Energy

1. Chapter 5Thermochemistry(The study of the relationships between chemical reactions and energy changes involving heat)&Chapter 8 Section 8Strength of Covalent Bonds

2. Energy • The ability to do work or transfer heat.

3. Work • Energy used to move an object over some distance. • w = F d, where w is work, F is the force, and d is the distance over which the force is exerted.

4. Heat • Heat is energy transferred between substances due to a difference in temperature (from higher temp to lower temp)

5. Potential Energy Energy an object possesses by virtue of its position or chemical composition. PE = mgh (mass x gravity x height)

6. 1 KE =  mv2 2 Kinetic Energy Energy an object possesses by virtue of its motion. (½ x mass x velocity2) (true except at 0 K)

7. Transferal of Energy Example: • The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

8. Transferal of Energy • The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. • As the ball falls, its potential energy is converted to kinetic energy.

9. Transferal of Energy • The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall. • As the ball falls, its potential energy is converted to kinetic energy. • When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

10. kg m2 1 J = 1  s2 Units of Energy • The SI unit of energy is the joule (J). • An older, non-SI unit is still in widespread use: The calorie (cal). 1 cal = 4.184 J 1 food Calorie = 1kcal

11. System and Surroundings • System - the molecules we want to study (here, the hydrogen and oxygen molecules). • Surroundings - everything else (here, the cylinder and piston).

12. First Law of Thermodynamicsa.k.a. Law of Conservation of Energy • Energy is neither created nor destroyed. • The total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

13. Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. Usually we don’t measure E, we just measure the change in E (ΔE)

14. Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal − Einitial (We don’t actually use this equation….we just need to know the change is between two states!)

15. If E is positive (E > 0, Efinal > Einitial) • The system absorbed energy from the surroundings. • Energy of the system increases • Energy of the surroundings decreases

16. If E is negative (E < 0, Efinal < Einitial) • The system released energy to the surroundings. • Energy of the system decreases • Energy of the surroundings increases

17. Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). E = q + w.

18. E, q, w, and Their Signs

19. State Functions • Depend only on the present state of the system, not on the path by which the system arrived at that state. • In the system below, the water could have reached room temperature from either direction.

20. E depends only on Einitial and Efinaland is a state function.

21. State Functions • However, q and w are not state functions. • Whether the battery is shorted out or is discharged by running the fan, its E is the same. • But q and w are different in the two cases.

22. State Functions • Generally….. Quantities we look at the change in (E) are state functions Quantities we don’t look at changes in (q & w) are NOT state functions.

23. P-V Work When a chemical reaction occurs, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

24. PV Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston. w = −PV

25. Example #1 Hydrogen & Oxygen gas are in a cylinder with a movable piston. When ignited, the system loses 1150J of heat to the surroundings. The reaction also causes the piston to rise (from the expanding gases) and 480J of work is done as the piston is pushed against the atmosphere. What is the change in internal energy of the system?

26. Heat is transferred out of the system so q is negative (q=-1150J) Work is done by the system, so w is negative (w = -480J) ΔE = q + w = -1150J + -480J = -1630J 1630J of energy has been transferred from the system to the surroundings

27. Enthalpy at Constant Pressure • If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system. • Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV

28. Enthalpy at Constant Pressure • When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) • This can be written H = E + PV

29. Enthalpy at Constant Pressure • Since E = q + w and w = −PV, we can substitute these into the enthalpy expression: H = E + PV H = (q+w)− w H = qp (pmeans constant pressure) • So, at constant pressure the change in enthalpy isthe heat gained or lost.

30. Important! We use H to describe chemical reactions because most reactions happen at constant pressure, and at constant pressure H= q and q is (relatively) easy to find! Often for chemical reactions, PΔV is very small and ΔE ≈ ΔH

31. ΔH and sign: Endothermic – ΔH is positive (heat transferred to the system from the surroundings) Phase changes: s l  g Exothermic – ΔH is negative (heat transferred from the system to the surroundings) Phase changes: g l  s

32. Thermochemical Equations • Must specify physical states of products and reactants • Enthalpy depends on state! • Must specify enthalpy change Example: CaO(s) + CO2(g)  CaCO3(s) H298 = -178kJ H, is called the enthalpy of reaction, or the heat of reaction (HRXN). (Temp at which the reaction occurs is sometimes included) How to we figure out H?

33. Method #1 to find :Calorimetry When =q, you can calculate q like you did in your first year class when given the proper information! qp = m Tcp qp =ccT qlost = qgained Specific heat of water (and most aqueous sol’ns): 4.184J/gºC or 4.184J/g K *On AP Equation Sheet! (q=mc T)

34. Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, we can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

35. Constant Volume (Bomb) Calorimetry Reactions can be carried out in a sealed “bomb” calorimeter, such as this one, to measure the heat absorbed by the water.

