1. Unit 12 Thermodynamics Chapter 16

2. Thermodynamics Definition A study of heat transfer that accompanies chemical changes Concerned with overall chemical changes Chemical Change involves: A change in energy A degree of disorder

3. 16.1 Energy The Nature of Energy Energy is the ability to do work or produce heat. It exists in two basic forms, potential energy and kinetic energy. Potential energy is energy due to the composition or position of an object. A macroscopic example of potential energy of position is water stored behind a dam above the turbines of a hydroelectric generating plant. When the dam gates are opened, the water rushes down and does work by turning the turbines to produce electrical energy. The energy stored in a substance because of its composition is called chemical potential energy. Chemical potential energy plays an important role in chemical reactions.

4. 16.1 Energy The Nature of Energy Kinetic energy is energy of motion. You can observe kinetic energy in the motion of people and objects all around you. The potential energy of the dammed water is converted to kinetic energy as the dam gates are opened and the water flows out. Chemical systems contain both kinetic energy and potential energy. As temperature increases, the motion of submicroscopic particles increases. The potential energy of a substance depends upon its composition: the type of atoms in the substance, the number and type of chemical bonds joining the atoms, and the particular way the atoms are arranged.

5. 16.2 Heat in Chemical Reactions & Processes Chemical Potential Energy Heat, which is represented by the symbol q, is energy that is in the process of flowing from a warmer object to a cooler object. When the warmer object loses heat, its temperature decreases. When the cooler object absorbs heat, its temperature rises.

6. 16.2 Heat in Chemical Reactions & Processes Measuring Heat The flow of energy and the resulting change in temperature are clues to how heat is measured. In the metric system of units, the amount of heat required to raise the temperature of one gram of pure water by one degree Celsius (1°C) is defined as a calorie (cal). The SI unit of heat and energy is the joule (J). One joule is the equivalent of 0.2390 calories, or one calorie equals 4.184 joules.

7. Reactions Endothermic Exothermic

8. Exothermic Reactions

9. Potential Energy converts to Kinetic Energy as you release the energy to the surroundings. Where is the Potential energy stored? How do we know there is a shift to Kinetic energy? Most reactions are exothermic and spontaneous

10. Endothermic Reactions

11. Energy content of products is greater than reactants If the products have more energy why are the surroundings cold? (Hint: PE↑) The surroundings feel cold because the bonds absorb the heat energy from the surroundings, so Kinetic energy converts to Potential energy

12. Comparing reactions… 2 H 2 + O 2 ----> 2H 2 O + 136 Kcal 2 C + 2 H 2 + 52 kJ ----> C 2 H 4 If the energy appears on the product side, it is negative. If the energy appears on the reactant side, it is positive.

13. In Thermodynamics… System refers to the reaction itself Surroundings refers to everything else Standard Conditions 25°C (298 K) 1 atmosphere 1 molar solution

14. 16.4 Calculating Enthalpy Change Chemical Energy and the Universe Thermochemistry is the study of heat changes that accompany chemical reactions and phase changes. In thermochemistry, the system is the specific part of the universe that contains the reaction or process you wish to study. Everything in the universe other than the system is considered the surroundings. Therefore, the universe is defined as the system plus the surroundings. universe = system + surroundings

15. Enthalpy Symbol:H Measure of heat content (energy) of a system at constant pressure Can’t be measure directly, can only measure the change in enthalpy. We call this the Heat of Reaction, ΔH Measure of the heat released or absorbed in a chemical reaction! ΔH rxn = ΔH products - ΔH reactants

16. Standard Heat of Formation Change in enthalpy from the formation of 1 mol of a compound, in its standard state, from its elements. Symbol: “  ” refers to standard conditions Units: kJ/mol Example: S (s) + O 2(g) → SO 2(g) -297 kJ/mol * Table 16-7 on page 510 lists Standard H f

17. Chemical Reactions Compare and contrast the following graphs: What did you notice? How would you write these reactions in standard format?

18. Standard Heat of Formation What can we tell from values? Positive value means? Negative value means? Thermochemical equations 4Fe (s) + 3O 2(g) → 2Fe 2 O 3(s) + 1625 kJ NH 4 NO 3(s) + 27 kJ → NH 4 + (aq) + NO 3 - (aq)

19. When you state the height of a mountain, it is relative to another point (usually sea level). In the same way enthalpies of formation are stated based on the following arbitrary standard: Every free element in its standard state has a value of exactly 0.0 kJ. That way when the heat of formation is negative the system has lost heat, when positive the system has gained heat! Where do Standard Heats of formation come from?

