Energy and Chemical Change

1. Energy and Chemical Change

2. Energy: ability to do work/produce heat http://www.solpass.org/5s/images/roller.gif • Potential energy: based on object's position or composition • Kinetic energy: based on object’s motion • Law of Conservation of Energy • In any chemical reaction or physical process • Energy can be converted from one form to another • But is neither created nor destroyed

3. Chemical potential energy: energy stored in chemicals due to their composition • Heat (q) • Form of energy • Always flows from warmer object to cooler object • Transfer of energy due to temperature differences • Temperature: property directly proportional to KE of substance • In metric system, heat is measured in • calorie (cal): amount of heat required to raise T of 1 g of pure water by 1oC • Calories: amount of heat required to raise T of 1 kg H2O 1oC

4. SI unit of heat and energy-joule (J)1 Calorie = 1 kcal = 1000 cal = 4184 J1J = 0.239 cal1 cal = 4.184 J a. 1.69 Joules to calories b. 0.3587 J to cal c. 820.1 J to kilocalories d. 68 calories to kilocalories e. 423 calories to kilocalories f. 20.0 calories to Joules g. 252 cal to J h. 2.45 kilocalories to calorie i. 556 kilocalories to cal j. 6.78 kilocalories to kilojoules k. 59.6 calories to kcal l. 449.6 joules to kilojoules m. 9.806 kJ to J n. 5.567 cal to J o. 5467.9 kcal to J

5. amount of heat per unit mass required to raise T by one degree Celsius

6. 1oC = 1K in units (cal/g-oC or cal/g-K) • Units of in SI system are J/g-K • Specific heat for each substance is different because each has different compositions • Specific heat of water is 1 cal/g·°C = 4.186 joule/g·°C which is higher than any other common substance • Why water plays important role in temperature regulation

7. How many degrees will T of a 63-gram sample of a substance with a specific heat of 1.672 J/g * K rise if 9000 joules are applied? 1.672 J/g*K = 9000 J/63 g * ∆T ∆T = 85.44 K What is specific heat of element if 10 kJ heats 214-g sample 24 K? Specific heat = 10000 J / (214 g * 24 K) = 1.947 J/g * K How much water is in beaker if 130 kilojoules heats water by 76 K? 4.184 J/g * K=130000 J / (X grams * 76 K) mass= 408.826 g

8. How many joules are required to heat a 1-kg bar of Al (specific heat = 0.902 J/g * K) by 250 K? 0.902J/g*K=X joules/(1000 g * 250 K) J=225500 or 225.5 kJ How much energy does it take to raise T of 50g Cu by 10oC? 0.385 J/g*C = heat/50 g * 10oC heat = 19.3 J If add 30 J heat to 10g of Al, by how much will its T increase? If the initial T was 20o, what would the new T be? 0.902 J/g*C = 30 J/10 g * ∆T ∆T= 3.33oC/23.3oC

9. If a gold ring with a mass of 5.5 grams changes temperature from 25.0oC to 28.0oC, how much energy (in joules) has it absorbed?  0.129 J/g*C = heat/5.5 g * 3oC heat = 2.13 J The temperature of a sample of 250 g of water is changed from 25.0oC to 30.0oC. How much energy was transferred into the water to cause this change? Calculate answer in joules and in calories.  4/184 J/g*C = heat/250 g * 5oC heat = 5230 J Convert to calories: 5230 J X 1 calorie/4.184 J = 1250 C

10. The T of 335 g of water changed from 24.5oC to 26.4oC. How much heat did this sample absorb? Calculate in kilojoules/kilocalories. 4.184 J/g*C = heat/335 g * 1.9oC heat = 2663.12 J Convert to kJ:  2663.12J X 1 kJ/1000 Joules = 2.66 kJ To kcal/Cal: 2663.12J X 1C/4.184J X1 kcal/1000 C=0.64 kcal If 10.2 kJ of heat is used to warm 250. g of iron (specific heat = 0.451 J/gK) at 35.0oC, what will be the final temperature of iron? 0.451 J/g*K = 10.2 kJ/250. g * ∆T ∆T = 90.5 Final T = 125.5oC(since warmed, must go up from initial)

11. Measuring heat • How much energy must be absorbed by 20.0 g of water to increase its temperature from 283.0 °C to 303.0 °C? • When 15.0 g of steam drops in temperature from 275.0 °C to 250.0 °C, how much heat energy is released? • How much energy is required to heat 120.0 g of water from 2.0 °C to 24.0 °C?

