4 Hess’s Law - Ex. Gold is solid at 1atm 25°C, Mercury is liquid at 1atm 25°C, Chlorine is in gaseous state at 1atm 25°C, 6.5 Standard Enthalpies of formation Sodium is solid at 1atm 25°C - [The standard state for oxygen is O2(g) at a pressure WHAT IS A THERMOCHEMISTRY? of 1 atmosphere, the standard state for sodium is - The branch of chemistry with the quantities of heat Na(s), the standard state for mercury is Hg(l); and so evolved or absorbed during chemical reactions on.] Enthalpy, H - quantity of heat transferred into and out of a Some examples of standard enthalpies… system in a constant-pressure process (usually, 1 atm) - an extensive property ENTHALPY OF REACTION, ΔH - the difference between the enthalpies of the products and the enthalpies of the reactants Table 7.2 Standard Enthalpy of Formation of Several Compounds at 25°C This is called a thermochemical equation: We use the standard enthalpies of formation (ΔHf°) to calculate the standard enthalpy of reaction (ΔH°rxn) - Theoretical method of finding the heat of a reaction Standard Enthalpy of Reaction, ΔH°rxn Where ΔH is calculated as difference in enthalpy between - the enthalpy of a reaction carried out at 1 atm products and reactants Let’s have an example… WHERE DO WE GET THE ENTHALPY VALUES OF THE REACTANTS AND PRODUCTS? Standard Molar Enthalpy of Formation, ΔHf° - enthalpy change for the formation of one mole of a compound from its elements in their standard states (normal/ambiant condition) - this indicates standard conditions, 1 atm, 25 °C - Higher temperature, higher energy Where do we get the enthalpy values of the reactants and products?  - ENTHALPY is an extensive property, based on amount of matter present, this value is used to determine how much product is used from a chemical reaction 6.6 Calorimetry WHAT IS A CALORIMETRY? Calorimetry: measure of heat changes by observing - Balance the equation temperature change - Find the heat of formation of each compound involved Calorimeter: a device used to measure heat changes in a in the reaction (table) chemical reactions - Plug in all the values to the formula (check number of HOW IS HEAT MEASURED? moles) - The relationship between heat flow (q) and - (negative answer is exothermic) temperature change (ΔT): What are the characteristics of enthalpy change? - If a reaction is reversed, the sign of ΔH is also reversed. - The magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction. If Heat capacity amount of heat required to raise the the coefficients in a balanced reaction are multiplied temperature of a given substance (system) by 1°C by an integer, the value of ΔH is multiplied by the - Simply, it is the amount of heat absorbed per °C (unit: same integer. J/ C) - Enthalpy is an extensive property (dependent on the - Heat can be measured as a function of change in mass) temperature. COMBINING THERMOCHEMICAL EQUATIONS: HESS’ LAW For a pure substance, heat capacity is given per gram of - the enthalpy for a reaction is the same whether it the substance: occurs by one step or by any series of steps C = ms where m = mass of the substance Specific heat capacity (s) - the amount of heat absorbed by 1g of a substance per °C (unit: J/g°C) - If the heat capacity is given per mole of the substance, it is called molar heat capacity. q = C x ΔT For a pure substance, the equation becomes: For example… q = ms x ΔT Specific heat of some substance: - Since CH3CH2OH is a product, we can reverse the first reaction and change the sign of ΔH - Taking the sum of the two equations…  - Metals are good conductors of heat. Why is there a negative sign? - Water is often used in cooling systems. - the negative sign indicates flow of heat: o when the system release energy (-), the surrounding absorbs it (+) o when the system absorbs energy (+), the surrounding released it (-) EXAMPLES: 1. 50.0 mL of water at 41.0 °C is added to a calorimeter containing 50.0 mL of water at 17.4 °C. After waiting for the system to equilibrate, the final temperature reached is 28.3 °C. Calculate the heat capacity of the calorimeter. (sp_heat of water = 4.184 J/g×°C) Let’s have an example… 1. How much heat is absorbed by 255 g of H2O if its temperature is raised from 20.0 to 80.0 °C? (s = 4.184 J/g °C) o Note: heat is absorbed, q must be positive (endothermic) 2. When 1.00 g of NH4NO3 is dissolved in 50.0 g of water in a coffee-cup calorimeter that has a heat 2. How much heat is released by a 5.0 g Cu metal as it capacity of 34.7 J/°C: cools from 350 to 40°C? (s = 0.382 J/g °C) o Note: heat is released, q must be negative (exothermic) o the temperature drops from 25.0 to 23.3°C. Calculate the heat change for the dissolution. Useful formulas: HOW IS CALORIMETRY DONE? Constant Pressure Calorimetry - used to measure enthalpy change for aqueous reactions o Note: “temperature drops” indicates that the system absorbed heat from the surrounding (q is positive). 3. When 1.00 g of NH4NO3 is dissolved in 50.0 g of water in a coffee-cup calorimeter that has a heat capacity of 35 J/°C: the temperature drops from 25.0 to 23.3°C. Remember system and surroundings? - system: aqueous reaction - surrounding: calorimeter (including water) LAW OF CONSERVATION OF ENERGY - heat is exchanged between system and surrounding  4. Calculate the heat change when 50.0 mL of 1M HCl and 50.0 mL of 1M NaOH are mixed in a coffee-cup calorimeter that has a heat capacity of 35 J/°C and the temperature increases from 21.0 to 27.5 °C. o Why 100.0 g of water? It is assumed that 50 mL of 1M HCl contains 50 g of water, 50 mL of 1 M NaOH also contains 50 g of water, NOTES: hence total to 100 g. ENERGY — driving force that allows you to move from one CONSTANT VOLUME CALORIMETRY point to another - allows to measure the heat of combustion of a - used for reaction to take place or produced as a substance product of chemical