AP Thermochemistry AP Thermodynamics

1. AP ThermochemistryAP Thermodynamics Mack

2. Thermochemistry vs. Thermodynamics • Thermochemistry – relationships between chemical reactions and energy changes that involve heat • Thermodynamics – study of energy and its transformations

3. In Thermo. What specifically are we studying?! • The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). • The surroundings are everything else (here, the cylinder and piston).

4. Systems • An open system allows for energy and mass to change, like an open cup of coffee or beaker. • A closed system allows for changes in energy only, like a closed coffee cup or beaker. • An isolated system doesn’t allow any change in temperature or heat to occur, like a perfect calorimeter.

5. Energy • Energy – ability to do work or produce heat • Sum of all potential and kinetic energy in a system is the internal energy of the system • Potential Energy – energy by virtue of position • Energy stored in chemical bonds • Kinetic Energy – energy of motion • Translational, rotational, & vibrational (we will learn about this more in-depth later)

7. Law of Conservation of Energy • 1st law of Thermodynamics – Energy is never created nor destroyed • Any change in energy of a system must be balanced by the transfer of energy either into or out of the system • Energy of the universe is constant

8. Heat vs. Temperature • Heat (q) – flow of energy from an area of warmer energy to an area of lower energy • Warmer  cooler • Thermal Equilibrium – when systems with two different heat contents (typically two different temperatures) come in contact with each other, they will exchange heat content until they reach thermal equilibrium with each other (i.e. same temperature) • Temperature (T) – the measure of the average kinetic energy of the particles • Heat ‘em up, speed ‘em up

10. Heat (q) • Endothermic: Energy goes into the system, the reaction is not spontaneous (low stability), positive value. • Exothermic: Energy comes out of the system, the reaction is spontaneous (high stability), negative value.

11. Reaction Energy Diagrams

12. Energy and Work • Recall: Energy is the ability to do work or produce heat • E = q + w • Translation: Change in energy = heat + work • Heat can either be + or – • + q means heat is absorbed (endothermic) • - q means heat is released (exothermic) • Work can either be + or – • + w means work done on the system (compression) • - w means work done by the system (expansion) • Work = -P V

13. Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). • That is, E = q + w.

14. E, q, w, and Their Signs

15. Change in Energy Example • Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system. • E = q + w • E = 15.6 kJ + 1.4 kJ • 17 kJ • 17000 J

16. Work Examples • Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. • Work = -P V • W = - (15 atm)(64 L - 46 L) • W = 270 atm x L

17. Energy and Work Example • A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the volume expands against a constant pressure of 1.0 atm, calculate the change in energy for the process. • Need to use both formulas but I can plug and chug to make one E = q + wWork = -P V E = q + -P V (1.3 x 108 J) + - (1.00 atm)(4.50 x 106 L– 4.00 x 106 L) 1 L *atm = 101.3 J (1.3 x 108 J) + - (455850000 J – 405200000 J) 79350000 J 7.935 x 107 J 8.0 x 107 J

18. Enthalpy • Enthalpy (H) • State Function • Function that is independent of the pathway taken to get the answer • Only the answer matters • H = q at constant pressure • Enthalpy can be calculated multiple ways • Stoichiometry • Calorimetry • ∑HProducts – ∑HReactants • Hess’s Law • Bond Energies

19. Stoichiometry • When you are looking at a value of H, it normally is reported in units of kJ/mole. • This assumes you have one mole (i.e. kJ/1 mole) • But what happens if you do not have just one mole?? • You need to account for that utilizing stoichiometry! (i.e. dimensional analysis / conversions)

20. Calculation of H through stoichiometry, example 1 • When 1 mole of methane (CH4) is burned at constant pressure 890 kJ/mol of energy is released as heat. Calculate H for a process in which a 5.8 gram sample of methane is burned at constant pressure. CH4 + O2 CO2 + 2 H2O + (-890 kJ) How many moles is 5.8 grams of methane? = 0.36155 mole How much heat is released when you only have 0.36155 moles? = -321.7795 kJ/molereaction -320 kJ/molereaction

21. Calorimetry • Calorimetry - The process of measuring heat based on observing the temperature change when the system absorbs or discharges heat • Since we cannot measure the exact enthalpy of the reactants and products of a reaction, we can measure the change in enthalpy through a calorimeter

