1. Chapter 6 Thermochemistry

2. The Nature of Energy  Energy- the capacity to do work or produce heat  Law of conservation of energy- energy can be converted but not created or destroyed. Energy of universe is constant 6.1 The Nature of Energy

3. Types of Energy  potential energy- (PE) due to position or composition Attractive or repulsive forces Attractive or repulsive forces  kinetic energy- (KE) due to motion of the object KE = ½ mv 2 :depends on mass and volume KE = ½ mv 2 :depends on mass and volume

4. Types of Energy (a): PE A > PE B (b):  ball A has rolled down the hill  has lost PE to PE in ball B  and to friction

5. Transfer of Energy  Temperature: measure of the average kinetic energy of the particles. It a property that reflects random motion of particles in substance  Two Ways to Transfer Energy: Heat- (q) transfer of energy between two objects because of a temperature difference Heat- (q) transfer of energy between two objects because of a temperature difference Work- (w) force acting over a distance Work- (w) force acting over a distance

6. Pathway  the specific conditions of energy transfer  energy change is independent of pathway thus it is a state function (state property) (state property)  State function: a property of a system that depends only on its present state (it does not depend on system ’ s past or future, i.e, it does not depend on how the system arrived at the present state)  work and heat depend on pathway so they  work and heat depend on pathway so they are not state functions

7. Chemical Energy  System - part of the universe you are focused on  Surroundings- everything else in the universe  CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat  usually system: what is inside the container. Reactants or products of a chemical reaction system: what is inside the container. Reactants or products of a chemical reaction surroundings: container,room, etc. surroundings: container,room, etc.

8. Transfer of Energy Exothermic process energy is produced in reaction energy is produced in reaction energy flows out of system energy flows out of system container becomes hot when touched container becomes hot when touched CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat  Endothermic process energy is consumed by the reaction energy is consumed by the reaction energy flows into the system energy flows into the system container becomes cold when touched container becomes cold when touched N 2 (g) + O 2 (g) + Energy 2NO(g) N 2 (g) + O 2 (g) + Energy 2NO(g)

9. Transfer of Energy Combustion of Methane Gas is exothermic CH 4 (g) + 2O 2 (g) CO 2 + 2H 2 O(g) + energy (heat)

10. Transfer of Energy Reaction between nitrogen and oxygen is endothermic N 2 (g) + 2O 2 (g) energy (heat) CO 2 + 2H 2 O(g) +

11.  Where does the energy, released as heat come from in an exothermic reaction?  Energy gained by surrounding must be equal to energy lost by the system  The heat flow to the surroundings results from a lowering of potential energy of the reaction system.  Thus, in any exothermic reaction, some of the potential energy stored in the chemical bonds is being converted into thermal energy

12. Transfer of Energy  the energy comes from the potential difference between the reactants and products  energy produced (or absorbed) by reaction must equal the energy absorbed (or produced) by surroundings  usually the molecules with higher potential energy have weaker bonds than molecules with lower potential energy

13. Thermodynamics  Thermodynamics: study of energy and its interconversions (transfers)  First Law of Thermodynamics Energy of universe is constant (Law of conservation of energy) (Law of conservation of energy)

14. Internal Energy, E  Internal energy, E, is the sum of potential and kinetic energy of all particles in the system in system E = PE + KE E = PE + KE  E can be changed (- or + ∆E) by a flow of work, heat, or both ∆E = q + w ∆E = q + w

15. Signs

16. Signs  Signs are very important in thermodynamic quantities  Signs will always reflect the  Signs will always reflect the system ’ s point of view unless otherwise stated ∆Eqw change in internal energy heatwork Exothermic --- Endothermic +++

17. Work  common types of work Expansion- work done by gas Expansion- work done by gas Compression- work done on gas Compression- work done on gasexpansion+∆V-wcompression-∆V+w P is external pressure – not internal like we normally refer to Work is force applied over distance

