1. MYP/Honors Chemistry Introduction to Thermochemistry

2. Thermochemistry: Study of energy changes that occur during chemical reactions and changes in state

4. Energy Many types of energy  must obey law of conservation of energy Energy cannot be created or destroyed, it can only change forms energy – capacity to do work Work = Force x distance Kinetic energy, potential energy, radiant (solar) energy, thermal energy (heat), chemical energy (stored in bonds, also called chemical potential energy)

5. Systems & Surroundings In thermodynamics, the world is divided into a system and its surroundings A system is the part of the world we want to study (e.g. a reaction mixture in a flask) The surroundings consist of everything else outside the system SYSTEM CLOSED OPEN ISOLATED

6. OPEN SYSTEM: can exchange both matter and energy with the surroundings (e.g. open reaction flask, rocket engine) CLOSED SYSTEM: can exchange only energy with the surroundings (matter remains fixed) e.g. a sealed reaction flask ISOLATED SYSTEM: can exchange neither energy nor matter with its surroundings (e.g. a thermos flask)

7. Heat Energy that transfers from one object to another because of a temperature difference between them ALWAYS flows from a warmer object to a cooler object Represented by q

8. EXOTHERMIC & ENDOTHERMIC REACTIONS Exothermic Process: a change (e.g. a chemical reaction) that releases heat; heat is lost as the surroundings heat up During Exothermic Process: heat < 0 (at constant pressure) Burning fossil fuels is an exothermic reaction

9. Endothermic Process: a change (e.g. a chemical reaction) that requires (or absorbs) heat; heat is gained as the surroundings cool down During Endothermic Process: heat > 0 (at constant pressure) Photosynthesis is an endothermic reaction (requires energy input from sun) Forming Na + and Cl - ions from NaCl is an endothermic process

10. Measuring Heat reaction Exothermic reaction, heat given off & temperature of water rises Endothermic reaction, heat taken in & temperature of water drops

11. Conceptual Problem: On a sunny winter day, the snow on a rooftop begins to melt. As the melting water drips from the roof, it refreezes into icicles. Describe the direction of heat flow as the water freezes. Is this process exothermic or endothermic?

13. Units for Measuring Heat Flow calorie-the quantity of heat needed to raise the temperature of 1g of pure water 1° C calorie with a small c is the unit used in thermochemistry (if it is Calorie with a capital C, then it is referring to the energy contained in food) 1 Calorie = 1 kilocalorie = 1000 calories joule-the SI unit for energy; the amount of energy required to raise the temperature of 1 g of pure water 0.2390°C 1 J = 0.2390 cal 1 cal = 4.184 J 1000 J = 1 kJ (kJ represents kilojoules)

14. Heat Capacity and Specific Heat Heat capacity: The amount of heat needed to increase the temperature of an object exactly by one degree Celsius. A cup of water would have a much bigger heat capacity than a drop of water.

15. Heat Capacity and Specific Heat Specific Heat: The amount of heat needed to raise the temperature of one gram of the substance by one degree Celsius. Water has a very high specific heat value: 1.00 cal/g-C or 4.184 J/g-C. q=heat m=mass in grams ∆T=T f - T i

16. Example Problem:

18. Calorimetry Calorimetry - the measurement of the heat into or out of a system for chemical and physical processes. Based on the fact that the heat released = the heat absorbed The device used to measure the absorption or release of heat in chemical or physical processes is called a “Calorimeter”

19. Constant-Pressure Calorimeters Foam cups are excellent heat insulators, and are commonly used as simple calorimeters under constant pressure. For systems at constant pressure, the “heat content” is the same as a property called Enthalpy (H) of the system

20. Enthalpy & ∆H The symbol for enthalpy is H Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV At constant pressure:  H =  E = q So at constant pressure, which is what we will deal with:  H = heat lost or gained by the system Therefore,  H = q