12. If 720.0 g of steam at 400.0 °C absorbs 800.0 kJ of heat energy, what will be its increase in temperature? • How much heat (in kJ) is given out when 85.0 g of lead cools from 200.0 °C to 10.0 °C? (Cp of lead = 0.129 J/g °C) • If it takes 41.72 joules to heat a piece of gold weighing 18.69 g from 10.0 °C to 27.0 °C, what is the specific heat of the gold? • It takes 333.51 joules to melt exactly 1 gram of H2O. What is the molar heat of fusion for water, from this data? • A certain mass of water was heated with 41,840 Joules, raising its temperature from 22.0 °C to 28.5 °C. Find the mass of water.

13. Homework: Read 16.1, pp. 488-495 Q pg. 495, #8 Q pg. 525, #75-77

14. Calorimeter: insulated device used for measuring amount of heat absorbed or released during chemical or physical process 'bomb' calorimeter

15. Thermochemistry: study of E changes during chemical reactions and phase changes determines direction (sign) of heat flow Boundary reaction/process studied Includes reactants/products, container, air or other substances, thermal (heat) in contact with system immediate environment that will exchange energy w/system

16. Enthalpy(H): heat content of system at constant pressure, used to measure/study energy changes that accompany reactions • Can’t measure actual energy or enthalpy of substance, only change in enthalpy for reaction • Enthalpy (heat) of reaction (∆Hrxn) • ∆Hrxn : difference between enthalpy of substances that exist at end of reaction and enthalpy of substances present at start • ∆Hrxn = Hfinal – Hinitial • ∆Hrxn = Hproducts –Hreactants

17. Exothermic Reactions • Bond forming releases more energy than was required in bond breaking • Energy released to surroundings • Reaction vessel becomes warmer Ereactants > Eproducts A + B  C + D + heat ∆H = Hproducts – Hreactants = (-ve)

18. Heats of Dilution and Solution • Place Styrofoam cup inside each of three beakers. • Add 50 mL of water to each cup and record the temperature. • Add 5 mL of 18 M sulfuric acid to 1st cup, stir with glass rod, and record the highest temperature reached. • Repeat with 10 mL and 15 mL of sulfuric acid, recording highest temperatures.

19. Place Styrofoam cup inside each of three beakers. • Add 50 mL to each cup and record the temperature. • Add 5 g calcium chloride into 1st cup, stir and record highest temperature. • Add 5 g sodium chloride to the 2nd, and 5 g lithium chloride to the 3rd. Record the temperatures. • Determine the heat of solution for each of these. • Is the amount of heat produced in the dilution of sulfuric acid related to the amount of sulfuric acid diluted? Explain. • A reaction is exothermic if the energy released by bond formation exceeds the energy required to break bonds. Explain. • Which of the chemicals tested released heat upon dissolving (exo)? Which consumed heat? (endo) • Write the reaction for the dissolving of lithium chloride in water.

20. Endothermic Reactions • Bond forming releases less energy than required in bond breaking • Energy absorbed from surroundings • Reaction vessel becomes cooler Ereactants < Eproducts A + B + heat  C + D ∆H = Hproducts – Hreactants = (+ve)

21. Endothermic Reactions • Break a cold pack. • Pour 100 mL of tap water into a beaker and record the temperature. • Add 50 g of ammonium nitrate, stir with a glass rod and record the coldest temperature. • Repeat with 50 g of table salt (NaCl) and 50 g of calcium chloride. • Approximately 326 joules of heat are consumed per gram of ammonium nitrate in this reaction. Where does this heat come from since there are no burners or heaters involved? Explain. • What change in temperature occurs when ammonium nitrate is dissolved. When sodium chloride is dissolved? When calcium chloride is dissolved? • Are all dissolution reactions endothermic? Explain. • Would any of the chemicals used in this experiment serve well for a “hot pack”? Explain.