22. Heat Capacity, Specific Heat, & Molar Heat Capacity • We define specific heat capacity (or simply specific heat, c) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (1C). • The higher the specific heat, the more energy required to raise the temperature • The amount of energy required to raise the temperature of a substance with a fixed mass (such as the calorimeter) by 1 K is its heat capacity, C. • Molar heat capacity is the SAME thing BUT it is specific to one mole of substance

23. Calorimetry continued • The specific heat of water is 4.18 J/g*°C • This is a value that you MUST know • When using this value, your units of q will be in J • When we are talking about a calorimeter, we will be using the equation q = mc𝚫T • Heat lost by one substance is gained by another • - q = + q

24. Calorimetry Examples • A 466 g sample of water is heated from 8.5 °C to 74.6 °C. Calculate the heat change. • A 869 g sample of Fe (with c = 0.46 J/g C) cools from 84 °C to 5 °C. Calculate the heat change. q = mc𝚫T = (466)(4.18)(66.1) = 128754.868 J = 130000 J or 1.3 x 105 J q = mc𝚫T = (869)(0.46)(-79) = -31579 J = -30000 J or -3 x 104 J

25. Calorimetry Example • 7.85g of Cu is heated to 70.0 °C and placed in a calorimeter with 5.00 g of water at 10.0 °C. At equilibrium, the temperature equals 20.0 °C. Calculate the specific heat of the Cu. -q = +q -mc𝚫T = mc𝚫T - (7.85)(x)(20 – 70) = (5.00)(4.18)(20 – 10) 392.5 x = 209 x = .532 J/g* °C

26. Molar heat of combustion example • 1.435 g of decyne (C10H18) is burned in a calorimeter and the temperature rose from 20.17 °C to 25.84 °C. The mass of the water equals 2000 g and the heat capacity of the calorimeter equals 1.80 kJ/C. Calculate the molar heat of combustion. = 0.010380 moles = 598.43 J/mole = .598 kJ/mole qreaction = qwater + qcalorimeter (2000)(4.18)(5.67)+(1800)(5.67) = 57652.56 J

27. What about the heating/cooling curve? • We can only use the formula q = mc𝚫T for a temperature change, but what is we are just undergoing a phase change?

28. Heat of Fusion q = mHf • Used when no temperature change has taken place, but the substance turns from solid to liquid or liquid to solid. • Hf is specific to the substance like specific heat (c). • The Hf values have been experimentally determined • The value of Hf Relates to IMF (intermolecular forces) • Solid  liquid (+ Hf) • You are putting energy into the system • Liquid  solid (- Hf) • You are removing energy from the system

29. Heat of fusion example • Calculate the heat absorbed when 5 grams of ice melts (Hf=334 J/g). q = mHf (5)(334) q = 1670 J

30. Heat of vaporization q = mHv • Used when no temperature change has taken place, but the substance turns from gas to liquid or liquid to gas. • Hf is specific to the substance like specific heat (c). • The Hf values have been experimentally determined • The value of Hf Relates to IMF (intermolecular forces) • liquid  gas (+ Hv) • You are putting energy into the system • gas  liquid (- Hv) • You are removing energy from the system

31. Heat of vaporization example • Calculate the heat absorbed when 5 grams of water vaporizes (Hv=2260 J/g). q = mHv(5)(2260) q = 11300 J

32. Putting it all together example

33. Enthalpies of Formation(∑HProducts – ∑Hreactants) • A standard enthalpy of formation, Hf°, is defined as the enthalpy change for the reaction in which one mole of a compound is made from its elements in their elemental forms at 1 atm and 25 °C/298 K (these are standard conditions). Elements are zero if the element is at it’s standard state. •  Hrxn° is the standard heat of a reaction at 1atm and 25C. This is obtained by  Hf°(P-R) • These values will normally be worked into a problem for you

34. Enthalpy of Formation Example Given the values below, calculate the ΔHrxn° of: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) C3H8 ΔHf° = -103.85 kJ/mol CO2ΔHf° = -393.5 kJ/mol H2O ΔHf° = -285.8 kJ/mol [3(-393.5) + 4(-285.8)] - [-103.85 + 5(0)] = -2220.0 kJ/mol This tells us only that the reaction is exothermic and may be favorable (spontaneous).