20. Check for understanding! Do elements in their standard states possess zero energy? Why are elements in their standard states assigned enthalpies of zero? What does the +33.2 on the graph tell you? What does the -396 on the graph tell you?

21. Enthalpy change from Standard Heat of Formation Use standard heat of formation to calculate ΔH rxn  for the combustion of methane CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) We can summarize Hess’s Law into the following equation: ΔH rxn  = ΣΔH f  (products) - ΣΔH f  (reactants) The symbol Σ means “to take the sum of the terms.”

22. Enthalpy change from Standard Heat of Formation Use standard heat of formation to calculate ΔH rxn  for the combustion of methane CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) First: look up ΔH f  values Second: Use the formula and multiply each term by the coefficient of the substance in the balanced chemical equation Third: Do the math

23. Enthalpy change from Standard Heat of Formation Use standard heat of formation to calculate ΔH rxn  for the combustion of methane CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) ΔH f  (CO 2 ) = -394 kJΔH f  (H 2 O) = -286 kJ ΔH f  (CH 4 ) = -75 kJΔH f  (O 2 ) = 0.0 kJ ΔH rxn  = [(-394 kJ) + (2)(-286 kJ)] – [(-75 kJ) + (2)(0.0 kJ)] products- reactants ΔH rxn  = [-966 kJ] – [-75 kJ] = -891 kJ Is this reaction endothermic or exothermic?

24. Reaction Spontaneity What happens when you leave a iron nail outside for a few months? What happens when you light a gas stove? Do these reactions take place spontaneously? (without outside intervention) Will the reverse of these reactions take place spontaneously? What about ice melting at room temperature? There is something more than ΔH determining spontaneity!

25. Entropy Symbol: S Measure of the disorder or randomness of the particles that make up the system Molecules are more likely to exist in a high state of disorder than in a low state of disorder. Change in entropy is similar to change in enthalpy ΔS system =S products - S reactants

26. 16.5 Reaction Spontaneity Dissolving of a gas in a solvent: When a gas is dissolved in a liquid or solid solvent, the motion and randomness of the particles are limited and the entropy of the gas decreases. Change in the number of gaseous particles: When the number of gaseous particles increases, the entropy of the system usually increases because more random arrangements are possible. Dissolving of a solid or liquid to form a solution: When solute particles become dispersed in a solvent, the disorder of the particles and the entropy of the system usually increase. Change in temperature: A temperature increase results in increased disorder of the particles and an increase in entropy.

27. Entropy Do you expect the ΔS of the phase change shown below to be positive or negative? S products > S reactants ΔS system positive S products < S reactants ΔS system negative

28. Entropy What happens to molecules when you increase their temperature? Do you think this will increase or decrease their entropy? ΔS positive = more entropy, ie more disorder ΔS negative = less entropy, ie less disorder Reactions tend to go spontaneous towards increased entropy.

29. Entropy Practice Predict the ΔS for the following changes: 1. H 2 O(l) → H 2 O(g) 2. CO 2 (g) → CO 2 (aq) 3. 2SO 3 (g) → 2SO 2 (g) + O 2 (g) 4. NaCl(s) → Na + (aq) + Cl - (aq) 5. CH 3 OH(l) → CH 3 OH(aq)

30. Gibbs Free Energy By calculating free energy (energy that is available to do work) we can determine if a reaction is spontaneous. Just like Enthalpy and Entropy we can only measure the free energy as a change. ΔG system = ΔH system – TΔS system The sign of ΔG system tells you if the reaction is spontaneous: Negative = spontaneous, will occur Postive = nonspontaneous, will NOT occur

31. Gibbs Free Energy How do the enthalpies and entropies affect reaction spontaneity? What is happening if ΔG system = 0? -ΔH system +ΔH system +ΔS system Always spontaneous, -ΔG system Spontaneous only at high Temperatures, + or - ΔG system -ΔS system Spontaneous only at low Temperatures, + or - ΔG system Never spontaneous +ΔG system

32. Review of Symbols ΔH – Tells us if a reaction is endothermic or exothermic (measure of change in energy) Postive = endothermic Negative = exothermic ΔS – Tells us if the reaction is more or less ordered (randomness of particles) Positive = more disordered Negative = more ordered ΔG – Tells us if the reaction is spontaneous by determining the amount of energy available to do work. Positive = nonspontaneous Negative = spontaneous

