Thermochemistry (Production of Heat).pptx

Thermochemistry (Production of Heat).pptx

• 1.
  Thermochemistry Chapter 6 AP Chemistry
• 2.
  Thermodynamics: the studyof energy and its transformations Thermochemistry: the sub discipline involving chemical reactions and energy changes
• 3.
  Energy •Energy is definedas the capacity to do work or to produce heat. •The 1st Law of Thermodynamics …. Energy is conserved as it is converted between one form and another, it will be neither created or destroyed, but simply change form…thus making the energy of the universe constant! •Energy can be classified as either Kinetic or Potential
• 4.
  Kinetic energy: energyassociated with the motion of atoms and molecules in a system. KE = ½ mv2 Temperature is a measure of the average KE of a collection of particles in a system. RECALL…….
• 5.
  Heat vs Energy •Thermalenergy is the energy of the object and is not in the process of being transferred or moved. •Heat is kinetic energy being transferred –It moves from a hotter object towards cooler object until the temperatures are the same. –At this point the KE’s are at equilibrium. –It is not a property of the substance.
• 6.
  System: the partof the universe we are studying Surroundings: everything else Usually, energy is transferred to... (1) (2) Change an object’s state of motion like fuel in a vehicle. Cause a temperature change like a furnace warming a house. In chemistry the system is the reaction that we are interested in and the surroundings could be the container that the reaction takes place in.
• 7.
  Units of energyare either joules (J) kilojoules (kJ) CONVERSIONS: Divide by 1000 to convert from J to KJ Multiply by 1000 to convert from KJ to J James Prescott Joule (1818-1889) 4184 J = 4.184 kJ
• 8.
  In endothermic processes,heat is _________ by the system. absorbed melting boiling sublimation released freezing condensation deposition In exothermic processes, heat is ________ by the system.
• 9.
  Water Phase ChangeDiagram
• 10.
  • Exothermic processis any process that gives off heat – The energy will be listed as a product. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat is required by the system. The energy is listed as a reactant. energy + 2HgO (s) 2Hg (l) + O2 (g) energy + H2O (s) H2O (l)
• 12.
  ΔH = Hproducts– Hreactants When ΔH is +, the system... has gained heat. When ΔH is –, the system... has lost heat. (ENDO) (EXO) Enthalpy is used to measure the heat that is either gained or lost by a system that is at constant pressure. • Enthalpy is an extensive property, meaning that…the amount of material affects its value
• 13.
  Enthalpy (H)
• 14.
  H2O (s) H2O(l) ΔH = 6.01 kJ/mol ΔH = 6.01 kJ Thermochemical Equations If you reverse a reaction, the sign of ΔH changes H2O (l) H2O (s) ΔH =- 6.01 kJ If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n. 2H2O (s) 2H2O (l) ΔH = 2 mol x 6.01 kJ/mol = 12.0 kJ
• 15.
  H2O (s) H2O(l) ΔH = 6.01KJ The physical states of all reactants and products must be specified in thermochemical equations. Thermochemical Equations H2O (l) H2O (g) ΔH = 44.0 KJ Practice Question: How much heat is evolved when 266 g of white phosphorus (P4) burns in air? ΔHreaction = -3013 kJ
• 16.
  2 H2(g) +O2(g) → 2 H2O(g) ΔH = – 483.6 kJ What is the enthalpy change when 178 g of H2O(g) are produced? The space shuttle was powered by the reaction above.
• 17.
  Calorimetry: the measurementof heat flow A Calorimeter is used to measure the heat changes molar heat (capacity): amt. of heat needed to raise temp. of 1 mol of a substance J/ C *mol or J/ K*mol specific heat (capacity): amt. of heat needed to raise temp. of 1 g of a substance J/ C *g or J/ K*g
• 18.
  cX = heatof fusion (s/l) or heat of vaporization (l/g) We calculate the heat a substance loses or gains using: q = heat m = amount of substance c = substance’s heat capacity ΔT = temperature change q = m c ΔT (for within a given state of matter) AND q = m cX (for between two states of matter when temp is constant)
• 19.
  Heat capacities ofmetals are very low when compared to water or other substances.
• 20.
  In an experimentit was determined that 59.8 J was required to change the temperature of 25.0 g of ethylene glycol (a compound used as antifreeze in automobile engines) by 10.0 C. Calculate the specific heat capacity of ethylene glycol. q = m c ΔT
• 21.
