General Chemistry 2 Midterm Learning Package PDF

**GENERAL CHEMISTRY 2** **SECOND SEMESTER -- MIDTERMS A. Y. 2024-2025** **LESSON 1. INTERMOLECULAR FORCES:** **Liquids and Solids** **Kinetic molecular Theory-** this explains how gases behave. The word kinetic comes from the word **"kinetic"** comes form the Greek word ***kinein***, which means...

**GENERAL CHEMISTRY 2** **SECOND SEMESTER -- MIDTERMS A. Y. 2024-2025** **LESSON 1. INTERMOLECULAR FORCES:** **Liquids and Solids** **Kinetic molecular Theory-** this explains how gases behave. The word kinetic comes from the word **"kinetic"** comes form the Greek word ***kinein***, which means "to move". *Gas* is made up of molecules that move randomly and are widely separated from one another. Because of their wide separation, the forces that attract the molecule together are negligible in ideal gas molecules. In addition, when the absolute temperature is increased, the average kinetic energy of the gas molecules also increase. These may result in an increase in the number of collisions, with energy transferred between collisions. The **Kinetic Molecular Theory** allows us to explain the existence of the three phases of matter: solid, liquid, and gas. In addition, it helps explain the physical characteristics of each phase and how phases change from one to another. The Kinetic Molecular Theory is essential for the explanations of gas pressure, compressibility, diffusion, and mixing. Our explanations for reaction rates and equilibrium also rest on the concepts of the Kinetic Molecular Theory. **Table 1.1 Particle description of each phase of Matter** **Criterion** **Solid** **Liquid** **Gas** ----------------------- --------------------------- ------------------------------------------------------------------- ------------------------------------------ Molecular arrangement Closely packed together Slightly far from one another Freely moving; very far from one another Volume/ Shape Definite volume and shape Definite volume; assumes shape of an occupied part of a container Indefinite volume and shape Density High High Low Motion of Molecules Vibration in place Random Fast random fig 1.2. Phase Transition of a varying temperature **Intermolecular Forces-** pertains to forces that hold individual particles such as atoms, molecules, or ions together. The strength in the intermolecular forces of attraction is dependent on the arrangement of the particles, proximity of the particles relative to one another, and the nature of the interacting particles. The intermolecular forces of attraction influence the resulting properties of solids, liquids, and gases. In solids for example, the intermolecular forces of attraction directly affect its melting point and heat or fusion. *Generally, the stronger the intermolecular forces pf attraction, the greater is the required amount of energy to overcome these forces*. For example, compounds with high boiling points are those that possess strong intermolecular forces of attraction. On the other hand, are those responsible for interactions within a molecule, such as covalent and ionic bonds. One example would be different amino acids, present in a certain protein. The functional group in each amino acid interact with one another, enabling the protein to have a folded structure. *Generally, intramolecular forces of attraction are stronger than intermolecular attractions.* **Types of Intermolecular Forces of Attraction** 1. **London dispersion force-** Named after the German born physicist Fristz London, this is the weakest among the intermolecular forces. This dispersion is caused by polarization or distortion of electron cloud brought about by the presence of a highly charged particle. In this case, the electron cloud of one atom is attracted to the positively charged nucleus of another atom. London dispersion force happens when the molecule with a temporary dipole exerts a weak attractive force on another molecule, and because it is caused by polarization, the strength of this force depends on the number of electrons present. 2. **Dipole-dipole Interaction-** It is an intermolecular force of attraction that occurs between partially positive and partially negative ends. This interaction is observed in polar covalent molecules such as amino acids, wherein the electrons are shared both by oxygen and carbon atoms. The dipole-dipole interaction is effective over a short distance only, as it is still weak, containing only 1% of the strength of ionic bonds. Moreover, an increase in temperature diminishes the strength of a dipole-dipole interaction, which may explain the observed volatility of certain polar covalent compounds such as sulfur dioxide. 3. **Ion -dipole Interaction-** This arises form the interaction between an ion and a polar molecule. If the molecule is *anion*, it will be attracted to the partially positive end of the polar molecule; however, if the molecule is *cation,* it will be attracted to a partially negative end of the polar molecule. The energy of the strength of the ion-dipole interaction is about 15 kilojoules perm mole for a 500 parts per million distance. The ion-dipole interaction is responsible for the formation of cations in a solution. Cations are formed from ionic solids such as calcium dichloride when these are dissolved in water to yield aqueous solution. Elements belonging to group I and II of the periodic table such as calcium, lose electron easily, and thus form cations. 4. **Hydrogen Bond-** Is a special kind of dipole-dipole interaction, which is formed when hydrogen bonds with fluorine, oxygen, or nitrogen. Typically, the distance needed to form a hydrogen bond is about 2X10 \^ -10m. Generally, hydrogen bonds are still weaker than ionic or covalent bonds, but they are the strongest intermolecular force of attraction (when the hydrogen bond is present between two atoms of two different molecules). 