22. Magnitude of ∆H dependent upon amounts of reactants consumed Doubling reactants, doubles enthalpy Reversing chemical reaction results in same magnitude of enthalpy but with opposite sign Splitting 2 mol H2O to produce 2 mol H2 and 1 mol O2 requires input of +483.6 kJ of energy Combining 2 mol H2 and 1 mol O2 to get 2 mol H2O gives off -483.6 kJ of energy Enthalpy change depends upon state of reactants and products States (g, l, s or aq) must be specified

23. Thermochemical calculations • Thermochemical calculations: direction of heat flow between chemical reaction/surroundings from point of view of system • Thermochemical equation: balanced chemical equation that includes physical states of all reactants/ products and energy change (change in enthalpy, ∆H) • Chemical potential energy: energy available from chemical reaction • Reaction: • Series of bond-breaking followed by bond-forming steps • Chemical reactions involve absorption/release of heat • Bond breaking always requires energy • Bond making releases energy

24. Heat and change of state • Sensibleheat (heat raising T of system can be sensed) • Latentheat (heat entering system without changing T, only phase change)

25. 5 major steps in phase change of H2O • T ice rises from -10.0 to 0.00°C • Ice melts at 0.00 °C. • T liquid rises from 0 to 100.0°C • Liquid water boils at 100.0°C • T steam  from 100.0 to 120°C http://college.hmco.com/chemistry/shared/media/animations/ucchangesofstate.html

26. Symbols used: • ∆ t = change in T from start to finish in °C • m = mass of substance in g • Cp = specific heat in J/g°C (p = measured at constant pressure) • q = amount of heat involved, in J or kJ • mol = moles of substance (must also know molar mass) • ∆ Hfus=molar heat of fusion • ∆ Hvap=molar heat of vaporization

27. Step 1: Solid ice rises in temperature • Add heat-T  until arrives at normal MP of 0oC (∆t = 10.0oC) • Ice has not melted yet • At end of step have solid ice at 0o

28. Step 2: Solid ice melts • Continue to add energy • Ice begins to melt but temperature does not change (remains at OoC) • Energy overcomes water molecules' attraction for each other, destroying ice’s 3-D structure

29. Heat of Fusion • Energy required to change gram of substance from solid to liquid state without changing its T • Each mole H2O requires constant amount of energy to melt (molar heat of fusion) • Energy breaks solid bonds, but not liquid bonds • Unit is kJ/mol (older is kcal/mol) • 4.184 J = 1cal

30. Molar heat of fusion of water = 6.01 kJ / mol Equation: q = ∆Hfus (mass) (molar mass) 31.5 g of H2O is being melted at its melting point of 0 °C. How many kJ is required? q = (6.01 kJ / mol) (31.5 g / 18.0 g/mol) 53.1 g of H2O exists as a liquid at 0 °C. How many kJ must be removed to turn water into a solid at 0 °C molar heat of fusion value used at S-L phase change, regardless of direction-melting/freezing). q = (6.01 kJ / mol) (53.1 g / 18.0 g/mol)

31. Calculate the heat of fusion for water in J/g • molar heat of fusion (J) x mass 1 mol H2O. • (6010 J/mol)/(18.0 g/mol) = 334.44 J/g-heat of fusion • Using the heat of fusion for water in J/g, calculate the energy needed to melt 50.0 g of water at its melting point of 0 °C. • heat of fusion (J/g) x mass of water involved. • (334.44 J/g) (50.0 g) = 16722.2 J = 16.72 kJ • Given ∆H CaCO3(s)  CaO(s) + CO2 (g) is 178.3 kJ, calculate ∆H for 3 CO2 (g) + 3CaO(s)  3CaCO3 (s). • Reverse reaction: -178.3 kJ x 3 = -534.9 kJ