reaction - no work is done at constant volume - ex. Heat energy - heat comes from surface - Heat of combustion transferred to you - heat measurement is done using a bomb calorimeter - Or your heat (energy) can transfer to the surface - Yes, it’s a bomb calorimeter. Don’t worry, it does not - bond formed - energy released; break bond - req. explode. energy System - part of universe wherein we are focusing the attention Surroundings - anything around the system Barrier - there is a barrier between both TYPES OF SYSTEM Open system - transfer between system and surroundings - thermal equilibrium Closed system - transfer of heat only Isolated system - aqua flask, isolator is air, keep the heat or cold intact EXAMPLES: - in example, cup is the barrier, coffee is the system, 1. Octane is a component of gasoline that is very and anything else is the surrounding volatile and flammable. A 2.425 mL octane sample HEAT was placed inside a bomb calorimeter that contains - We feel cold because our body is releasing heat into 675.0 mL of water. the surroundings, jacket is our barrier (insulator) Igniting the octane caused the temperature of the - Joules - unit of energy and work; equal to Na water to increase from 28.4°C to 57.3°C. - Extensive property - dependent on mass Calculate the molar heat of combustion of octane, - Temperature - intensive property; not dependent on assuming that the bomb calorimeter has negligible quantity heat capacity. Octane has a molar mass of 114.23 and density of 0.703 g/mL. o Note: If “negligible heat capacity”, we assume that Ccal = 0. (Only happens in bomb calorimeter) o Note: Remember, density of water is 1 g/mL 2 WAYS HEAT WLL FLOW —> THERMODYNAMICS BCM 621 LABORATORY - heat flow between system and surroundings Exothermic —> heat is released from system going out to CALORIMETRY - PRE-LAB surroundings; system —> surroundings ENERGY - releases energy (-) - can only be known by its effects and not by Endothermic —> heat is absorbed by the system; from the perceptible characteristics surroundings to the system; surroundings —> system - Measured in Joules - requires energy (+) - The best way to describe energy is thru its forms THERMAL EQUILIBRIUM o Radiant energy – energy that travels - Energy - from this energy emanates everything, in the through space (light from the sun or beginning there was just energy (God) commonly known as solar energy) - Heat is exchanged between system and surroundings o Chemical energy– Energy generated until it reaches this (system = surrounding) through bond formation Kinetic energy - energy in motion o Thermal energy– associated with the Potential energy - energy at rest random motion of atom and molecules INTERNAL ENERGY THERMAL ENERGY - Exothermic —> work to system; work done on - This form of energy is our main concern, that is to system measure the energy changes in chemical reaction known as thermochemistry - This is important for understanding the natural physical, chemical and biological processes. ENERGY CONSERVATION - Endothermic —> system work on you; work done by - The law of the conservation of energy states that system; heat absorbed by system energy is not created nor destroyed but only transformed to another form of energy Chemical → electrical → heat → light DIFFERENT TYPES OF THERMOCHEMICAL SYSTEM System – a particular part of the universe in which we are Heat is required —> Endothermic interested Heat is produced —> Exothermic - if you feel the heat coming Surrounding – refers to the rest of the universe system out, the system is releasing heat surrounding ENTHALPY - heat involved in the chemical reaction - what is directly in contact with the system SYSTEM MAY BE CLASSIFIED AS: Open – There is exchange of mass and heat between the system and surroundings Closed – There is only exchange of heat between the system and surroundings Isolated – There is no exchange of both mass and heat between the system and surroundings THERMAL PROCESSES H2O(l) → H2O (g) Endothermic process – energy is absorbed by the system form the surroundings - Heat is absorbed by system - We are measuring the surrounding H2O (l) → H2O(s) Exothermic process – energy is released by the system to the surroundings MEASURING ENERGY CHANGES CALORIMETRIC METHOD - To measure the change in heat in a chemical reaction - Calorimetric methods are used to measure the heat the method needed to be employed is calorimetry absorbed or released by the system, by measuring using an instrument called calorimeter. changes in temperature of the surrounding. - With measured amounts of hot and cold water, the heat capacity of the calorimeter can calculated: A CALORIMETER IS AN ISOLATED SYSTEM - this means that there is no exchange mass and heat between the system and the surroundings - it can measure the change of heat during chemical HEAT OF SOLUTION reaction by the increase or decrease in temperature - it is the amount of heat energy that is released or HEAT CAPACITY VS SPECIFIC HEAT absorbed when a solution is formed. Heat capacity or thermal capacity (c) – a physical property of Enthalpy of solution matter, defined as the amount of heat to raise the temperature by 1oC - intensive property - More insulation, less heat capacity 1. The breaking of bonds within the solute, such as the electrostatic attraction between two ions (endothermic) 2. The breaking of intermolecular attractive forces within the solvent, such as hydrogen bonds (endothermic) 3. The formation of new attractive solute-solvent bonds Specific heat (s) – the quantity of heat required to raise the in solution (exothermic) temperature of one gram of a substance by one Celsius degree HEAT OF NEUTRALIZATION - When molecules and ions absorb the heat, their - the heat evolved when an acid and a base react to thermal energy increase and causes increase in form a salt plus water temperature - Extensive property - This enables us to measure the amount of heat absorbed or released (q) in a calorimetric - the ∆H of neutralization will include experiment of a known mass (m) of sample by virtue o ionizing the acid and base of the change in temperature (Δt). o reaction between the hydrogen ions and q = msΔt hydroxide ions. (-)q Exothermic (+)q Endothermic CONSTRUCTING A STYRO-CUP (SC) CALORIMETER OBJECTIVE - Measure the heat capacity of a coffee-cup calorimeter - Determine the heat of reaction between an acid and a base - Determine the heat of solution of two solids - Differentiate endothermic from exothermic reactions BASED ON THE LAW OF CONSERVATION OF ENERGY - heat is exchanged between the system and the surrounding - A system that loses heat to the surrounding is 1. Insert sc 1 into sc 2 exothermic, while a system that absorbs heat from the 2. Cut sc 3 along cross section surrounding is endothermic. 3. Make a smal hole on the bottom of Styro 3 enough to - Thus: insert your thermometer q (system) + q (surrounding) = 0 4. Inset thermometer q (system) = -q (surrounding) MEASURING THE HEAT CAPACITY OF CALORIMETER CHAPTER 8 PART 1 LIQUIDS AND SOLIDS 8.1 The Liquid State 8.2 The Solid State 8.3 Types of Crystalline Solids 8.4 Heating Curves 8.5 Phase Changes and Phase Diagram INTENDED LEARNING OUTCOMES 1. explain the properties of liquids and solids 2. interpret the phase diagram of a substance 1.Place tap water in sc 2 about 1/3 a cup (80 ml) and 3. Illustrate the different types of crystalline solids measure the temperature A REVIEW OF THE STATES OF MATTER 2. Place sc 2 with the cold water into sc 1 3. Measure the same amount of hot water (40°C) and immediately pour into the cold water 4. Cover the calorimeter and swirl gently 5. Measure the temperature after 10 swirls HEAT CAPACITY OF THE CALORIMETER HEAT OF REACTION BETWEEN NAOH AND HCL STATES OF MATTER IN TERMS OF IMF - attractive forces between individual particles - gas < liquid < solid PROPERTIES OF THE LIQUID STATE SURFACE TENSION SOLUTIONS - the resistance of a liquid to an HEAT OF SOLUTION increase in its surface area - surface tension allows this insect to walk on water and a paper clip to float on water CAPILLARY ACTION Cohesion – attraction among the molecules of the liquid - this is why water form droplets Adhesion – attraction between a liquid and a surface or container - this is why water sticks on the surface of a glass - Surface outside is involved We learned about capillary action in grade school science  - the concave - There are more molecules that can overcome meniscus of water atmospheric pressure shows that its - the vapor pressure of a liquid increases with adhesive force with temperature glass is stronger than its cohesive forces - the convex meniscus of mercury shows that its cohesive force is stronger than its adhesive force to the glass VISCOSITY - the resistance of a liquid to flow - liquids with strong IMF are more viscous - Cohesive force - As we increase temperature, vapor pressure increases Linear equation: y=mx+b - When we draw a graph, we PREDICT - Each liquid has different profiles, but when made linear, they are the same A linear relationship between vapor pressure (Pvap) and temperature (T in K) is given by the formula: WHY DO LIQUIDS EVAPORATE? - some molecules on the surface of a liquid have enough energy to escape into the gas phase - molecules with weaker IMF can evaporate more easily - This equation is the Clausius-Clapeyron equation named after the German physicist Rudolf Clausius (1822 - 1888) and the French engineer Emil Clapeyron (1799 - 1864). CLAUSIUS-CLAPEYRON EQUATION The exponential behavior of vapor pressure (P) as a function of temperature (T) is given by the exponential function: VAPOR PRESSURE - molecules that escape into the gas phase exert - A in the expression is an experimental constant that pressure can be related to the normal boiling point, ΔHvap is - molecules with weaker IMF can evaporate more the heat of vaporization of the liquid, and R is the gas easily; they are more volatile and has higher vapor constant, R = 8.314 J/(K mol). pressure - It is not 1 is to 1, it is exponential (wider range) - Vapor pressure is the property of the liquid, force that To remove an exponent from an expression the log exists in the liquid due to the cohesive force of the function (log or ln) can be used. Taking the natural log (ln) molecules of both sides of equation. - Atmospheric pressure – pushes it down, molecules resists this force - Pressure is a force - Due to cohesive force - Overcome pressure and form into gasses due to - We can predict certain temperature s at certain vapor pressure conditions - Weaker IMF, higher vapor pressure - Initial and final points A simple relationship can be found by integrating between what is the equation? two pressure-temperature endpoints: can be rearranged to: - This is important to determine the changes in the vapor pressure of a liquid - Altitude changes boiling points because atmospheric pressure changes where: BOILING - boiling point is the temperature (degree of hotness or coldness) when the vapor pressure of a liquid becomes equal to the atmospheric pressure - Atmospheric pressure = atmospheric pressure plug in the values: solving for T2: T2 = 362 K = 89 °C HEATING CURVE - A plot of temperature versus time for a process where energy is added at a constant rate o Melting Point o Boiling Point o Heat of Fusion o Heat of Vaporization WHAT INFORMATION CAN WE GET FROM A HEATING CURVE? WHAT HAPPENS DURING BOILING? - more volatile liquids have lower boiling point - as the temperature increase, the rate of evaporation and the vapor pressure increases - at the boiling point, the vapor pressure becomes equal to the atmospheric pressure - the temperature of the ice increases as it absorbs heat as a function of its