Thermodynamics & Thermochemistry

1. Thermodynamics & Thermochemistry Unit 10

2. Overview • Energy • Units • Heat /temperature • System/surrounding • Calorimeters • 1st Law Thermodynamics • Heat Capacity • Enthalpy • ∆H and ∆H° • Endothermic/exothermic • Heat of formation • Stoichiometry • Bond energy • Hess’s Law • Entropy • ∆S and ∆S° • Spontaneous reactions • Entropy for phase changes • 2nd Law of Thermodynamics • Gibbs Free Energy • ∆G and ∆G° • ∆G = ∆H - T∆S • Spontaneous reactions • State functions

3. What is Thermodynamics? What is Thermochemistry? The study of the energy and its transformations The study of the transfers of energy as heat that accompany chemical reactions and physical changes

4. Energy • The capacity to do work or to transfer heat • Work: the energy used to cause an object with mass to move against a force • Heat: the energy used to cause the temperature change in an object • Units of Energy • Joule(J) = 1 kg∙m2/s2 • Calorie (cal) = 4.184 J • In nutrition 1 calorie (Cal) = 1000 cal • (notice difference is that nutrition calorie is capitalized) • In chemistry 1000 cal = 1 kcal

5. What is the difference between heat and temperature?

6. What’s the difference? HEAT TEMPERATURE Measure of the average kinetic energy of the particles in a sample of matter Directly measureable (affected by heat transfer) Celsius (°C) or Kelvin (K) Higher kinetic energy = higher temperature • Energy transferred between samples of matter because of a difference in their temperatures • Cannot be measured directly (determined by temperature change) • Joules (J) = 1 kg∙m2/s2 • Moves from high to low temperature

7. How do we measure energy? Calorimeter Measures energy absorbed or released as heat in a chemical/physical change Constant Pressure Example “coffee-cup” calorimeter uses unsealed calorimeter where reaction occurs under atmospheric pressure

8. How do we measure energy? Constant Volume • Example “bomb” calorimeter uses pressurized sealed vessel called “bomb” submerged in water

9. System and Surroundings • System: portion we single out for studying • Open system: matter and energy exchanged with surroundings • Closed system: energy exchanged within system • Surroundings: everything else that is not included in the system

10. Work E = q + w E= change in internal energy of a system q= heat flowing into or out of the system -qif energy is leavingthe system +q if energy is enteringthe system w= work done by, or on, the system -w if work is done bythe system on the surroundings +w if work is done onthe system by thesurroundings

11. Pressure-Volume Work • Most common type of work. • Example: A sample of gas is held by a pressure of 2.4 atm. The pressure is then changed to 1.3 atm. The gas expands. How much work is done? • Note that because the gas is expanding, the system is doing work on the surroundings so work must be negative

12. Pressure-Volume Work When the volume (V) of a system increases by an amount of ΔV against an external pressure (p), the system pushes back, and thus does pV work on the surroundings w = -PΔV

13. Pressure-Volume Work Compression Expansion +V(increase) -V(decrease) -wresults +wresults Esystemdecreases Esystemincreases Work has been done on the system by the surroundings Work has been done by the system on the surroundings

14. FIRST LAW OF THERMODYNAMICS Energy is neither created nor destroyed (conservation of energy)

15. FIRST LAW OF THERMODYNAMICS • Energy is transferred but not lost • Energy lost by a system must be gained by surroundings and vice versa

16. What determines how much heat is transferred during a temperature change?

17. Amount depends on… • Nature of material changing temperature. • Mass of material changing temperature. • Size of the temperature change.

18. Example 1

19. Example 2 One gram of each of two metals (iron and silver) are heated to 100°C and cooled to 50°C in a beaker of water. Iron transfers 22.5 J of energy. Silver transfers 11.8 J of energy. What makes the difference?

20. Heat Capacity • Heat Capacity (Cp) – temperature change experienced by an object when it absorbs a certain amount of heat • The greater the heat capacity, the greater the heat required to produce given increase in temperature

21. Heat Capacity Cp = Cp = heat capacity ∆H = heat added (J or cal) ∆T = temperature change (°C or K) ∆H ∆T

22. Heat Capacity (Pure Substances) • Molar Heat Capacity (Cm): Amount of energy required to raise the temperature of 1 mole of a substance by 1 degrees – Celsius or Kelvin • Specific Heat (Cs): Amount of energy required to raise the temperature of 1 gram of a substance by 1 degrees – Celsius or Kelvin • Higher specific heat means… • Higher heat capacity • Heats up slower • Transfers more heat

23. Specific Heat

24. Explain in terms of specific heat

25. Calculations q = mc∆T q = heat added (J or cal) m = mass of substance (g or kg) c = specific heat ∆T = temperature change (°C or K)

26. Specific Heat (Units) J/g°C or J/gK

27. Example 1 A 4.0 g sample of glass was heated from 274 K to 314 K and was found to have absorbed 32 J of heat. • What is the specific heat of this type of glass? • How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

28. Example 2 A piece of copper alloy with a mass of 85.0 g is heated from 30.°C to 45°C. In the process, it absorbs 523 J of energy as heat. • What is the specific heat of this copper alloy? • How much energy will the same sample lost if it is cooled from 45°C to 25°C?