1. **Viscosity-** This property refers to the measure of a liquid's resistance to flow. The viscosity of a liquid is strongly dependent on the strength of the intermolecular forces in play. Highly viscous liquids are those exhibiting the strongest intermolecular forces, example is the glycerol which has a viscosity of 1.7 pascal-seconds at 293K (20°C). 2. **Surface Tension --** This is referred to as the amount of resistance needed to increase the surface area of liquids. Surface free energy is the work required to increase the surface area of a liquid by a unit area. The surface tension of water is 78.8 millinewtons per meter and has a surface energy of 72.8 millijoules per square meter. Surface tension decreases with decreasing polarity of the liquid as seen in table 1.2. **Liquid** **Surface Tension (mN/m)** ------------ ---------------------------- Water **72.8** Chloroform **27.1** Ethanol **22** Benzene **28.9** 3. **Vapor Pressure-** *Vaporization* is the transformation of a substance form a liquid to gas. It is an endothermic process as it requires absorption of energy to break the intermolecular forces of attraction into a liquid. Consider a liquid inside an enclosed container. Initially, especially when heated, liquid evaporate by escaping form the liquid's surface and are transformed into gas molecules above the liquid **(evaporation).** After some time, the number of gaseous molecules increases. When this happens, some of these molecules will have the tendency to return to their liquid phase. This process is called **condensation.** The **vapor pressure of a liquid** is defined as the pressure of the vapor present at equilibrium. Liquids with high vapor pressure have weak intermolecular forces of attraction. Liquids exhibit **volatility**, which is the ability to readily evaporate from an open vessel. Liquids that have strong intermolecular forces of attraction have relatively lower vapor pressure so they require a large amount of energy. Distillation- is an effective method in separating the components of a mixture or in the removal of impurities in solvents. Distillation is performed by heating the mixture at a known temperature. 4. **Boiling Point-** a liquid boils when its vapor pressure is equal to the prevailing atmospheric pressure. The normal boiling point of a liquid is the temperature at which its pressure is equal to the standard pressure of 101 325 Pa. The normal boiling point of any liquid is directly influenced by the existing intermolecular forces of attraction. So if a stronger intermolecular force of attraction is in place, the liquid will have a relatively higher boiling point. **Importance of Water** Water is an extremely important molecule. It actively participates in many chemical reactions, particularly those involving aqueous processes. For example, photosynthesis is based on the oxidation of water to yield oxygen as by-product. Photosynthesis is highly useful as this process convert the sun's energy into chemical energy that organism need. Water also facilitates the transport of molecules, nutrients, an products of reactions within and between cells. Furthermore, the behavior of properties of biomolecules may be influenced by the presence of water. **Physical and Chemical Properties of Water** \*\*Pure water is an odorless, tasteless and colorless substance \*\*Water molecule can form hydrogen bond with many other molecule. \*\*A water molecule is also polar due to the lone pair of electron in the oxygen atom. \*\*Water is considered as the universal solvent because it has great dissolving power. \*\*Each water molecule is surrounded by four other neighboring water molecules through hydrogen bonding. -An **acid** is a molecule that contains a hydrogen proton and a **base** is a molecule that contains hydroxide according to Arrhenius definition. Water may act as both acid and base, enabling it to undergo autoprotolysis to give hydronium and OH ions. Buffer- is the ability to resist extreme pH changes. The buffering capacity of water us so important form an environmental perspective. Any accidental spill of substances with extreme pH such as strong bases and acids will not immediately induce significant pH changes in natural bodies of water. Water also exhibits high heat capacity which prevents oceans or river systems from generating large changes in their temperature conditions. **Solids-** solids have a definite shape and volume due to the compact arrangement of their particles. They can be broadly classified as crystalline and amorphous. **Crystalline Solids** **Amorphous Solids** --------------------------------------------- -------------------------------------------------- Regular Have as random and disordered arrangement Highly ordered arrangement Gradually soften when heated Examples are amethyst, fluorite, and pyrite Melt at a wide range of temperature Examples are glass, charcoal, plastic containers **Space lattice-**refers to the three dimensional pattern formed by the point s representing the location of these particles; it defines the basic structure of the crystal. A **unit cell** is the smallest unit of the lattice. Each unit is stacked together repeatedly to resemble the whole.  