specific heat (2.03 J/g°C) - At this point, the temperature remains constant as the ice absorbs heat and melts as a function of its enthalpy of fusion (6.02 kJ/mol) At high altitude, does water boil higher or lower than - the temperature of the water increases as it absorbs 100°C? heat as a function of its specific heat (4.184 J/g°C) EXAMPLE: In Breckenridge, Colorado, the typical atmospheric - At this point, the temperature remains constant as the pressure is 520 torr. What is the boiling point of water in water absorbs heat and evaporate as a function of its Breckenridge? (ΔHvap of water is 40.7 kJ/mol) enthalpy of vaporization (40.7 kJ/mol) what is given? - the temperature of the steam increases as it absorbs Pvap = 520 torr heat as a function of its specific heat (2.04 J/g°C) R = 8.314 J/molK ΔHvap = 40.7 kJ/mol what is being asked? T=? what else do we know? at 760 torr, water boils at 373 K 1atm = 760torr 760mHg = 760torr 760 = 373K boiling, Pvap = Patm reminder: during When does “SUPER COOLING” happen? Let’s have an example… - When a liquid is cooled below its freezing point What quantity of energy does it take to convert 0.500 kg ice at without solidifying -20. °C to steam at 250. °C? What happens to the molecules and the energy of the Specific heat capacities: system when it is super cooled? ice, 2.03 J/g °C; liquid, 4.2 J/g °C; steam, 2.0 J/g °C - Crystallization ∆Hvap = 40.7 kJ/mol; ∆Hfus = 6.02 kJ/mol When does “super heating” happen? - happens when a liquid is heated past its boiling point without boiling 1. first, how much heat is used to increase the WHAT INFORMATION CAN WE GET FROM A PHASE temperature of 0.500 kg ice from -20 to 0 °C DIAGRAM? q1 = ms∆T - Phase diagram is a convenient way of representing = (500 g)(2.03 J/g°C)(20 °C) the phases of a substance in a closed system as a = 20,300 J function of temperature and pressure q1 = 20 kJ - At the triple point, all three phases co-exist 2. second, how much heat is absorbed as 0.500 kg ice HOW DO WE INTERPRET THIS PHASE DIAGRAM FOR melts to water WATER? q2 = ∆Hfus x (m/MM) = (6020 J/mol)(500 g)/(18 g/mol) = 167,222 J q1 = 20 kJ q2 = 167 kJ 3. third, how much heat is used to increase the temperature of 0.500 kg water from 0 to 100 °C q3 = ms∆T = (500 g)(4.18 J/g°C)(100 °C) = 209,000 J q1 = 20 kJ q2 = 167 kJ q3 = 209 kJ 4. fourth, how much heat is absorbed as 0.500 kg water evaporates q2 = ∆Hvap x (m/MM) = (40,700 J/mol)(500 g)/(18 g/mol) = 1,130,556 J Experiment 1: q1 = 20 kJ - At a pressure of 1 atm, ice melts at this temperature q2 = 167 kJ (273 K) water evaporates at this temperature (373 K). q3 = 209 kJ Experiment 2: q4 = 1,131 kJ - At this 5. fifth, how much heat is used to increase the pressure, ice temperature of 0.500 kg steam from 100 to 250 °C is converted q3 = ms∆T to vapor = (500 g)(2.0 J/g°C)(150 °C) (sublimation) = 150,000 J at this q5 = 150 kJ temperature q1 = 20 kJ Experiment 3: q2 = 167 kJ - At this q3 = 209 kJ pressure, the q4 = 1,131 kJ 3 phases of water co-exist at this temperature Try to find out what Experiment 4 is about.  - There are different ways by which atoms, ions or molecules can be arranged in a crystal lattice, depending on their sizes - We will study more about this in Inorganic Chemistry X-RAY DIFFRACTION - scattering of x-rays by the unit cells of the crystalline solid - It is a powerful tool for determining the structure of crystals - Intensity of light will give what kind of atoms are present (how they react) THE SOLID STATE DIFFERENT TYPES OF CRYSTALS CRYSTALLINE SOLID ATOMIC CRYSTALS: - rigid with long range order 1. Network Covalent - atoms, molecules or ions occupy specific positions in o 3D network of covalent bonds of non-metals the crystal lattice (rigid or proper arrangement of o high mp, poor conductor, hard to soft molecules) o diamond and graphite are not compounds; - Each portion of the lattice is called a crystal cell they are different forms of carbon held - A unit cell is the repeating unit of a crystalline solid together by covalent bonding A. diamond - one of the hardest substances; o mp = 3550 °C o all C atoms are tetrahedral (sp3) B. graphite - soft, layers slide past one another, o mp = 3652 °C o all C atoms are trigonal planar (sp2) 2. Metallic crystals o held by delocalized electrons between metals o varied hardness and mp o high electrical conductivity (due to sea of electrons) o localized sea of electrons model  3. Ionic crystals o composed of ions of ionic compounds; hard, TO SUMMARIZE… brittle o have very high mp ▪ NaCl: 801°C ▪ CaO: 2927 °C o poor conductor of heat and electricity o There are many kinds of ionic crystals o The kind depends on the size of the cations and anions o Held together by attraction between cations and anions o Ductility, malleability HOW DO AMORPHOUS SOLIDS DIFFER FROM CRYSTALS? - lacking in 3D ordered arrangement of atoms - crystals have long range of order - amorphous solids lack 3D ordered arrangement Quartz crystal and glass: - both of them are made of SiO2 4. Molecular crystals o held together by dispersion, dipole-dipole, H- bond o soft, relatively low mp o examples: naphthalene, glucose, I2, P4, S8 o exhibited by elements or compounds that are molecules o If the crystalline is caused by IMF it is molecular crystals Brittle edges – crystalline solid o Soft edges – amorphous solid PERCENTAGES OF VARIOUS COMPON ATOMIC CRYSTALS, IONIC CRYSTALS, MOLECULAR CRYSTALS - glass is amorphous LAB AND LEC – GAS LAWS The average kinetic energy of a gas is directly proportional to the absolute temperature (in K). CHAPTER 7 PART 1: GASES o When you increase heat, kinetic energy also 7.1 Properties of Gases increases 7.2 Kinetic Molecular Theory of Gases WHAT ARE THE GAS LAWS? 7.3 Gas Laws and Ideal Gas Equation 1. BOYLE’S LAW - the pressure (P) of a gas 7.4 Chemistry of the Atmosphere is inversely proportional to its