21. Measuring Enthalpy Change Changes in enthalpy =  H q =  H (These terms will be used interchangeably) Thus, q surr =  H = m x C x  T q sys =  H = -q surr = -m x C x  T  H is negative for an exothermic reaction  H is positive for an endothermic reaction

22. Constant-Volume Calorimeters Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system Used by nutritionists to measure energy content of food

23. Example Problem When 25.0 mL of water containing 0.025 mol HCl at 25.0°C is added to 25.0 mL of water containing 0.025 mol of NaOH at 25.0°C in a foam cup calorimeter, a reaction occurs. Calculate the enthalpy change (kJ) during this reaction if the highest temperature observed is 32.0°C. Assume the densities of the solutions are 1.00 g/mL.

24. Solution

25. 25 C + O 2 → CO 2 Energy ReactantsProducts  C + O 2 CO 2 395kJ given off + 395 kJ

26. 26 Exothermic The products are lower in energy than the reactants Thus, energy is released. ΔH = -395 kJ The negative sign does not mean negative energy, but instead that energy is lost.

27. 27 CaCO 3 → CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 176 kJ absorbed CaCO 3 + 176 kJ → CaO + CO 2

28. 28 Endothermic The products are higher in energy than the reactants Thus, energy is absorbed. ΔH = +176 kJ The positive sign means energy is absorbed

29. 29 Chemistry Happens in MOLES! An equation that includes energy is called a thermochemical equation CH 4 + 2O 2  CO 2 + 2H 2 O + 802.2 kJ 1 mole of CH 4 releases 802.2 kJ of energy. When you make 802.2 kJ you also make 2 moles of water

30. Thermochemical Equations The heat of reaction is the enthalpy(heat) change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions (SC) for the reaction is 101.3 kPa (1 atm.) and 25 o C (different from STP)

31. 31 CH 4(g) + 2 O 2(g)  CO 2(g) + 2 H 2 O (l) + 802.2 kJ If 10. 3 grams of CH 4 are burned completely, how much heat will be produced? 10. 3 g CH 4 16.05 g CH 4 1 mol CH 4 802.2 kJ = 514 kJ ΔH = -514 kJ, which means the heat is released for the reaction of 10.3 grams CH 4 Ratio from balanced equation 1 Start with known value Convert to moles Convert moles to desired unit

32. Another Example Problem

33. Summary So Far… Enthalpy: The heat content a substance has at a given temperature and pressure Can’t be measured directly because there is no set starting point The reactants start with a heat content The products end up with a heat content So we can measure how much enthalpy changes

34. Symbol is H Change in enthalpy is ∆H (delta H) If heat is released, the heat content of the products is lower ∆ H is negative (exothermic) If heat is absorbed, the heat content of the products is higher ∆ H is positive (endothermic)

35. 35 Energy ReactantsProducts  Change is down ΔH is <0 = Exothermic (heat is given off)

36. 36 Energy ReactantsProducts  Change is up ΔH is > 0 = Endothermic (heat is absorbed)

37. Heat of Reaction The heat that is released or absorbed in a chemical reaction Equivalent to ∆H C + O 2 (g)  CO 2 (g) + 393.5 kJ C + O 2 (g)  CO 2 (g) ∆H = -393.5 kJ In thermochemical equation, it is important to indicate the physical state a) H 2 (g) + 1/2O 2 (g)  H 2 O(g) ∆ H = -241.8 kJ b) H 2 (g) + 1/2O 2 (g)  H 2 O(l) ∆ H = -285.8 kJ

38. Heat of Combustion The heat from the reaction that completely burns 1 mole of a substance: C + O 2 (g)  CO 2 (g) + 393.5 kJ Could also be written as: C + O 2 (g)  CO 2 (g) ∆H = -393.5 kJ Some common Heats of Combustion at 25°C: Hydrogen = -286 ∆HPropane = -2220 ∆H Methane = -890 ∆HGlucose = -2808 ∆H Ethanol = -1368 ∆HCarbon = -394 ∆H

