2.  Section 1 – Thermochemistry  Section 2 – Driving Force of Reactions

3.  Define temperature and state the units in which it is measured.  Define heat and state its units.  Perform specific-heat calculations.  Explain enthalpy change, enthalpy of reaction, enthalpy of formation, and enthalpy of combustion.  Solve problems involving enthalpies of reaction, enthalpies of formation, and enthalpies of combustion.

4.  Virtually every chemical reaction is accompanied by a change in energy. ◦ What two ways can energy be involved?  Thermochemistry: study of the transfer of energy as heat that accompanies chemical reactions and physical changes.

5.  Calorimeter: used to measure heat absorbed or released.  Energy released from the reaction is measured from the temp change of the water

6.  measure of the average kinetic energy of the particles in a sample of matter. ◦ The greater the kinetic energy of the particles in a sample, the hotter it feels.  Remember: How do we convert Celsius to Kelvin?

7.  Joule: SI unit of heat and energy  Heat: the energy transferred between samples of matter because of a difference in their temperatures.  Energy transferred as heat moves spontaneously  from matter at a higher temp  lower temp

8.  Energy transferred depends on: ◦ The material ◦ The mass  Specific Heat: specific heat of a substance is the amount of energy required to raise the temperature of one gram by one Celsius degree (1°C) or one kelvin (1 K). ◦ Units: J/(g°C) or J/(gK) ◦ The temperature difference as measured in either Celsius degrees or kelvins is the same.

10. Heat released Specific heat: or absorbed: c p = specific heat at a given pressure q = energy lost or gained m = mass of the sample ∆T = difference between the initial and final temperatures.

11.  A 4.0 g sample of glass was heated from 274 K to 314 K, and was found to have absorbed 32 J of energy as heat. a. What is the specific heat of this type of glass? b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

12. Given: m = 4.0 g ∆ T = 314 – 274 = 40. K q = 32 J Unknown:a. c p in J/(gK) b. q for ∆ T of 314 K → 344 K Solution: a.

13. b.

14.   Please note how each part of the curve is numbered. These numbers correspond to the steps in our problem-solving.

15.  Assume we have 85 grams of ice at -10 °C. How much energy is needed to convert it to water vapor at 120 o C?  Step 1 – We are below 0°C ◦ What is the phase? ◦ What is happening to the particles as heat is added? ◦ General formula: q=(m)(cp)(∆T) ◦ Answer = 1751 J or 1.8kJ  Step 2 - We are at 0 °C ◦ What is happening to the phase? ◦ Notice how there is no increase in temp. ◦ Where is this energy going? ◦ General formula: q=(∆H fusion)(mol) ◦ Answer = 28 kJ

16.  Step 3 – We are above 0 °C but below 100 °C ◦ What is the phase? ◦ What is happening to the particles as we heat? ◦ General formula: q=(m)(cp)(∆T) ◦ Answer = 36 kJ  Step 4 - We are at 100 °C ◦ What is happening to the phase? ◦ Notice how there is no increase in temp. ◦ Where is this energy going? ◦ General formula: q=(∆H vap)(mol) ◦ Answer = 190 kJ  Step 5 – We are above 100 °C ◦ What is the phase? ◦ What is happening to the particles as we heat? ◦ General formula = (m)(cp)(∆T) ◦ Answer = 3.2kJ

17.  Now, add up all the steps for total heat. ◦ Answer = 259 kJ  Summary:  When the substance is heating up and is in a single phase, the formula used is : q=(m)(cp)(∆T)  When a phase change is occurring all the energy is involved with the intermolecular forces and because of this we do not see a temperature change. To calculate we must multiply the moles by molar heat; ◦ q=(∆H fusion)(mol) OR ◦ q=(∆H vap)(mol)

18.  Enthalpy(∆H): “change in enthalpy” energy absorbed as heat during a chemical reaction at constant pressure

19.  The difference between the enthalpies of products and reactants. ∆H = H products – H reactants

20.  Equation that includes the quantity of energy released or absorbed as heat 2H 2 (g) + O 2 (g) → 2H 2 O(g) + 483.6 kJ  States of matter are ALWAYS shown because it can influence the energy of released or absorbed.  Coefficients are always viewed as number of moles in the reaction.  If 2x as many reactants were provide, 2x as many moles of water would be produced and 2x as many kJ of energy would be released.

21.  chemical reaction that releases energy the energy of the products is less than the energy of the reactants ◦ example: 2H 2 (g) + O 2 (g) → 2H 2 O(g) + 483.6 kJ

22.  chemical reaction that absorbs energy the products have a larger enthalpy than the reactants 2H 2 O(g) + 483.6 kJ → 2H 2 (g) + O 2 (g)

23.  Enthalpy is often not included in the reaction expression, instead as a ∆ H value ◦ Example: 2H 2 (g) + O 2 (g) → 2H 2 O(g) ∆ H = –483.6 kJ ◦ Negative ∆H is exothermic – releases heat ◦ Positive ∆H is endothermic – absorbs heat

24. Energy released, ∆ H is negative.

25. Energy is absorbed, ∆ H is positive.

26.  molar enthalpy of formation: enthalpy change that occurs when one mole of a compound is formed from its elements in their standard state at 25°C and 1 atm. ◦ Enthalpies of formation are given for a standard temperature and pressure so that comparisons between compounds are meaningful. standard states heat of FORMATION change in enthalpy

27.  In your textbook!  The values are given as the enthalpy of formation for one mole of the compound from its elements in their standard states.