volume (V) at INTENDED LEARNING OUTCOMES a given temperature (T) 1) associate the kinetic molecular theory to the gas laws o Robert Boyle 2) perform calculations and predict changes in pressure, volume and temperature based on the gas laws GASES EXERT PRESSURE UNITS OF PRESSURE ▪ 760 mm Hg - 760 torr ▪ 1 atmosphere - 101.325 kPa (N/m2) o decreasing the volume increases ▪ 1.01325 bar - 14.7 psi the frequency of collision, hence HOW DO WE MEASURE THE increasing the pressure PRESSURE OF GASES? o Higher pressure, Lower volume STATE VARIABLES when temperature is constant In terms of: o Natural law ▪ Volume – space matter will occupy (area is part of o Decrease in space, increase in this) pressure ▪ Pressure – force applied to the walls of the molecules against the walls of the container o One state variable o Caused by frequency of collision ▪ Temperature – if you increase temperature, molecules will move faster o State variable Barometer is used to measure atmospheric pressure. A. When you plot pressure on the y-axis and volume on - The barometer was invented the x-axis, you will get a gaph that appears like by Evangelista Toricelli - Used to measure the pressure of a specific environment - Experiment: In a pool of mercury, how far can mercury rise up in a tube that doesn’t contain anything inside. It rose 760mmHg/torr/atm. - Property of mercury is affected by the mass. P = lgh B. When you plot pressure on the y-axis and inverse of P = F/A —> N/m2 * kPa volume (1/volume) on the x-axis, you will get a graph 760mmHg = torr = atm —> limits pressure that appears like - Surface – part of liquid being pushed downward by the atmosphere - Capillary action Manometer is used to measure pressure of a gas in a container THE KINETIC MOLECULAR THEORY OF GASES ASSUMES THAT… The volume occupied by gas particles are negligible compared to the distance between them. Let’s have an example... o Volume of gas can expand limitless A cylinder with a moveable piston contains 15 L of gas with a o Diffusibility – one of the properties of gases pressure of 650 mm Hg. If the gas is compressed to a volume Gas particles are in constant motion; the pressure of 11 L, what will be the resulting pressure of the gas inside the exerted by a gas is due to its movement. cylinder? o It is a continuous motion first, identify what are given: o Reason why it can expand and convert o V1 =15L Gas particles are assumed not to attract or repel each o P1 = 650 mmHg other (collisions are perfectly inelastic). o V2 =11L o Velocity of gas moving towards each other causes collision  next, what is being asked? what equation are you going to use? o P2 = ? Which gas law is this about? o Boyle’s what equation are you going to use? Plug in the values and compute.. Plug in the values and compute.. o Pressure is increased o Volume decreases 2. CHARLES’S LAW - the volume of a gas at constant 3. GAY-LUSSAC’S LAW - the pressure of a gas of pressure increases linearly with temperature constant volume increases linearly with temperature o Jacques o Joseph Gay-Lussac Charles o increasing the temperature increases the o The constant is kinetic energy of the gases, increasing the pressure frequency of collisions at constant o increasing the volume temperature o Volume is constant increases the o Temperature increases kinetic energy of the gases, causing the o When pressure decreases, volume to expand with constant temperature decreases pressure o When gas in tank is used, it gets cold because pressure decreses o You need higher volume to maintain the pressure o When you decrease pressure, molecules slow down and compress o Volume increases, temperature increases, 4. AVOGADRO’S LAW - equal volumes of gases at the pressure decreases same T and P contain the same number of particles o You always need to convert temperature in o the volume is directly proportional to the Kelvin scale number of moles (n) at constant T and P o Amedeo Avogadro o If you want to maintain a specific temperature, you have to keep the same number of moles (moles Let’s have an example... A very elastic balloon contains 2.5 L of gas at 28 °C. If this balloon was placed in a cold room with a temperature of 4.0 °C under the same atmospheric pressure, what will happen to the volume of the balloon? are number of particles) (assuming the elasticity does not affect the pressure) first, identify what are given: o V1 = 2.5 L WHAT HAPPENS WHEN WE COMBINE THE GAS o T1 = 28 °C + 273 = 301 K LAWS? o T2 = 4.0 °C + 273 = 277 K Let’s have an example... next, what is being asked? A 10.0 mL syringe contains of 0.016 g of NO2 gas (46.00 o V2 = ? g/mol) at 28 °C with a pressure of 0.85 atm. If the plunger is Which gas law is this about? pushed releasing 4.5 mL of gas, what mass of NO2 will remain o Charles’s in the syringe? first, identify what are given: o V1 = 10.0 mL o T2 = 28 °C + 273 = 301 K o m1 = 0.016 g NO2  o V2 = 10.0 – 4.5 = 5.5 mL (remaining) what equation are you going to use? next, what is being asked? o m2 = ? Which gas law is this about? o Avogadro’s what equation are you going to use? Plug in the values and compute. First, the number of moles of 0.016 g NO2 o N = mass/molar mass Plug in the values and compute. ANOTHER GAS LAW DALTON’S LAW OF PARTIAL PRESSURES - for a mixture of gases in a container, the total pressure exerted is the sum of the individual pressures of each of the gases Then the remaining mass. - Total pressure is due to the total number of moles - John Dalton PT = P1 + P2 + P3 +... IDEAL GAS LAW They can be combined into one equation: We can also summarize the gas laws into this equation: - The partial pressure of each gas in a mixture of gases in a container depends on the number of moles of that gas. The total pressure is the sum of the partial pressures and depends on the total moles of gas particles present, no matter what they are. Note