  HEAT Temp. s s/l l l/g g heat added (+q)→ ← heat removed (–q) Typical Heating Curve
• 22.
  Constant Pressure Calorimetry •Commonly called “COFFEE CUP” calorimetry • It’s used to determine any changes in enthalpy for reactions occurring in solution. • Atmospheric pressure remains constant during the reaction . Calorimetry: The measurement of heat flow
• 23.
  Practice Problem A lead(Pb) pellet having a mass of 26.47 g at 89.98°C was placed in a constant-pressure calorimeter 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?
• 24.
  A sketch ofthe initial and final situation is as follows: We know the masses of water and the lead pellet as well as the initial and final temperatures. Assuming no heat is lost to the surroundings, we can equate the heat lost by the lead pellet to the heat gained by the water. Knowing the specific heat of water, we can then calculate the specific heat of lead.
• 25.
  Because the heatlost by the lead pellet is equal to the heat gained by the water, qPb = −280.3 J.
• 26.
  Combustion reactions are studiedusing constant volume calorimetry. It requires a BOMB CALORIMETER.
• 27.
  We assume thatno energy escapes into the surroundings, so that the heat absorbed by the bomb calorimeter equals the heat given off by the reaction.
• 28.
  Hess’ Law 1840 •Thechange of enthalpy in a chemical reaction is independent of the route by which the chemical change occurs. •This is true because enthalpy is a state function , which is a value that does not depend on the path taken
• 29.
  ● The ΔHrxnshave been calculated and tabulated for many basic reactions. ● Hess’s law allows us to put these simple reactions together like puzzle pieces so that they can add up to a more complicated reaction. ● By adding or subtracting the ΔHrxns, we can determine the ΔHrxn of the more complicated reaction. How Hess’s Law works
• 30.
  Hess's Law issaying: If you convert reactants A into products B, the overall enthalpy change will be exactly the same whether you do it in one step or two steps or however many steps.
• 31.
  Important things toremember when using Hess’s Law: ● If a reaction is reversed, the sign of H is also reversed. ● The size of H is directly related to the quantities of reactants and products ● If the coefficients in a balanced reaction are multiplied by an integer, the value of H is multiplied by the same integer.
• 32.
  Calculate the enthalpyfor this reaction 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1299.5 kJ C(s) + O2(g) ---> CO2(g) ΔH° = -393.5 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ
• 33.
  1) Determine whatmust be done to the given equations to get the target equation: a) first eq: flip it so as to put C2H2 on the product side b) second eq: multiply it by two to get 2C c) third eq: do nothing. We need one H2 on the reactant side and that's what we have. 2CO2(g) + H2O(ℓ) ---> C2H2(g) + (5/2)O2(g) ΔH° = +1299.5 kJ 2C(s) + 2O2(g) ---> 2CO2(g) ΔH° = -787.0 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -285.8 kJ Notice that the ΔH values changed as well Add up ΔH values for our answer: +1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ
• 34.
  Standard enthalpy offormation (ΔHf 0 ) is the heat change that results when one mole of a compound is formed from its elements at STP. Whenever a standard enthalpy change is quoted, standard conditions are assumed. The standard enthalpy of formation of any element in its most stable form is zero. ΔHf 0 (O2) = 0 ΔH0 (O3) = 142kJ/mol f ΔHf 0 (C, graphite) = 0 ΔHf 0 (C, diamond) = 1.90 kJ/mol
• 35.
  Some other importanttypes of enthalpy changes Standard enthalpy change of combustion, ΔH°c The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one mole of the compound is burned completely in oxygen at STP . The enthalpy change of solution (ΔH soln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent at STP.
• 36.
  Standard enthalpy ofa reaction (ΔHo rxn): Using Hess’s law, we can easily calculate ΔHo rxn from the ΔHf o of all reactants and products by using the following equation: ΔHo rxn = Σ (ΔH fproducts) – Σ (ΔH f reactants)
• 37.
