1 / 42

Thermochemistry

Thermochemistry. Chapter 6. 화학반응의 에너지 관계. 에너지의 본성과 에너지의 형태 화학반응에서의 에너지 변화 , Δ E 열역학 . 계와 상태함수 열역학 제 1 법칙 , 일 ( w ) 과 열 ( q ) 화학반응에서의 엔탈피 , H 열계량법 표준생성 ( Δ H f 0 ) 및 반응엔탈피 ( Δ H ). Energy is the capacity to do work

lark
Télécharger la présentation

Thermochemistry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Thermochemistry Chapter 6

  2. 화학반응의 에너지 관계 • 에너지의 본성과 에너지의 형태 • 화학반응에서의 에너지 변화, ΔE • 열역학. 계와 상태함수 열역학 제1법칙, 일(w)과 열(q) • 화학반응에서의 엔탈피, H • 열계량법 • 표준생성(ΔHf0) 및 반응엔탈피(ΔH)

  3. Energyis the capacity to do work • Radiant energy comes from the sun and is earth’s primary energy source • Thermal energy is the energy associated with the random motion of atoms and molecules • Chemical energy is the energy stored within the bonds of chemical substances • Nuclear energy is the energy stored within the collection of neutrons and protons in the atom • Potential energy is the energy available by virtue of an object’s position

  4. Temperature = Thermal Energy 900C 400C Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. greater thermal energy

  5. Thermodynamics is the study of heat and its transformations. Thermochemistry is a branch of thermodynamics that deals with the heat involved with chemical and physical changes. Fundamental premise When energy is transferred from one object to another, it appears as work and/or as heat. For our work we must define a system to study; everything else then becomes the surroundings. The system is composed of particles with their own internal energies (E or U). Therefore the system has an internal energy. When a change occurs, the internal energy changes.

  6. Various Systems

  7. Various Systems open

  8. Various Systems closed

  9. Various Systems isolated

  10. 열역학의용어들 우주 = 계 + 주위 Universe= System + Surroundings 상태2 경로(path) q, w 과정(process) 단열, 정압, 정적, 가역, 비가역, 발열, 흡열, … 상태1 (state) State functions p, V, T, S, H, U, m, n, …

  11. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

  12. Exothermic Endothermic

  13. Δp = pfinal - pinitial ΔV = Vfinal - Vinitial ΔT = Tfinal - Tinitial Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature ΔU = Ufinal - Uinitial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

  14. The First Law of Thermodynamics -- Heat and work are forms of energy transfer and energy is conserved. U = q + w change in total internal energy work done on the system heat added to the system State Function Process Functions

  15. Another form of the first law for ΔUsys ΔU = q + w ΔU is the change in internal energy of a system q is the heat exchange between the system and the surroundings w is the work done on (or by) the system w = -pΔVwhen a gas expands against a constant external pressure

  16. Increase in volume, dV Work, and the expansion(p-V) work Work = force • distance pex=F/A force acting on the piston = pex × Area distance when expand = dy F F +y dW (p-V work)

  17. (a) (b) ΔV = 5.4 L – 1.6 L = 3.8 L ΔV = 5.4 L – 1.6 L = 3.8 L 101.3 J = -1430 J w = -14.1 L•atm × 1L•atm • Q. A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. • What is the work done in joules if the gas expands • against a vacuum and • (b) against a constant pressure of 3.7 atm? w = -p ΔV p = 0 atm w = -0 atm × 3.8 L = 0 L•atm = 0 joules p = 3.7 atm w = -3.7 atm × 3.8 L = -14.1 L•atm

  18. Chemistry in Action: Making Snow ΔU = q + w q = 0 w < 0, ΔU < 0 ΔU = C ΔT ΔT < 0, SNOW!

  19. Enthalpy and the First Law of Thermodynamics ΔU = q + w At constant pressure: q = ΔH and w = -p ΔV ΔU = ΔH - p ΔV ΔH = ΔE + p ΔV

  20. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. ΔH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts > Hreactants Hproducts < Hreactants ΔH > 0 ΔH < 0

  21. H2O (s) H2O (l) ΔH = 6.01 kJ Thermochemical Equations Is ΔH negative or positive? System absorbs heat Endothermic ΔH > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

  22. DH = -890.4 kJ CH4(g) + 2O2(g) CO2(g) + 2H2O (l) Thermochemical Equations Is DH negative or positive? System gives off heat Exothermic DH < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

  23. Thermochemical Equations • The stoichiometric coefficients always refer to the number of moles of a substance ΔH = 6.01 kJ H2O (s) → H2O (l) • If you reverse a reaction, the sign of ΔH changes ΔH = -6.01 kJ H2O (l) → H2O (s) • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n. ΔH = 2 × 6.01= 12.0 kJ 2H2O (s) → 2H2O (l)

  24. × H2O (l) H2O (g) H2O (s) H2O (l) 3013 kJ 1 mol P4 × ΔH = 6.01 kJ ΔH = 44.0 kJ 1 mol P4 123.9 g P4 Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations. Q. How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4(s) + 5O2(g) → P4O10(s)ΔH = -3013 kJ = 6470 kJ 266 g P4