18. Work

19. Example 1  Find the ∆E for endothermic process where 15.6 kJ of heat flows and 1.4 kJ of work is done on system Since it is endothermic, q is + and w is + Since it is endothermic, q is + and w is +

20. Example 2  Calculate the work of expansion of a gas from 46 L to 64 L at a constant pressure of 15 atm. Since it is an expansion, ∆V is + and w is - Since it is an expansion, ∆V is + and w is -

21. Example 3  A balloon was inflated from 4.00 x 10 6 L to 4.50 x 10 6 L by the addition of 1.3 x 10 8 J of heat. Assuming the pressure is 1.0 atm, find the ∆E in Joules. (1 L∙atm=101.3 J) Since it is an expansion, ∆V is + and w is - Since it is an expansion, ∆V is + and w is -

22. 6.2 Enthalpy and Calorimetry

23.  The term enthalpy is composed of the prefix en-, meaning "to put into" and the Greek word -thalpein, meaning "to heat", that is  The term enthalpy is composed of the prefix en-, meaning "to put into" and the Greek word -thalpein, meaning "to heat", that is “to put heat into” Greek  This is the heat change which occurs when 1 mol of a substance reacts completely with oxygen to form products at 298 K and 1 atm.  the enthalpy or heat content (denoted as H or ΔH,) is a quotient or description of thermodynamic potential of a system, which can be used to calculate the "useful" work obtainable from a closed thermodynamic system under constant pressure.

24. Enthalpy  Enthalpy is a property of a system  definition: H = E + PV  since E, P and V are all state functions, then H is also a state function  Consider a process carried out at constant P and the only type of work allowed is P-V work (w= -P ∆V )  Under these conditions, the expression: ∆E = q P + w ∆E = q P - P∆V ∆E = q P + w becomes ∆E = q P - P∆V OR q P = ∆ E +P∆V OR q P = ∆ E +P∆V

25. Relationship between q P and ∆H  By definition: H = E + PV  Therefore ∆H = ∆E + ∆ (PV)  Since P is constant: ∆H = ∆E + P ∆ V  Since q P = ∆ E +P∆V then ∆H = q P then ∆H = q P ∆H At constant P, ∆H of a system is equal to the energy flow as heat

26.  Thus, heat of reaction and change in enthalpy are used interchangeably for a reaction at constant P  For a chemical reaction: ∆H = H products - H reactants ∆H = H products - H reactants For endothermic reaction : ∆H is +ve Heat is absorbed For exothermic reaction : ∆H is – ve For exothermic reaction : ∆H is – ve Heat is released Heat is released This means that for a reaction carried out at constant P, the flow of heat is a measure of the change in enthalpy of the system

27. Calorimetry  The science of measuring heat  calorimeter- device used to experimentally find the heat associated with a chemical reaction  substances respond differently when heated ( to raise T for two substances by 1 degree, they require different amount of heat)

28. Heat Capacity  (C) how much heat it takes to raise a substance ’ s T by one °C or K  The amount of energy depends on the amount of substance

29. Heat Capacity  Specific heat capacity (s) heat capacity per gram (s) heat capacity per gram in J/°C*g or J/K*g in J/°C*g or J/K*g  Molar heat capacity heat capacity per mole heat capacity per mole in J/°C*mol or J/K*mol in J/°C*mol or J/K*mol

30. Constant-Pressure Calorimetry  uses simplest calorimeter (like coffee-cup calorimeter) since it is open to air  used to find changes in enthalpy (heats of reaction) for reactions occurring in a solution since q P = ∆H  heat of reaction is an extensive property, so we usually write them per mole so they are easier to use

31. Constant-Pressure Calorimetry  when 2 reactants are mixed and T increases, the chemical reaction must be releasing heat so it is exothermic  the released energy from the reaction increases the motion of molecules, which in turn increases the T

32. Constant-Pressure Calorimetry  If we assume that the calorimeter did not leak energy or absorb any itself (that all the energy was used to increase the T), we can find the energy released by the reaction: E released by rxn = E absorbed by soln ∆H = q P = s P x m x ∆T ∆H = q P = s P x m x ∆T

