1 / 43

Thermochemistry & Thermodynamics

Thermochemistry & Thermodynamics. AP Chemistry. Brief Overview. Chemical reactions involve changes in energy Thermochemistry Study of relationships of chemical reactions and energy changes Thermodynamics Study of energy and its transformations. Energy.

katina
Télécharger la présentation

Thermochemistry & Thermodynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Thermochemistry & Thermodynamics AP Chemistry

  2. Brief Overview • Chemical reactions involve changes in energy • Thermochemistry • Study of relationships of chemical reactions and energy changes • Thermodynamics • Study of energy and its transformations

  3. Energy • Energy is the ability to do work or produce heat and is the sum of all potential and kinetic energy in a system • Potential energy is energy possessed due to relative position to other objects • Commonly in chemistry this is energy stored in bonds • Kinetic energy is the energy of motion, usually of particles

  4. Work & Heat • Heat (q) transfer of energy from object with higher KE to object with lower KE • Hence heat flows from hot to cold • Heat transfers because of temperature (KE) difference, but temperature is nota measure of heat directly • Work (w) is a force acting over a distance • Usually used when discussing gases • When related to gases, work is a function of pressure and volume:

  5. Energy, Work, and Heat • Energy is a state function while heat and work are not • State function is a property that is independent of past or future behavior, only on current conditions (temperature, pressure, etc) • Example: Many roads from your home to school, but you have the same starting point and ending point

  6. Signs of Work & Heat • Signs of q • +q means heat is absorbed (endothermic) • -q means heat is released (exothermic) • Signs of w • +w if work done on the system (i.e. compression) • -w if work done by the system (i.e. expansion)

  7. First Law of Thermodynamics • Energy cannot be created or destroyed, only conserved • Law of Conservation of Energy • Energy of universe is constant • Energy may only transfers between system (the experiment) and surroundings (universe)

  8. Units of Energy • SI unit of energy is Joule • 1 Joule = 1 kg*m2/s2 • Commonly reported in kilojoules • Non-SI unit of energy is calories • 1 cal = 4.184 J • Nutrional Calorie (Cal) = 1 kcal = 1000 calories

  9. Enthalpy (∆H) • Describes heat gained or lost at constant pressure H = qp • Following the derivation, E = H + P∆V or H = E - P∆V • Enthalpy is a state function • Can be calculated from multiple sources: • Calorimetry • Stoichiometry • Heats of formation (tables of standard values) • Hess’ Law • Bond Energies

  10. Enthalpy of Reaction • Enthalpy of a reaction can be utilized stoichiometrically based on coefficients • For endothermic processes ∆H is positive and should be included as a reactant in the rxn • For exothermic processes ∆H is negative and should be included as a product in the rxn

  11. Enthalpy of Reaction • Example Upon adding solid potassium hydroxide pellets to water the following reaction takes place: Answer the following questions regarding the addition of 14.0 g of KOH to water: • Does the beaker get warmer or colder? • Is the reaction endothermic or exothermic? • What is the enthalpy change for the dissolution of the 14.0 g of KOH?

  12. Thermochemical Equations • Guidelines • Enthalpy is an extensive property • Enthalpy of the reverse reaction is equal in magnitude but opposite in sign • Enthalpy change depends on physical states of reactants and products • If the thermochemical equation is multiplied by a factor of n, then ∆H must also change by the same factor

  13. Calorimetry • Process of measuring heat based on observing the temperature change when a body absorbs or discharges energy as heat • Assumption is that no heat is lost to surroundings (aka closed system) • 2 types of Calorimetry: • Constant pressure(coffee-cup setup) • Constant volume (bomb calorimeter) • Used in industry for determining food calories

  14. Calorimetry • Commonly used in lab to determine specific heats of metals • Can also measure the heat released/absorbed of many types of reactions: • Neutralizations • Ionization (Dissolution/Dissolving) • Reaction (ppt, combustion, etc.)

