Thermodynamics I Overview of Energy

Thermodynamics I Overview of Energy • 1st Law of Thermodynamics: Energy is conserved. • Thermodynamic viewpoint: • Energy (heat) can be transferred from one object to another until a thermal equilibrium is established. • A change in temperature (∆T) measures the transfer of energy from one object to another. • Heat flows between objects because of difference in temperature. • Classifications of Energy: • Potential energy (PE: = mgh )is stored energy. • Kinetic energy (KE: ½ mv2) is the energy of movement (motion). • Chemists view energy as the ability to do work or transfer heat. Thermodynamic Terminology • System • Object understudy such as a chemical reaction or a phase change • Surroundings • Everything outside of the system • Universe • Combination of the system & the surroundings Reference credit: catalog.flatworldknowledge.com Temperature is not Heat • Temperature is defined as the measure of hotness or coldness of an object (degrees Celsius, or degrees Fahrenheit, or kelvins). • Whereas heat is a form of energy that flows between two objects of different temperatures. • Energy flows from HOT to COLD • Directionality of heat transfer (2nd Law of Thermodynamics) • Temperature is result of the movement of matter. • It is related to the average translational kinetic energy per molecule in a substance. • Cold: slow movement of particles • Hot: fast movement of particle Reference credit: ces.fau.edu Heat Transfer: Directionality Endothermic • Energy enters the system from the surroundings. • DH is positive (+) • Energy of system goes up. • Temperature of the system rises; therefore, the Temperature of the surroundings decreases. Exothermic • Energy leaves the system & goes into the surroundings. • DH is negative (-) • Energy of system goes down. • Temperature of the system decreases; therefore, the Temperature of the surroundings rises. Reference crediteasynotecards.com Common Energy Units • Joule (J): • 1 joule of energy is the amount of energy needed to move a 1 kg mass at a speed of 1 m/s. • 4.184 J = 1 calorie • Calorie (cal): • A calorie is the amount of energy needed to raise the temperature of one gram of water 1 °C. • 1 kcal = energy needed to raise 1000 g of water 1.0 0C • A food Calorie (Cal) is 1000 calories. • 1 kcal is equivalent to ONE food Calorie . Thermal Properties of Matter Thermal properties are EXTENSIVE physical properties of matter. • Heat Capacity: • The amount of energy (heat as q) required to raise an object’s Temperature by 1 ˚C. • Specific Heat: • The amount of energy (heat as q) required to raise ONE gram of matter’s temperature by 1 oC. • Molar Heat Capacity: • The amount of energy (heat as q) required to raise ONE mole of matter’s temperature by 1 oC. Calorimetry: Measuring Heat Transfer • Technique: • Calorimetry is a technique used to quantitatively determine the amount of heat exchanged between objects. • The equipment used to carry out this measurement is referred to as a calorimeter. • Calorimeters can be simple pieces ((a) coffee cup calorimeter) or very sophisticated instruments ((b) bomb calorimeter). (a) (b) Reference credit: OpenStax2017 Calorimetry: Measuring Heat Transfer • Calorimetric Calculations • The heat (q) “lost” or “gained” is related to the: • sample mass (grams) • change in Temperature (DT) • Where DT is Tinitial - Tfinal • specific heat (c) of substance • Specific heat (c) units • cal/goC or J/goC • The equation: • q = c x mass x DT (a) heat, q, is transferred from the hot metal, M, to the cool water, W, until (b) both are at the same temperature. Reference credit: OpenStax2017 Problem#1: A 348 g piece of metal absorbs 6.64 kJ of heat. The initial temperature of the metal was 22.4 oC and after the absorbing 6.64 kJ its final temperature was 43.6 oC. • Determine the identify of the metal from its specific heat. metal Specific Heat (c) (J/goC) Aluminum (Al) 0.897 Copper (Cu) 0.385 Iron (Fe) 0.449 Lead (Pb) 0.130 Gold (Au) 0.712 Problem#2: • Into a beaker containing 390.0 mL of water (T=22.6oC) a heated a 248 g piece of copper is added. • The final temperature of the water was 39.9oC. • Given: • Assume that all the heat transferred from the heated copper metal was absorbed by the water. • The density of water is 0.9875 g/mL • Specific heat for copper is 0.385 J/goC & for water 4.184 J/goC Determine the initial temperature of the copper metal. Why A Steam Burns Hurt! Scenario: • After stopping at their favorite hamburger drive-thru place, Grandma places her “HOT” cup of coffee on dashboard in order for her to put her change from the purchase back into her purse. Just as Grandma was closing her purse, her grandson speeds off from the drive-thru window resulting in the “hot” cup of coffee spilling all over Grandma. • Because of the severe burns resulting from the spilled coffee, Grandma and her grandson sue the hamburger establishment for damages incurred from the accident. Problem#3: How much energy did Grandma’s hand absorbed when the hot coffee spilled on her? Information: • The coffee cup contains 100.0 grams of hot water (100.0 oC) • Grandma’s initial hand (skin) temperature was 37.0 oC. • D Hvap = 540 cal/g & Specific heat (c) of water is 1.00 cal/goC Problem#4: • In lab, 100.0 mL of 0.200 M NaCl (aq) solution is mixed with 75.0 mL 0.200 M Pb(NO3)2 (aq). A precipitate forms. • The initial temperature of both solutions was 21.9 oC and after mixing the temperature of the mixture was 23.5oC. 1. 