33. Constant-Volume Calorimetry  uses a bomb calorimeter  weighed reactants are placed inside a rigid, steel container and ignited  water surrounds the reactant container so the T of it and other parts are measured before and after reaction

34. Constant-Volume Calorimetry  Here, the ∆V = 0 so -P∆V = w = 0 ∆E = q + w = q V for constant volume E released by rxn = ∆T x C calorimeter

35. Example 1  When 1 mol of CH 4 is burned at constant P, 890 kJ of heat is released. Find ∆H for burning of 5.8 g of CH 4 at constant P.  ∆H = q p  890 kJ is released per mole of CH 4

36. Example 2  When 1.00 L of 1.00 M Ba(NO 3 ) 2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na 2 SO 4 solution at 25.0°C in a coffee-cup calorimeter, solid BaSO 4 forms and the T increases to 28.1°C. The specific heat capacity of the solution is 4.18 J/g.°C and the density is 1.0 g/mL. Find the enthalpy change per mole of BaSO 4 formed.

37. Example 2  Write the net ionic equation for the reaction: Ba 2+ (aq) + SO 4 2- (aq)  BaSO 4 (s)  Is the energy released or absorbed? What does that mean about ∆H and q? exothermic: -∆H and – q P  How can we calculate ∆H or heat? heat = q p = s P x m x ∆T  How can we find the m? use density and volume

38. Example 2  Find the mass:  Find the change in T:  Calculate the heat evolved

39. Example 2  since it is a one-to-one ratio and the moles of reactants are the same, there is no limiting reactant  1.0 mol of solid BaSO 4 is made so ∆H= -2.6x10 4 J/mol = -26 kJ/mol

40. Example 3  Compare the energy released in the combustion of H 2 and CH 4 carried out in a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T change of 7.3°C while the combustion of 1.15 g of hydrogen produced a T change of 14.3°C. Find the energy of combustion per gram for each.

41. Example 3  methane: CH 4  hydrogen: H 2  The energy released by H 2 is about 2.5 times the energy released by CH 4

42. 6.3 Hess ’ Law

43. Hess ’ Law  since H is a state function, the change in H is independent of pathway  Hess ’ Law- when going from a set of reactants to a set of products, the ∆H is the same whether it happens in one step or a series of steps

44. Example 1

45.  N 2 (g) + 2O 2 (g)  2NO 2 (g)∆H = 68 kJ OR  N 2 (g) + O 2 (g)  2NO(g)∆H = 180 kJ 2NO(g) + O 2 (g)  2NO 2 (g)∆H = -112 kJ N 2 (g) + 2O 2 (g)  2NO 2 (g)∆H = 68 kJ

46. Rules 1. If a reaction is reversed, the sign of ∆H must be reversed as well. because the sign tells us the direction of heat flow at constant P because the sign tells us the direction of heat flow at constant P 2. The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction. If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way. because ∆H is an extensive property because ∆H is an extensive property

47. Example 2  Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396 kJ/mol), find the ∆H for the conversion of graphite to diamond. C graphite (s)  C diamond (s)∆H=?

48. Example 2 (1) C graphite (s) + O 2 (g)  CO 2 (g) ∆H=-394kJ/mol (2) C diamond (s) + O 2 (g)  CO 2 (g) ∆H=-396kJ/mol  to get the desired equation, we must reverse 2 nd equation: (1) C graphite (s) + O 2 (g)  CO 2 (g) ∆H=-394kJ/mol -(2) CO 2 (g)  C diamond (s) + O 2 (g) ∆H=396kJ/mol C graphite (s)  C diamond (s)∆H=-394 + 396 ∆H=2 kJ/mol ∆H=2 kJ/mol

