1. DO NOW: CaCO 3 (s)  CaO (s) + CO 2 (g) Is the change in entropy ∆S positive or negative for the above reaction, and why?

2. DO NOW: CaCO 3 (s)  CaO (s) + CO 2 (g) Is the change in entropy ∆S positive or negative for the above reaction, and why?

3. VUONG-SCHMICK THERMODYNAMICS AND STANDARD STATES

4. DEMONSTRATION : Dehydration of sugar Ice cube Bomb calorimeter

5. REVIEW: Heat q = ∆H when??? ∆H reported for a reaction is the amount of h___ r___ or a____ when reactants are converted to products at the same t______ and in the m____ amounts represented by c_______ in the balanced equation. Example: Combustion of propane 1 mole release 2043 kJ Now if we have 0.5 mol then what?

6. Note that physical states of reactants and products must be specified as s, l, aq, or g when enthalpy changes are reported. Why does it matter? Write equation for propane to release: Water gas versus water liquid -2043 kJ versus -2219 kJ Why is there a difference of 176 kJ?

7. The difference of 176 kJ between the two arise because the conversion of liquid water to gas requires energy. If water liquid is produced: larger enthalpy (more negative) If water gas is produced :smaller enthalpy (less negative) Because 44.0 kJ/mol is needed for vaporization.

8. Besides physical states, it is important to specify pressure and temperature Which leads to Thermodynamic Standard State ∆H° Most stable form of substance at 1 atm pressure and 25 °C, 1M for all substances in solution Mention the difference between STP*** Standard state condition versus standard temperature pressure.

9. EXAMPLE:

10. The reaction between H 2 and O 2 yield water vapor and has a change of heat = - 484kJ. How much PV work is done and what is the value of the change of E in kJ for the reaction of 0.50 mol of H 2 with 0.25 mol O 2 amd atmospheric pressure of 1.00 atm, the volume change is -5.61L?

11. ENTHALPIES OF PHYSICAL AND CHEMICAL CHANGE Almost every change in a system involves either g___ or l___ of e____. The change can either be physical (melting) or chemical (burning)…

12. ENTHALPIES OF PHYSICAL CHANGE Explain a block of ice melt start from let say -10°C Draw graph of melting point… Explain enthalpy of fusion or heat of fusion ∆H fusion Value look up in a table Explain enthalpy of vaporization or heat of vaporization ∆H vap Sublimation Heat of sublimation equals sum of fusion and vaporization

14. ENTHALPY OF CHEMICAL CHANGE Review endo and exo Why hot water freezes faster than cold water… If the reaction is reverse, the sign in enthalpy changes. Enthalpy given in a chemical reaction is depend on the coefficient given in the equation.

15. EXAMPLE:

16. DO NOW:

17. CALORIMETRY AND HEAT CAPACITY Calorimeter device Heat evolved or absorbed is calculated from temperature change. Keep pressure constant. Calculate enthalpy change during reaction Bomb calorimeter: Reaction takes place at constant volume, not constant pressure So the measurement provides value of ∆E rather than ∆H

18. How to calculate ∆H: When calorimeter and its contents absorb a given amount of heat, the temp rise results depends on calorimeter’s heat capacity Heat Capacity: Amount of heat required to raise temperature of object or substance a given amount: C = q/ ∆T Greater heat capacity, greater amount of heat needed to produce a given temperature change Bathtub full of water has greater heat capacity than coffee cup full Therefore takes far more heat to warm than cup

19. Q = C X ∆T Depends on size of object and its composition Leads to “ specific heat ” Heat necessary to raise temp of 1 gram of substance by 1 degree C Q = (specific heat) x (mass of substance) x ∆T Just as you can raise 1 gram of substance, you can raise one mol, so we have molar heat capacity

20. WHY LEARN SPECIFIC HEAT How human survive Specific heat is explain how you is alive We are able to maintain a steady internal temperature under changing outside condition Since water specific heat is so high