36. Bomb Calorimetry • Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. • For most reactions, the difference is very small.

37. Temperature changes are converted to heat energy by using heat capacities. Heat Capacity (C)- The amount of energy required to raise the temperature of a substance by 1 K (J/K) Molar Heat Capacity (Cm)– The amount of energy required to raise the temperature of 1 mole of a substance by 1 K (J/molK) Specific Heat Capacity (or simply Specific Heat) (Cs)- The amount of energy required to raise the temperature of 1 g of a substance by 1 K. (J/gK) Calorimeter Constant (cc) – The heat capacity of a calorimeter(J/K) *temperatures may also be in °C

38. Example #2 The specific heat of ammonia is 4.381 J/gK. Calculate the heat required to raise the temperature of 1.50g of ammonia from 213.0K to 218.0K. q = mΔTcp q = (1.50g)(218.0K – 213.0K )(4.381J/gK ) q = 33 J

39. Example #3 2.15 g of methane is combusted (burned) to form CO2 and H2O. All of the heat evolved is used to raise the temperature of 100.0 g of water by 17.8 K. Calculate the heat (q) of combustion per molof methane. q = (100.0 g) (17.8 K) (4.184 J/g K) = 7450 J = 7.45 kJ 7.45 kJ ---------- 2.15 g CH4 16.0 g CH4 X ----------- 1 mole CH4 = 55.4 kJ/mol

40. Example # 4 50.00 mL of 0.750 M HCl solution and 50.00 mL of 0.875 M KOH solution are combined in a calorimeter. The temperature of the calorimeter and both solutions goes from 25.0 oC to 29.6 oC. Assume that the specific heat of each solution is the same as water, 4.184 J/g oC and that the density of each solution is 1.00 g/mL. The heat capacity of the calorimeter is 125 J/oC. Calculate the heat of reaction per mole of water formed.

41. HCl(aq) + KOH(aq) H2O(l) + KCl(aq) Hrxn= q= energy change of solutions + energy change of calorimeter = (m)( T)(cp) + (cc)( T) (100.00g)(4.6oC)(4.184 J/goC) + (125 J/oC)(4.6oC) 1900 J + 580 J 2500 J or 2.5 kJ since this reaction is exothermic: Hrxn= - 2500J or - 2.5 kJ We want the heat of reaction per mole of water!

42.   Do an excess-limiting problem: .0500L .0500L .750M .875M ? mol HCl(aq) + KOH(aq) H2O(l) + KCl(aq) 0.0375 moles of H2O are formed in the reaction - 2.5 kJ -------------- = - 67 kJ / mol H2O 0.0375 mol

43. Some more about Enthalpy… • Enthalpy is an extensive property (it depends on amount). • H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. • H for a reaction depends on the state of the products and the state of the reactants. (Enthalpy: gases > liquids > solids)

44. ΔH & Stoichiometry Enthalpy is an extensive property so changing the stoichiometry changes the value of ΔH CaO(s) + CO2(g)  CaCO3(s) H= -178kJ 2CaO(s) + 2CO2(g)  2CaCO3(s) H= -356kJ ½ CaO(s) + ½ CO2(g)  ½ CaCO3(s) H= -89.0kJ

45. Example #5 How much heat is released when 4.50g of methane gas is burned in a constant pressure system? CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔH = -802 kJ For every 1 mol of methane burned 802 kJ is released

46. Method #2 to find : Hess’s Law Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

47. Example #6 Given: C(s) + O2(g)  CO2(g) ΔH = -393.5 kJ CO(g) + ½ O2(g)  CO2(g) ΔH = -283.0 kJ Find ΔH for the following: C(s) + ½ O2(g)  CO(g)

48. ½ C(s) + O2(g)  CO2(g) ΔH = -393.5 kJ CO(g) + ½ O2(g)  CO2(g) ΔH = -283.0 kJ C(s) + O2(g)  CO2(g) ΔH = -393.5 kJ + CO2(g)  CO(g) + ½ O2(g) ΔH = +283.0 kJ C(s) + ½ O2(g)  CO(g) ΔH = -110.5 kJ

49. Example #7 Given: C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(l) ΔH = - 1299.6 kJ C(s) + O2(g)  CO2(g) ΔH = - 393.5 kJ H2(g) + ½ O2(g)  H2O(l) ΔH = - 285.9 kJ Find: 2 C(s) + H2(g)  C2H2(g)

50. 2 CO2(g) + H2O(l)  C2H2(g) + 5/2 O2(g) ΔH = + 1299.6 kJ 2 C(s) + 2 O2(g)  2 CO2(g) ΔH = - 787.0 kJ H2(g) + ½ O2(g)  H2O(l) ΔH = - 285.9 kJ 2C(s) + H2(g)  C2H2(g) ΔH = 226.7 kJ