1 / 66

TOPIC 8: THERMOCHEMISTRY

TOPIC 8: THERMOCHEMISTRY. Thermochemistry is the branch of chemistry dealing with determining quantities of heat by measurement and calculation. Some of these calculations will allow us to establish indirectly, a quantity of heat that would be difficult to measure directly.

ursa
Download Presentation

TOPIC 8: 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. TOPIC 8: THERMOCHEMISTRY Thermochemistry is the branch of chemistry dealing with determining quantities of heat by measurement and calculation. Some of these calculations will allow us to establish indirectly, a quantity of heat that would be difficult to measure directly.

  2. Some Main Concepts in Thermochemistry System: The part of the universe we choose to study. Surroundings: The parts of the universe with which the system interacts. Systems are classified into 3 groups: Open system: Free transfer of energy and matter between the system and its surroundings. Closed system: Only free transfer of energy between the system and its surroundings Isolated system: No possible transfer of energy and or matter between the system and its surroundings. Closed system Isolated system Open system

  3. Some main Concepts in Thermochemistry Energy transfer can occur as heat (q) or in several other forms, known as work (w). Heat: The form of energy which is transfered from a system with higher temperature to a system with low temperature due to the temperature difference. Work: The effect that causes the change in position of a body by exerting a force on it. Energy: The capacity of a system to perform a work. Energy transfers occuring as heat and work affect the total amount of energy contained within a system, its internal energy (E). The components of internal energy of special interest to us are thermal energy and chemical energy. Thermal energy: Energy associated with random molecular motion. Chemical energy: Energy associated with chemical bonds and intermolecular forces.

  4. Energy Kinetic energy (KE): The energy of a moving object. m: mass, v: velocity Energy Unit: Kinetic energy Potential energy: The stored energy or «energy of position» associated with forces of attraction or repulsion between objects. The stored energy arises from the kinetic energy of molecules and atoms. Potential Energy: g: acceleration of gravity, 9,81 m/s2 h: height of the matter, m m: mass of the matter,kg

  5. Heat transfer betweenthesystemanditssurroundingsoccurs as a result of temperaturedifference. Heatmovesfromthe hot environmenttothecoldenvironment. Temperature is variable. Change of thestatemayoccur Heat The heat flux occuring at constant temperature is called ISOTHERMAL PROCESS.

  6. Units of Heat The quantity of heat energy ,q , depends upon: How much the temperature is to be changed The quantity of substance The nature of substance (type of atoms or molecules) • Calorie (cal) • The quantity of heat required to change the temperature of one gram of water by one degree Celcius. • Joule (J) • SI unit of work and energy 1 cal = 4.184 J

  7. q = q = mcT CT Heat Capacity • The quantity of heat required to change the temperature of a system by one degree . • Molar heat capacity: • The System is one mole of substance • Specific heat (capacity), c. • The system is 1 g of substance • Heat capacity, C • Massx specific heat.

  8. Conservation of Energy • In interactions between a system and its surroundings, the total energy remains constant- energy is neither created nor destroyed. • Applied to the exchange of heat, this means : qsystem + qsurroundings = 0 qsystem = -qsurroundings Thus, heat lost by a system is gained by its surroundings, and vice versa.

  9. Experimental Determination of Specific Heats Lead The transfer of energy, as heat, from the lead to the cooler water causes the temperature of the lead decrease and that of the water increase, to the point where both the lead and water are at the same temperature. If we consider lead to be the system and water as the surroundings, we can write : qlead= -qwater

  10. Experimental Determination of Specific Heats qlead= -qwater q water= mcT = (50.0 g water)(4.184 J/g water°C)(28.8 - 22.0)°C q water= 1.4x103 J q lead= -1.4x103 J = mcT = (150.0 g lead)(c)(28.8 - 100.0)°C c lead= 0.13 Jg-1°C-1

  11. Heats of Reaction and Calorimetry • Chemical Energy. • Type of energy related to the internal energy of the system. • The energy which occurs as a result of chemical reactions. The batteries and accumulators are the vehicles which convert the chemical energy into electrical energy. The accumulation of electrical energy in batteries is performed by chemical methods. Chemical energy can be also converted into the mechanical, heat and light energy. • Heat of reaction, qrxn. • The quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within a system at constant temperature and pressure.

