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CHEMISTRY 161 Energy and Chemical Change Chapter 7

CHEMISTRY 161 Energy and Chemical Change Chapter 7. Energy an Chemical Change. Forms of Energy SI Unit of Energy Energy in Atoms and Molecules Thermodynamics Calculation of Heat and Energy Changes Measuring Heat and Energy Changes. 1. Forms of Energy. 1. Kinetic energy.

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CHEMISTRY 161 Energy and Chemical Change Chapter 7

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  1. CHEMISTRY 161 Energy and Chemical Change Chapter 7

  2. Energy an Chemical Change • Forms of Energy • SI Unit of Energy • Energy in Atoms and Molecules • Thermodynamics • Calculation of Heat and Energy Changes • Measuring Heat and Energy Changes

  3. 1. Forms of Energy 1. Kinetic energy energy of a moving microscopic or macroscopic object E = ½ m v2 2. Radiant energy energy in form of photons (‘light’) (solar energy) E = h  (h = Planck’s constant) (Chapter 8) 3. Potential energy energy by changing object’s position in height E = m g h (h = height)

  4. 4. Thermal Energy energy associated with random motion of atoms and molecules Ekin = ½ M v2 = 3/2 R T (Chapter 7) M = m  Na 5. Chemical Energy EXP1 energy stored in chemical bonds of substances (Chapter 7) LAW OF CONSERVATION OF ENERGY THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT

  5. 2. SI Unit of Energy 1 Joule = 1 J 1 cal = 4.184 J 1 J = 1 Nm = 1 kg m2 s-2 Ekin = ½  m  v2 macroscopic versus microscopic 1 J vs. 1 kJ mol-1

  6. 3. Energy in Atoms and Molecules Atoms – Kinetic and Thermal Energy gases are constantly in motion and hold a kinetic energy Ekin = ½ M v2 = 3/2 R T EXP2

  7. Molecules – Kinetic, Thermal, & Potential Energy (N2) molecules have different ‘internal’ (vibrational) energy when bond distances are changed EXP3/4 different bonds have different bond strength (stabilities) (H2 vs. N2)

  8. 4. Thermodynamics reactants  products (different energies) THERMODYNAMICS HEAT CHANGE study of the energy associated with change

  9. THERMOCHEMISTRY study of the energy associated with chemical change 2 H2(g) + O2(g) → 2 H2O(l) + energy Hindenburg 1937 Challenger 1986

  10. Surrounding Surrounding System System heat heat ENDOTHERMIC EXOTHERMIC EXP 2 HgO(s) → O2(g) + 2 Hg(l) 2 H2(g) + O2(g) → 2 H2O(l)

  11. Energy NH4NO3 (aq) 2 H2(g) + O2(g) Exothermic (heat given off by system) Endothermic (heat absorbed by system) NH4NO3(s) + H2O (l) 2 H2O(l)

  12. QUANTIFICATION Enthalpy of Reaction Enthalpy is the heat release at a constant pressure (mostly atmospheric pressure) DH = Hfinal - Hinitial DH = Hproducts - Hreactants board Hfinal > Hinitial : DH > 0 ENDOTHERMIC Hfinal < Hinitial : DH < 0 EXOTHERMIC

  13. Energy NH4NO3(aq) 2 H2(g) + O2(g) Hfinal > Hinitial Hfinal < Hinitial Exothermic Endothermic NH4NO3(s) + H2O(l) 2 H2O(l)

  14. Energy H2O(l) Hfinal > Hinitial DH = Hfinal – Hinitial Endothermic H2O(s) H2O(s) → H2O(l) ΔH = + 6.01 kJ mol-1

  15. Energy H2O(l) Hfinal < Hinitial DH = Hfinal – Hinitial Exothermic H2O(s) H2O(l) → H2O(s) ΔH = - 6.01 kJ mol-1

  16. THERMOCHEMICAL EQUATIONS H2O(l) → H2O(s) ΔH = - 6.01 kJ mol-1 CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g) ΔH=-890.4 kJ mol-1 Calculate the heat evolved when combusting 24.0 g of methane gas.

  17. 5. Calculation of Heat and Enthalpy Changes DHm= Hm,products – Hm,reactants molar REFERENCE SYSTEM e.g. oxidation numbers of elements are zero

  18. Standard Enthalpy of Formation DHfO heat change when 1 mole of a compound is formed from its elements at a pressure of 1 atm (T = 298 K) DHfO (element) = 0 kJ/mol DHfO (graphite) = 0 kJ/mol DHfO (diamond) = 1.9 kJ/mol

  19. C(s, graphite) + O2(g) Hreactants 0 ENTHALPY, H DHf0 = - 393.51 kJ mol-1 Hproducts -393.51 CO2(g)

  20. Standard Enthalpy of Formation C(s, graphite) + O2(g) CO2(g) DHf0 = - 393.51 kJ mol-1 CH4(g) DHf0 = - 74.81 kJ mol-1 C(s, graphite) + 2H2(g) DHf0 = - 46.11 kJ mol-1 ½ N2(g) + 3/2 H2(g) NH3(g) NO(g) DHf0 = + 33.18 kJ mol-1 (1/2) N2(g) + (1/2) O2(g)

