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Chapter 8

Chapter 8. Thermochemistry: Chemical Energy. 8.1 Energy and Its Conservation. Conservation of Energy Law : Energy cannot be created or destroyed; it can only be converted from one form to another. Energy and Its Conservation. Energy : The capacity to supply heat or do work.

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Chapter 8

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  1. Chapter 8 Thermochemistry: Chemical Energy

  2. 8.1 Energy and Its Conservation Conservation of Energy Law: Energy cannot be created or destroyed; it can only be converted from one form to another.

  3. Energy and Its Conservation Energy: The capacity to supply heat or do work. Kinetic Energy (EK): The energy of motion. Potential Energy (EP): Stored energy. Thermal Energy: The kinetic energy of molecular motion and is measured by finding the temperature of an object Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two 1 cal = 4.184 J (exactly) Units: 1 Cal = 1000 cal = 1 kcal

  4. 8.2 Internal Energy and State Functions First Law of Thermodynamics: The total internal energy E of an isolated system is constant - The sum of all the kinetic and potential energies of particles making up a substance DE = Efinal - Einitial

  5. Internal Energy and State Functions State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state

  6. CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 802 kJ energy Internal Energy and State Functions DE = Efinal - Einitial = -802 kJ 802 kJ is released when 1 mole of methane, CH4, reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and two moles of water.

  7. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 8.3 Expansion Work Expansion Work: Work done as the result of a volume change in the system w = F x d = - P∆V 7 mol of gas 6 mol of gas

  8. 8.4 Energy and Enthalpy DE = q + w q = heat transferred w = work = -PDV q = DE + PDV Constant Volume (DV = 0): qV = DE Constant Pressure: qP = DE + PDV

  9. Energy and Enthalpy qP = DE + PDV = H Enthalpy change or Heat of reaction (at constant pressure) DH = Hfinal - Hinitial Enthalpy is a state function whose value depends only on the current state of the system, not on the path taken to arrive at that state. = Hproducts - Hreactants

  10. C3H8(g) + 5O2(g) C3H8(g) + 5O2(g) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 3CO2(g) + 4H2O(l) 3CO2(g) + 4H2O(g) 8.6The Thermodynamic Standard State DH= -2043 kJ DH = -2219 kJ Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution. Standard enthalpy of reaction is indicated by the symbol ΔHo DH° = -2043 kJ

  11. Enthalpies of Physical and Chemical Change

  12. Enthalpies of Physical and Chemical Change Enthalpy of Fusion (DHfusion): The amount of heat necessary to melt a substance without changing its temperature Enthalpy of Vaporization (DHvap): The amount of heat required to vaporize a substance without changing its temperature Enthalpy of Sublimation (DHsubl): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase

  13. 2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s) 2Fe(s) + Al2O3(s) 2Al(s) + Fe2O3(s) Enthalpies of Physical and Chemical Change DHo = -852 kJ exothermic The reverse reaction DHo = +852 kJ endothermic

  14. Applying Stoichiometry to Heats of Reaction • A propellant for rockets is obtained by mixing the liquids hydrazine, N2H4, and dinitrogen tetroxide, N2O4. These compounds react to give gaseous nitrogen, N2 and water vapor, evolving 1049 kJ of heat at constant pressure when 1 mol N2O4 reacts. • Write the thermochemical equation for this reaction • Write the thermochemical equation for the reverse of the reaction • How much heat evolves when 10.0 g of hydrazine reacts according to the reaction described in (a) ?

  15. Example • An LP gas tank in a home barbeque contains 13.2 kg of propane, C3H8. Calculate the heat (in kJ) associated with the complete combustion of all of the propane in the tank C3H8(g) + CO2(g)  3 CO2(g) + 4 H2O(g) ΔHo = -2044kJ

  16. 8.6Calorimetry and Heat Capacity Measure the heat flow at constant pressure (DH).

  17. Calorimetry and Heat Capacity Measure the heat flow at constant volume (DE).

  18. Calorimetry and Heat Capacity Heat Capacity (C): The amount of heat required to raise the temperature of an object or substance a given amount. q = C x DT Molar Heat Capacity (Cm): The amount of heat required to raise the temperature of 1 mol of a substance by 1 °C. q = (Cm) x (moles of substance) x DT Specific Heat: The amount of heat required to raise the temperature of 1 g of a substance by 1 °C. q = (specific heat) x (mass of substance) x DT

  19. Calorimetry and Heat Capacity

  20. Example • Iron metal has a specific heat of 0.449 J/goC. How much heat (in kJ) is transferred to a 5.00g piece of iron, initally at 20.0oC, when it is placed in a pot of boiling water? Assume that the temperature of the water is 100.0oC and that the water remains at this temperature, which is the final temperature of iron

  21. Example • When 1.045 g of CaO is added to 50.0 mL of water at 25.0oC in a calorimeter, the temperature of the water increases to 32.2 oC. Assuming that the specific heat of the solution is 4.18 J/goC and that the calorimeter itself absorbes a negligible amount of heat, calculate ∆H in kilojoules for the reaction CaO(s) + H2O(l)  Ca(OH)2(aq)

  22. N2H4(g) + H2(g) 2H2(g) + N2(g) 3H2(g) + N2(g) 3H2(g) + N2(g) N2H4(g) 2NH3(g) 2NH3(g) 2NH3(g) 8.9 Hess’s Law Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Haber Process: DH°total = ??? Multiple-Step Process - Given DH°1 = 95.4 Kj DH°2 = -187.6 kJ DH°total = DH°1 + DH°2

  23. Hess’s Law

  24. Example • Find ΔHorxn for the following reaction: 3H2(g) + O3(g)  3 H2O(g) ΔHorxn = ?? Use the following reactions with known ΔH’s 2H2(g) + O2(g)  2 H2O(g) Δ Ho = -483.6 kJ 3O2(g)  2 O3 (g) Δ Ho = +285.4 kJ

  25. Example • Find ΔHorxn for the following reaction C(s) + H2O(g)  CO(g) + H2(g) Horxn = ? Use the following reactions with known H’s C(s) + O2(g)  CO2(g) ΔHo = -393.5 kJ 2CO(g) + O2(g)  2CO2(g) Δ Ho = -566.0kJ 2H2(g) + O2(g)  2H2O(g) Δ Ho = -483.6 kJ

  26. C(s) + 2H2(g) CH4(g) Standard Heats of Formation Standard Heat of Formation (DHof ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states Standard states DHof = -74.8 kJ 1 mol of 1 substance

  27. Standard Heats of Formation

  28. aA + bB cC + dD Standard Heats of Formation Ho = Hof (Products) - Hof (Reactants) Ho = [c Hof (C) + d Hof (D)] - [a Hof (A) + b Hof (B)] Reactants Products

  29. 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g) Standard Heats of Formation Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O. DHo = ?

  30. Example • Calculate the enthalpy change for the following reaction 3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) *Use standard enthalpies of formation

  31. An Introduction to Entropy Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence

  32. An Introduction to Entropy Entropy (S): The amount of molecular randomness in a system

  33. An Introduction to Entropy Spontaneous processes are • favored by a decrease in H (negative DH). • favored by an increase in S (positive DS). Nonspontaneous processes are • favored by an increase in H (positive DH). • favored by a decrease in S (negative DS).

  34. Example • Predict whether ΔSo is likely to be positive or negative for each of the following reactions • H2C=CH2(g) + Br2(g)  BrCH2CH2Br(l) • Consider the following figures

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