  12. Heat of Reaction CaO(s) + H2O(l) → Ca(OH)2 (aq) • Exothermicreaction. • Gives off heat to the surroundings, qrxn < 0. • Temperature increase in the system (a) • Endothermic reaction. • Gain of heat from the surroundings, qrxn > 0. • Temperature decrease in the system (b) • Calorimeter • Device for measuring quantities of heat Ba(OH)2·8H2O + 2NH4Cl(s)→BaCl2(s) + 2 NH3(aq) + 8 H2O(l)

  13. qcal = miciT = CcalT heat Heat capacity of calorimeter Bomb Calorimetry The type of calorimeter shown in the figure is called bomb calorimetry. It is ideally suited for measuring the heat evolved in combustion reaction. It is an isolated system from its surroundings When the combustion reaction occurs, chemical energy is converted to thermal energy and temperature of the system rises.The heat of reaction qrxn is the quantity of heat that the system would have to lose to its surroundings to be restored to its initial temperature and pressure. qrxn = -qcal qcal = qbomb + qwater + qwire+…

  14. Example • Determination of heat of reaction by the bomb calorimeter • The combustion of 1,010 g sucrose, C12 H22 O11 , in a bomb calorimeter causes the temperature rise from 24,92˚ to 28,33˚C. The heat capacity of the calorimeter assembly is 4,90kJ/ ˚C • What is the heat of combustion of sucrose (kJ/mol) ? • Verify the claim of sugar producers that one teaspoon of sugar (4,8 g) contains only 19 calories .

  15. Calculate qrxn qrxn = -qcal = -16,7 kJ Solution Example 7-3 Calculate qcal qcal = CT = (4,90 kJ/°C)(28,33-24,92)°C = (4,90)(3,41) kJ = 16,7 kJ in 1,01 g sample of sugar

  16. 343,3 g = -16,5 kJ/g qrxn 1,00 mol (a) = -5,65 x 103 kJ/mol 1,00 kal 4,8 g )= -19 cal/teaspoon qrxn = (-16,5 kJ/g)( )( (b) 4,184 J 1 ts Solution Example 7-3 Conversion of the unit ofqrxn as kJ/mol : -16,7 kJ qrxn = -qcal = = -16,5 kJ/g 1,010 g (b) Verification of the claim that a teaspon of sugar(4,8 g) has 19 Calories. For a teaspoon

  17. The Coffee-Cup Calorimeter • Easy to handle . • An isolated system composed of styrofoam cup. • Reactants are mixed up in the cup • Temperature difference is measured at the end of reaction. • System is at constant pressure. qrxn = -qcal qrxn = - CcalT

  18. WORK In thermodynamics work means the transfer of energy between the system and its surroundings due to an external makroscopic force. • Apart from transfer of heat, some chemical processes may do work.(the expansion or compression of gases) • The gas pushes the atmosphere. • The volume changes. • Pressure-volume work

  19. Pressure-Volume Work Unit of work Since work is done by the system , it has a negative sign - . If the volume increases, work has - sign.

  20. Example Example 7-3 Calculation of Pressure-Volume Work. What is the work done in joules, when 0,100 mol He(g) at 298Kexpands from an initial pressure of Pi= 2,4 atm to a final pressure Pf= 1,3 atm? We assume that He behaves as an ideal gas: Vi = nRT/Pi = (0,100 mol)(0,08201 L atm mol-1 K-1)(298K)/(2,40 atm) = 1,02 L Vf = nRT/Pf= 1,88 L V = 1.88-1.02 L = 0.86 L

  21. Example Example 7-3 Calculation of work done by the system w= -PV = -(1.30 atm)(0.86 L)( = -1.1 x 102 J 101 J ) 1 L atm Conversion factor: 8.3145 J/mol K ≡ 0.082057 L atm/mol K 1 ≡ 101.33 J/L atm

  22. Pressure-Volume Work Example: What is the quantity of work, in joules, done by the gas in the figure next if it expands against a constant pressure of 0,980 atm and the change in Volume(ΔV) is 25 L .

  23. First Law of Thermodynamics • Internal Energy, U. • The TOTAL energy of the system (potential and kinetic). • Translational kinetic energy. • Rotations. • Bond vibrations. • Intermolecular attractions • Chemical bonds. • Electrons.