The number of particles present in a unit cell is systematically determined to quantify the **coordination number** or the number of the nearest neighboring particles, as well as to identify the physical properties such as density and conductivity. In determining the number of particles of each type in a unit cell, particle that are outside the cell, such as those occupying vertices and faces, are shared by neighboring particles. Here are some guidelines in determining the number of particles in a unit cell: 1. A particle I the body of a cell belongs entirely to the cell and is counted as one. 2. A particle located on the face is shared by two unit cells and is counted as one-half of a unit cell. 3. A particle occupying the edge is counted as one- fourth of a unit cell and is counted as one. 4. A particle situated at the vertex is counted as one -eight of a unit cell, as is it shared by unit cell. **Polymorphus-** Substances that crystallize in several arrangement **Polytypes-** are crystals that are the same in two dimensions but different in the third. It can be either hexagonal or cubic close-packed. **Types of Crystalline Solid** **Crystalline solids** are classified according to the nature of bonding or interactions present among their particles. Solids are classified as *ionic, molecular, network, or metallic*. **Ionic Solids-** are hard, brittle, and poor conductor of heat and electricity. However, when in their molten state, ionic solid are strong electrolytes because the ions are in their free mobile forms. **Molecular Solids**-the intermolecular forces of attraction that exist between particles in a molecular solid are relatively weak, resulting in low melting points, ranging from 1to 673K (from -272 to 400°C.) Moreover. Molecular solids are soft and are also poor conductor of heat and electricity. Examples of molecular solids include CH₄, P₄, O₂, CO₂ and fullerenes. **Network or covalent Solids --** these are large or "giant" molecules in which atoms are covalently bonded in highly cross- linked, rigid network. Polymeric materials, diamond, and quartz are classified as network covalent solids. Because of the strong covalent bond between particles, network solids are very hard and have high melting points, ranging from 1473 to 4273 K (1200- 4000°C). Network solids are also poor thermal and electrical conductors at normal temperature, because their electrons are localized in covalent bonds and thus, they are not free to move. Diamond and graphite are allotropes of carbon. Allotropes are different physical forms of the same element in the same physical state. **Metallic Solids --** are bound by metallic bonding, a type of bond in which the metal atoms "swims" in a sea of electrons. Thus , intermolecular attraction exists between the nucleus of the metal atom and the negatively charged electrons. Metals are excellent thermal and electrical conductors. Metals also possess malleability, ductility and luster and hardness. The melting of metal cover a wide range form 234 to 3673K (form -39 to 3400°C). Cupper, nickel, and chromium are a few examples of metals. **Phase changes --** involve the transition form one phase to another. For example, a solid changes to a liquid during melting. A phase change entails either absorption or release of heat, represented by a change in the latent heat of fusion. other phase changes includes Vaporization, condensation, sublimation. **Melting** -- solid may turn into liquid. Take ice as an example. The energy that flows into the ice causes the random motion of water molecules. This random motion disrupts the intermolecular forces of attraction between the water molecules. Thus, water molecules break lose from the solid lattice of the ice. The temperature at which a solid changes into liquid is called melting point. It can also be defined as the temperature at which the melting rate of the solid is the same as its freezing rate at a given pressure. This implies that solid and liquid phases of the solid are in equilibrium. **Vaporization --** Continuing the heating process will eventually cause the liquid to boil and eventually vaporize, or turn to gas. The temperature further increases and the liquid boils at 373 K (100 °C). this temperature remains constant until the liquid has been completely vaporized. The melting and boiling points of the substances are affected by their vapor pressure, with the vapor pressure of ice increases rapidly with an increase in temperature., as compared to the vapor pressure of liquid water. Thus, at melting point, liquids and solids have the same vapor pressures. **Sublimation --** is the process at which solid is vaporized at atmospheric pressure without transforming to a liquid. Solids with high vapor pressure easily sublime. Household products such as mothballs and deodorizers undergo sublimation. Occurrences of such a phase change is marked by the distinct odor these product gives off. Note that all phases changes are physical changes, which means that no chemical bonds are broken during these process. **LESSON 2. PHYSICAL PROPERTIES OF SOLUTIONS** **SOLUTIONS** Solutions are classified according to the following: A. **PHYSICAL STATE** **Physical properties of solutions** Technically, any mixture that is homogeneous on a microscopic level is a solution. According to that definition, air is a solution, as are alloys such as bronze, brass and steel. However, when most people use the word solution, they are usually referring to a homogeneous liquid mixture. A homogeneous liquid mixture has one primary component which is a liquid as well as one or more additional ingredients that are present in smaller amounts. 1. Solutions are classified according to physical state after solutes are dissolved in solvents. 2. Solid solution such as jewelry, coins, bronze, brass, etc. 3. Liquid solution such as soft drinks, coffee, juices, mineral water, etc. 4. Gaseous solution such as air **Components of a solution** In a solution, the substance that used to dissolve another substance is called the **solvent** while the substance being dissolved is the **solute**. For a solution composed of two substances in the same state suck as liquid-liquid solution, it is difficult to establish which substance is the solute and which is the solvent. In such case, the component present in relatively large quantity is the solvent. The solute may be gas, liquid or solid. ***Types of Solutions*** \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ State of State of State of Solvent Solute Solution Examples \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ Gas Gas Gas Air, natural gas Liquid Liquid Liquid Alcoholic beverages, Antifreeze solution; Liquid solid liquid seawater, sugar solution Liquid gas liquid carbonated water (soda) Ammonia solution; Solid solid solid metal alloys: brass, bronze,... Solid gas solid hydrogen in platinum B. **CONCENTRATION:** Classified according to their general concentration or ratio of solute and solvent. 