  25. A Comparison of ΔH and ΔU 2Na (s) + 2H2O (l) → 2NaOH (aq) + H2 (g) ΔH = -367.5 kJ/mol ΔU = ΔH - p ΔV At 25 0C, 1 mole H2 = 24.5 L at 1 atm p ΔV = 1 atm × 24.5 L = 2.5 kJ ΔU = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol

  26. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. C = m × s Heat (q) absorbed or released: q = m × s ×ΔT q = C ×ΔT ΔT = Tfinal - Tinitial

  27. Q. How much heat is given off when an 869 g iron bar cools from 950C to 50C? s of Fe = 0.444 J/g •0C ΔT = Tf – Ti = 50C – 950C = -900C = -34,000 J q = ms ΔT = 869 g × 0.444 J/g •0C × (–900C) = -34725 J = 3.4 × 104 J (34 kJ) 단원 김홍도 작품, 대장간

  28. DH = qrxn Constant-Volume Calorimetry qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = m ×s ×ΔT qbomb = Cbomb ×ΔT Reaction at Constant V DH~qrxn No heat enters or leaves!

  29. Constant-Pressure Calorimetry 온도계 qsys = qwater + qcal + qrxn 막대젓개 뚜껑 qsys = 0 qrxn = - (qwater + qcal) qwater = m ×s ×Dt qcal = Ccal ×Dt 이중보온 컵 Reaction at Constant P DH = qrxn No heat enters or leaves!

  30. Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l) ΔH = -2801 kJ/mol 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J(큰칼로리로 식품에 사용하던 단위)

  31. Q. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? Establish an arbitrary scale with the standard enthalpy of formation (ΔHfo) as a reference point for all enthalpy expressions. Standard enthalpy of formation (ΔHfo) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. The standard enthalpy of formation of any element in its most stable form is zero. ΔHfo(O2) = 0 ΔHfo(C, graphite) = 0 ΔHfo(O3) = 142 kJ/mol ΔHfo(C, diamond) = 1.90 kJ/mol

  32. The standard enthalpy of reaction (ΔHrxno ) is the enthalpy of a reaction carried out at 1 atm. aA + bB → cC + dD ΔHrxno = [c ΔHfo (C) +d ΔHfo (D)] - [a ΔHfo (A) +b ΔHfo (B)] ΔHrxno = S nΔHfo (products) - S mΔHfo (reactants) Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

  33. C (graphite) + 1/2O2(g) CO (g) CO (g) + 1/2O2(g) CO2(g) C (graphite) + O2(g) CO2(g)

  34. C(graphite) + 2S(rhombic) CS2 (l) + Q. Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2(g) → CO2(g)ΔHrxno= -393.5 kJ S(rhombic) + O2(g) → SO2(g)ΔHrxno = -296.1 kJ CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)ΔHrxno = -1072 kJ 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) → CS2 (l) 2. Add the given rxns so that the result is the desired rxn. C(graphite) + O2(g) → CO2(g)ΔHrxno = -393.5 kJ 2S(rhombic) + 2O2(g) → 2SO2(g)ΔHrxno = -296.1×2 kJ CO2(g) + 2SO2(g) → CS2(l) + 3O2(g)ΔHrxno = +1072 kJ ΔHrxno = -393.5 + 2×(-296.1) + 1072 = 86.3 kJ

  35. = - 2973 kJ/mol C6H6 -5946 kJ 2 mol Q. Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O (l) ΔHrxno = S nΔHfo (products) - S mΔHfo (reactants) ΔHrxno = [ 12×ΔHfo (CO2) + 6×ΔHfo (H2O) ] - [ΔHfo (2×ΔHfo (C6H6) )] = [ 12×(–393.5) + 6 ×(– 187.6) ] – [ 2×49.04 ] = -5946 kJ

  36. Chemistry in Action: Bombardier Beetle Defense C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O (l) ΔHo = ? C6H4(OH)2(aq) → C6H4O2(aq) + H2 (g) ΔHo = 177.2 kJ/mol H2O2(aq) → H2O (l) + ½O2 (g) ΔHo = -94.5 kJ/mol H2(g)+ ½O2(g) → H2O (l) ΔHo = -285.5 kJ/mol 폭탄먼지벌레 ΔHo = 177.2 - 94.5 – 285.5 = -202.8 kJ/mol Exothermic! benzene-1,4-diol hydroquinone quinol

  37. Hydrogen Peroxide Hydroquinone Enzymes – Catalase and Peroxidase exit at boiling point and vaporize one-fifth of it! Figure 2. Bombardier beetle discharge apparatus. Novid Beheshti and Andy C Mcintosh, Bioinsp. Biomim. 2 (2007) 57–64 doi:10.1088/1748-3182/2/4/001

  38. Energy Calculation: Bombardier Beetle Defense C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O(l) ΔH0 = -202.8 kJ Aqueous solution of reactants is stored in a reservoir, and is composed of hydroquinone at a concentration of 25% and hydrogen peroxideat concentration of 10% 79kJ= 0.4 mol HQ = 25 g/100mL water Hydroquinone Solubility water 20g 100℃ vapor 100g 25℃ water 100g 100℃ water 31 kJ /100g water 48kJ /20g vapor

  39. The enthalpy of solution (ΔHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. ΔHsoln = Hsoln - Hcomponents Q. Which substance(s) could be used for melting ice? Q. Which substance(s) could be used for a cold pack?

  40. The Solution Process for NaCl DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

More Related