28.  Compounds with a large negative enthalpy of formation are very stable.  Compounds with positive values of enthalpies of formation are typically unstable.

29.  ∆H c : enthalpy change that occurs during the complete combustion of one mole of a substance **Enthalpy of combustion is defined in terms of one mole of reactant **Enthalpy of formation is defined in terms of one mole of product

31.  Hess’s Law: the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process.  This means that the energy difference between reactants and products is independent of the route taken to get from one to the other.

32.  If you know the reaction enthalpies of individual steps in an overall reaction, you can calculate the overall enthalpy without having to measure it experimentally. What is the heat of formation of methane? C(s) + 2H 2 (g) → CH 4 (g)

33. H 2 (g) + ½O 2 (g) → H 2 O(l) C(s) + O 2 (g) → CO 2 (g) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l)  The component reactions in this case are the combustion reactions of carbon, hydrogen, and methane: This reaction must be reversed! The Δ H is changed to positive! CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ∆ H 0 = +890.8 kJ 2 [ H 2 (g) + ½O 2 (g) → H 2 O(l) ] This reaction must x2! The Δ H is x2! Goal Reaction: C(s) + 2H 2 (g) → CH 4 (g)

34. 2H 2 (g) + O 2 (g) → 2H 2 O(l) C(s) + O 2 (g) → CO 2 (g) C(s) + 2H 2 (g) → CH 4 (g) CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g)

35. ∆ H 0 = Σ [( ΔH 0 f of products) × (mol of products)] – Σ [( ΔH 0 f of reactants) × (mol of reactants)]

36.  Explain the relationship between enthalpy change and the tendency of a reaction to occur.  Explain the relationship between entropy change and the tendency of a reaction to occur.  Discuss the concept of free energy, and explain how the value of this quantity is calculated and interpreted.  Describe the use of free energy change to determine the tendency of a reaction to occur.

37.  Majority of spontaneous reactions are exothermic ◦ Reactions proceed in a direction that leads to a lower energy state  Some spontaneous reactions are endothermic ◦ Example: melting (particles have more energy in the liquid state)

38.  Entropy: ◦ Increases when a system goes from one state to another without an enthalpy change 2NH 4 NO 3 (s)  2N 2 (g) + 4H 2 O(l) + O 2 (g)  The arrangement of particles on the right- hand side of the equation is more random than the arrangement on the left side and hence is less ordered.

39.  Entropy (S) : the degree of randomness or disorder of the particles in a system  There is a tendency of nature to go from order to disorder ◦ Think about your bedroom, does it always stay neat and organized? Or does it get messier and messier? ◦ Increase in disorder  Increase in entropy ◦ The entropy of the universe is ALWAYS increasing

41.  As you go from a solid  liquid  gas What happens to the entropy?  As you go from a gas  liquid  solid What happens to the entropy?

43.  Unit: kJ/(molK)  Change in Entropy ΔS ◦ Difference between the entropy of the products and the reactants  Increase in Entropy  Positive Value  Decrease in Entropy  Negative Value

44.  Spontaneous Reactions: ◦ Least enthalpy ◦ Greatest entropy  Gibb’s Free Energy (G): ◦ Combined enthalpy-entropy function

45.  Only the change in free energy can be measured.  At a constant pressure and temperature, the free-energy change, ∆G, of a system is defined as the difference between the change in enthalpy, ∆H, and the product of the Kelvin temperature and the entropy change, which is defined as T∆S: ∆G 0 = ∆H 0 – T ∆ S 0

47.  The expression for free energy change is for substances in their standard states. Unit: kJ/mol  If ∆G < 0  the reaction is spontaneous.

49. For the reaction NH 4 Cl (s) → NH 3(g) + HCl (g) ∆H 0 = 176 kJ/mol and ∆S 0 = 0.285 kJ/(molK) at 298.15 K. Calculate ∆G 0, and tell whether this reaction is spontaneous in the forward direction at 298.15 K.

50. Given: ∆ H 0 = 176 kJ/mol at 298.15 K ∆ S 0 = 0.285 kJ/(molK) at 298.15 K Unknown: ∆ G 0 at 298.15 K Solution: The value of ∆ G 0 can be calculated according to the following equation: ∆G 0 = ∆ H 0 – T ∆ S 0 ∆G 0 = 176 kJ/mol – 298 K [0.285kJ/(molK)] ∆G 0 = 176 kJ/mol – 84.9 kJ/mol ∆G 0 = 91 kJ/mol