that the volume remains constant. o Henri Victor Regnault o where K= R is the universal gas constant equal to 0.08206 L-atm/mol K or 8.3145 J/ mol K o At STP, 1 mole of gas will take up 22.4 L of the volume of the container. o Volume is inversely proportional to pressure o Volume is directly proportional to temperature o Volume is directly proportional to number of moles in the gas o K is universal gas constant —> STP Let’s have an example... o To maintain A mixture of 5.00 g N2 and 6.00 g O2 is placed in a 6.0-L container at 27 °C. Calculate the partial pressure of each gas Let’s have an example... and the total pressure. What volume will 75 g of chloroform gas (CHCl3, MM = 119.4 first, identify what are given: g/mol) occupy at atmospheric pressure if its temperature is 32 o 5.00 g N2 °C? o 6.00 g O2 - If pressure is given, use 0.0821 o 6.0 L container - If Joules is given, use 8.314 o 27 °C - Should be in meters (m), grams (g), and Kelvin (K) next, what is being asked? first, identify what are given: o partial pressure of each gas total pressure o 75 g CHCl3 Which gas law is this about? o P = 1.0 atm o Dalton’s o MM = 119.4 g/mol To get the partial pressure, we use the ideal gas o T = 32 °C + 273 = 305 K equation next, what is being asked? o V=? Which gas law is this about? o Ideal Gas equation  Plug in the values and compute.. WHAT ARE THE DIFFERENT TYPES OF SOLUTIONS? - An unsaturated solution may be described as “dilute”. - A saturated solution may be described as “concentrated”. - A supersaturated solution may be described as more than the maximum amount of solute that is capable of being To get the total pressure: dissolved at a given temperature o Once the equilibrium is disrupted, crystallization occurs (precipitate) If we dissolve 30 g of NaCl in 100 mL H2O We produce an unsaturated solution; it has not Plug in the values and compute. reached the limit of solubility CHAPTER 9 PART 1: PHYSICAL PROPERTIES OF SOLUTIONS 9.1 Types of Solutions 9.2 Factors that Affect Solubility 9.3 Concentration Units 9.4 Colligative Properties If we dissolve 40 g of NaCl in 100 mL INTENDED LEARNING OUTCOMES H2O 1. differentiate among types of solutions We produce a saturated 2. explain the factors that affect solubility solution; it has reached the 3. perform calculations involving concentrations units limit of solubility (36 g/100 mL) 4. perform calculations involving colligative properties The excess 4 g will not 5. relate colligative properties to real samples dissolve anymore WHAT IS A SOLUTION? Standard temperature: 25 - A solution is a homogenous mixture of degrees C 2 or more substances. WHAT IS SOLUBILITY? - The solute is the substance present in - the maximum amount of solute the smaller amount. that can be dissolved in a - The solvent is the substance present in given amount of solvent at a the larger amount. specific temperature WHAT ARE THE FACTORS THAT AFFECT SOLUBILITY? Solids: temperature o generally, the solubility of solids increase with increasing temperature Gases: temperature o the solubility of gases decrease with increasing The phase of a solution takes the phase of the solvent: temperature o Solubility of gases can happen at low temperature WHAT ARE THE FACTORS THAT AFFECT SOLUBILITY? Gases: pressure o Henry’s Law: the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution o Pressure changes have little or no effect on solubility of liquids and  solids in liquids because Liquids and solids HOW DOES A COLLOID DIFFER FROM A SOLUTION are not compressible AND A SUSPENSION? o Increasing pressure liquifies gases Tyndall Effect Effervescence - a beam of light is scattered by the dispersed phase; o evolution of gas as a result of a used to differentiate colloids from decrease in pressure over the solutions solution (not as a result of a HOW DOES TYNDALL EFFECT WORK? chemical reaction) - Suspension – has larger particle size o Result of decrease in pressure of the dispersed phase; the particles not a chemical reaction settle after standing Example: - The suspended particles of a colloid Let’s say we dissolved 100 g of a solid in 250 mL of reflects the light that passes through H2O it, showing the path of the light, unlike After thorough stirring, only 91 g dissolved. What is solutions the solubility of the solid (g/100 mL)? HOW DO WE EXPRESS SOLUTION COMPOSITION? Heating increases the solubility of the solid in the - Solute identifies the characteristics of a solution water; now all 100 g dissolved (that’s why we forcus on solutes) Supersaturated is when the solution exceeds its How much of the solute is present in proportion to the normal solubility solvent or solution as a whole? The excess 9 g will come out again as limited by its There are different units of concentration: solubility o percent, % Compare these two o mole fraction, x o molarity, M o molality, m o normality, N % OF SOLUTE (m/m, m/v, v/v) - always part over the whole (part/whole —> solute/solution) - Amount of solute/amount of solution —> ratio —> proportion - the percentage of the solute in the solution, by mass or volume WHAT ARE COLLOIDS? - an intermediate type of mixture that has a particle size between those of true solutions and suspensions - dispersion of particles of one substance throughout a dispersing medium made of another substance - the particles do not settle out of the solution but they Let’s have an example… make the solution cloudy or opaque How many grams of NaCl is present in 500.0 mL of a normal - Dispersion saline solution (NSS, 0.90 % m/v)? - Milk is a dispersion as it makes the solution cloudy or what is the formula? (note: m/v) opaque (powdered milk) How many grams of isopropyl alcohol is present in 250 mL of a 70.0 % (v/v) solution? (density of isopropyl alcohol = 0.786 g/mL) what is the formula? (note: v/v) mole fraction, X MOLE FRACTION, X Let’s make it