Engineering Chemistry Preparatory Year 2023/2024 PDF

Engineering Chemistry Preparatory year Faculty of Engineering , Zagazig University 2023 / 2024 Chapter (2) States of Matter  Chapter contents: Types of Phase Transitions Phase diagram C...

Engineering Chemistry Preparatory year Faculty of Engineering , Zagazig University 2023 / 2024 Chapter (2) States of Matter  Chapter contents: Types of Phase Transitions Phase diagram Cooling and Heating Curves Characteristics of Liquid Characteristics of Gases Characteristics of Solid 10/5/2023 State (Phase): a physical form of a substance, such as the solid, liquid, and gaseous states over a range of pressure or temperature. 10/5/2023  Change of State and Phase Transitions Change of State (Phase Transition): The change of a substance from one state to another. Solid Sublimation Gas Liquid Gas Vaporization Heating Heating Condensation Condensation (deposition) Cooling Cooling CO2 CO2 Liquid Solid 1 atm, -78 0C Freezing Solid Gas Cooling Dry Ice – CO2 sublimation: Solid Melting (Fusing) carbon dioxide CO2 at -78 0C is converted directly into CO2 gas without passing through the liquid state. Heating 10/5/2023  Types of Phase Transitions 1. Melting (fusion): The change of a solid to the liquid state (melting is also referred to as fusion) Solid Liquid 𝐻2 𝑂(𝑠)(𝐼𝑐𝑒) → 𝐻2 𝑂(𝑙)(𝑤𝑎𝑡𝑒𝑟) 2. Freezing: The change of a liquid to the solid state 2 2 3. Vaporization: The change of a liquid to the vapor 𝐿𝑖𝑞𝑢𝑖𝑑 → 𝐺𝑎𝑠 𝐻2 𝑂(𝐿)(𝑤𝑎𝑡𝑒𝑟) → 𝐻2 𝑂(𝑔)(𝑠𝑡𝑒𝑎𝑚) 4. Condensation: The change of a gas to either liquid or solid state 2 2 5. Sublimation: The change of a solid directly to the vapor without passing through the liquid state 2 2 6. Deposition: The change of a gas directly to a solid without passing through the liquid state H O (steam) H O (ice) Gas Solid 2 2 10/5/2023  How Do We Predict The States or Phases of Matter? The Answer is: Using The Phase Diagram Phase Diagram is a graphical way (between pressure and temperature) to summarize the conditions under which the different states of a substance are stable. ‫غازية عند درجات الحرارة و الضغط‬, ‫ سائلة‬, ‫هو طريقة بيانية لتوضيح حالة المادة من حيث كونها صلبة‬ Triple point (Point A): It is a point on a phase diagram representing the temperature and pressure at which the three phases of a substance coexist. 2.0 Critical Point (Point C): The temperature and pressure above which the liquid state of a substance no longer exists regardless of the pressure -20 ‫هي درجة الحرارة التي فوقها مباشرة تختفي الحالة السائلة للمادة‬ 10/5/2023.