  21. Standard Enthalpy of Reaction a A + b B → c C + d D a A + b B a × DHfO (A) + b × ΔHfO(B) Hreactants ENTHALPY, H DHOrxn = ΣΔHf0(prod) – ΣΔHf0(react) Hproducts c × DHfO(C) + d × ΔHfO(D) c C + d D

  22. Standard Enthalpy of Reaction DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(react) CaO(s) + CO2(g) → CaCO3(s) [kJ/mol] -393.5 -1206.9 -635.6 DHOrxn = -177.8 kJ/mol

  23. Standard Enthalpy of Reaction CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) DHOrxn = ΣnΔHf0(prod) – ΣmΔHf0(react) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) 2 H2O(g) → 2 H2O(l) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

  24. Hess’s Law The overall reaction enthalpy is the sum of the reaction enthalpies of the steps in which the reaction can be divided

  25. CH4(g) + 2O2(g) Reactants - 802 kJ - 890 kJ ENTHALPY, H CO2(g) + 2H2O(g) - 88 kJ CO2(g) + 2H2O(l) Products

  26. S solid Indirect Path +O 2 direct path DH1 = + 3/2 O2 -320.5 kJ DH3 = -395.7 kJ SO gas 2 SO3 gas + 1/2 O 2 DH2 = -75.2 kJ DHrxn for S(s) + 3/2 O2(g) SO3(g) S(s) + O2(g)  SO2(g) DH1 = -320.5 kJ SO2(g) + 1/2 O2(g) SO3(g) DH2 = -75.2 kJ

  27. 6. State Functions THERMODYNAMICS quantitative study of heat and energy changes of a system CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) the state (condition) of a system is defined by T, p, n, V, E

  28. the state (condition) of a system is defined by T, p, n, V, E STATE FUNCTIONS properties which depend only on the initial and final state, but not on the way how this condition was achieved Hess Law

  29. ΔV = Vfinal – Vinitial Δp = pfinal – pinitial ΔT = Tfinal – Tinitial ΔE = Efinal – Einitial

  30. Energy is a STATE FUNCTION ΔE = m g Δh IT DOES NOT MATTER WHICH PATH YOU TAKE

  31. Hess Law CH4(g) + 2O2(g) Reactants - 802 kJ - 890 kJ ENTHALPY, H CO2(g) + 2H2O(g) - 88 kJ CO2(g) + 2H2O(l) Products

  32. Applications Zeroth Law of Thermodynamics a system at thermodynamical equilibrium has a constant temperature heat is spontaneous transfer of thermal energy two bodies at different temperatures T1 > T2 spontaneous T2 T1 EXP LN2/Metal/H2O

  33. First Law of Thermodynamics energy can be converted from one form to another, but cannot be created or destroyed CONSERVATION OF ENERGY

  34. SURROUNDINGS + - SYSTEM THE TOTAL ENERGY OF THE UNIVERSE IS CONSTANT

  35. First Law of Thermodynamics ΔEsystem = ΔQ + ΔW ΔQ heat change ΔW work done DQ > 0 ENDOTHERMIC ? DQ < 0 EXOTHERMIC

  36. M M mechanical work ΔW = - p ΔV ΔV < 0 ΔV > 0 the energy of gas goes up the energy of gas goes down

  37. Surrounding System heat 6. Measurement of Heat Changes temperature increase DH = ΔQ∞ ΔT (pressure is constant)

  38. Where does the ‘heat’ go?

  39. DH = ΔQ ∞ ΔT DH = ΔQ = const × ΔT DH =ΔQ = C ΔT temperature change enthalpy change heat capacity C = m  s s = specific heat capacity DH = ΔQ = m s ΔT EXP

  40. specific heat capacity capability of substances to store heat and energy s = J g-1 K-1 the J necessary to increase the temperature of 1 g of a compound by 1 K

  41. DH = ΔQ = m s ΔT • prepare two styrofoam cups 2. carry out chemical reaction in a compound with known s s (H2O) = 4.184 J g-1 K-1 3. measure temperature change 4. determine ΔH calorimeter

  42. 100 ml of 0.5 M HCl is mixed with 100 ml 0.5 M NaOH in a constant pressure calorimeter (scup = 335 J K-1). The initial temperature of the HCl and NaOH solutions are 22.5C, and the final temperature of the solution is 24.9C. Calculate the molar heat of neutralization assuming the specific heat of the solution is the same as for water. DH = ΔQ = C ΔT DH = ΔQ = (c1 + c2) ΔT • Neutralization reactions • Redox reactions • 3. Precipitation reactions

  43. Constant Volume Calorimeter ΔQ = (m s(H2O) + cbomb) ΔT

  44. Energy an Chemical Change • Forms of Energy • SI Unit of Energy • Energy in Atoms and Molecules • Thermodynamics • Calculation of Heat and Energy Changes • Measuring Heat and Energy Changes

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