39. Heat in Changes of State Molar Heat of Fusion (  H fus. ) = the heat absorbed by one mole of a substance in melting from a solid to a liquid q = mol x  H fus. (no temperature change) Molar Heat of Solidification (  H solid. ) = the heat lost when one mole of liquid solidifies (or freezes) to a solid q = mol x  H solid. (no temperature change)

40. 40 Heat in Changes of State Note: You may also have the value of these equations as: q = mass x  H This is because some textbooks give the value of  H as, kJ/gram instead of kJ/mol

41. Heat in Changes of State Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies Thus,  H fus. = -  H solid

42. Example Problem

43. Heats of Vaporization and Condensation When liquids absorb heat at their boiling points, they become vapors. Molar Heat of Vaporization (  H vap. ) = the amount of heat necessary to vaporize one mole of a given liquid. q = mol x  H vap. (no temperature change)

44. Heats of Vaporization and Condensation Condensation is the opposite of vaporization. Molar Heat of Condensation (  H cond. ) = amount of heat released when one mole of vapor condenses to a liquid q = mol x  H cond. (no temperature change)  H vap. = -  H cond.

45. The large values for water  H vap. and  H cond. (40.7kJ/mol) is the reason hot vapors such as steam are very dangerous! You can receive a scalding burn from steam when the heat of condensation is released! H 2 0 (g)  H 2 0 (l)  H cond. = - 40.7kJ/mol

46. Example Problem

47. 47 The solid temperature is rising from -20 to 0 o C (use q = mol x ΔT x C) The solid is melting at 0 o C; no temperature change (use q = mol x ΔH fus. ) The liquid temperature is rising from 0 to 100 o C (use q = mol x ΔT x C) The liquid is boiling at 100 o C; no temperature change (use q = mol x ΔH vap. ) The gas temperature is rising from 100 to 120 o C (use q = mol x ΔT x C) The Heat Curve for Water 120

48. Heat of Solution Heat changes can also occur when a solute dissolves in a solvent. Molar Heat of Solution (  H soln. ) = heat change caused by dissolution of one mole of substance q = mol x  H soln. (no temperature change) Sodium hydroxide provides a good example of an exothermic molar heat of solution (next slide)

49. NaOH (s)  Na 1+ (aq) + OH 1- (aq)  H soln. = -445.1 kJ/mol The heat is released as the ions separate (by dissolving) and interact with water, releasing 445.1 kJ of heat as  H soln. thus becoming so hot it steams! H 2 O(l)

50. Example Problem H 2 O(l)

51. 51 Hess’s Law (developed in 1840) If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Called Hess’s Law of Heat Summation Germain Henri Hess (1802-1850)

52. How to add chemical equations: If two identical substances are on opposite sides of the arrow, they will cancel. If two identical substances are on same side of the arrow, add the coefficients together. Keep substances on the same side of the arrow in the final equation.

53. Example

55. 55 How Does It Work? 1)If you turn an equation around, you change the sign: If H 2 (g) + 1/2 O 2 (g)  H 2 O(g) ∆H=-285.5 kJ then the reverse is: H 2 O(g)  H 2 (g) + 1/2 O 2 (g) ∆H =+285.5 kJ 2)If you multiply the equation by a number, you multiply the heat by that number: 2 H 2 O(g)  2 H 2 (g) + O 2 (g) ∆H =+571.0 kJ 3)Or, you can just leave the equation “as is”

61. Hess’s Law - Procedure Options: 1. Use the equation as written 2. Reverse the equation (and change heat sign + to -, etc.) 3. Increase the coefficients in the equation (and increase heat by same amount)

62. Final Example: NO rxn Our reaction of interest is: N 2 (g) + 2O 2 (g)  2NO 2 (g) ΔH = 68 kJ This reaction can also be carried out in two steps: N 2 (g) + O 2 (g)  2NO(g) ΔH = 180 kJ 2NO (g) + O 2 (g)  2NO 2 (g) ΔH = -112 kJ

63. If we take the previous two reactions and add them, we get the original reaction of interest: N 2 (g) + O 2 (g)  2NO(g) ΔH = 180 kJ 2NO (g) + O 2 (g)  2NO 2 (g) ΔH = -112 kJ _________________________________ N 2 (g) + 2O 2 (g)  2NO 2 (g) ΔH = 68 kJ

64. Note the important things about this example, the sum of ΔH for the two reaction steps is equal to the ΔH for the reaction of interest. Big point: We can combine reactions of known ΔH to determine the ΔH for the “combined” reaction.