  24. The First Law of Thermodynamics • A system keeps the energy only in form of internal energy. • A system does not contain energy in the form of heat or work. • Heat and work are the means by which the system exchanges energy with its surroundings. • Heat and work only exist during a change. • Law of Conservation of energy • The energy of an isolated system is constant. U = q + w Uisolated = 0

  25. System The First Law of Thermodynamics Any energy entering the system carries + sign.Thus,if heat is absorbed by the system, q>0 .If work is done on the system ,w>0. Any energy leaving the system carries a – sign.Thus if heat is given off by the system , q<0. If work is done by the system, w<0. If, on balance more energy enters the system than leaves, ΔU is positive.If more energy leaves than enters, ΔU is negative.

  26. Functions of State • Any property having a unique value when the state of the system is defined is called function of state. • Example: • The state of water at 293,15 K and 1,00 atm is specific • In this state d = 0.99820 g/mL • Density just depends on the state of the system. • It does not depend on how it is reached.

  27. State 2 Internal energy State 1 tot Functions of State • U is a function of state. • Can not be measured. • We do not need to know the real value. • The U between U2 and U1 have a unique value. • It can be measured easily. The value of the internal energy is the value of heat energy given off from the surroundings to the system to reach from the state U1 to U2

  28. Functions depending on the path to be followed • Heat and work are not functions of state! • The values of heat and work depend upon the path to be followed for a change in the system. • 0,1 mol He 298 K, 2,40 atm (State 1) (1,02 L) A↓ 298 K, 1,30 atm (State 2) (1,88 L) B 298 K, 1,80 atm (1,36 L) C wBC = (-1,80 atm)(1,36-1,02)L – (1,30 atm)(1,88-1,36)L = -0,61 L atm – 0,68 L atm = -1,3 L atm = -1,3 x 102 Jin contrast; wA = -1,1 x 102 J

  29. Heats of Reaction and Entalpy Change: U and H Reactants→ Products Ui Uf U = Uf- Ui U = qrxn + w In a system at a constant volume(Bomb calorimeter): w = -PV= 0 U = qrxn+ 0 = qrxn = qv However, lots of chemical reactions occur in the earth under constant pressure! What is the relationship between qpand qv?

  30. Heat of Reactions First state First state qV = qP + w Internal energy Final state Final state

  31. Heat of reactions qV = qP + w w = - PV and U = qv: U = qP - PVqP = U + PV P,V,U are functions of state Enthalpy H = U + PV H = Hf – Hi = U + PV At constant pressure and temperature: H = U + PV = qP

  32. heat Constant volume Constant pressure heat Comparison of heat of reactions 2 CO(g) + O2(g) → 2CO2(g) qP = -566 kJ/mol = H PV = P(Vf – Vi) = RT(nf – ni) = -2,5 kJ U = H - PV = -563,5 kJ/mol = qV

  33. Example • Theheat of combustion at constantvolume of CH4 (g) is measured in a bombcalorimeter at 25˚C and is foundto be – 885,389 J/mol. What is theenthalpychange,ΔH ? • Solution: CH4 (g) + 2 O2 (g) CO2(g) + 2 H2O (l) ΔU= - 885,389 J Δn= Σnproducts- Σnreactants 1-(2+1)= - 2 (Notethat, themolenumber of H2O is not included in thecalculation!!!Since volumechange of liquidsandsolidsaretoosmall, it is here neglected.

  34. Example • ΔH= ΔU + PΔV • ΔH= ΔU + ΔnRT • ΔH= - 885,389 -2x 8,314 J/molK x 298,15K • ΔH=- 885,389 kJ- 4,957 kJ • ΔH=- 890,346 kJ Notethatthevaluemeasuredbythebombcalorimeter is equaltoΔU!!!

  35. Example • Forthereaction ; B2H6(g) + 3 O2(g) B2O3(s) + 3 H2O(l) ΔU is – 2143,2 kJ a) CalculateΔH forthereaction at 25 ˚C b) Determinethevalue of standardenthalpy of formation of B2H6(g). For B2O3 (s) ΔHf˚=-1264,0 kJ/molandfor H2O(l) ΔHf˚=-285,9 kJ/mol Solvethe problem byyourself!

