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- 1. Thermochemistry: Calorimetry and Stoichiometry Lecture 17
- 2. Calories everywhere.
- 3. Energy content of chocolate
- 4. However, how do we measure the heat released by a chocolate bar?
- 5. We cannot measure the enthalpy of a system just because we do not have a starting point with which to compare it.
- 6. We can measure the change in enthalpy of a system. But not directly.
- 7. To measure q P (equals ΔH), we: <ul><li>construct “surroundings” that retain the heat; </li></ul><ul><li>observe the temperature change; </li></ul><ul><li>relate the quantity of heat released (or absorbed) to that temperature change through the specific heat capacity. </li></ul>
- 8. The more we heat an object, the higher its temperature. The quantity of heat (q) absorbed by an object is proportional to its temperature change: q ~ ΔT or q = constant x ΔT or q/ΔT = constant
- 9. Heat capacity of an object equals quantity of heat required to change its temperature by 1 K. Q /ΔT = constant Heat capacity = q /ΔT [in units J/K]
- 10. Specific heat capacity (c) of a substance equals quantity of heat required to change the temperature of 1 gram of this substance by 1 K. Q /ΔT = constant Specific heat capacity (c) = q / (mass x ΔT) [in units J/g*K]
- 11. Specific heat capacity
- 12. Specific heat capacities
- 13. Same specific heat capacity, different heat capacities:
- 14. Same specific heat capacity, different heat capacities:
- 15. How to find q P (water)
- 16. Molar heat capacity (C) of a substance equals quantity of heat required to change the temperature of 1 mole of this substance by 1 K. Q /ΔT = constant Molar heat capacity = q / (moles x ΔT) [in units J/mol*K]
- 17. The specific heat capacity of liquid water c(H 2 O) = 4.181 J/g*K So molar heat capacity of liquid water C(H 2 O) = 4.184 J/(g*K) x 18.02 g/mol = = 75.40 J/(mol*K)
- 18. A sample problem on finding the quantity of heat from specific heat capacity.
- 19. Calorimeter is a device used to measure the heat released (or absorbed) by a chemical or physical process.
- 20. Constant-pressure calorimetry uses “coffee-cup” calorimeter. The processes open to atmosphere.
- 21. Coffee-cup calorimeter
- 22. A sample problem on determining the heat of a reaction.
- 23. Constant-volume calorimetry uses “bomb” calorimeter. The processes are closed to atmosphere.
- 24. Bomb calorimeter
- 25. A sample problem on calculating the heat of combustion.
- 26. Thermochemical equation is a balanced equation that includes the heat of reaction (ΔH rxn ).
- 27. ΔH rxn value refers to the amounts of substances and their states of matter in that specific equation.
- 28. The enthalpy change has a sign. If reaction is exothermic, ΔH rxn is negative : S 8(s) + O 2(g) 8 SO 2(g) ΔH rxn < 0 If reaction is endothermic, ΔH rxn is positive : N 2(g) + O 2(g) 2N O (g) ΔH rxn > 0
- 29. The enthalpy change has a sign. A forward reaction has the opposite sign of the reverse reaction: 2H 2 O (l) 2H 2 O (g) + O 2(g) ΔH rxn = 572 kJ 2H 2 O (g) + O 2(g) 2H 2 O (l) ΔH rxn = -572 kJ
- 30. The enthalpy change has a magnitude. It is proportional to the amount of substance reacting: 2 H 2 O (g) + 1 O 2(g) 2 H 2 O (l) ΔH rxn = -572 kJ 1 H 2 O (g) + 1/2 O 2(g) 1 H 2 O (l) ΔH rxn = -286 kJ
- 31. Fractional coefficients are common in thermochemistry: 2 H 2 O (g) + 1 O 2(g) 2 H 2 O (l) ΔH rxn = -572 kJ 1 H 2 O (g) + 1/2 O 2(g) 1 H 2 O (l) ΔH rxn = -286 kJ
- 32. molar ratio from balanced equation H rxn (kJ/mol) Figure 6.11 Summary of the relationship between amount (mol) of substance and the heat (kJ) transferred during a reaction. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. AMOUNT (mol) of compound A AMOUNT (mol) of compound B HEAT (kJ) gained or lost
- 33. A sample problem using the heat of reaction to find mass of substance.
- 34. A sample problem using the mass of substance to find the heat of reaction.
- 35. <ul><li>Problem: When 1 mol of NO (g) forms from its elements, 90.29 kJ of heat is absorbed. How much heat is involved when 1.50 g of NO decomposes to its elements? </li></ul><ul><li>Plan : forward balanced equation reverse balanced equation ; grams of NO decomposing mols of NO decomposing heat (kJ). </li></ul><ul><li>Solution: N 2(g) + O 2(g) 2NO (g) </li></ul><ul><li>1/2N 2(g) + 1/2O 2(g) NO (g) ΔH rxn = 90.29 kJ </li></ul><ul><li>NO (g) 1/2N 2(g) + 1/2O 2(g) ΔH rxn = -90.29 kJ </li></ul><ul><li>Amount of NO = mass of NO / molar mass of NO </li></ul><ul><li>ΔH = - 90.29 kJ/mol x 1.50 g / (14+16) g/mol = -4.515 kJ </li></ul>
- 36. THE END

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