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CHEMISTRY 161 Chapter 6 chem.hawaii/Bil301/welcome.html

CHEMISTRY 161 Chapter 6 www.chem.hawaii.edu/Bil301/welcome.html. Thermochemistry. Enthalpy of Reaction. heat released or absorbed by the system at a constant pressure. D H = H products - H reactants. H products > H reactants : D H > 0 ENDOTHERMIC.

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CHEMISTRY 161 Chapter 6 chem.hawaii/Bil301/welcome.html

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  1. CHEMISTRY 161 Chapter 6 www.chem.hawaii.edu/Bil301/welcome.html

  2. Thermochemistry Enthalpy of Reaction heat released or absorbed by the system at a constant pressure DH = Hproducts - Hreactants Hproducts > Hreactants : DH > 0 ENDOTHERMIC Hproducts < Hreactants : DH < 0 EXOTHERMIC

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

  4. DH = ΔQ ∞ ΔT DH = ΔQ = const × ΔT DHm =ΔQm = cmpΔT temperature change molar enthalpy change molar specific heat capacity DHm = ΔQm = cmpΔT

  5. molar specific heat capacity capability of substances to store heat and energy [cmp] = J mol-1 K-1 the J necessary to increase the temperature of 1 mol of a compound by 1 K cmp(H2O) = 75.3 J mol-1 K-1

  6. DHm = ΔQm = cmpΔT • prepare two styrofoam cups 2. carry out chemical reaction in a compound with known cmp cmp(H2O) = 75.3 J mol-1 K-1 3. measure temperature change 4. determine ΔHm calorimeter

  7. The Real Experiment DHm = ΔQm = cmpΔT / × n DH = ΔQ = (cmp × n) ΔT DH = ΔQ = (cmp × n + ccup) ΔT

  8. 100 ml of 0.5 M HCl is mixed with 100 ml 0.5 M NaOH in a constant pressure calorimeter (ccup = 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. cmp(H2O) = 75.3 J mol-1 K-1 • Neutralization reactions • Redox reactions • 3. Precipitation reactions

  9. Where does the ‘heat’ go?

  10. Constant Volume Calorimeter ΔQ = (cmv × n + cbomb) ΔT cmv(H2O) ≈ cmp(H2O)

  11. Constant Volume Calorimeter DH = ΔQ = cpΔT ΔE = ΔQ + ΔW ΔQ = cvΔT ΔW = - p ΔV

  12. CHANGING THE INTERNAL ENERGY DE = ΔQ + Δw C8H18(l) + (25/2) O2(g) ENERGY Einitial E lost as HEAT E lost as HEAT and WORK 8 CO2(g) + 9H2O(g) Efinal

  13. CHANGING THE INTERNAL ENERGY DE = ΔQ + Δw different combinations of q and w INITIAL STATE Q  0 w  0 Q  0 w = 0 INTERNAL ENERGY, E FINAL STATE

  14. SUMMARY measurement of heat changes DHm = ΔQm = cmpΔT ΔE = ΔQ + ΔW cmp(H2O) = 75.3 J mol-1 K-1

  15. Homework Chapter 6, p. 212-216 problems

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