35. Hess’s Law • H° is well known for many reactions, and it is inconvenient to measure H° for every reaction in which we are interested. • However, we can estimate H° using other H° values that are published. • Hess’s law states that “If a reaction is carried out in a series of steps, H° for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” • What you do to the equation, you have to do to the H • Multiply the equation by 2, multiply the H by two • Flip the equation, flip the sign of H

36. Example Given: H2O(s)  H2O(l) H°=6.01KJ CH4(g)+2O2(g)CO2(g)+2H2O(l) H°=-890.4KJ Calculate the heat of reaction of: 2CH4(g)+4O2(g)2CO2(g)+ 4H2O(s) Multiply equation 2 by 2: 2CH4(g)+ 4O2(g)2CO2(g)+4H2O(l) H=2(-890.4KJ) Reverse quadruple equation 1: 4H2O(l)  4H2O(s) H=-4(6.01KJ) Combine new equations : H = 2(-890.4)+-4(6.01) = -1804.84kJ

37. Bond Energies (Enthalpies) • Breaking bonds is endothermic • You need to put energy in • Making bonds is exothermic • Energy is released • If we have a reaction that uses and makes only covalent compounds we can find H using bond energies • ∑Hbonds broken – ∑Hbonds formed • ∑Hreactants – ∑Hproducts • THIS IS THE ONLY EQUATION THAT IS REACTANTS - PRODUCTS

38. Average Bond Enthalpies Table • For your information: These values are only general estimations—not every single Carbon—Hydrogen bond in the world is the same as all the others. Our DH will be an approximation.

39. Bond Enthalpy Reaction Example 1 CH4 + 2O2 CO2 + 2H2O ∑Hreactants – ∑Hproducts Reactants – Products = [4(413) + 2(495)] – [(4 x 463) + 2 (799)] = 2642 - 3450 = -808 kJ/mole

40. Bond Enthalpy Reaction Example 2 CH3Br + HI CH3I + HBr ∑Hreactants – ∑Hproducts Reactants – Products = [3(413) + 276 + 299] – [(3 x 413) + 240 + 366] = 1814 - 1845 = - 31 kJ/mole

41. What drives a reaction to be thermodynamically favored? • Enthalpy (H) • Heat exchange • Exothermic reactions are generally favored • Entropy (S) • Dispersal (disorder) • The more disorder/chaos is favored • Thermodynamically favored processes are reactions that involve both a decrease in internal energy of the system (𝚫H < 0) and an increase in entropy (𝚫S > 0) • These are thermodynamically favored (i.e. spontaneous) • Spontaneous reactions depend on both H and S

42. Spontaneous Processes • Spontaneous processes are those that can proceed without any outside intervention. • The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously change.

43. Spontaneous Processes • Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures. • Above 0C it is spontaneous for ice to melt. • Below 0C the reverse process is spontaneous.

44. Second Law of Thermodynamics • The second law of thermodynamics states that the entropy of the universe increases for spontaneous processes.

45. Entropy is dependent on: • The greater the dispersal of matter and/or energy, the larger the entropy • Solids < liquids < gases • When a solid or liquid dissolves in a solvent (becomes aq), the entropy increases • When a gas molecule escapes from a solvent, the entropy increases • Think carbonated beverages • Entropy increases with increasing molecular complexity • KCl vs CaCl2 • Reactions increasing the number of moles of particles increase entropy

46. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

47. Standard Entropy • These are molar entropy values of substances in their standard states. • Standard entropies tend to increase with increasing molar mass.

48. Standard Entropies Larger and more complex molecules have greater entropies.

49. Examples • Which has a greater entropy? • NaCl(s) or NaCl(aq) • CO2(s) or CO2(g) • KI(aq) at 45 °C or 100 °C • Is the entropy positive (increasing) or negative (decreasing)? • 2H2(g) + O2(g)  2H2O(l) becoming more ordered • NH4Cl(s)  NH3(g) + HCl(g) becoming less ordered

50. Entropy • Like total energy and enthalpy, entropy is a state function (dependant on only final and initial conditions, not the changes it takes in between). • Entropy changes for a reaction can be estimated in the same manner by which H is estimated: Srxn° = Sf°(products) - Sf°(reactants)