‫مهما زادت قيمة الضغط‬ Phase Diagram of Water  Phase Diagram of Carbon Dioxide The phase diagram of CO2 indicates the following: Normal Boiling Point: Not Applicable because there is no liquid state at atmospheric pressure Normal Sublimation Point: P = 1 atm and T = -78 0C Triple Point: P = 5.1 atm and T = - 57 0C Critical Point: P = 73 atm and T = 31 0C Phase diagram Carbon dioxide 10/5/2023  Applications of Phase Diagram Many Gases Cannot Be Liquefied at Room Temperature Nitrogen, for example, has a critical temperature of -147oC Oxygen has a critical temperature of -118.6 oC 10/5/2023  Applications of Phase Diagram Separation of Air Gases by Fractional Distillation Cooling air to −200 °C, the liquified air is fed into a vessel that is slightly warmer at the bottom (−185 °C) than it is at the top (−190 °C). Oxygen liquifies at −183 °C, so it flows out of the flask through a tube in the bottom. Nitrogen turns back into a gas, however, because N2 boiling point is −196 °C. 10/5/2023  Applications of Phase Diagram Application of Vaporization and Sublimation: Food conservation & Freeze-Drying  Drying removes the moisture from the food so that bacteria and yeasts cannot grow and spoil the food.  Freeze-drying of a food is accomplished by placing the frozen food in a vacuum (below 0.00603 atm) so that the ice in it sublimes. Food vacuum freeze dryers are often used 10/5/2023 in vegetables, meat, aquatic products. Heat of Phase Transition: State change involves the addition or removal of energy as heat..‫كمية الطاقة الحرارية الﻼزمة لتغيير المادة من حالة الي حالة اخري‬ ‫و يتم حسابها للمول الواحد من المادة و هي المحتوي الحراري الﻼزم لتحول مول من المادة الي حالة اخري‬ ‫المحتوي الحراري الﻼزم لتحويل مول صلب الي سائل‬ ∆H fusion Heat of fusion (or enthalpy of fusion) and is denoted ∆H fusion the heat needed for the melting of a solid. 𝐻2 𝑂(𝑠) ⇌ 𝐻2 𝑂(𝑙); ∆𝐻𝑓𝑢𝑠 = 6.01 𝑘𝐽/𝑚𝑜𝑙 Heat of vaporization and is denoted ∆H vap the heat needed for the vaporization of a liquid. 𝐻2 𝑂(𝑙 ) ⇌ 𝐻2 𝑂(𝑔); ∆𝐻𝑣𝑎𝑝 = 40.7 𝑘𝐽/𝑚𝑜𝑙 10/5/2023  Cooling and Heating Curves 10/5/2023  Temperature 𝟏𝟐𝟎 Q5 𝐋→𝑮 𝟏𝟎𝟎 Q3 Q 4= n x ∆H vap 𝐒→𝑳 𝟎 Q 2 = n x ∆H fusion −𝟒𝟎 Time Q 1=m s ∆t Q added at constant rate Q total = Q 1 + Q 2 + Q 3 + Q 4 +Q 5 10/5/2023  Q 1 = m s ∆t Q 2 = n x ∆H fusion Where: m = mass (g), s = specific heat (j/g.OC), ∆t=t final – t initial n = moles (mass /Mm) , ∆H fusion = heat of fusion (j/mole) q = q heat ice + q fusion + q heat water + q vaporization + q heat steam q = (m×S×ΔT) Ice + n× ∆𝐻 + (m×S×ΔT) water + n× ∆𝐻 + (m×S×ΔT) steam 10/5/2023 Example: Calculate the heat absorbed when 46 g of ice at –10ºC is converted to liquid water at 85ºC. The specific heat of ice is 2.03 J/(g ºC), the molar heat of fusion of ice is 6010 J/mol, and the specific