32. Step 3: Liquid water rises in temperature • Once ice totally melted, T begins to rise • Rises until normal BP of 100.0 °C (∆t is 100) • Liquid has not boiled yet • Have liquid water at 100o

33. Step 4: Liquid water boils • Add more energy • Water boils-T doesn’t change (100oC) • Energy overcomes molecules' attraction-they move from close together (liquid) to far apart (gas state) • Each mol H2O requires constant amount of E to boil • Molar heat of vaporization (each substance has own value) • E required to boil 1 mol substance at normal BP • Water = 40.7 kJ/mol = 2261 J/g or 2.26 kJ/g.

34. Heat of Vaporization • Energy required to change gram of liquid into gaseous state at boiling point • Energy breaks down intermolecular attractive forces of liquid and provides energy necessary to expand gas

35. Molar heat of vaporization for water = 40.7 kJ/mol Equation: q = ∆Hvap (mass) (molar mass) 49.5 g H2O boiled at BP=100 °C. How many kJ is required? q = (40.7 kJ / mol) (49.5 g / 18.0 g/mol) 80.1 g of H2O exists as a gas at 100 °C. How many kJ must be removed to turn water into liquid at 100 °C? molar heat of vaporization value used at S-L phase change regardless of direction-boiling/condensing q = (40.7 kJ / mol) (80.1 g / 18.0 g/mol)

36. Calculate heat of vaporization for water in J/g • molar heat of vaporization (in J)/mass 1 mol H2O • (40700 J/mol)/(18.0 g/mol) = 2261.11 J/g-heat of vaporization • Using heat of vaporization for water in J/g, calculate energy needed to boil 50.0 g of water at its boiling point of 100 °C. • heat of vaporization (J/g) x mass of water involved • (2261.11 J/g) (50.0 g) = 113055.6 J = 113.0 kJ

37. Step 5: Steam rises in temperature • Once water completely changed to steam, T rises again • Continues to rise until we stop adding energy • Since T went from 100 to 120, ∆t is 20

40. ∆ Hfus Molar Heat of Fusion Heat energy absorbed when 1 mole of substance changes from solid to liquid (at constant temp) ∆ Hsol Molar Heat of Solidification Heat energy released when 1 mole of substance changes from liquid to solid (at constant temp) ∆ Hvap Molar Heat of Vaporization Heat energy absorbed when 1 mole of substance changes from liquid to gas (at constant temp) ∆ Hcond Molar Heat of Condensation Heat energy released when 1 mole of substance changes from gas to liquid (at constant temp) All measured in kJ / mole

42. Standard enthalpy changes-∆Ho-zero superscript tells you the reactions were carried out under standard conditions-1 atm/298K • Molar heat of reaction: quantity of heat released or absorbed during chemical reaction (same as ∆H) • If energy given off, T of system rises- exothermic, heat of reaction is negative (-) • If absorbed, T falls, endothermic (+) • Molar heat of formation: heat released or absorbed when 1 mol of compound formed by combination of elements

43. Molar heat of combustion-heat of reaction released by complete combustion (burning) of 1 mol of substance (J/mol) Calculating the heat change for a reaction: ∆H =m x C x ∆ T CH4(g) + 2O2(g) CO2 (g) + 2H2O(g) ∆H = -802 kJ How much heat energy is released when 4.5 grams of methane is burned (in a constant pressure system)? 4.5 g CH4 x __1 mole__ x –802kJ__ = -225.5 kJ 16.004 g CH4 1 mol CH4 Given N2 (g) + 3H2 (g)  2NH3 (g) ∆ H = +10 kJ, calculate mass of NH3 produced when 300 J of energy is absorbed. 10 kJ absorbed when 2 mol of NH3 is produced. (∆ H is for # moles of reactants/products in equation) That means heat per mole of NH3, ∆HRxn[NH3], is 5 kJ. 300 J x 1 kJ x 17.04 g/mol NH3 = 1.02 g 5 kJ/mol 1000 J 1 mol NH3

