2. Thermochemistry Part 4: Phase Changes & Enthalpies of Formation

3. Specific Heat  Specific heat: The amount of heat that must be added to a stated mass of a substance to raise it’s temperature, with no change in state.

4. Example: How much heat is released by 250.0 g of H 2 O as it cools from 85.0 o C to 40.0 o C? (Remember, specific heat of water = 4.18 J/g o C) q = mcT q = (250.0 g)(4.18 J/g o C)(40.0-85.0) q = -47,025 J = -47.0 kJ

5. But what if there is a phase change?

6. LATENT HEAT OF FUSION,  H fus  Definition: the enthalpy change (energy absorbed) when a compound is converted from a solid to a liquid without a change in temperature.  “Latent” means hidden; the heat absorbed/released during a phase change does not cause the temperature to change.  Note: H fus for water is 334 J/g

7. LATENT HEAT OF FUSION,  H fus A = solid B = melting (solid + liquid) C = liquid D = boiling (liquid + gas) E = gas

8. LATENT HEAT OF VAPORIZATION,  H vap  Definition: the enthalpy change (energy absorbed) when one mole of the compound is converted from a liquid to a gas without a change in temperature.  Note: for water H vap is 2260 J/g

9. A = solid B = melting (solid + liquid) C = liquid D = boiling (liquid + gas) E = gas LATENT HEAT OF VAPORIZATION,  H vap

10. Example 1: How much heat is released by 250.0 g of H 2 O as it cools from 125.0  C to -40.0  C? Five steps… 1. Cool the steam m∙c steam ∙T 2. Condense m(-H vap ) 3. Cool the liquid waterm∙c water ∙T 4. Freezem(-H fus ) 5. Cool the solid icem∙c ice ∙T

11. Example 1: How much heat is released by 250.0 g of H 2 O as it cools from 125.0  C to -40.0  C? When substances change state, they often have different specific heats: c ice = 2.09 J/g o C c water = 4.18 J/g o C c steam = 2.03 J/g o C

12. Example 1: How much heat is released by 250.0 g of H 2 O as it cools from 125.0  C to -40.0  C? q total = -787,000J q ice = mc  T = (250.0g)(2.09J/g o C)(-40.0-0) = -20,900 J q fus = mH fus = (250.0g)(-334J/g) = -83,500 J q water = mc  T = (250.0g)(4.18J/g o C)(0-100) = -105,000 J q vap = mH vap = (250.0g)(-2260J/g) = -565,000 J -787 kJ q steam = mc  T = (250.0g)(2.03J/g o C)(100.0–125.0) = -12,700 J cooling = exothermic → negative heat values

13. Now YOU try a few…

14. Example 2: How much heat energy is required to bring 135.5 g of water at 55.0 o C to it’s boiling point (100. o C) and then vaporize it? 1. Water must be heated to it’s boiling point. q = mcT = (135.5 g)(4.18J/g o C)(100-55.0) q = 25,500 J 2. Water must be vaporized: q = mH vap = (135.5 g)(2260 J/g) q = 306,000 J Add them together: q = 25,500 J + 306,000 J = 331,500 J

15. Example 3: How much energy is required to convert 15.0 g of ice at -12.5 o C to steam at 123.0 o C? 1. Heat the ice (from -12.5 C  0 C) 2. Melt the ice (at 0 C) 3. Heat the water (from 0 C  100 C) 4. Vaporize the water (at 100 C) 5. Heat the steam. (from 100 C  123 C)

16. q ice = mc  T = (15.0g)(2.09J/g o C)(0.0 - -12.5) = 392 J q total = 392 J + 5,010 J + 6,270 J + 33,900 J + 700. J = 46,300J q steam = mc  T = (15.0g)(2.03J/g o C)(123.0-100.0) = 700. J q vap = mH vap = (15.0g)(2260J/g) = 33,900 J q water = mc  T = (15.0g)(4.18J/g o C)(100-0) = 6,270 J q fus = mH fus = (15.0g)(334J/g) = 5,010 J 46.3 kJ Example 3: How much energy is required to convert 15.0 g of ice at -12.5 o C to steam at 123.0 o C?

17. Enthalpies of Formation enthalpy change (delta) standard conditions formation

18. Enthalpies of Formation  usually exothermic  see table for H f  value  enthalpy of formation of an element in its stable state = 0  these can be used to calculate H for a reaction

19. Standard Enthalpy Change Standard enthalpy change, H, for a given thermochemical equation is = to the sum of the standard enthalpies of formation of the product – the standard enthalpies of formation of the reactants. “sum of” (sigma)

20. Standard Enthalpy Change  elements in their standard states can be omitted: 2 Al(s) + Fe 2 O 3 (s)  2 Fe(s) + Al 2 O 3 (s) ΔH rxn =  (ΔH f  products ) -  (ΔH f  reactants ) ΔH rxn = ΔH f  Al 2 O 3 - ΔH f  Fe 2 O 3 ΔH rxn = (-1676.0 kJ) – (-822.1 kJ) ΔH rxn = -853.9 kJ

21. Standard Enthalpy Change  the coefficient of the products and reactants in the thermochemical equation must be taken into account: 2 Al(s) + 3 Cu 2+ (aq)  2 Al 3+ (aq) + 3 Cu(s) ΔH rxn =  (ΔH f  products ) -  (ΔH f  reactants ) ΔH rxn = 2ΔH f  Al 3+ - 3ΔH f  Cu 2+ ΔH rxn = 2(-531.0 kJ) – 3(64.8 kJ) ΔH rxn = -1256.4 kJ

22. Standard Enthalpy Change  Example: Calculate H for the combustion of one mole of propane: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(l) ΔH rxn =  (ΔH f  products ) -  (ΔH f  reactants ) ΔH rxn = [3ΔH f  CO 2 +4ΔHfH 2 O] - ΔH f  C 3 H 8 ΔH rxn = [3(-393.5kJ)+4(-285.8kJ)]–(-103.8 kJ) ΔH rxn = -2219.9 kJ

23.  Example: The thermochemical equation for the combustion of benzene, C 6 H 6, is: C 6 H 6 (l) + 15/2 O 2 (g)  6CO 2 (g) + 3H 2 O(l) H = -3267.4 kJ Calculate the standard heat of formation of benzene. -3267.4kJ = [6(-393.5kJ)+3(-285.8kJ)]–ΔH f  C 6 H 6 ΔH f  C 6 H 6 = +49.0 kJ -3267.4kJ = -3218.4–ΔH f  C 6 H 6 -49.0kJ = –ΔH f  C 6 H 6

24. Standard Enthalpy Change  Example: When hydrochloric acid is added to a solution of sodium carbonate, carbon dioxide gas is formed. The equation for the reaction is: 2H + (aq) + CO 3 2- (aq)  CO 2 (g) + H 2 O(l) Calculate H for this thermochemical equation. ΔH = [(-393.5kJ)+(-285.8kJ)]–[2(0 kJ)+(-677.1kJ) ΔH = (-679.3kJ)–(-677.1kJ) ΔH = -2.2 kJ