heat of water is 4.18 J/(g ºC). Solution: Three-step process of 1- warming ice from –10ºC to 0ºC, 2- melting the ice, and 3- heating liquid water to 85ºC: q = q heat ice + q fusion + q heat water q = (m × S × ΔT) heat ice + n × ∆H fusion + (m × S × ΔT) heat water q = (46 × 2.03 × 10) heat ice + × 6010+ (46 × 4.18 × 85) heat water q = 934 J + (1.53 × 104 J) + (1.63 × 104 J) q = 3.25 × 104 J 10/5/2023  Properties of Liquids Vapor Pressure (a) before the evaporation begins and (b) at equilibrium, when no further change. In (b) the number of molecules leaving the liquid is equal to the number of molecules returning to the liquid. The difference in the mercury levels (h) gives the equilibrium vapor pressure of the liquid at the specified temperature. Vapor pressure: The partial pressure of the vapor over the liquid, at equilibrium at a given temperature. ‫ هو الضغط الجزئي للبخار فوق سطح السائل عند اﻻتزان ودرجة حرارة معينة‬: ‫ضغط البخار‬ 2 2 Note: Vapour Pressure is a function of temperature 10/5/2023  Measurement of The Liquid Vapor Pressure Using the Barometer Barometer  Plot of vapour pressure  Plot of the logarithm of versus temperature vapour pressure versus 1/T  The relation follows a  The relation is 10/5/2023 CURVE LINEAR Clausius Equation: ‫ كلما زادت درجة حرارة السائل زاد ضغط بخار‬: ‫العﻼقة بين ضغط البخار و درجة حرارة السائل‬ ‫السائل و بالتالي زادت كمية السائل التي تتحول الي بخار‬ 𝒗𝒂𝒑 𝒗𝒂𝒑 𝑷𝒗𝒂𝒑 = vapor pressure, 𝒗𝒂𝒑 = heat of vaporization (J/mole) T = OK, R = 8.31 J/ OK mole If the temperature changes from T1 to T2 the vapor pressure changes from P1 to P2: ∆𝑯𝒗𝒂𝒑 ∆𝑯𝒗𝒂𝒑 𝒍𝒏 𝑷𝟏 = 𝑹𝑻𝟏 +𝑪 𝒍𝒏 𝑷𝟐 = 𝑹𝑻𝟐 +𝑪 If you subtract the first equation from the second, you get −∆𝑯𝒗𝒂𝒑 ∆𝑯𝒗𝒂𝒑 𝒍𝒏 𝑷𝟐 − 𝒍𝒏 𝑷𝟏 = +𝑪+ −𝑪 𝑹𝑻𝟐 𝑹𝑻𝟏 𝑷𝟐 ∆𝑯𝒗𝒂𝒑 𝟏 𝟏 10/5/2023 𝑷𝟏 𝑹 𝑻𝟏 𝑻𝟐  Vapor pressure and boiling temperature  When the vapor pressure equals the atmospheric pressure, the liquid begins to boil.  The temperature of a boiling liquid remains constant until all the liquid has vaporized.  The temperature at which the vapor pressure of a liquid equals the pressure exerted on the liquid is called the boiling point of the liquid. Pressure cooking is the process of cooking food under high pressure steam,  Water cannot be heated above its boiling point because the added energy is used to evaporate the water.  Liquid under a pressure higher than atmospheric pressure can be heated above its normal boiling point.  Since water boils at a higher temperature under such conditions, the food cooks more rapidly. 