44. If 150.0 grams of iron at 95.0 °C, is placed in an insulated container containing 500.0 grams of water at 25.0 °C, and both are allowed to come to the same temperature, what will that temperature be? The specific heat of water is 4.18 J/g °C and the specific heat of iron is 0.444 J/g °C) • In this problem no phase changes are involved, so only the specific heat equation will be involved. Also, it should be apparent that the lead will go down in temperature as it loses energy to the colder water, which will go up in temperature. • The first key to the solution lies in thinking about the relationship of the starting temperatures to the final temperature. So let's start by calling the final, ending temperature 'x.' Keep in mind that BOTH the water and the lead will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the Dt, but the FINAL temperature.

45. The lead goes down from to 95 to x, so this means its Dt equals 95 minus x. The water goes up in temperature, so its Dt equals x minus 25. • The second key is to see that the energy amount going out of the lead is equal to the energy amount going into the water. This means qlost = qgain. Remember, q stands for the amount of heat involved. • So, by substitution, we have: • (150) (95 - x) (0.444) = (500) (x - 25) (4.18) • Solve for x

46. When 80.0 grams of a certain metal at 90.0 °C was mixed with 100.0 grams of water at 30.0 °C, the final equilibrium temperature of the mixture was 36.0 °C. What is the specific heat (Joules gram¯1 °C¯1) of the metal? • The heat lost by the metal is equal to the heat gained by the water. When put as an equation, it is: • qlost = qgain • By substitution, we then have: • (mass) (Dt) (Cp) = (mass) (Dt) (Cp) • with the metal values on the left and the water values on the right. • Plugging in the numbers gives: • (80.0 g) (54.0 °C) (x) = (100.0 g) (6.0 °C) (4.184 J / g °C) • You may finish the solution.

47. Calculate the specific heat of a metal if a 55.0 g sample of an unknown metal at 99.0 °C causes a 1.7 °C temperature rise when added to 225.0 g of water at 22.0 °C. • Heat energy is transferred from the metal to the water. The amount lost is equal to the amount gained, so we can set the equations equal to each other • (55.0 g) (75.3 °C) (x) = (225 g) (1.7 °C) (4.184 J / g °C) • The 75.3°C Dt results from it going from 99 to 23.7 (NOT 22).

48. Homework: Read 16.2-16.3, pp. 496-505 Q pg. 500, #15 Q pg. 524-525, #53, 56, 78, 80, 82, 86 (extra credit)

49. Hess’s Law: • If you add two or more thermochemical equations to produce final equation for a reaction, sum of enthalpy changes for individual reactions is enthalpy change for final reaction

50. S(s) + O2(g)  SO2 (g)           H°rxn = –296.1 kJH2 (g) + ½ O2 (g)  H2O(l)            H°rxn = –285.8 kJ H2S(g) + 3/2O2 (g)  SO2 (g) + H2O(l)     H°rxn = –561.7 kJ These equations can be arranged so that their sum is the formation reaction of hydrogen sulfide. First one is unchanged: S(s) + O2 (g)  SO2 (g)Second one is unchanged: H2 (g) + ½ O2 (g)  H2O(l)Third one is reversed: SO2(g) + H2O(l)  H2S(g) +  O2 (g)Their sum is overall equation: S(s) + H2 (g) H2S(g) We must reverse the sign of the enthalpy change of the third reaction before summing the enthalpy changes. S(s) + O2 (g)  SO2 (g)            H°rxn = –296.1 kJ H2 (g) + O2 (g)  H2O(l)           H°rxn= –285.8 kJ SO2 (g) + H2O(l)  H2S(g) + 3/2O2 (g) H°rxn= 561.7 kJS(s) + H2 (g)  H2S(g)           H°f = –20.2 kJ Summation of the three reaction steps yields the net reaction, and summation of the H° values yields the enthalpy change of the net reaction.