65. Standard Heats of Formation The ∆H for a reaction that produces (or forms) 1 mol of a compound from its elements at standard conditions Standard conditions: 25°C and 1 atm. Symbol is: The standard heat of formation of an element in it’s standard state is arbitrarily set at “ 0” This includes the diatomic elements

66. The heat of a reaction can be calculated by: subtracting the heats of formation of the reactants from the products Remember, from balanced equation: Products - Reactants  H o = ( (Products) - Reactants)

68. Example Problem What is the standard heat of reaction (∆H o ) for the reaction of CO(g) with O 2 (g) to form CO 2 ? O 2 (g) = 0 kJ/mol (free element) CO(g) = -110.5 kJ/mol CO2(g) = -393.5 kJ/mol First we need to balance the equation: 2CO(g) + O 2  2CO 2 (g)

70. 70 Another Example CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = - 74.86 kJ/mol O 2 (g) = 0 kJ/molCO 2 (g) = - 393.5 kJ/mol H 2 O(g) = - 241.8 kJ/mol  ∆Hº= [-393.5 + 2(-241.8)] - [-74.86 +2 (0)]  ∆Hº= -802.24 kJ (endothermic or exothermic?) (Because it is an element )

71. Thermodynamics Thermodynamics – scientific study of the interconversions of heat and energy Thermochemistry is a state function Thermochem depends only on where you started and where you ended, does NOT matter how you get there

72. First Law of Thermodynamics Conservation of energy, energy cannot be created or destroyed but only change forms  E = E f – E i If we consider a reaction, we can consider the products and the reactants S (s) + O 2(g)  SO 2(g)  E = E products - E reactants Also, since we know that energy must be conserved, the energy lost (or gained) by the system (the reaction), must be equal to the energy gained (or lost) by the surroundings 0 =  E system +  E surrounding OR  E sys = -  E surr

73. Relating  E to Heat(q) and Work(w) Energy cannot be created or destroyed. Energy of (system + surroundings) is constant. Any energy transferred from a system must be transferred to the surroundings (and vice versa). We normally don’t care about the surroundings, we just care about the work done (or by) the surroundings on the system From the first law of thermodynamics: The First Law of Thermodynamics

75. First Law of Thermodynamics Calculate the energy change for a system undergoing an exothermic process in which 15.4 kJ of heat flows and where 6.3 kJ of work is done on the system. Work/energy/heat (q) is almost ALWAYS given in Joules (or kJ), sometimes calories  E = q + w

77. Work and Heat The concept of work can be applied to gases Pistons move gases up and down If gases are expanded – work is negative If gases contrast – work is positive Work done on gases is in atm (usually), we need a conversion factor to convert to Joules 1 L*atm = 101.3 J

78. Example 6.1 A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 1.2 atm (a) w = -P  V w = -(0 atm)(6 – 2 L) = 0 J No work done (b) w = -P  V w = -(1.2 atm) (6.0 – 2.0 L) = -4.8 L*atm -4.8 L*atm x (101.3 J/1 L*atm) = -4.9 x 10 2 J Work is negative, gas expanded This example shows work (or heat) is NOT a state function, beginning and ending volume are the same, but the external pressure affects it

79. Example 6.2 The work done when a gas is compressed in a cylinder is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the energy for this process.  E = q + w q = - 128 J (exothermic) w = + 462 J (compressed) YOU MUST FIRST ALWAYS FIND THE SIGNS! Now plug it in  E = -128 J + 462 J = 334 J