10/5/2023 Example Estimate the vapor pressure of water at 85 oC. Note that the normal boiling point of water is 100 oC and that its heat of vaporization is 40.7 kJ/mol. Solution The normal boiling point is at vapor pressure equals 760 mmHg. P1 = 760 mmHg and T1 = 373 K (100 oC). P2 = ?? and T2 = 358 K (85 oC). Hvap, = 40.7 x 103 J/mol. 𝑷𝟐 ∆𝑯𝒗𝒂𝒑 𝟏 𝟏 𝑷𝟐 𝟒𝟎𝟕𝟎𝟎 𝟏 𝟏 𝑷𝟏 𝑹 𝑻𝟏 𝑻𝟐 𝟕𝟔𝟎 𝟖.𝟑𝟏 𝟑𝟕𝟑 𝟑𝟓𝟖  Therefore, P2 = 0.577 x760 mmHg = 438 mmHg 10/5/2023 Properties of Gases T n V P (mole) m3 = 103 dm3 (L) = 106 cm3 = 109 mm3 1 atm = 1.013 x 105 Pa = 100 kPa o K= oC+273 = 1bar = 760 mmHg (torr) STP (standard temperature and pressure) is defined as a temperature of 273 K (0 °C) and an absolute pressure of 100 kPa By holding any two of these physical properties constant, it is possible to show a simple relationship between the other properties ‫يتم وصف عينة من الغاز بتحديد عدد الموﻻت – الحجم – درجة الحرارة – والضغط‬ 10/5/2023  ‫‪Empirical Gas Laws‬‬ ‫‪1- Boyle's Law‬‬ ‫‪V-P‬‬ ‫حجم كمية من الغاز يتناسب عكسيا مع ضغطة عند ثبوت درجة حرارته‬ ‫‪At constant (T):‬‬ ‫‪2-Charles's Law V-T‬‬ ‫حجم كمية من الغاز يتناسب طرديا مع درجة الحرارة المطلقة عند ثبوت الضغط‬ ‫‪At constant (P):‬‬ ‫‪3- Avogadro's Law‬‬ ‫‪V-n‬‬ ‫يشغل المول من اي غاز حجم ثابت عند ثبوت درجة الحرارة و الضغط‬ ‫‪At constant P and T:‬‬ ‫‪10/5/2023‬‬  The ideal gas law The ideal gas law could be deduced from ‫استنتاج القانون العام‬ Boyle’s law, Charles’s law and Avogadro's Law, 1 𝑛𝑇 𝑛𝑇 𝑉∝ 𝑉∝𝑇 𝑉∝ 𝑉=𝑅 𝑃 𝑃 𝑃 Calculating R: , Combined gas law equation (When Properties Change) Another useful equation, derived from the ideal gas equation, can be used to calculate changes in the properties of a gas. 10/5/2023  Example: Light bulbs “burn out” because their tungsten filament evaporates, weakening the thin wire until it breaks. Argon gas is added inside the bulbs to reduce the rate of evaporation. (Argon is chosen because, as a noble gas, it will not react with the components of the bulb, and because it is easy to obtain in significant quantities). What is the pressure in atmospheres of 3.4×10-3 moles of argon gas in a 75 mL light bulb at 20 °C? Solution: P=? n = 3.4 × 10-3 mol V = 75 mL = 0.075 L T = 20 °C + 273 = 293 K We use PV = nRT 𝒏𝑹𝑻 𝟑.𝟒×𝟏𝟎 𝟑 ×𝟖.𝟑𝟏×𝟐𝟗𝟑 𝑷= 𝑽 = 𝟎.𝟎𝟕𝟓 = 𝟏𝟏𝟎. 𝟒 𝒌𝒑𝒂 𝟏𝟏𝟎. 𝟒 𝐏= = 𝟏. 𝟏 𝐚𝐭𝐦 𝟏𝟎𝟎 10/5/2023  Example: Neon gas in luminous tubes radiates red light—the original “neon light.” The standard gas containers used to fill the tubes have a volume of 1.0 L and store neon gas at a pressure of 101 kPa at 22 °C. A typical luminous neon tube contains enough neon gas to exert a pressure of 1.3 kPa at 19 °C. If all the gas from a standard container is allowed to expand until it exerts a pressure of 1.3 kPa at 19 °C, what will its final volume be? If that volume of Neon is added to luminous tubes that have an average