49. Example 3  Find ∆H for the synthesis of B 2 H 6, diborane: 2B(s) + 3H 2 (g)  B 2 H 6 (g) Given: (1) 2B(s) + 3/2O 2 (g)  B 2 O 3 (s) ∆H 1 =-1273kJ (2) B 2 H 6 (g) + 3O 2 (g)  B 2 O 3 (s) + 3H 2 O(g) ∆H 2 =-2035kJ (3) H 2 (g) + ½ O 2 (g)  H 2 O(l) ∆H 3 =-286kJ (4) H 2 O(l)  H 2 O(g) ∆H 4 =44 kJ

50. 2B(s) + 3/2O 2 (g)  B 2 O 3 (s) ∆H 1 =-1273kJ (-1) B 2 O 3 (s) + 3H 2 O(g)  B 2 H 6 (g) + 3O 2 (g) -∆H 2 =-(-2035kJ) (x3) 3H 2 (g) + 3/2O 2 (g)  3H 2 O(l) 3∆H 3 =3(-286kJ) (x3)3H 2 O(l)  3H 2 O(g) 3∆H 4 =3(44 kJ) 2B(s) + 3H 2 (g)  B 2 H 6 (g) 2B(s) + 3H 2 (g)  B 2 H 6 (g) ∆H = -1273 + -(-2035) + 3(-286) + 3(44) = 36kJ

51. 6.4 Standard Enthalpies of Formation

52. Standard Enthalpy of Formation  ∆H f °  change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states  ° means that the process happened under standard conditions so we can compare more easily

53. Standard States  For a :  For a compound: for gas: P = 1 atm for gas: P = 1 atm For pure substances, it is a pure liquid or pure solid state For pure substances, it is a pure liquid or pure solid state in solution: concentration is 1 M in solution: concentration is 1 M  For an :  For an element: form that it exists in at 1 atm and 25°C form that it exists in at 1 atm and 25°C O: O 2 (g)K: K(s)Br: Br 2 (l)

54. Writing Formation Equations  always write equation where 1 mole of compound is formed ( )  always write equation where 1 mole of compound is formed (even if you must use non-integer coefficients) NO 2 (g): ½ N 2 (g) + O 2 (g)  NO 2 (g) ∆H f °= 34 kJ/mol CH 3 OH(l): C(s) + 2H 2 (g) + O 2 (g)  CH 3 OH(l) ∆H f °= -239 kJ/mol ∆H f °= -239 kJ/mol

55. Using Standard Enthalpies of Formation where n = number of moles of products/reactants ∑ means “ sum of ” ∆H f ° is the standard enthalpy of formation for reactants or products  zero  ∆H f ° for any element in standard state is zero so elements are not included in the summation

56. Using Standard Enthalpies of Formation  since ∆H is a state function, we can use any pathway to calculate it  one convenient pathway is to break reactants into elements and then recombine them into products

57. Using Standard Enthalpies of Formation

58. Example 1  Calculate the standard enthalpy change for the reaction that occurs when ammonia is burned in air to make nitrogen dioxide and water 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(l)  break them apart into elements and then recombine them into products

59. Example 1

60.  can be solved using Hess ’ Law: (1) 4NH 3 (g)  2N 2 (g) + 6H 2 (g)-4∆H f ° NH3 (2) 7O 2 (g)  7O 2 (g)0 (3) 2N 2 (g) + 4O 2 (g)  4NO 2 (g)4 ∆H f ° NO2 (4) 6H 2 (g) + 3O 2 (g)  6H 2 O(l)6 ∆H f ° H2O

61. Example 1  can also be solved using enthalpy of formation equation:  values are in Appendix 4

62. Example 2  Calculate the standard enthalpy change for the following reaction: 2Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s)

63. Example 3  Compare the standard enthalpy of combustion per gram of methanol with per gram of gasoline (it is C 8 H 18 ).  Write equations: 2CH 3 OH(l) + 3O 2 (g)  2CO 2 (g) + 4H 2 O(l) 2C 8 H 18 (l) + 25O 2 (g)  16CO 2 (g) + 18H 2 O(l)

64. Example 3  Calculate the enthalpy of combustion per mole:

65. Example 3  Convert to per gram using molar mass:  so octane is about 2x more effective