more challenging… - the proportion of the number of moles of the solute Calculate the molarity of an aqueous solution that is 10.0% over the total number of moles (m/m) glucose, C6H12O6. The density of the solution is 1.04 - take note, this is a fraction, not percentage; mole g/mL. fraction is always less than 1 1 mol C6H12O6 = 180 g Note: - 1 kg of the solution contains 100 g glucose - 1 kg of the solution is 0.96 L (see density) Let’s have an example… What are the mole fractions of glucose (180.0 g/mol) and water (18.0 g/mol) in a 10.0 % (m/m) glucose solution? Note: 100 g of the solution will contain 10 g of glucose and 90 g of water Some more challenge… Calculate the percentage by mass of HCl in a 12.0 M (concentrated) solution. The density of the solution is 1.19 g/mL. 1 mol of HCl = 36.45 g Note: - 12 M solution consists of: o 12 moles or 437 g HCl (12 x 36.45) o 1 L solution is equal to 1,190 g (see density) Another example… MOLALITY, M - number of moles of the solute per kilogram of the solvent - Ratio of mass of solute and mass of solvent MOLARITY, M - number of moles of the solute per liter of the solution - How does molality differ from molarity? o Molarity – proportion of solute against solution in liters o Molality – Amount of solute against the mass of the solvent in kg Let’s have an example… Calculate the molality of a solution that contains 7.25 g of Let’s have an example… benzoic acid C6H5COOH, in 200.0 mL of benzene, C6H6. The Calculate the molarity of an aqueous solution that contains density of benzene is 0.879 g/mL. 55.0 g of H2SO4 in 2.50 L of solution. (MM H2SO4 = 98.0 1 mol C6H5COOH = 122 g g/mol) what’s the formula? what’s the formula? first, the kg of the solvent (convert gram to kg): next, plug in the values: Let’s have an example… HOW DO WE KNOW THE NUMBER OF REACTING Calculate the molality of a 0.90% (m/m) NaCl solution. UNITS? 0.90% NaCl is called normal saline solution (NSS). 1 mol NaCl = 58.45 g what’s the formula? Note: - NSS contains 9.0 g NaCl per 1.0 kg solution; o If you have 1kg solution it means that you have 0.991 kg water if you are using 0.9 grams of NSS - that means it has 0.991 kg of water - Simply put, N = M x h Let’s have an example… Calculate the mass of H2SO4 that is present in 750 mL of a 6.0 N solution? what’s the formula? what is the equivalent mass of H2SO4? (98.0 g/mol)/2 = 49.0 g/eq plug in the values: NORMALITY, N - number of equivalents of the solute per kilogram of the solvent Answer exercises 31 – 40 of chapter 10 for your practice. WHAT DO WE MEAN BY COLLIGATIVE PROPERTIES? - These are properties of a solution that depend on the amount of the solute, and not on the nature of the solute. - It is applicable to dilute solutions only. - Extending - Dependent on amount of solute. - Colligative properties are extensive properties VAPOR PRESSURE - It is a force that allows the liquid to vaporize or go to liquid face - the pressure exerted over the surface of a liquid as molecules - IMF is attributed to pressure, cohesive force that moulds all these together - escape into the gas phase HAPPENS DURING BOILING? - Vapor pressure is equal to atmospheris pressure WHY IS COOKING OIL USEFUL FOR FRYING? - It has a high boiling poi t or vapor pressure COLLIGATIVE PROPERTIES OF NON-ELECTROLYTE SOLUTIONS Vapor Pressure Lowering - a non-volatile solute lowers the vapor pressure of a WHAT IS EQUIVALENTS? solvent - equivalents is mass divided by the equivalent mass - the presence of a nonvolatile solute inhibits the (EM) escape of solvent molecules from the liquid and so WHAT IS EQUIVALENT MASS (EM)? lowers the vapor pressure of the solvent. - equivalent mass is molar mass divided by the reacting species, h - where for acids/bases, h is the number of H+/OH- Raoult’s Law Kf – molal freezing point depression constant the solution has a - the partial pressure of a solvent over a solution, P1, I s lower vp than pure solvent, hence will freeze at lower temp given by the vapor pressure of the pure solvent, P1o, - aqueous solutions will always have a freezing point times the mole fraction of the solvent in the solution, below 0°C (Tf = 0 - ΔTf) X1 P1 = X1P1o - In terms of mole fraction of the solute X2 since X1 + X2 = 1 ΔP = X2P1o where P1 = P1o - ΔP - ΔP is the vapor pressure lowering, how much the vapor pressure decreased Let’s have an example… Calculate the vapor pressure (in mm Hg) of the solvent in a solution at 20oC where 20.0 g of glucose (MM=180 g/mol) was dissolved in 250 g of water (18.0 g/mol). The vapor pressure of pure water at this temp is 17.25 mm Hg. - try to memorize Kb and Kf of water so you won’t need - what are given? to refer to the table every time o mass of solute = 20.0 g Let’s have an example… o mass of solvent = 250 g Ethylene glycol is used both as antifreeze and as coolant in o vp of pure water at 20°C = 17.25 mm Hg radiators. Calculate the bp and fp of a solution containing 45.8g - what do we need? of ethylene glycol (MW=62.0 g/mol) in 3202 g of water. o mole fraction of glucose, X2 - what do we need first? - what is asked? o the concentration of the solution in molality, o P1 m - what is the formula? - the vapor pressure lowering is: - the vapor pressure of the solvent over the solution is: - what is the formula for bp elevation? o ΔTb = Kbm o ΔTb = (0.51)(0.231) = 0.12 - what is the bp of the solution? Another approach in terms of mole fraction of solvent, X1 o Tb = 100 + ΔTb = 100.12 °C Let’s have an example… Ethylene glycol is used both as antifreeze and as coolant in radiators. Calculate the bp and fp of a solution containing 45.8 g of ethylene glycol (MW=62.0 g/mol) in 3202 g of water. m = 0.231 COLLIGATIVE PROPERTIES OF NON-ELECTROLYTE - what is the formula for bp elevation? SOLUTIONS o ΔTb = Kbm Boiling Point Elevation o ΔTb = (0.51)(0.231) = 0.12 - a non-volatile solute elevates the boiling