volume of 500 mL, what is the approximate number of tubes can be filled? Solution: V1 = 1 L P1 = 101 kPa T1 = 22 °C + 273 = 295 K V2 = ? P2 = 1.3 kPa T2 = 19 °C + 273 = 292 K 𝟏 𝟏 𝟐 𝟐 n1=n2 So 𝟏 𝟏 𝟐 𝟐 = 77 L 𝟏 𝟏 𝟐 𝟐 𝟐 𝟏 𝟐 𝟏𝟎𝟑 No of tubes = 10/5/2023 𝟓𝟎𝟎  Gas Mixtures O2 H2 O2+H2 O2 V=1L n1 n2 n = n1+ n2 H2 T = cons. P1 P2 P = P1+ P2 Mixture P = P1+P2 Consider two flasks each with 1L volume, -One filled with O2 (g) , n1 moles and P1 at a given T. -The other with H2 has n2 moles and P2 at the same T. Suppose all of the O2 and H2 are put in 1L flask. After the gases are mixed in one flask, each gas occupies a volume of one liter, and had the same T. 10/5/2023  Dalton’s law of partial pressures Dalton’s law, the sum of the partial pressure of all the gases in a mixture is equal to the total pressure of the mixture. ‫مجموع الضغوط الجزئية للغازات المختلفة في الخليط يساوي الضغط الكلي للخليط‬ 𝑡𝑜𝑡𝑎𝑙 1 2 3 The ideal gas law for mixture 𝑡𝑜𝑡𝑎𝑙 𝑡𝑜𝑡𝑎𝑙 The ideal gas law for gas (1) 1 1 The ideal gas law for gas (2) 2 2 The ideal gas law for gas (3) 3 3 10/5/2023  Mole fractions 1 1 𝑡𝑜𝑡𝑎𝑙 𝑡𝑜𝑡𝑎𝑙 For gas mixture V and T constants So Pα n The composition of a gas mixture is often described in terms of the mole fractions of component gases. The mole fraction of a component gas is simply the fraction of moles of that component in the total moles of gas mixture. Because the pressure of a gas is proportional to moles, for fixed volume and temperature; the mole fraction also equals the partial pressure divided by total pressure. 10/5/2023 Example: If an 85 mL light bulb contains 0.140 grams of argon and 0.011 grams of nitrogen at 20 °C, what is the total pressure of the mixture of gases? (Ar = 40 , N=14) Solution: V = 85 mL =0.085L T = 20 °C + 273 = 293 K n total = nAr + nN2 𝑡𝑜𝑡𝑎𝑙 𝑡𝑜𝑡𝑎𝑙 nAr = mol Ar = 0.14 g / 40 = 0.0035 mol nN2 = mol N2 = 0.011 g /28 g = 0.00039 mol nAr + nN20 0.0035 + 10/5/2023  Properties of Solid Reasons of differences : 1- Type of attraction forces 2- Arrangement of atoms 10/5/2023  Types of Solids Crystalline solid Solids that have highly regular arrangement of their components (its atoms, molecules, or ions), i.e. occupy specific positions such as ice or sodium chloride. ‫الذرات داخل المادة مرصوصة في توزيع منتظم الشكل‬ Amorphous Solids: Solids that have disordered structure such as glass. ‫الذرات داخل المادة مرصوصة بشكل عشوائي‬ Unit cell: The repeating structural unit of a crystalline solid, it is the smallest collection of atoms that displays the features of this structure. ‫اصغر وحدة يتضح فيها شكل و توزيع الذرات داخل المادة‬ 10/5/2023  Cubic unit cell a :‫مهم جدا معرفة‬ 2r ‫عدد الذرات داخل المكعب‬-1 ‫ العﻼقة بين نصف قطر‬-2 ‫الذرة و ضلع المكعب‬ SC: Number of atoms BCC: Number of atoms FCC: Number of atoms in the unit cell = 1 in the unit cell = 2 in the unit cell = 4 10/5/2023  How to Calculate the Number of Atoms in a Unit Cell  The atom at the corner of a cube shares by 1/8 atom in the unit cell Simple Cube (SC) 𝟏 𝟏 𝑵 =Number of atoms at the corners × = 𝟖 × = 𝟏 𝟖 𝟖  For body centred unit cell (BCC) 𝟏 𝑵 = 𝑵𝒄𝒐𝒓𝒏𝒆𝒓𝒔 × + 𝑵𝒄𝒆𝒏𝒕𝒓𝒆 × 𝟏 𝟖 𝟏 =𝟖× +𝟏×𝟏=𝟏+𝟏=𝟐 𝟖  The atom at the canter of each face shares by 1/2 atom in the unit cell Face-centred Cube (FCC) 𝟏 𝟏 𝟏 𝟏 𝑵 = 𝑵𝒄𝒐𝒓𝒏𝒆𝒓𝒔 × + 𝑵𝑭𝒂𝒄𝒆𝒔 × = 𝟖 × + 𝟔 × = 𝟏 + 𝟑 = 𝟒 10/5/2023 𝟖 𝟐 𝟖 𝟐 Relationship between the atomic radius r and the edge length a of a cubic cell 10/5/2023 Packing efficiency: is the percentage of space that is occupied in a given arrangement Volume of the atoms 𝑽 𝑵𝒐 𝒐𝒇 𝒂𝒕𝒐𝒎𝒔 ×𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒂𝒕𝒐𝒎 Packing efficiency = Volume of unit cell = 𝑽𝒂𝒕𝒐𝒎𝒔 = 𝑽𝒄𝒆𝒍𝒍 𝒄𝒆𝒍𝒍 Example : Packing efficiency for Face-centered cubic (fcc), 𝟒 𝟒 𝑽𝒂𝒕𝒐𝒎𝒔 𝟒× × 𝝅 × 𝒓𝟑 𝟒 × × 𝝅 × 𝒓𝟑 Packing efficiency = = 𝟑 = 𝟑 = 𝟎. 𝟕𝟒 𝑽𝒄𝒆𝒍𝒍 𝒂 𝟑 𝟑 𝟖×𝒓 Simple cubic (scc), The packing efficiency 52%. Body-centered cubic (bcc), The packing efficiency 68%. Face-centered cubic (fcc), The packing efficiency 74%. 10/5/2023 Example: Silver crystallizes in a face-centered cubic lattice with all atoms at the lattice points. The length of an edge of the unit cell was determined by x-ray diffraction to be 408.6 pm. The density of silver is 10.50 g/cm3. Calculate the mass of a silver atom. Then, using the known value of the atomic mass, calculate Avogadro’s number. Solution 𝑽𝒄𝒆𝒍𝒍 = (𝟒. 𝟎𝟖𝟔 × 𝟏𝟎 𝟏𝟎 𝒎)𝟑 = 𝟔. 𝟖𝟐𝟐 × 𝟏𝟎 𝟐𝟗 𝒎𝟑 The density, 𝝆, of silver in grams per cubic meter is 𝟑 𝒈 𝟏𝒄𝒎 𝝆 = 𝟏𝟎. 𝟓 𝟑 × 𝟐 = 𝟏. 𝟎𝟓 × 𝟏𝟎𝟕 𝒈/𝒎𝟑 𝒄𝒎 𝟏𝟎 𝒎 the mass of a unit cell 𝒎 = 𝑽𝒄𝒆𝒍𝒍 × 𝝆 = 𝟔. 𝟖𝟐𝟐 × 𝟏𝟎 𝟐𝟗 × 𝟏. 𝟎𝟓 × 𝟏𝟎𝟕 = 𝟕. 𝟏𝟔𝟑 × 𝟏𝟎 𝟐𝟐 𝒈 There are four atoms in a face-centered unit cell, the mass of a silver atom is 𝟏 𝑴𝒂𝒔𝒔 𝒐𝒇 𝟏 𝑨𝒈 𝒂𝒕𝒐𝒎 = × 𝟕. 𝟏𝟔𝟑 × 𝟏𝟎 𝟐𝟐 = 𝟏. 𝟕𝟗𝟏 × 𝟏𝟎 𝟐𝟐 𝒈 𝟒 𝟏𝟎𝟕. 𝟖𝟕 𝑵𝑨 = 𝟐𝟐 = 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝟏. 𝟕𝟗𝟏 × 𝟏𝟎 10/5/2023  Example: Silver crystallizes in a cubic closest packed structure (face-centered cubic cell). The radius of a silver atom is 144 pm. Calculate the density of a solid silver, atomic mass = 107.87 Solution Since r = 144 pm for a silver atom The volume of the unit cell is: 𝟑 𝟑 𝟕 𝟑 To convert pm3 to cm3, multiply by 10-10 𝟕 𝟏𝟎 𝟑 𝟐𝟑 𝟑 The total mass of atoms in the unit cell = number of atoms in the unit cell × the mass of one atom 𝒂𝒕𝒐𝒎𝒊𝒄 𝒎𝒂𝒔𝒔 𝟏𝟎𝟕.𝟖𝟕 𝟐𝟑 =4 𝑵𝒂 𝟔.𝟎𝟐×𝟏𝟎𝟐𝟑 𝒎𝒂𝒔𝒔 𝟕𝟏.𝟔𝟕𝟒×𝟏𝟎 𝟐𝟑 𝟑 Density = 10/5/2023 𝒗𝒐𝒍𝒖𝒎𝒆 𝟔.𝟕𝟓𝟕×𝟏𝟎 𝟐𝟑 10/5/2023