29. Enthalpy

30. What is Enthalpy? The amount of energy absorbed by a system as heat during a process at constant pressure. H

31. How does it apply to reactions? Enthalpy Change Difference between enthalpies of products and reactants H = Hproducts - Hreactants Enthalpy of Reaction Quantity of energy transferred as heat during chemical reaction (“heat of reaction”)

32. What’s the difference? Endothermic Exothermic Heat released Energy lost • Heat absorbed • Energy gained

33. When you gamble and lose your money, do you end up with less money or more? ….do you have a “positive or negative” amount of money in your pocket after? When you gamble and win money, do you end up with less money or more? ….do you have a “positive or negative” amount of money in your pocket after?

34. What’s the difference? Endothermic Exothermic Heat released Energy lost H = negative (-) H2 + O2  H2O + 483.6 kJ H2 + O2 H2O H = -483.6 kJ • Heat absorbed • Energy gained • H = positive (+) • H2O + 483.6 kJ  H2 + O2 • H2O  H2 + O2H = +483.6 kJ

35. Energy Diagrams • Activation Energy – energy needed for reactants to transition to products (become activated complex) • At activated complex, all bonds have been broken but no bonds formed (highest energy/lowest stability) • Exothermic diagram  energy of products are lower than energy of reactants (∆H is negative)

36. Energy Diagrams • Activation Energy – energy needed for reactants to transition to products (become activated complex) • At activated complex, all bonds have been broken but no bonds formed (highest energy/lowest stability) • Endothermic diagram  energy of products are higher than energy of reactants (∆H is positive)

37. Most reactions in nature are exothermic… Hf = negative

38. When Hfpositive or slightly positive, the compound is unstable… …very positive Hf can even be explosive For example: Mercury fulminate is used as a detonator in explosives. HgC2N2O2Hf = + 270 kJ/mol

39. Standard State Conditions • All gases are at 1 atm • All liquids are pure • All solids are pure • All solutions are at 1 M concentration • Temperature is room temperature (25°C or 298K) unless told otherwise • Energy of formation of element in its normal state = 0 • Standard state conditions indicated by superscript 0 • Example: ∆H° means standard state conditions

40. Formation of Compounds Enthalpy of Formation (Hf) – Heat of Formation Enthalpy change that occurs when 1 mole of a compound is formed from its elements (sometimes called molar enthalpy of formation) To represent standard state useHf0 Elements in their standard state have Hf0= 0

41. Heat of Formation • To form compounds from elements or compounds in their standard states break apart compounds into individual components • Balance by using fractions on standard state elements/compounds rather than whole numbers Example 1: Equation for formation of NaCl(s) Na(s) + ½ Cl2(g)  NaCl(s) Example 2: Equation for formation of CaCO3(s) Ca(s) + C(s) + 3/2 O2(g)  CaCO3(s)

42. Heat of Formation Hf0 = ∑ Hf0 products - ∑ Hf0 reactants Example: 2CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g) Hf0 = [2(-394) + 4(-242)] – [2(-201) + 3(0)] Hf0 = [-1756] – [-402] Hf0 = -1354 kJ

43. Enthalpy Stoichiometry How much heat is released when 5.40 g of methane gas is burned in a constant-pressure system? CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ∆H = -890 kJ 4.50 g CH4 × × = - 250 kJ 1 mol CH4 16.0 g CH4 -890 kJ 1 mol CH4

44. Bond Energy • Energy required to break a bond. • Breaking a bond requires energy (endothermic) • Bond energy is positive • Forming a bond releases energy (exothermic) • Bond energy is negative H0 = ∑ Bond Energies of bonds broken – ∑Bond Energies of bonds formed H0 = ∑ Bond Energyreactants– ∑ Bond Energyproducts

45. Bond Energy Example: 2H2(g) + O2(g)  2H2O(g) H0 = [2(H-H) + 1(O-O)] – [4(O-H)] H0 = [2(436) + 1(499)] – [4(463)] H0 = -481 kJ/mol H0 = ∑ Bond Energies of bonds broken – ∑Bond Energies of bonds formed

46. How can we find Hf for these types of reactions? Sometimes finding the enthalpy change in reactions is difficult or even impossible due to the reaction.

47. Hess’s Law The overall enthalpy change in a reaction equals the sum of the enthalpy changes for individual steps in the process. Can be used to find the Hf for reactions that are difficult or impossible to measure.

48. Example C(s) + 2H2(g) CH4(g)Hf0 = ? We can use the following to help… C(s) + O2(g) CO2(g)Hf0 = -393.5 kJ H2(g) + ½O2(g)  H2O(l)Hf0 = -285.8 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)Hf0 = -890.8 kJ

49. Example Rules to remember: • When a reaction is reversed, the sign of Hf must be reversed (+/-). • When coefficients are multiplied by a number, Hf must be multiplied by the same number.

50. Example C(s)+ O2(g) CO2(g)Hf0 = -393.5 kJ H2(g) + ½O2(g)  H2O(l)Hf0 = -285.8 kJ CH4(g)+ 2O2(g)  CO2(g) + 2H2O(l)Hf0 = -890.8 kJ C(s)+ 2H2(g) CH4(g)Hf0 = ?