1. Dilute -- contains relatively small amount of solute in a given solvent. 2. Concentrated -- contains large amount of solute in a given solvent. C. **SATURATION** ***Solubility and Equilibrium in Solution*** The ***solubility*** of a substance is the amount of solute (in grams) that is dissolved in a given quantity of solvent to give a *saturated* solution at a particular temperature. 1. A ***saturated solution*** is *one that contains the maximum quantity of dissolved solute that is normally possible at a given temperature*, where a state of dynamic equilibrium exists between dissolution and crystallization. Solubility is temperature dependent. For example, the solubility of KNO~3~ is about 30. g per 100 g water at 20 ^o^ C and 63 g per 100 g water at 40 ^o^ C. When a solution that is almost saturated at a higher temperature is cooled to a lower temperature where the solubility is lower, the excess solute would normally precipitate out to give a saturated solution at the lower temperature. However, if the solution is cooled down too rapidly and the solute fails to seed, precipitates will not form and the resulting solution would contain more dissolved solute than it would be in a normal saturated solution. The resulting solution is said to be [*supersaturated*.] 2. A ***supersaturated solution*** is one that contains more dissolved solute than it normally possible under normal condition. The solution is unstable and crystallization will occur readily by seeding (introducing particles that provide nuclei for precipitation). The solution contains the maximum amount of solute capable of being dissolved. 3. An ***unsaturated solution*** is a solution that contains less than the maximum amount of solute that is capable of being dissolved. If more solute is added and it does not dissolve, then the original solution is saturated. If the added solute dissolves, then the original solution was unsaturated. The amount of dissolved solute is less than the saturation point of the solvent. ***Solution process*** When a solute is dissolved in a solvent, the molecule or ions that make up the solute separate from one another as they become surrounded by molecules of the solvent. A molecular substance when dissolved in water will mingle with the water molecules and occupy up between the water molecules. For ionic substances to dissolve in water, the cations and anions of the solute separate from one another and the ions will be surrounded by water molecules. A molecule or ion surrounded by solvent molecules is said to be solvated: if the solvent is water it is hydrated. The formation of solutions of sucrose and water, and of NaCl and water followed very similar processes: forces of attraction between solvent molecules are broken; forces of attraction between solute particles are also broken, while forces of attraction are formed between solute and solvent particles are formed. These processes can happen with any pair of solute and solvent and solute-solvent in the solution process, and the energy involved in their breaking or formation. The major determining factor in solution formation is the relative strength of intermolecular forces between and among solute and solvent particles. The extent to which one substance is able to dissolve in another depends on the relative magnitudes of interactions between: solute-solute, solvent-solvent, and solute-solvent in the solution process, and the energy involved in their breaking or formation. **Solution Composition** Concentration is a general term that expresses the quantity of solute contained in a given amount of solution. Various ways of expressing concentration are in use; the choice is usually a matter of convenience in a particular application. **Different Ways of Expressing the Concentration of Solutions** 1. **Parts-per concentration** - Defined as a part of solute per 100 parts of solution - Parts of a solution may be expressed either in mass or volume or both - The solution consists of both solute and solvent In the consumer and industrial world, the most common method of expressing the concentration is based on the quantity of solute in a fixed quantity of solution. The "quantities" referred to here can be expressed in weight, in volume, or both (i.e., the weight of solute in a given volume of solution.) In order to distinguish among these possibilities, the abbreviations (w/w), (v/v) and (w/v) are used. Percent means parts per 100; we can also use parts per thousand (ppt) for expressing concentrations in grams of solute per kilogram of solution. For more dilute solutions, parts per million (ppm) and parts per billion (109; ppb) are used. These terms are widely employed to express the amounts of trace pollutants in the environment. a. **Mass/Mass Percent Concentration, % by mass or % (w/w)--** refers to the amount of solute (in grams) in 100 grams solution **% (w/w) = [Mass of solute\_] x 100%** **Note: mass of solution = mass of solute + mass of solvent** In most applied fields of Chemistry, (w/w) measure is often used, and is commonly expressed as weight-percent concentration, or simply \"percent concentration\". For example, a solution made by dissolving 10 g of salt with 200 g of water contains \"1 part of salt per 20 g of water\". **Sample Problem:** Describe how you would prepare 30 g of a 20 percent (w/w) solution of KCl in water. **Given:** 30 g = mass of solution, 20 % m/m **Required**: mass of salt and mass of water **Solution:**  **NOTE**: A 20 % by mass means there are 20 g KCl in 30g of a solution. \% (w/w) = [Mass of solute\_] x 100% Mass of solution Derived formula using cross multiplication: mass of KCl = [20 % x 30 g kCl] = **[6 g KCl]** Thus, the remainder of the solution (30 -- 6 = 24) g consists of water. So, you would dissolve 6.0 g of KCl in **24 g of water**. b. **Volume/Volume Percent Concentration, % by volume or % (v/v) -** Most practical way to express