  15. Terms to know (and love!) • Heat capacity– energy required to raise the temperature of an object by 1 degree C (J/°C) • Specific heat capacity– same as above but specific to 1 gram of the substance (J/ g °C) • Molar heat capacity – Same as specifc heat capacity, but specific to one mole of a substance (J/mol K or J/mol °C)

  16. Relationships of Heat • Heat capacities are extensive properties, while specific heat is an intensive property • Heat capacity can be determined by multiplying a substances mass by its specific heat: J/°C = J/g°C x g • Specific heat of water is 4.184 J/g°C and is the same for all dilute aqueous solutions • Heat of substance + heat of solution = 0 • Therefore, qsubstance = -qsoln

  17. Calculating Specific Heat q=mCp∆T • q= heat transferred (J) • Recall: at constant pressure q=∆H • m = mass (g) • Cp = specific heat of material at constant pressure • ∆T = Tf – Ti (final – initial)

  18. Calculating Specific Heat • Example How much heat is needed to raise 10.0 grams of aluminum from 22.0 °C to 42.0 °C? (Specific heat of aluminum is 0.90 J/ g K) Also, what is the molar heat capacity of aluminum?

  19. Constant Pressure Calorimetry • Example (Specific Heat of Metal) A lead (Pb) pellet having a mass of 26.47 g at 89.98C was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50 °C to 23.17 °C. What is the specific heat of the lead pellet? 0.158 J/g˚C

  20. Constant Pressure Calorimetry • Example (Molar Heat of Neutralization) A quantity of 1.00 x 102mL of 0.500 M HCl was mixed with 1.00 x 102mL of 0.500 M NaOH in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solutions was the same, 22.50 °C, and the final temperature of the mixed solution was 25.86 °C. Calculate the heat change for the neutralization reaction on a molar basis (that is, the molar heat of neutralizaton) -56.2 kJ/mol

  21. Constant Volume Calorimetry • In order to perform calculations using a constant volume calorimeter, the bomb calorimeter must be calibrated so that the heat capacity of the calorimeter is known qcal = Ccal∆T (where Ccal is the heat capacity of the calorimeter) • The Ccal is determined by burning a substance with an accurately known heat of combustion • This is a constant that is generally given! Finally, qcal= -qrxn

  22. Constant Volume Calorimetry • Example (Molar Heat of Combustion) A quantity of 1.435 g of naphthalene (C10H8) was burned in a bomb calorimeter. The temperature of the water rose from 20.28 °C to 25.95 °C. If the heat capacity of the bomb plus water is 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis. -5.151x103 kJ/mol

  23. Constant Volume Calorimetry • Example (Heat Capacity of Bomb Calorimeter) Camphor (C10H17O) has a heat of combustion of 5903 kJ/mol. When a sample of camphor with mass of 0.1204 g is burned in a bomb calorimeter, the temperature increases by 2.28 °C. Calculate the heat capacity of the calorimeter

  24. Standard Enthalpy of Formation & Reaction • Heat (or enthalpy) of formation, ∆Hf, is the heat required to form the elements • Standard enthalpies of formation (∆H°f) can be used as reference points for determining the standard enthalpy of a reaction (∆H°rxn) • Standard state conditions are 1 atm and 25 °C • Standard enthalpies of formation of any element in its most stable form is zero • The greater the heat of formation of a molecule, the less stable the molecule • Higher ∆H°fimplies more energy to form the bonds

  25. Standard Enthalpy of Formation & Reaction • Change of enthalpy that occurs in a chemical reaction can be given by: • Where n and m represent coefficients • Two methods exist for determining the ∆H°rxnfor a reaction • Direct Method (using ∆H°f) • Indirect Method (using Hess’ Law)

  26. Standard Enthalpy of Formation & Reaction • When to use Direct Method? • If all of the standard heats of formation are known for each participant in a chemical equation, then plug them into the equation on the last slide • When to use Indirect Method? • The heat of a reaction is independent of the steps it takes to get there (enthalpy is a state function) • If only heats of reaction for a series of reactions is known, then these can be manipulated to determine the heat of the reaction • This is known as Hess’s Law

  27. Direct Method • Example (Enthalpy of Reaction) • Calculate the ∆H°rxn for the following: Given the following values: • 3 Al (s) + 3 NH4ClO4 (s) → Al2O3 (s) + AlCl3 (s) + 3 NO (g) + 6 H2O (g) Answer: -2680 kJ/mol (exo)