point of a - what is the bp of the solution? solvent a solution has a higher boiling point (Tb) than o Tb = 100 + ΔTb = 100.12 °C the pure solvent (To) - what is the formula for fp depression? ΔTb = Kbm o ΔTf = Kfm where: Tb = To + ΔTb o ΔTf = (1.86)(0.231) = 0.43 - ΔTb is the boiling point elevation, how much the bp - what is the fp of the solution? increased o Tf = 0 - ΔTf = -0.43 °C Kb – molal boiling point elevation constant Osmotic Pressure m - molality if the solvent is water, Tb = 100 + ΔTb - since a solution has a lower vp than the pure solvent, it will boil at a higher temp Freezing Point Depression - the freezing point of a solution is lower than that of the pure solvent a solution has a lower freezing point (Tf) than the pure solvent (To) ΔTf = Kfm where: ΔTf = To– Tf ΔTf is the freezing point depression, how much the fp decreased  - osmosis: selective passage of solvent molecules - Dialysis through a porous membrane from a dilute solution to a more concentrated one - semipermeable membrane: allows the passage of solvent molecules but blocks solute molecules - osmotic pressure: the pressure required to stop osmosis - the minimum pressure required to stop osmosis - water tends to move to the more concentrated solution until equilibrium is reached Let’s have an example… - Reverse Osmosis Calculate the osmotic pressure against a solution prepared by dissolving 135 g of glucose (180 g/mol) to make 250 mL of solution at 25oC. - what do we need first? o the concentration of the solution in molarity, M - what is the formula? What if the solute is a non-volatile electrolyte? - We multiply the colligative property with the van’t Hoff factor, i - To compute for osmotic pressure: APPLICATIONS OF OSMOTIC PRESSURE Some terms: hypertonic solution – more concentrated, contains more solute hypotonic solution – more dilute, contains less solute - It is assumed that electrolytes completely dissolve into isotonic - same solute concentrations its component ions, contributing to the colligative hemolysis – the cell becomes swollen when placed property in a hypotonic solution How do we know the van’t Hoff factor? crenation – the cell becomes shriveled when placed - It is assumed that electrolytes completely dissolve into in a hypertonic solution its component ions, contributing to the colligative property - *non-electrolytes do not dissociate, hence i=1 - IV fluids must be isotonic with blood Let’s have an example… Calculate the bp and fp of a solution prepared by dissolving PERIODIC PROPERTIES 13.8 g of NaCl in 550 g of water. (MM=58.45 g/mol) - Those properties which are determined by the - what do we need first? electronic configuration of elements or which depend on electronic configuration of elements - The periodic law states: "When elements are arranged in order of increasing atomic number, there is a periodic repetition of their chemical and physical - what is the van’t Hoff factor? properties - These trends can be predicted merely by examing the periodic table and can be explained and understood by analyzing the electron configurations of the - what do we need next? elements. ATOMIC RADIUS - The atomic radius is the distance from the nucleus to the outermost electron. - Trend Across a Period: o Atomic radius decreases from left to right across a period. o This happens because the increasing number of protons pulls electrons closer to the nucleus. Let’s have an example… - Trend Down a Group: Calculate the osmotic pressure of a solution prepared by o Atomic radius increases down a group. dissolving 23.0 g of MgCl2 (95.21 g/mol) in water to make - This is due to the addition of electron shell 625mL of the solution at 20.0oC? - Atomic size is influenced by nuclear charge and - what do we need first? electron shielding - what is the van’t Hoff factor? - what do we need next? PERIODIC PROPERTIES AND TRENDS IONIZATION ENERGY THE PERIODIC TABLE - Ionization energy is the energy required to remove an - The periodic table arranges elements by increasing electron from an atom in the gas phase. atomic number. - Trend Across a Period: - Periodic trends refer to the patterns observed across o Ionization energy increases across a period. periods (rows) and groups (columns) in the table. o This is because the atoms are more tightly - Trends include atomic radius, ionization energy, held as atomic size decreases. electron affinity, and electronegativity. - Trend Down a Group: o Ionization energy decreases down a group. o Outer electrons are farther from the nucleus and experience more shielding. - Higher ionization energy indicates stronger attraction between the nucleus and electrons ELECTRON AFFINITY - Electron affinity is the energy change that occurs when an electron is added to a neutral atom. - Trend Across a Period: o Electron affinity becomes more negative across a period (atoms release more energy when gaining an electron).  - Trend Down a Group: o Electron affinity becomes less negative (atoms tend to gain electrons less readily down a group) ELECTRONEGATIVITY - Electronegativity is the ability of an atom to attract electrons in a chemical bond. - Trend Across a Period: o Electronegativity increases across a period. o This is because atoms have more protons and a greater ability to attract electrons. - Trend Down a Group: o Electronegativity decreases down a group. o Larger atomic size means electrons are farther from the nucleus and harder to attract. APPLICATIONS OF PERIODIC TRENDS - Periodic trends help us understand and predict the properties of elements as well reactivity of elements - These trends are based on the atomic structure and the arrangement of electrons in an atom. - Understanding the formation of chemical bonds - The periodic table is a powerful tool for understanding chemistry and explaining the behavior of elements in different chemical reactions