the method of preparation of a solution of liquids in liquids. The amount of both solute and solvent are expressed in volume units. **% (v/v) = [Volume of Solute\_] x 100%** **Volume of Solution** **Sample Problem:** A certain gin is labeled 80 proof. What is the percent by volume? How many ml alcohol is contained in250 ml of gin? **Given: 80 =** proof, 250 ml = volume of solution **Required:** % v/v, volume of solute **Solution:** **% (v/v) = [proof\_]** **2** \% (v/v) = [80 proof\_] = 40% 2 Volume Percent % (v/v) = [Volume of Solute\_] x 100% Volume of Solution 40 % (v/v) = [Volume of Solute\_] x 100% 250 ml Volume of solute = [250 ml x 40 %] = **[100 ml]** 100% c. **Mass/Volume Concentration, % by mass-volume % (w/v) --** this concentration refers to the number of grams of solute in 100.0 ml of solution. **% by mass-volume = [mass of solute] x 100%** **Volume of solution** **Sample Problem:** How much CaO is needed to prepare 500 ml of a 7.5% mass --volume solution? **Given:** Concentration of solution = 7.5% Volume of solution = 500 ml **Required:** mass of CaO (mass of solute) **Solution:** \% by mass-volume = [mass of solute] x 100%, manipulate the formula to derive the formula for mass of solute Volume of solution Mass of solute = % by mass -- volume x volume of solution/ 100 % = 7.5 x 500 /100 = **[37.5 g CaO]** 2. **Molarity (*M*) = mass of solute** **Liter of solution** This is the method most used by chemists to express concentration, and it is the one most important for you to master. Molar concentration (molarity) is the number of moles of solute per liter of solution. The important point to remember is that the volume of the solution is different from the volume of the solvent; the latter quantity can be found from the molarity only if the densities of both the solution and of the pure solvent are known. **Sample Example:** How would you make 120 mL of a 0.10 M solution of potassium hydroxide in water? KOH has a molecular weight of 56 g/mol. **Given:** 120 ml = volume of solution, 0.10 M= Concentration of solution, 56 g/mol = MW of KOH **Required**: mass of KOH **Solution:** Molarity (*M*) = mass of solute 0.10 (*M*) = [mass of KOH] 56 g/mol x 120 ml Mass of KOH = 0.10 mol/L (56 g/mol) (120 ml) NOTE: The amount of KOH required is (0.120 L) × (0.10 mol L--1) = 0.012 mol. THUS: Mass of KOH = 0.10 mol/L (56 g/mol) (0.012 L) **[Mass of KOH = 0.0672 g]** 3. ***Molality* (*m*) = [No. of *moles* of solute]** A 1-molal solution contains one mole of solute per 1 kg of solvent. Molality is a hybrid concentration unit, retaining the convenience of mole measure for the solute, but expressing it in relation to a temperature-independent mass rather than a volume. Molality, like mole fraction, is used in applications dealing with certain physical properties of solutions; we will see some of these in the next lesson. **Sample Problem:** Calculate the molality of a 60-% (w/w) solution of ethanol in water. **Solution:**  From the above problems, we know that one liter of this solution contains 11.6 mol of ethanol in (893.7 -- 536.2) = 357.5 g of water. The molarity of ethanol in the solution is therefore (11.6 mol) / (0.3575 kg) = **[32.4 mol/Kg.]** 4. **Normality (*N*) = No. [of *equivalent* of solute]** **Liter of solution** Normality is a now-obsolete concentration measure based on the number of equivalents per liter of solution. Although the latter term is now also officially obsolete, it still finds some use in clinical- and environmental chemistry and in electrochemistry. The equivalent weight of an acid is its molecular weight divided by the number of titratable hydrogens it carries. Thus, for sulfuric acid H~2~SO~4~, one mole has a mass of 98 g, but because both hydrogens can be neutralized by strong base, its equivalent weight is 98/2 = 49 g. A solution of 49 g of H~2~SO~4~ per liter of water is 0.5 molar, but also \"1 normal\" (1N = 1 eq/L). Such a solution is \"equivalent\" to a 1M solution of HCl in the sense that each can be neutralized by 1 mol of strong base. 5. **Mole Fraction** - of a solution is the ration of the number of moles of the component to the total number of moles of all the components(solute and solvent) of the solution. **Mole fraction, *X~i~* = [Mole of a component]\_\_\_\_\_\_\_\_\_\_** **Total moles of components in solution** a. **Mole fraction of solute (X ~solute~) = [mole of solute] or [n solute]** **mole of solute + mole of solvent n solution** b. **Mole fraction of solvent (X ~solvent~) = [mole of solvent] or [n solvent]** **mole of solute + mole of solvent n solution** **Sample Problem:** What is the mole fraction of the solute in a 40% by mass ethanol (C~2~H~6~O) solution in water? **Given:** 40% by mass ethanol solution, MM ethanol = 46 g/mole, MM water = 18 g/mole **Required**: mole fraction of solute: X ~ethanol~ **Solute -- ethanol** **Solvent - water** **Solution:** Step1: In converting concentration units based on the mass or moles of a solute and solvent or mass percentage, it is useful to assume a certain total mass of solution. Assume there is exactly 100 grams of solution. Because the solution is 40% ethanol (C~2~H~6~O), it contains 40 grams of ethanol and 60 grams of water. Step 2: Change the masses of the components ethanol and water to number of moles. mole ethanol = [40 g] = 0.87 mol 46 g/mol mole water = [60 g] = 3.33 mol 18 g/mol Step 3: Substitute the values obtained in the formula and solve for the mole fraction of the solute ethanol, and the solvent water. \(x) mole fraction ethanol = [mole ethanol] mole ethanol + mole water X~ethanol~ = [0.87 mole] = **[0.21]** 0.87 mol + 3.33 mol The mole fraction of water can be solved using the formula: X~water~ = [mole water] = [3.33mol water] = **[0.79]** mole ethanol + mole water 0.87 mol + 3.33 mol +-----------------------------------+-----------------------------------+ | **FORMATIVE ASSESSMENT 2B** | Solve each of the following | | | problems on concentrations of | | | solutions. (30 