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# Lecture17221

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### Lecture17221

1. 1. Thermochemistry: Calorimetry and Stoichiometry Lecture 17
2. 2. Calories everywhere.
3. 3. Energy content of chocolate
4. 4. However, how do we measure the heat released by a chocolate bar?
5. 5. We cannot measure the enthalpy of a system just because we do not have a starting point with which to compare it.
6. 6. We can measure the change in enthalpy of a system. But not directly.
7. 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. 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. 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. 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. 11. Specific heat capacity
12. 12. Specific heat capacities
13. 13. Same specific heat capacity, different heat capacities:
14. 14. Same specific heat capacity, different heat capacities:
15. 15. How to find q P (water)
16. 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. 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. 18. A sample problem on finding the quantity of heat from specific heat capacity.
19. 19. Calorimeter is a device used to measure the heat released (or absorbed) by a chemical or physical process.
20. 20. Constant-pressure calorimetry uses “coffee-cup” calorimeter. The processes open to atmosphere.
21. 21. Coffee-cup calorimeter
22. 22. A sample problem on determining the heat of a reaction.
23. 23. Constant-volume calorimetry uses “bomb” calorimeter. The processes are closed to atmosphere.
24. 24. Bomb calorimeter
25. 25. A sample problem on calculating the heat of combustion.
26. 26. Thermochemical equation is a balanced equation that includes the heat of reaction (ΔH rxn ).
27. 27. ΔH rxn value refers to the amounts of substances and their states of matter in that specific equation.
28. 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. 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. 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. 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. 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. 33. A sample problem using the heat of reaction to find mass of substance.
34. 34. A sample problem using the mass of substance to find the heat of reaction.
35. 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. 36. THE END