1. Thermochemistry Chapter 5

2. First Law of Thermodynamics states that energy is conserved.Energy that is lost by a system must be gained by the surroundings or vice versa.

3. Enthalpy (H) accounts for heat flow in chemical changes occurring at constant pressure

4. Enthalpy (H) State function: H=H final -H initial

5. Enthalpy (H) When  H > 0, the system has gained heat from surroundings (endothermic) When  H < 0, the system has released heat to the surroundings (exothermic)

6. Enthalpy (H)

7. Enthalpies of Reaction Thermochemical equations are balanced chemical equations that show the associated enthalpy change

8. Enthalpies of Reaction D H = H(products) – H(reactants)

9. Guidelines (1) H is an extensive property meaning it depends on the amount of reactant consumed in the process

10. Guidelines (2) The  H for a reaction is equal in magnitude, but opposite in sign, to the reverse reaction

11. Guidelines (2)

12. Guidelines (3) The  H for a reaction depends on the state of the reactants and products and we assume they are at constant T

13. Practice 5.5 How much heat is released when 4.50 g of methane gas is burned in a constant pressure system? Given: CH 4 (g) + 2O 2 (g)  CO 2 (g) +2H 2 O(l)  H = -890kJ

14.  H can be determined experimentally by measuring the heat flow accompanying a reaction at constant pressure which is done by measuring Temperature changes Enthalpies of Reaction

15. Calorimetry is the measurement of heat flow A calorimeter will measure this heat flow

16. Heat Capacity the amount of heat required to raise the temperature of an object by 1K or 1 o C.

17. Molar Heat Capacity heat capacity of one mole of a substance

18. Specific Heat Capacity heat capacity of 1g of substance

19. Specific Heat Capacity

20. Enthalpies of Reaction Heat = specific heat x grams of substance x  T q=cm  T

21. Sample Exercise 5.5 Relating Heat, Temperature Change, and Heat Capacity (a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 °C (about room temperature) to near its boiling point, 98 °C? The specific heat of water is 4.18 J/g-K. (b) What is the molar heat capacity of water?

22. Constant- Pressure Calorimetry

23. (1)Assume the calorimeter prevents loss or gain of heat from the solution to its surroundings

24. Constant- Pressure Calorimetry (2a) For an exothermic rxn, heat is lost by rxn and gained by soln so the T of soln rises

25. Constant- Pressure Calorimetry (2b) For an endothermic rxn, heat is gained by rxn and lost by soln so the T of soln goes down

26. Constant- Pressure Calorimetry (3) q soln = specific heat of solution x gram of solution x  T = -q rxn

27. Sample Exercise 5.6 Measuring ΔH Using a Coffee-Cup Calorimeter When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g-K.

28. Bomb Calorimetry Constant-Volume CalorimetryConstant-Volume Calorimetry Combustion reactions are studied in bomb calorimetersCombustion reactions are studied in bomb calorimeters

29. Bomb Calorimetry

30. The heat released in combustion is absorbed by the calorimeter contents, causing a rise in the temperature of the waterThe heat released in combustion is absorbed by the calorimeter contents, causing a rise in the temperature of the water

31. Bomb Calorimetry To calculate  H combustion from the measured temperature increase, we must know the heat capacity of the bomb calorimeter, CcalTo calculate  H combustion from the measured temperature increase, we must know the heat capacity of the bomb calorimeter, Ccal

32. Bomb Calorimetry q rxn = -Ccal x  Tq rxn = -Ccal x  T

33. Sample Exercise 5.7 Measuring qrxn Using a Bomb Calorimeter Methylhydrazine (CH6N2) is used as a liquid rocket fuel. The combustion of methylhydrazine with oxygen produces N2(g), CO2(g), and H2O(l): 2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l) When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °C to 39.50 °C. In a separate experiment the heat capacity of the calorimeter is measured to be 7.794 kJ/°C. Calculate the heat of reaction for the combustion of a mole of CH6N2.

34. Hess’ Law It is possible to calculate  H from tabulated values for other reactionsIt is possible to calculate  H from tabulated values for other reactions

35. Hess’ Law If a reaction is carried out in a series of steps,  Hrxn will equal the sum of  H for the individual steps.If a reaction is carried out in a series of steps,  Hrxn will equal the sum of  H for the individual steps.  H rxn =  H 1 +  H 2 + …

36. Hess’ Law The overall enthalpy change is independent of the number of steps by which the reaction is carried outThe overall enthalpy change is independent of the number of steps by which the reaction is carried out

37. Hess’ Law Hess’ law provides ways to calculate energy changes that are difficult to measure directly.Hess’ law provides ways to calculate energy changes that are difficult to measure directly.

38. Sample Exercise 5.8 Using Hess’s Law to Calculate ΔH

39. Hess’ Law

40. Hess’ Law (5.9)

41. Practice Exercise

42. Standard enthalpy of a reaction is the enthalpy change when all reactants and products are in their standard states,  H o rxnis the enthalpy change when all reactants and products are in their standard states,  H o rxn

43. Standard States Standard state is when a substance is in its pure form at 1 atm and 298K.Standard state is when a substance is in its pure form at 1 atm and 298K.

44. Standard enthalpy of formation  H o f,, reported in kj/mol, is the change in enthalpy for the reaction that forms 1 mole of the compound from its elements, with all substances in their standard state.

45. Standard enthalpy of a formation  H f o of the most stable form of an element is zero because there is no formation reaction needed for an element in its standard state.  H f o of the most stable form of an element is zero because there is no formation reaction needed for an element in its standard state.

46. Standard enthalpy of a formation

47. Standard enthalpy of a formation (5.10)

48. Enthalpy of Reaction  H o rxn =  n  H o f (products) -  m  H o f (reactants), where n and m are the coefficients in the balanced chemical equation  H o rxn =  n  H o f (products) -  m  H o f (reactants), where n and m are the coefficients in the balanced chemical equation

49. 5.11 Enthalpy of Reaction Calculate the standard enthalpy change for the combustion of 1 mol of benzene(l), to form CO2(g) and H2O(l).

50. 5.12 Enthalpy of Reaction The standard enthalpy change for the decomposition reaction of CaCO 3 is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) and CO 2 (g), calculate the standard enthalpy of formation of CaCO 3 (s).