2. 3. Write a balanced net ionic equation for this reaction. Determine the amount of heat produced during the reaction. Determine the J/mol (molar heat capacity) for this reaction. • The density of the solution are 1.05 g/mL and the specific heat of is 4.184 J/goC. Internal Energy & Enthalpy Connection Internal Energy: Defined as the energy a substance possess. • PE + KE = Internal energy (U) A system’s internal energy (U) can change due to the flow of heat (q) and/or work (work or PDV). • DU = q(heat) + w (work or PDV) • Heat (qin) flowing into the system or work (won) being done on the system, causes its internal energy to increase, ΔU < 0. • Heat (qout ) flowing out of the system or work (wby)being done by the system causes the internal energy to decrease, ΔU > 0 When a system is under or at constant pressure, the work component (PDV) is constant and mathematically, the above relationship can now be expressed as • DU = qp (heat) • qp (heat) is said to equals DH; Enthalpy • Where DH denotes the change in energy (heat) of a system at constant pressure. Enthalpy (DH): Heat For a chemical reaction: • The DHrxn is the S DH along one path (products) - S DH along another path(reactants) • The above mathematical relationship is known as Hess’s Law (direct) • Enthalpy is a state function of matter, meaning it is independent of pathway. Thus: ∆H = Hfinal – Hinitial or ∆H = Hproducts – Hreactants • Direction of heat flow • EXOTHERMIC • If Hfinal < Hinitial then ∆H is negative; indicating energy (heat) left the system. • ENDOTHERMIC • If Hfinal > Hinitial then ∆H is positive: indicating energy (heat) has entered the system Problem: • Gummy bears (C12H22O11, sucrose) when mixed with potassium chlorate (KCl03) react violently. • When 2.67 g sucrose is mixed with 7.19 g of potassium chlorate 43.7 kJ of energy is produced. C12H22O11 (aq) + 8 KClO3 (aq) → + 8 KCl (aq) + 12 CO2 (g) + 11 H2O (l) 1. Determine the change in enthalpy for this reaction when 2.67 grams sucrose & 7.19 grams of potassium chlorate are reacted. 2. Determine the change in enthalpy for the balanced reactions 3. Indicate if this process is exothermic or endothermic and give the DH value. Justify answer. https://www.youtube.com/watch?v=OMtl52pBEKY Diagram of the Formation of CO2 from its Elements Reference credit: OpenStax2017 • The formation of CO2(g) from its elements occurs in two steps, which sum to the overall reaction, as described by Hess’s law. • The horizontal blue lines represent enthalpies. • For an exothermic process, the products are at lower enthalpy than are the reactants. Standard Enthalpy of Formation (DHof) & Hess Law • Standard Enthalpy of Formation, DHof. • Enthalpy calculations under standard state conditions (i.e., 1 mole, 1 atm, 298K). • The DHof values have units of kJ/mol • Values are listed in Thermodynamic Tables according to state (phase) • The DHof value (kJ/mol) for all elements in their standard states is ZERO kJ/mol (e.g., C = graphite and O2 = gas). • Use Hess’ Law to calculate the DHorxn from the reaction’s products and reactants’ DHof values. • DHorxn = S DHof products - S DHof reactants Problem1: substance 1. Determine the DHorxn for the following reaction: sulfur (s) + oxygen (g) → sulfur trioxide (g) 2. Is this reaction endothermic or exothermic? DHof (kJ/mol) Sulfur (s) 0 Sulfur trioxide (g) -395.7 Oxygen (g) 0 3. Would the system's temperature increase or decrease in temperature? Problem2: The combustion of ammonia is given by the following chemical reaction. 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) Determine the DHof for ammonia if the DHorxn is -904.8 kJ substance DHof (kJ/mol) NO (g) + 90.4 H2O (l) - 286.0 H2O (g) - 241.8 Calculating DHorxn not Using Hess’ Law Equation • The DHorxn of a reaction can be determined when individual tables DHof values are not given. • Criteria. • Question/problem ask to calculate DHorxn for a specific reaction. • Information available to solve problem is several chemical reactions with known DHorxn values . • You DO NOT USE the Hess Law equation. DHorxn = S DHof products - S DHof reactants Problem#1: Determine the DHorxn for 0 (g) + H (g) →OH (g). Given : O2 (g) + H2 (g) → 2 OH (g) O2 (g) → 2 O (g) H2 (g) → 2 H (g) DHo + 77.9 kJ + 49.5 kJ + 43.6 kJ Problem#2: Determine the DHorxn for 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) Given : DHo HCl (g) → HCl (aq) H2 (g) + Cl2 (g) → 2 HCl (g) AlCl3 (s) → AlCl3 (aq) 2 Al (s) + 6 HCl (aq) → 2 AlCl3 (aq) + 3 H2 (g) - 74.8 - 185.0 - 323.0 - 1049.0 Bond Energies • Bond Energy: The energy required to separate atoms in a bond. • Influenced by the strength. ∆Hrxn = S energy required to break bonds (D) - S energy evolved when bonds are made (D) Where (D) is bond energies in kJ/mol. • Breaking bonds REQUIRES energy (when determining use the (+) value of bond energies). • Making bonds RELEASES energy (when determining use the (-) value of bond energies). • Exothermic: • Breakage of weak bonds and the formation of stronger bonds: exothermic reaction occurs. • Endothermic: • Breakage of strong bonds and the formation of weak bonds: endothermic reaction occurs. Problem: Using bond energies, determine DHrxn for CH4 (g) + Cl2 (g) → HCl (g) + CH3Cl (g) Bond Ave. Bond Energies (kJ/mol) H-H 436 Cl-Cl 242 H-Cl 432 H-C 414 Cl-C 339