  28. Direct Method • Example (Enthalpy of Formation) • Sometimes all values are not found in the table of thermodynamic data. For most substances it is impossible to go into a lab and directly synthesize a compound from its free elements. The heat of formation for the substance must be found by working backwards from its heat of combustion. Find the ∆Hf of C6H12O6 (s) from the following information: C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) + 2800 kJ Answer: -1276 kJ.mol for glucose

  29. Hess’s Law (of heat summation) • Guidelines • First decide how to rearrange equations so reactants and products are on appropriate sides of the arrows • Manipulate the equations using the thermochemical equation guidelines (mentioned previously) • Check to ensure that everything cancels out to give you the exact equation you want • Note: It is often helpful to begin your work backwards from the answer that you want!

  30. Hess’s Law • Given the following equations: H3BO3 (aq) → HBO2 (aq) + H2O (l) ∆Hrxn= -0.02 kJ/mol H2B4O7 (aq) + H2O (l) → 4 HBO2 (aq) ∆Hrxn= -11.3 kJ/mol H2B4O7 (aq) → 2 B2O3 (s) + H2O (l) ∆Hrxn= 17.5 kJ/mol Find the ∆Hfor this overall reaction: 2 H3BO3 (aq) → B2O3 (s) + 3 H2O (l) Answer: 14.4 kJ/mol (endothermic)

  31. Thermodynamics! • Study of energy changes in chemistry • Involves three major players: ∆H, ∆S, ∆G • One of the major objectives of thermodynamics is to predict whether or not a reaction will occur when reactants are brought together (spontaneous vs. nonspontaneous) • First Law of Thermodynamics (stated previously) is that energy cannot be created or destroyed, only conserved

  32. Thermodynamics! • Enthalpy can be used to predict spontaneity of a reaction • In general, lower energy states are preferred, meaning exothermic reactions are favorable (-∆H)

  33. Entropy (∆S) • Enthalpy (∆H) is not the only predictor of spontaneity • Entropy (∆S) can be used, and is the measurement of disorder in a system • Second Law of Thermodynamics states that the entropy (or disorder) of the universe is constantly increasing • Examples: Ice cube melting, your room at the end of a week. Nature tends towards chaos!

  34. Entropy (∆S) • + ∆S = more disorder in the system • Increase in disorder is favorable, indicates higher chance to be spont. • - ∆S = less disorder in the system • Decrease in disorder is not favorable, indicates lower chance to be spont. • Thermodynamics can predict spontaneity, but does NOT predict the rate (speed) of the reaction! • Talking state functions: we do not look at pathways, only beginning and end states!

  35. Twice is extra nice! • Entropy, like enthalpy, is a state function • Entropy can be calculated for reactions exactly like enthalpy: • Units are J/K*mol (where ∆H is kJ/mol)

  36. Changes in Entropy • Entropy increases as one goes from solid to a liquid Ssolid  Sliquid  Sgas

  37. Changes in Entropy • Entropy increases as a substance divides into parts • Dissolving solids into liquids (exception: carbonates!)

  38. Changes In Entropy • Entropy will increase in reactions in which number of product molecules are greater • Entropy increases with temperature

  39. Third Law of Thermodynamics • States that the entropy of a perfect crystal at 0 K is zero • This implies no movement, and no randomness exists • Not many perfect crystals out there, so entropy values rarely ever zero • For determination of standard values of entropy (as in ∆S°), this allows an absolute standard to exist which all other values can be based on

  40. Free Energy • To ultimately decide the spontaneity of a reaction, we use Gibb’s free energy (G) • For a constant temperature process, change in free energy can be given by: • The most important thermodynamic equation, and one of the most beneficial equations in chemistry!

  41. Free Energy & Spontaneity • Free energy is the energy available to do work • - ∆G means forward reaction is spontaneous • + ∆G means forward reaction is non-spontaneous • Reverse reaction would be spontaneous! • ∆G = 0 means reaction is at equilibrium (more on this later)

  42. Third Time’s the Charm! • Standard free enthalpies of reaction can be calculated using the following equation: • ∆G has units of energy (J or kJ) • ∆G is a state function (like ∆H and ∆S)

  43. Predicting Reaction Spontaneity • Both enthalpy and entropy must be known to predict spontaneous reactions

More Related