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UNIT 3 - THERMOCHEMISTRY

Chapters 5 and 19. UNIT 3 - THERMOCHEMISTRY. Energy = capacity to do work Kinetic = energy of motion Potential = energy of position relative to other objects Work = energy used to cause an object with mass to move against a force Force = push or pull exerted on an object

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UNIT 3 - THERMOCHEMISTRY

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  1. Chapters 5 and 19 UNIT 3 - THERMOCHEMISTRY

  2. Energy = capacity to do work • Kinetic = energy of motion • Potential = energy of position relative to other objects • Work = energy used to cause an object with mass to move against a force • Force = push or pull exerted on an object • Heat = energy used to cause temp. to increase The Nature of energy

  3. Coulomb’s Law F = the electric force between 2 charged objects k = Coulomb’s constant = 8.99 x 109 J-m/C2 Q1 and Q2 are the electrical charges r is the distance of separation between the two objects

  4. System = portion singled out for study • Open • Closed • Isolated • Surroundings = everything else System and surroundings

  5. Work is energy used to cause an object to move against a force • w = F x d [Eq. 5.3] • Heat is energy transferred from a hotter object to a colder one Work and Heat

  6. Energy is conserved. • Internal energy • Sum of all kinetic and potential energies of the system’s components • Change in energy is what we can know • DE = Efinal – Einitial • Magnitude of the change is important • Direction of change is important (+ or -) First law of thermodynamics

  7. Shows the direction and magnitude of the energy change for a chemical reaction Energy diagrams

  8. DE is positive when energy is added to the system and negative when energy exits • DE = q + w (q is heat and w is work) • Endothermic • System absorbs heat • Exothermic • System releases heat Relating DE to heat and work

  9. Depends only on the present state of the system, and not the path it took to get there Energy of a system STATE FUNCTIONS Final

  10. P-V work = pressure-volume work, which is involved in the expansion or compression of gases • w = -PDV • Enthalpy = heat flow in processes @ constant P where no other work is done except P-V work • Internal energy + P*V • DH = DE + PDV • Change in enthalpy = heat gained/lost @ constant P Enthalpy

  11. Enthalpy change that accompanies a reaction Heat of reaction (DHrxn) DH= DHproducts – DHreactants Enthalpy of Reaction

  12. Enthalpy is an extensive property. • Directly proportional to the amount of reactant consumed in the reaction. • Enthalpy change for a rxn is equal in magnitude but opposite in sign to DH for the reverse rxn. • The enthalpy change for a rxn depends on the state of the reactants and products. Enthalpy

  13. Thermochemical Equations: show enthalpy changes as well as the mass relationships 2H2O (s) 2H2O (l) H2O (s) H2O (l) H2O (l) H2O (s) DH = -6.01 kJ DH = 6.01 kJ DH = 2 x6.01= 12.0 kJ • The stoichiometric coefficients always refer to the number of moles of a substance • If you reverse a reaction, the sign of DH changes • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.

  14. P4(s) + 5O2(g) P4O10(s)DH = -3013 kJ x H2O (l) H2O (g) H2O (s) H2O (l) -3013 kJ 1 mol P4 x DH = 6.01 kJ DH = 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. How much heat is evolved when 266 g of white phosphorus (P4) burn in air? = -6470 kJ 266 g P4

  15. Hydrogen peroxide can decompose to water and oxygen by the following reaction: • 2H2O2 (l) → 2H2O (l) + O2 (g) DH = -196 kJ Calculate the value of q when 5.00 g of hydrogen peroxide decomposes at constant pressure. More Practice

  16. Measurement of heat flow • Calorimeter = device used to measure magnitude of temp. change that the heat flow produces • Heat capacity = C = amount of heat required to raise the temp. by 1K (or 1°C). • Extensive property; units are J/K or J/°C Calorimetry

  17. Molar heat capacity (Cm)= heat capacity of one mole of a substance • Specific heat (Cs)= heat capacity of one gram of a substance • Intensive property • q = m * Cs * DT Calorimetry

  18. How much heat is given off when an 869 g iron bar cools from 940C to 50C? Cs of Fe = 0.444 J/g •0C Dt = tfinal – tinitial = 50C – 940C = -890C q = m*Cs*Dt = 869 g *0.444 J/g •0C * –890C = -34,000 J

  19. Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the temperature change 50.0 kg of rocks would undergo if they emitted 450. kJ of heat. Calorimetry Practice

  20. Coffee cup calorimeter Pressure of the system = atmospheric pressure Assume that the calorimeter contains all heat generated in the reaction Use q = m*Cs * DT Constant-Pressure Calorimetry DH = qrxn No heat enters or leaves!

  21. When 50.0 mL of 0.100 M silver nitrate and 50.0 mL of 0.100 M hydrochloric acid are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30°C to 23.11°C. The temperature increase is caused by the following reaction: AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3 (aq) • Calculate DH for this reaction in kJ/mol AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g-°C. Constant-Pressure Calorimetry Practice

  22. Bomb calorimetry Bomb calorimeter = used to study combustion reactions qrxn = -Ccal * DT Constant-Volume Calorimetry

  23. DH = qrxn Constant-Volume Calorimetry Measured in a constant-volume bomb calorimeter qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = m x Cs x Dt qbomb = Cbomb xDt Reaction at Constant V DH ~ qrxn No heat enters or leaves!

  24. A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10°C to 24.95°C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole. Constant-Volume Calorimetry Practice

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