33. States of Matter - Phase Changes During phase changes, there is no change in kinetic energy - only potential energy increases!!! Liquid H 2 O at 0 o C has more kinetic energy than solid H 2 O at 0 o C Gas H 2 O at 100 o C has more kinetic energy than liquid H 2 O at 100 o C Coke on ice - shows 3 phases of matter

34. Changes of State Heat of Vaporization: Heat required to vaporize one mole of a liquid Heat of Fusion: Heat required to melt one mole of a solid

35. Changes of State

36. 16.3 Thermochemical Equations Specific Heat You’ve learned that one calorie, or 4.184 J, is required to raise the temperature of one gram of pure water by one degree Celsius (1°C). That quantity, 4.184 J/(g∙°C), is defined as the specific heat (c) of water. The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius. Because different substances have different compositions, each substance has its own specific heat.

37. The heat absorbed or released by a substance during a change in temperature depends not only upon the specific heat of the substance, but also upon the mass of the substance and the amount by which the temperature changes. You can express these relationships in an equation. In the equation, q = the heat absorbed or released, c = the specific heat of the substance, m = the mass of the sample in grams, and ∆T is the change in temperature in °C. ∆T is the difference between the final temperature and the initial temperature or, T final – T initial. 16.3 Thermochemical Equations Calculating Heat Evolved and Absorbed

38. 16.3 Thermochemical Equations Calculating Specific Heat In the construction of bridges and skyscrapers, gaps must be left between adjoining steel beams to allow for the expansion and contraction of the metal due to heating and cooling. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 114 J heat. What is the specific heat of iron? You are given the mass of the sample, the initial and final temperatures, and the quantity of heat released. The specific heat of iron is to be calculated. The equation that relates these variables can be rearranged to solve for c. Known joules of energy released = 114 J ∆T= 50.4°C – 25.0°C = 25.4°C mass of iron = 10.0g Fe

39. 16.3 Thermochemical Equations Calculating Specific Heat Unknown specific heat of iron, c = ? J/(g∙°C) Rearrange the equation q = c x m x ∆T to isolate c by dividing each side of the equation by m and ∆T. Solve the equation using the known values.

40. 16.3 Thermochemical Equations Measuring Heat Heat changes that occur during chemical and physical processes can be measured accurately and precisely using a calorimeter. A calorimeter is an insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.

41. 16.3 Thermochemical Equations Determining Specific Heat You can use a calorimeter to determine the specific heat of an unknown metal. Suppose you put 125 g of water into a foam-cup calorimeter and find that its initial temperature is 25.6°C. Then, you heat a 50.0-g sample of the unknown metal to a temperature of 115.0°C and put the metal sample into the water. Heat flows from the hot metal to the cooler water and the temperature of the water rises. The flow of heat stops only when the temperature of the metal and the water are equal.

42. 16.3 Thermochemical Equations Determining Specific Heat Both water and metal have attained a final temperature of 29.3°C. Assuming no heat is lost to the surroundings, the heat gained by the water is equal to the heat lost by the metal. This quantity of heat can be calculated using the equation you learned, q = c x m x ΔT. First, calculate the heat gained by the water. For this you need the specific heat of water, 4.184 J/(g∙°C). The heat gained by the water, 1900 J, equals the heat lost by the metal, qmetal, so you can write this equation.

43. 16.3 Thermochemical Equations Determining Specific Heat Now, solve the equation for the specific heat of the metal, cmetal, by dividing both sides of the equation by m x ∆T. The change in temperature for the metal, ∆T, is the difference between the final temperature of the water and the initial temperature of the metal (115.0°C – 29.3°C = 85.7 °C ). Substitute the known values of m and ∆T (50.0 g and 85.7 °C) into the equation and solve.

44. 16.3 Thermochemical Equations Determining Specific Heat The unknown metal has a specific heat of 0.44 J/(g·°C). From the table, you can infer that the metal could be iron.

45. 16.3 Thermochemical Equations Using Data from Calorimetry A piece of metal with a mass of 4.68 g absorbs 256 J of heat when its temperature increases by 182°C. What is the specific heat of the metal? Known mass of metal = 4.68 g metal quantity of heat absorbed, q = 256 J ∆T = 182°C Unknown specific heat, c = ? J/(g·°C) Solve for c by dividing both sides of the equation by m x ∆T.

46. Using Data from Calorimetry Substitute the known values into the equation. The calculated specific heat is the same as that of strontium. 16.3 Thermochemical Equations