points) | +===================================+===================================+ | 1. Find the concentration of | | | solution in terms of weight | | | percent if 20 grams of common | | | salt is dissolved in 50 grams | | | of water. (5 points) | | | | | | 2. What is the mole fraction of | | | the salt solute in a 1.00m | | | aqueous solution? (5 points) | | | | | | 3. A solution was prepared by | | | dissolving 66.0 g of urea | | | (NH~2~)~2~CO in 950 g of | | | water had a density of 1.018 | | | g mL. Express the | | | concentration of urea in: | | | | | | a\. weight-percent c. molarity | | | | | | b\. mole fraction d. molality | | | | | | **( 20 points)** | | +-----------------------------------+-----------------------------------+ ***Factors Affecting Solubility*** ***Structural Effects*** There is a correlation between molecular structure and solubility. A solution will form if both solute and solvent have similar molecular structures, polarity and types of interactions -- "*like dissolves like*". For example, nonpolar substances like grease and oil are soluble in nonpolar organic solvent such as hexane (C~6~H~14~) and gasoline (C~8~H~18~). Ionic and polar compounds such as salts, mineral acids, and sugar are soluble in polar solvents such as water and methanol. ***Effect of Temperature:*** The solubility of solids generally increases as temperature increases, which imply that dissolving solid substances is generally an endothermic process (*H*~soln~ \> 0). The Le Chatelier\'s principle states that an endothermic process favors high temperature. The solubility of some ionic compounds that contain SO~3~^2-^, SO~4~^2-^, SeO~4~^2-^, AsO~4~^3-^ and PO~4~^3-^, are found to decrease as the temperature increases. These compounds have negative enthalpy of solution (*H*~soln~ \< 0). For gases dissolving in liquid solvents, their solubility decreases as the temperature increases; the dissolution of gaseous solutes is an exothermic process - when a gas dissolves heat is given off. Gases are more soluble at low temperature than at high temperature. This temperature effect on the solubility of gaseous substances has important environmental implications as a result of the widespread uses of water from lakes and rivers in the industrial cooling system. The discharged water from the cooling system back into the lakes or rivers is significantly warmer than the natural ambient temperature, which causes ***thermal pollution***. The rise in ambient temperature of lake or river water lowers the concentration of dissolved oxygen, which can be detrimental to the survival of aquatic lives. ***Effect of Pressure:*** Pressure has little effect on the solubility of solids or liquids, but significantly increases the solubility of gases. According to ***Henry\'s Law***, *the solubility of gases is directly proportional to the partial pressure of the gas above the solution*: ***C* = *kP***, where C = concentration of gas in solution, P = gas pressure above the solution, and *k* = Henry's constant characteristic of the gas and solvent. The solubility of gases is normally expressed in mL of gas at STP per liter of solvent. Henry's law is obeyed most accurately for very dilute solutions of gases that do not dissociate or react with the solvent. For example, Henry's law is obeyed by oxygen and nitrogen gases dissolving in water, but not by carbon dioxide or ammonia, because the latter gases react with water as follows: CO~2(g)~ + H~2~O~(*l*)~ H~2~CO~3(aq)~ H~3~O^+^~(aq)~ + HCO~3~^-^~(aq)~ NH~3(g)~ + H~2~O(*l*) NH~4~^+^~(aq)~ + OH^-^~(aq)~ **LESSON 2.2: COLLIGATIVE PROPERTIES** Colligative properties are properties of a solution that depend only on the number and not on the identity of the solute particles. Thus, these depend on the collective effect of the concentration of solute particles present in an ideal solution. Because of their direct relationship to the number of solute particles, the colligative properties are very useful for characterizing the nature of a solute after it is dissolved in a solvent and for determining the molar masses of substances. The Different Colligative properties of Solution \(1) vapor pressure lowering; \(2) boiling point elevation; and \(3) freezing point depression. ** Effect of solute concentration on the colligative properties of solutions.** The concentration or amount of non-volatile solute (i.e., a solute that does not have a vapour pressure of its own) in the solution has an effect on the colligative properties of solutions. The effect would depend on the ratio of the number of particles of solute and solvent in the solution and not on the identity of the solute. However, it is necessary to take into account whether the solute is an electrolyte or a nonelectrolyte. Effects of electrolyte and nonelectrolyte on colligative properties solutions. **1) Vapor Pressure Lowering** Vapor pressure is a direct measure of escaping tendency of molecules. A pure liquid (solvent) in a closed container will establish equilibrium with its vapor. And when that equilibrium is reached, the pressure exerted by the vapor is called the vapor pressure. A substance that has no measurable vapor pressure is non-volatile, while one that exhibits a vapor pressure is volatile. When a liquid evaporates easily, it will have a large number of its molecules in the gas phase resulting to a high vapor pressure. The addition of a nonvolatile solute resulted to a lowering of the vapor pressure of the solvent. The lowering of the vapor pressure depends on the number of solute particles that have been dissolved. The chemical nature of the solute is not considered because vapor pressure is merely a physical property of the solvent and does not undergo a chemical reaction with the solvent and does not itself escape into the gas phase. It is important to note that the reduction in the vapor pressure of a solution of this example is directly proportional to the fraction of the volatile molecules in the liquid, which is the mole fraction of the solvent. This