21. EXAMPLE:

22. DO WS 2 UP TO AND INCLUDING #8 Prelab due Thursday

23. HABER PROCESS: Hydrogen gas and nitrogen gas make ammonia talk about hydrazine N 2 H 4 Lead to Hess’ Law: _____________________________________

24. REVIEW: Hess Law: Overall enthalpy change for reaction is equal to sum of enthalpy change for individual steps in reaction (Haber- cycle) Enthalpy is state function Hess Law let us calculate final enthalpy change when some reaction doesn’t go cleanly Strategy: Combine individual reactions so that sum is desired reaction

27. HESS’S LAW PRACTICE PROBLEMS 1. Given the following equations and H o values, determine the heat of reaction (kJ) at 298 K for the reaction: B 2 H 6 (g) + 6 Cl 2 (g) 2 BCl 3 (g) + 6 HCl(g) BCl 3 (g) + 3 H 2 O(l) H 3 BO 3 (g) + 3 HCl(g)H o /kJ = -112.5 B 2 H 6 (g) + 6 H 2 O(l) 2 H 3 BO 3 (s) + 6 H 2 (g)H o /kJ = -493.4 1/2 H 2 (g) + 1/2 Cl 2 (g) HCl(g)H o /kJ = -92.3

28. 2. Given the following equations and H o values, determine the heat of reaction (kJ) at 298 K for the reaction: 2 OF 2 (g) + 2 S(s) SO 2 (g) + SF 4 (g) OF 2 (g) + H 2 O(l) O 2 (g) + 2 HF(g)H o /kJ = -276.6 SF 4 (g) + 2 H 2 O(l) 4 HF(g) + SO 2 (g)H o /kJ = -827.5 S(s) + O 2 (g) SO 2 (g)H o /kJ = -296.9

29. 3.Determine H o /kJ for the following reaction using the listed enthalpies of reaction: N 2 H 4 (l) + 2 H 2 O 2 (g) N 2 (g) + 4 H 2 O(l) N 2 H 4 (l) + O 2 (g) N 2 (g) + 2 H 2 O(l)H o /kJ = -622.3 kJ H 2 (g) + 1/2 O 2 (g) H 2 O(l)H o /kJ = -285.8 kJ H 2 (g) + O 2 (g) H 2 O 2 (l)H o /kJ = -187.8 kJ

30. ENTHALPY SUMMARY OF WHAT WE JUST LEARNED: Physical state is important Example: Propane reacts with oxygen gas Release water (g) ∆H = -2043 kJ Release water (l) ∆H = -2219 kJ The difference = 176 kJ due to conversion of liquid water to gaseous water requires energy Critical thinking: If liquid water is produced vs. gaseous water, which ∆H is more negative?

31. SUMMARY: It is important to write states of matter It is also important to report specify pressure and temperature To compare between reactions Thermodynamic standard state: Most stable form of a substance at 1 atm pressure and at specified temperature, usually 25C, 1M concentration for substance in solution Most data is recorded as bar: 1 bar = 0.986923 atm

32. ∆H 0 The superscript = standard conditions Practice: The reaction of nitrogen with hydrogen

33. PRACTICE : The reaction between hydrogen and oxygen yield water vapor has standard enthalpy of -484 kJ. 2H 2 (g) + O 2 (g)  2H 2 O (g) How much work is done, and what is the value of energy change (in kJ) for the reaction of 0.50 mol of hydrogen gas with 0.25 mole of oxygen gas at atmospheric pressure if the volume change is -5.6 L?

34. Convert: 101J/(atmL) =.57 kJ = -120 kJ

35. MORE PRACTICE: The explosion of 2.00 mol of solid trinitrotoluene (TNT) with a volume of approximately 274 mL produces gases with a volume of 448 L at room temp. How much PV work (in kJ) is done during the explosion? 2C 7 H 5 N 3 O 6 (s)  12CO (g) + 5H 2 (g) + 3N 2 (g) + 2C (s)

36. -45 kJ