Thermochemistry: Energy Changes in Chemical Reactions

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2. Chapter 9 Thermochemistry: Energy Changes in Chemical Reactions

3. Particulate Review • Which representation depicts the solid phase of water? The liquid? The gaseous? • Is energy added or released during the physical change from (a) to (b)? What intermolecular forces are involved? • Describe the energy changes that accompanies the physical change from (a) to (c)

4. Particulate Review (Cont. 1) • Which representation depicts the solid phase of water? The liquid? The gaseous? • Solid = (c) Liquid = (a) Gas = (b)

5. Particulate Review (Cont. 2) • Is energy added or released during the physical change from (a) to (b)? What intermolecular forces are involved? • Energy is added to go from liquid (a) to gas (b). Hydrogen bonding is the intermolecular force.

6. Particulate Review (Cont. 3) • Describe the energy changes that accompanies the physical change from (a) to (c). • Energy is released as the liquid (a) becomes a solid (c).

7. Particulate Preview Bonds and Functional Groups • What kind of bonds must be broken for calcium chloride to dissolve in water? Is energy absorbed or released in order to break these bonds? • Which color spheres represents the chloride ions? Label the polar covalent bonds in water using d+ and d-. • What intermolecular interactions from as the salt dissolves? Is energy absorbed or released as these attractions form?

8. Particulate Preview (Cont. 1) Bonds and Functional Groups • What kind of bonds must be broken for calcium chloride to dissolve in water? Is energy absorbed or released in order to break these bonds? Ionic bonds between the calcium and chloride must be broken for calcium chloride to dissolve in water. Energy is released when the bonds break in calcium chloride.

9. Particulate Preview (Cont. 2) Bonds and Functional Groups • Which color spheres represents the chloride ions? Label the polar covalent bonds in water using d+ and d-. The chloride ions are represented by gray spheres.

10. Particulate Preview (Cont. 3) Bonds and Functional Groups • What intermolecular interactions from as the salt dissolves? Is energy absorbed or released as these attractions form? As the salt dissolves the ions interact with the permanent dipoles of water and ion-dipole intermolecular interactions form. Energy is released as these attractions form.

11. Chapter Outline 9.1 Energy as a Reactant or Product 9.2 Transferring Heat and Doing Work 9.3 Enthalpy and Enthalpy Changes 9.4 Heating Curves and Heating Capacity 9.5 Enthalpies of Reaction and Calorimetry 9.6 Hess’s Law and Standard Enthalpies of Reaction 9.7 Enthalpies of Reaction from Enthalpies of Formation and Bond Energies 9.8 Energy Changes When Substances Dissolve 9.9 More Applications of Thermochemistry

12. Energy as a Reactant or Product • Combustion reactions produce most of the energy we consume. • Combustion of coal • Combustion of natural gas (CH4) • Combustion of gasoline • Fossil fuels: decomposed plant/animal remains • Origins of energy in fossil fuels from food chains that supplied nutrition to plants and animals which started with photosynthesis

13. Energy as a Reactant or Product (Cont. 1)

14. Forms of Energy • Chemical energy • Potential energy stored in chemical bonds • Internal Energy (E) • Sum of all kinetic and potential energy • Change in energy (DE): when the collective internal energies of reactants are different than products

15. Forms of Energy (Cont. 1) • Thermochemistry • The study of the changes in energy that accompany chemical reactions • Thermodynamics • The study of energy and its transformations

16. Forms of Energy (Cont. 2) • First Law of Thermodynamics • Energy cannot be created or destroyed only changed in form. • Total energy of the universe is constant. • System • The part of interest • Surroundings • Everything else

17. Forms of Energy (Cont. 3) Universe = system + surroundings DEuniv = DEsys + DEsurr = 0 DEsys= −DEsurr • Whatever energy the system loses (DEsys < 0) its surroundings gain (DEsurr< 0).

18. Forms of Energy (Cont. 4) • Work is a force exerted through a distance. w= F × d • If heat flows out from a system or if work is being done by the system, Esys decreases. • If heat flows into a system or work is done on a system, Esys increases. DE = q + w • Internal energy is a state function. • The change in internal energy is independent of how it was acquired. • DE = Efinal– Einitial

19. Forms of Energy (Cont. 5) • Heat is energy that is in the process of being transferred from a higher-temperature object to a lower-temperature object. • Thermal energy is determined by its temperature and its composition of material and its mass.

20. Transferring Heat and Doing Work • System needs to be identified and defined • Can be galaxies or laboratory apparatus • May be just the particles involved or include the vessel

21. Isolated, Closed, and Open Systems Isolated system • A system completely shut off from its surroundings so that neither matter nor energy can be transferred Closed • If only energy can flow between system and surroundings Open • If energy and matter can flow freely between system and surroundings

22. Exothermic and Endothermic Processes • Exothermic • Energy flow (usually in form of heat) from a system to its surroundings • Endothermic • Energy flow (usually in form of heat) from surroundings into system

23. Exothermic and Endothermic Processes

24. P-V Work Pressure-volume work

25. Practice 9.2 Calculating P-V work A tank of compressed helium is used to inflate 100 balloons for sale at a carnival on a day when the atmospheric pressure is 1.01 atm. If each balloon is inflated with 4.8 L, how much P-V work is done by the compressed helium? Express your answer in joules. • Collect and Organize • Analyze • Solve • Think About It

26. Collect and Organize • Our task is to determine how much P-V work is done by inflating the 100 balloons, each with 4.8 L of helium, and each expanding against an opposing pressure of 1.01 atm.

27. Analyze • We focus on the work done on the atmosphere (the surroundings) by the helium (the system) as the volume of the balloons increases. Inflating each of 100 balloons with about 5 L of helium against a pressure of 1.01 atm means that PDV will be about 500 liter–atmospheres and about 100 times that number of joules, or 50,000 J.