reduced vapor pressure can be determined using *Raoult's Law* (1886). Where: P=vapor pressure of solution X= mole fraction of solvent P^o^ = vapor pressure of pure solvent **2) Boiling Point Elevation** The addition of a nonvolatile solute lowers the vapor pressure of the solution; consequently the temperature must be raised to restore the vapor pressure of the solution to the value conforming to the pure solvent. Specifically, the temperature at which the vapor pressure is 1 atm will be higher than the normal boiling point by an amount known as the boiling point elevation. Figure 3 below shows the phase diagram of a solution and the effect that the lowered vapor pressure has on the boiling point of the solution compared to the solvent. In this case the sucrose solution has a higher boiling point than the pure solvent. Since the vapor of the solution is lower, more heat must be supplied to the solution to bring its vapor pressure up to the pressure of the external atmosphere. The *boiling* *point elevation* is the difference in temperature between the boiling point of the pure solvent and that of the solution. **3) Freezing Point Elevation** The freezing point of a substance is the temperature at which the solid and liquid forms can coexist indefinitely, at equilibrium. Under these conditions molecules pass between the 2 phases at equal rates because their escaping tendencies from the two phases are identical. The lowering of the vapor pressure in a solution causes the boiling point of the solution to be higher than pure solvent (darker line). As a result, the freezing point of a solvent decreases when any solute is dissolved into it. **LESSON 3: THERMOCHEMISTRY** ***The Flow of Energy*** **3.1: ENERGY CHANGES: Exothermic and Endothermic Processes** In thermochemical calculations the direction of the heat flow is given from the point of view of the system. A process that ***absorbs heat*** from the surroundings is called ***[an endothermic process.]*** A process that ***loses heat*** to the surroundings is called ***[an exothermic process.]*** **EXOTHERMIC REACTIONS** **An exothermic reaction is one which releases heat energy to the surroundings**. **The temperature of the surroundings increases**. Exothermic reactions give out energy. There is a temperature rise and ∆H is **[negative.]** **Examples:** - Burning reactions including the combustion of fuels. - Detonation of explosives. - Reaction of acids with metals. - Charcoal burning - A candle burning. - A firework exploding **ENDOTHERMIC REACTIONS** **An endothermic reaction is on which takes in heat energy from the surroundings. The temperature of the surroundings decreases** **Endothermic reactions take in energy. There is a temperature drop and ∆H is [positive.]** **Examples:** Melting ice cubes. **3.2: FIRST LAW OF THERMODYNAMICS** **THERMODYNAMICS -- is the study of energy and its transformations.** **FIRST LAW OF THERMODYNAMICS -- states that the energy of the universe is constant. This is also referred to as the law of conservation of energy, which states that energy is neither created nor destroyed.** **3.3: ENTHALPY** **-a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance** **you can measure an enthalpy CHANGE** **written as the symbol ∆H, "delta H"** **Enthalpy change (∆H) = Enthalpy of products - Enthalpy of reactants**  **∆H = - ive ∆H = + ive** **EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed** **[ENTHALPY CHANGE OF FORMATION]** **The enthalpy change when ONE MOLE of a compound is formed from its elements. ∆~f~H or ∆H~f~** **If you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements** **Enthalpy of formation tends to be an exothermic process** **Step 1 Energy is released as reactants** **are formed from their elements.** **Step 2 Energy is released as products** **are formed from their elements.** **∆H~r~ =** **- Step 1 + Step 2 or** **Step 2 - Step 1** **In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined enthalpy change, its value is subtracteD** **Standard Enthalpies of Formation Table** **NOTE: *Abbreviations in compounds*** ***(s)* -- solid *(l)* -- liquid *(g)* -- gas *(aq)* -- aqueous solution** ***Sample calculation*** **1.Calculate the standard enthalpy change for the following reaction, given that the standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286, +33 and -173 kJ mol^-1^ respectively; the value for oxygen is ZERO as it is an element.** 2H~2~O (l) + 4NO~2~ (g) + O~2~ (g) ---------\> 4HNO~3~ (l) **\[ 4 x ∆~f~H of HNO~3~ \] minus \[ (2 x ∆~f~H of H~2~O) + (4 x ∆~f~H of NO~2~) + (1 x ∆~f~H of O~2~) \]** **2. NaOH(s) + HCl(g) \-\-\--\> NaCl(s) + H~2~O(g)** **\[(-411.0) + (-241.8)\] -- \[(-426.7) + (-92.3)\] = -133.8 kJ** **3. 2 CO (g) + O~2~ (g) \-\--\> 2 CO~2~ (g) ** \[2(-393.5)\] -- \[2(-110.5)\] = [-566 kJ/mol] **[ENTHALPY CHANGE OF COMBUSTION]** **Step 1 Energy is released as reactants** **undergo combustion.** **Step 2 Energy is released as products** **undergo combustion.** **∆H~r~ =** **Step 1 -** Step **2** **Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the defined change of Enthalpy of Combustion, its value is subtracted.** ***Sample calculation*** **Calculate the standard enthalpy of formation of methane; the standard enthalpies of combustion of carbon, hydrogen and methane are -394, -286 and -74.8 kJ mol^-1^.** ***[REACTANTS]* *[PRODUCTS]*** **\[(1 x ∆~c~H of C) + (2 x ∆~c~H of H~2~)\] minus \[1 x ∆~c~H of CH~4~\]** **ANSWER = [- 891 kJ mol^-1^ or -891 kJ/mol]** **[BOND DISSOCIATION ENTHALPY -] During a reaction, all the bonds of reacting species are broken and the individual atoms join up again but in the form of products. The overall energy change will depend on the difference between the energy required to break the bonds and that released as bonds are made.