  36. Enthalpy during the phase transitions Molar enthalpy of vaporization: H2O (l) → H2O(g) H = 44,0 kJ ; 298 K Molar enthalpy of melting : H2O (s) → H2O(l) H = 6,01 kJ ; 273,15 K Heat energy during the phase transition: qp= n*Hphchange

  37. Solution: qP = mcH2OT + nHvap 50,0 g = (50,0 g)(4,184 J/g °C)(100-25,0)°C + x 44,0 kJ/mol 18,0 g/mol Example Example 7-3 Enthalpy change during phase transition Calculate the molar heat of vaporization of 50,0 g sample of the water whose temperature is risen from 25,0°C to 100°C. cwater= 4,184 J/g °C, Hvap = 44,0 kJ/mol Think about the solution in two steps: First step: Increase in the temperature of the water. Second step: The vaporization Hvap = 44,0 kJ/mol = 15,69 kJ + 122 kJ = 137,69 kJ

  38. Standard Enthalpies of Formation • Standardstate is thepuresubstance at 1 atmpressure at thetemperature of interest. • Standardmolarenthalpy of formation(molarheat of formation), Hf • Thedifference in enthalpybetweenonemole of a compound in itsstandardstateanditselements in theirmoststableformsandstandardstates. Wewilluseenthalpies of formationtoperform a variety of calculations. Thefirstthingwehaveto do is towrite a chemicalcalculationandthentosketch an enthalpydiagram Temperature values must be given with H° !!

  39. Enthalpy Diagrams Products Reactants Enthalpy Enthalpy Reactants Products Endothermic reactants Exothermic reactants

  40. Indirect Determination of H:Hess’s Law • His an Extensive Property. • Enthalpy change is directly proportional to the amounts of substances in a system. N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ ½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ (we divide all coeeficients and H value by two) • H changes sign when a process is reversed. NO(g)→ ½N2(g) + ½O2(g)H = -90.25 kJ

  41. ½N2(g) + ½O2(g) → NO(g)H = +90.25 kJ NO(g) + ½O2(g) → NO2(g)H = -57.07 kJ Hess’s Law • Hess’s Law of Constant Heat Summation: • If a process occurs in stages or steps (even if only hypothetically) the enthalpy change for the overall(net) process (Hnet )is the sum of the enthalpy changes for the individual steps (H). ½N2(g) + O2(g) → NO2(g)H = +33.18 kJ

  42. Hess’s Law ½N2(g) + ½O2(g) → NO(g) Enthalpy NO(g) + ½O2(g) → NO2(g) ½N2(g) + O2(g) → NO2(g)

  43. Standard Enthalpy of Formation Hf ° • Weassignenthalpies of zerototheelements in theirmoststableformswhen in thestandardstate. • Moststableforms of theelementsaretheonesindicatedbelow • Na(s), H2(g), N2(g), O2(g), C(graphite), Br2(l)

  44. ½N2(g) + ½O2(g) → NO(g)H = +90.25 kJ NO(g) + ½O2(g) → NO2(g)H = -57.07 kJ Hess’s Law • Hess’s Law of Constant Heat Summation: • If a process occurs in stages or steps (even if only hypothetically) the enthalpy change for the overall(net) process (Hnet )is the sum of the enthalpy changes for the individual steps (H). ½N2(g) + O2(g) → NO2(g)H = +33.18 kJ

  45. Hess’s Law ½N2(g) + ½O2(g) → NO(g) Enthalpy NO(g) + ½O2(g) → NO2(g) ½N2(g) + O2(g) → NO2(g)

  46. Standard Enthalpy of Formation Hf ° • Weassignenthalpies of zerototheelements in theirmoststableformswhen in thestandardstate. • Moststableforms of theelementsaretheonesindicatedbelow • Na(s), H2(g), N2(g), O2(g), C(graphite), Br2(l)

  47. Standard Enthalpies of Formation Enthalpy of Formation Formation Enthalpy (formaldehyde)

  48. Standard Enthalpies of Formation Pozitive Enthalpies of Formation Standard Enthalpies of Formation Negative Enthalpies of Formation

  49. Standard Enthalpies of Reaction • If the reactants and products of a reaction are all in their standard states, we call the enthalpy change for a reaction standard enthalpy of reaction. • ΔHrxn0 • Instead of standard enthalpy of reaction the term is also frequently used: • Standard heat of reaction

  50. Standard Enthalpy of Reaction

More Related