  –238.6 kJ /mol Approximate theenthalpy change for the combustion of 246 g of liquid methanol. CH3OH(l) O2(g) CO2(g) H2O(g) + + 2 X 2 2 4 –393.5 kJ /mol –241.8 kJ /mol 0 kJ /mol X 2 X 4 –477.2 kJ –1754.2 kJ ΔHo rxn = (–1754.2 kJ) – (–477.2) kJ) = –1277 kJ for 2 mol (i.e., 64 g) of CH3OH So… X = ΔH = –4910 kJ 3 (Look these up. See App. 4, P A19.)
• 38.
  Practice problem #1 Benzene(C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (g)
• 39.
  Practice Problem #2 What is the Δ Hrxn for the complete combustion of Butane, C4H10 (g)? 2 C4H10 (g) + 13 O 2(g) 8CO2 (g) + 10 H2O (g)
• 40.
  Chapter 17 SecondLaw of Thermodynamics
• 41.
  Think of thesecommonplace experiences: ● A Hot frying pans cool down when taken off the stove. ● Air in a high-pressure tire shoots out from even a small hole in its side to the lower pressure atmosphere. ● Ice cubes melt in a warm room. ● Iron exposed to oxygen and water will form rust.
• 42.
  What’s happening ineach of those processes? ● Energy of some kind is changing from being localized, concentrated, and contained to becoming more spread out and dispersed. ● Entropy is the measurement of disorder of a system and it is given the letter S, and it is temperature dependent.
• 43.
  43 Entropy – quantitativemeasure of disorder
• 44.
  Ssolid < Sliquid< Sgas • Entropy increases with dispersal of particles so, entropies of gases are larger than liquids and liquid entropies are larger than solids. Entropies are greater for : - more complex molecules - Increased temperatures (KE) - When volume increases for gases +ΔS ………. Entropy increases - ΔS ………. Entropy decreases
• 45.
  Which substance hasthe greater entropy? CO2 (s) or CO2(g) H2 (g) at 1 atm or H2 (g) at 1.0 x 10 -2 atm What will the overall change in entropy? ● Solid sugar is added to water to make a sugar solution. ● Water vapor condenses. ● Ice melts
• 46.
  Entropy Changes inthe System (ΔSsys) When gases are produced (or consumed) • If a reaction produces more gas molecules than it consumes, ΔS0 + • If the total number of gas molecules diminishes, ΔS0 - What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s) The total number of gas molecules goes down….. ΔS0 -
• 47.
  Chemical Thermodynamics Entropy on theMolecular Scale • Ludwig Boltzmann described the concept of entropy on the molecular level.
• 48.
  Chemical Thermodynamics Entropy on theMolecular Scale Molecules exhibit several types of motion: Translational: Movement of the entire molecule from one place to another. Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule on about an axis or rotation about bonds.
• 49.
  Chemical Thermodynamics Entropy on theMolecular Scale • Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the molecules. • He referred to this sampling as a microstate of the thermodynamic system.
• 50.
  Chemical Thermodynamics Entropy on theMolecular Scale Implications: • More Particles -> more states -> more entropy • Higher Temp -> more energy states -> more entropy • Less Structure (gas vs solid) -> more states -> more entropy
• 51.
  The entropy changefor a system(reaction) is calculated from the entropies of the products and the reactants ΔSo system = Σ[So (products)] - Σ [So ( reactants)] ΔSo system is Positive, then entropy increases ΔSo system is Negative,then entropy decreases Calculations 
• 52.
  Chemical Thermodynamics
• 53.
  Gibbs Free Energy GibbsFree Energy (G) The energy associated with a chemical reaction that can be used to do work.
• 55.
  What are SpontaneousProcesses ? Spontaneous processes are those that can proceed without any outside intervention. For example, the gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously revert to its original state.
• 56.
  Chemical Thermodynamics Spontaneous Processes Processes thatare spontaneous in one direction are nonspontaneous in the reverse direction.
• 57.
  For a constant-temperatureprocess: Gibbs free energy (G) ΔG - The reaction is favorable (spontaneous) in the forward direction. No outside energy is needed Product formation is favored ΔG + The reaction is unfavorable (non-spontaneous) as written. The reaction is favorable (spontaneous) in the reverse direction Reactant formation is favored ΔG = 0 The reaction is at equilibrium and reactant and product formation are equally favored
• 58.
  Recap: Signs ofThermodynamic Values Negative Positive Enthalpy (ΔH) Exothermic Endothermic Entropy (ΔS) Less disorder More disorder Gibbs Free Energy (ΔG) Favored Not favored