** Energy released making bonds \> energy used to break bonds - EXOTHERMIC Energy used to break bonds \> energy released making bonds - ENDOTHERMIC [ ] Step 1: Identify bonds broken. \... Step 2: Find total energy to break bonds. \... Step 3: Identify bonds formed. \... Step 4: Find total energy released to form new bonds. \... Step 5: Add up energy for bonds broken and formed **Average Bond Enthalpies Table**  EXAMPLE: Calculate the enthalpy change for the hydrogenation of ethene. hesscalc1g CH~2~=CH~2~ + H~2~ \-\-\-\-\-\-\-- CH~3~ CH~3~ ∆ H~=~ 1 x C=C bond @ 614 = 614 kJ ∆ H~=~ 1 x C-C bond @ 348 = 348 kJ 4 x C-H bonds @ 413 = 1652 kJ 6 x C-H bonds @ 413 = 2478 kJ 1 x H-H bond @ 436 = 436 kJ **Total energy to break bonds of products= 2826 kJ** **Total energy to break bonds of reactants= 2702 KJ** ***ANSWER***: ∆ **H =** ∆ **H~R~ --** ∆ **H~P~** **= (2702 -- 2826) = [-- 124 kJ/mol]** 1.Calculate **∆ H for the following reaction using average bond enthalpies, assuming all compounds are in their gaseous states. CH~4~ + 2O~2~ \-\-\-\-\-\-\-\-\-\-- O=C=O + 2 (H-O-H)** **Bonds of reactants Bonds of products** **4x C-H \@413 =1652 2 x C=O \@799 =1 598** **2 x O2 \@495 =990 4 x O-H \@463 =1 852** **Total energy to break bonds of reactants =2 642 kJ Total energy to break bonds of products =3 450 KJ** ** ∆ H = ∆ H~R~ -- ∆ H~P~** **= (2 642 -- 3 450) = [-- 808 kJ/mol]** **3.4: CALORIMETRY** A scientific way of determining the quantity of heat by measuring the changes in the temperature of a system. The system is isolated from the surroundings, such a that whatever response that the system generates as it is heated will be due to the system only. A **CALORIMETER** is used for such purpose; it is used to measure the amount of heat involved in a chemical reaction or other processes. A substance possesses the property of heat capacity, C, which is mathematically represented as **C =** [\$\\frac{heat\\ absorbed\\ (\\ i\\ n\\ joules)}{change\\ in\\ temperature\\ (\\mathrm{\\Delta}T,\\ in\\ Kelvin)}\$]{.math.inline} The heat capacity of a substance is dependent on both its mass and its chemical composition. Different substances manifest different heat capacities. For example, an iron bar may absorb heat more strongly than a piece of cloth. Also, a glass water will have a heat capacity higher than a tablespoon of water, because the mass of water in the glass is higher than that of the tablespoon. **If heat capacity is expressed as per gram of the substance, it is termed as *specific heat capacity.*** **Molar heat capacity, on the other hand, pertains to the heat capacity expressed per mole of the substance**.Thus, specific heat capacity has a unit of joule per gram-kelvin (J/g-K). liquid water has specific heat capacity of 4.18J/g- [℃]{.math.inline}, and solid water has a specific heat capacity of 2.03 J/g - [℃]{.math.inline}. **3.5: HESS'S LAW** The net chemical equation for a given reaction is generally obtained by getting the summation of all the individual reactions from one or more series of steps to generate the product. The net heat of reaction can also be obtained in this manner. This principle is based on **Hess's law**. Hess's law is useful in indirectly determining the heat formation of a reaction from two or more known thermochemical equations. Hess's law is a convenient way of deriving the change in enthalpy involved in a given reaction, as there are chemical reactions that are very difficult to carry out experimentally. Two fundamental guidelines must be followed in the application of Hess's law. The change in enthalpy indicates the direction of the flow of heat. Therefore, should the thermochemical equation describing a particular reaction be written in the reversed direction, then -[*ΔH*]{.math.inline} must be reversed. Because [*ΔH*]{.math.inline} is an extensive property, its magnitude is dependent on the amount of both reactants and products. Thus, the thermochemical reaction should be balanced and the appropriate coefficients reflecting the number of moles of reactants and products should be determined. Should the thermochemical equation be multiplied by an integer, then the change in enthalpy of the reaction must also be multiplied by the same integer. EXAMPLE: Calculate [*ΔH* ]{.math.inline}for the reaction **2C~(s)~ +** **H~2(g)~** [**→**]{.math.inline} **C~2~H~2(g~**~)~ using the following data: Solution: 1. Balance the given reaction. 2. Check the reactants and the products in the overall reaction. 3. Inspect the individual reactions, and determine which of these bears the reactants and which bears the products. Reaction numbers 1 and 2 bear the reactants, with C~(s)~ and H~2(g)~ also in the product side of each respective reaction. Thus, there is no need to reverse the two reactions. However, should these reactions contain the product of the overall reaction in the reactant side, then that reaction has to be reversed, such that C~2~H~2(g)~ will be in the product side. The [*ΔH*]{.math.inline} of the reaction must also change its sign to indicate the reverse direction of heat flow. The overall reaction indicates that there are two moles of C~(s)~ necessary to yield the product C~2~H~2(g)~. therefore, the second reaction must be multiplied by two, including its [*ΔH*]{.math.inline}. 2C~(s)~ + 2O~2(g)~ [→]{.math.inline} 2CO~2(g)~ [*ΔH*]{.math.inline}= -788kJ Get the summation of the individual reactions using one of two ways: 1. By canceling out those substances(with the same coefficients)found on the opposite sides of the equation, 2. By adding the components that are present in different amounts written on the same side of the equation and then subtracting them from the amount of the same component written on the opposite side. For example, H~2~O~(l)~ is cancelled out because in the first reaction, it is on the product side, whereas in the third reaction, it is found on the reactant side. Each has the same coefficient, which is 1. Thus, Then get the summation of [*ΔH*]{.math.inline}, which is equal to 226kJ. Thus, the [*ΔH*]{.math.inline} for the synthesis of 1 mole of ethyne(or acetylene, a gas that is used for soldering purposes) is 226 kJ Note that you cannot cancel out components that exist in different phases.