28. Solve • The work (w) done by the compressed helium is

29. Think About It • The internal energy of the compressed helium (the system) decreases when some of it expands to inflate the balloons because work is done by the system on its surroundings. Lower internal energy in this case corresponds to the lower kinetic energy (and lower temperature) of both the helium atoms in the balloons and those remaining in the tank.

30. Practice 9.3 Relating DE, q, and w The racing cars in Figure 9.12 are powered by V8 engines in which the motion of each piston in its cylinders displaces a volume of 0.733 L. If combustion of the mixture of gasoline vapor and air in one cylinder releases 1.68 kJ of energy, and if 33% of the energy does P-V work, how much pressure, on average, does the combustion reaction mixture exert on each piston? How much energy is transferred as heat from the reaction mixture to its surrounding? • Collect and Organize • Analyze • Solve • Think About It

31. Collect, Organize, and Analyze • We know the quantity of energy (1.68 kJ) released by a combustion reaction and that 33% of this energy does P-V work on its surroundings. The rest (100 − 33 = 67%) must be lost as heat as described by Equation 9.4: DE = q − PDVThe system loses energy; therefore, DE = −1.68 kJ and the value of q will be about 2/3 of −1.68 kJ or about −1 kJ.

32. Solve • Using the information give and Equation 9.4 to solve for q and P:

33. Think About It • We used a negative sign in the last equation because PDV contains positive values of pressure and volume changes but represent work done by the system and are part of the energy lost by the system (−1.68 kJ). The heat transferred has a negative value because it is thermal energy lost by the system. Its value is close to the one we estimated.

34. Enthalpy and Enthalpy Changes • Enthalpy (H) • Sum of the internal energy and pressure-volume product of a system, H = E + PV • Enthalpy change (DH) • Quantity of heat transferred into or out from a system during a chemical reaction or physical change at constant pressure, DH = DE + PDV DH = qP • Exothermic DH < 0 • Endothermic DH > 0

35. Enthalpy and Enthalpy Changes (Cont. 1) • Enthalpy of fusion (DHfus) • Energy required to convert one mole of a solid substance at its melting point into the liquid state • Enthalpy of vaporization (DHvap) • Energy required to convert one mole of a liquid substance at its boiling point into the vapor state

37. Heating Curves and Heat Capacity • We can track the changes in temperature and physical states as water goes from snow at −18°C into boiling water.

38. Heating Curves and Heat Capacity (Cont. 1) • Heat capacity (CP) • The energy required to raise the temperature of an object by 1°C at constant pressure • Specific heat (cP) • The energy required to raise the temperature of 1 gram of a substance by 1°C at constant pressure • Molar heat capacity (cP,n) • The energy required to raise the temperature of one mole of substance by 1°C at constant pressure

40. Hot Soup on a Cold Day Suppose a hiker decides to make soup starting with 275 g of snow. They put the snow with an initial temperature of −18.0°C into a pan and start to heat it. • Heat snow from −18.0°C to 0.0°C • Melt snow • Heat water from 0.0°C to 100.0°C • Boil water

41. Hot Soup on a Cold Day (Cont. 1) • Water has high specific heat value. • Water can absorb large quantities of thermal energy. • Heat sink • Matter that can absorb energy without changing phases or significantly changing its temperature • Weather and climate changes driven by Earth’s oceans which serve as enormous heat sinks • Hydrogen bonding

42. Practice 9.5 Calculating a Final Temperature from Energy Gain and Loss Suppose you have exactly 1 cup (237 g) of hot (100.0°C) brewed tea in an insulated mug and that you add to it 2.50 × 102 g of ice initially at −18°C. If all the ice melts, what is the final temperature of the tea? Assume the tea has the same thermal properties as water. • Collect and Organize • Analyze • Solve • Think About It

43. Collect and Organize • We know the mass of tea, its initial temperature, and its specific heat (cP value for liquid water). We know the mass of ice and its initial temperature. From Tables 9.2 and 9.3 we know the specific heats and molar heat capacities of ice and liquid water and the enthalpy of fusion of ice. Our task is to find the final temperature of the tea.

44. Analyze • The energy lost by the tea has the same absolute value as, but the opposite sign of, the energy gained by the ice cubes. In equation form: qice = −qtea. Based on the way the problem is presented, we have to assume that once all the ice has melted the 250 g of 0.0°C water produced will mix with and be warmed by the still warmer 237 g of tea. Together they reach the same final temperature.

45. Analyze (Cont. 1) • Therefore there are three parts to qice: the energy gained when the ice warms from −18.0°C to 0.0°C, the energy gained when the ice melts, and the energy gained when the temperature of the melted ice increases from 0.0°C to the final temperature (which we will call x°C). • We have to calculate the number of moles of H2O in the ice to use DHfus, so we’ll use the molar heat capacities of ice and water to calculate qice.

46. Analyze (Cont. 2) • The tea undergoes no phase change, so its only loss of energy is due to its temperature change from 100.0°C to x°C.

47. Solve • Converting grams of ice to moles: • The ice gains energy as it • Warms to 0.0°C • Melts:

48. Solve (Cont. 1) • Warms to final temperature x: • The energy lost by the tea (qtea) as it cools to x:

49. Solve (Cont. 2) • Balancing the loss and gain of energy:

50. Think About It • The answer is consistent with the assumption that after all of the ice melted the temperature of the liquid water thus produced would increase a little to reach the same temperature as the tea. There were no temperature units in the final values used to calculate x but remember x was assumed to have units of °C when we inserted it into the DT term in the first steps of the calculation.