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Thermochemistry

Thermochemistry. Thermodynamics Study of energy and its transformations Thermochemistry Study of thermodynamics between the relationship of chemical reactions and their energy changes. Use of energy Energy in the form of work must be used to move an object W = fd

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Thermochemistry

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  1. Thermochemistry

  2. Thermodynamics • Study of energy and its transformations • Thermochemistry • Study of thermodynamics between the relationship of chemical reactions and their energy changes

  3. Use of energy • Energy in the form of work must be used to move an object • W = fd • f: force- push or pull on an object • d: distance- measurement of how far moved • When work is done, heat is generated • Heat is the energy transferred from a hotter to a cooler object (an object with greater energy to one with lesser energy

  4. Therefore energy is the capacity to do work or to transfer heat • All objects possess energy • K.E. kinetic- motion • Magnitude of K.E. or EK of an object is dependent on its mass(m) and velocity (v) • EK = 1/2mv2 • Mass and velocity determine how much work an object can do

  5. Potential energy: stored energy as a result of the energy an object possesses by its position relative to other objects (e- in atoms energy levels) • Charged particles have P.E. due to attractive and repulsive forces(electrostatic forces) between them • So an e- has P.E. when near a p+(attraction) • P.E. and K.E. can be interchanged • When happens, work is done • Some of energy is dissipated as heat

  6. Need to relate these changes on the atomic or molecular scale • Chemical energy is due to P.E. stored in bonds of a substances atoms • Energy substances possess due to temperature (thermal energy) are due to K.E. • Heat is the transfer of thermal energy from substances with greater average K.E. to those with lower avg. K.E.

  7. SI unit of energy is Joule (J) • 1J = 1kg-m2/s2 • So a mass of 2kg moving at 1m/s possess a K.E. of 1J • EK = ½ mv2 = ½(2kg)(1m/s)2=1kg-m2/s2=1J • Since a J is a small amt. of energy, most reactions use kJ for the energy associated with them • Energy used to be expressed as calorie(cal) • 1cal = 4.184J

  8. System • Part of universe under study • Surroundings • Everything else • The chemicals studied in a reaction are the system • The container and everything else is surroundings • Reactions are studied in Closed Systems • Exchange energy but not matter with surroundings • Usually done in calorimeter (matter conserved)

  9. First Law of Thermodynamics :Law of Conservation of Energy • Energy is always conserved • Energy lost by a system must be gained by its surroundings, vice versa • Total energy of a system called Internal Energy • Sum of all K.E. and P.E. within system

  10. Example • If place H2 and O2 in a container • Energy includes motion of H2 and O2 • Interactions of H2 and O2 • Interactions of nuclei and e- of molecules • Due to large number of interactions it is not possible to determine the exact energy of the system • Can measures changes in internal energy that accompany a physical or chemical process (changes)

  11. Change internal energy = difference between internal energy of system at the completion of a process and that of the beginning • DE = Efinal – EinitialDE= Ef – Ei • Thermodynamic quantities DE has 3 parts • A number and unit giving magnitude of change • Sign giving direction of change

  12. Positive DE results when Ef > Ei • Shows system gained energy from surroundings • Negative DE obtained when Ef < Ei • Shows system loses energy to surroundings • In chemical reactions • Ei is reactants • Ef is products • So H2 + O2 H2O + Energy • The energy of the products is less than the reactants: DE is negative

  13. Shown using Energy Diagram

  14. A system can exchange energy with its surroundings through work or heat • Internal energy = heat energy + work done • DE = q + w • q is pos.- gains heat from surroundings q>0 • w pos.- work done on system by surroundings w>0 • q- neg.value heat lost to surroundings q<0 • w neg. work done on surroundings w<0

  15. State Function • A system has a fixed value of energy for a given set of conditions • Value depends only on present condition not how it got there

  16. Enthalpy • Heat absorbed or lost in a chem. reaction • DH = Hf – Hi = qp (heat gain or lost) • Hf- product Hi- reactants qp-constant pressure ( reactions always take place in gaseous state) • DH > 0 heat gained endothermic • DH < 0 heat lost exothermic • Ex. H2(g) + O2(g) H2O(g)DH= -483.6kJ

  17. 1) Enthalpy is an extensive property • Magnitude DH proportional to amt. of reactant consumed • Ex. Above reaction gives double heat if double reactants used • 2) enthalpy change for a reaction is equal in magnitude but opposite in sign to DH of the reverse reaction • 3) enthalpy change for a reaction depends on state of reactant and product • Ex. H2O(g) H2O(l)DH= -88kJ

  18. Reactions for which DH is large and negative tend to be spontaneous • Measure of heat flow is called calorimetry • Use a calorimeter to measure heat flow • Heat capacity • Amt. of heat required to raise the temp. of any object 1.0oC or 1.0oK • Specific Heat • Amount of energy required to raise the temperature of 1.0g of any substance 1.0oC • Cp = q/mDT q=mDTCpDH=q CH2O(l)=4.184kJ/g.oC

  19. Constant pressure calorimeter • Used for dilute (aq) solutions q=4.184J/goC(K) • qsoln = (Cp soln)(g soln)DT • Heat of reaction is equal in magnitude; opposite in sign from qsoln • qrxn = -qsoln • So: if temp h (posDT) reaction exothermic neg(rxn) • Coffee-cup style

  20. Bomb Calorimeter • Used for constant volume • More familiar • Ccalorimeter heat capacity must be known • Remember: has a measured amount of water • qrxn = -Ccalorimeter x DT

  21. Hess’s Law • If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy changes for the individual steps • Therefore we can calculate DH for a reaction by “adding” up known DH of reactions • Used to obtain energy changes that are difficult to measure or predetermine

  22. Ex. • Calculate DH for : • 2C(s) + H2(g) C2H2(g) given the following: • Correct mols and divide so function of coefficients mols for wanted product is one • C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(l) DH = -1299.6kJ • C(s) + O2(g) CO2(g)DH = -393.5kJ • H2(g) + 1/2O2(g)  H2O(l)DH = -285.9kJ • Rearrange the equations so similar to algebra:

  23. Calculate DH for : • NO(g) + O(g) NO2(g) given: • NO(g) + O3(g)  NO2(g) + O2(g) DH = -198.9kJ • O3(g)  3/2O2(g)DH = -142.3kJ • O2(g)  2O(g)DH = 495.0kJ

  24. Enthalpies of Formation • DHf indicates a substance is formed from its elements • Enthalpy of vaporization DHvap- liquid to gas • Enthalpy of fusion DHfus- melting solids • Enthalpy of combustion DHcmb- combusting with oxygen • Standard Enthalpy of a reaction is defined as enthalpy change when all reactants and products are in the Standard State[1atm, 298K] • Denoted as DHo

  25. Standard Enthalpy of Formation DHfo • Formation of one mol of cmpd. from its constituent elements at 1atm, 298K • Represented in kJ/mol • DHfo of most stable form of an element is 0 • Reason: no formation reaction needed when element is already in std. state • Ex. At 1atm, 298K the most stable form of oxygen is O2 not O DHfo for O2(g) = 0

  26. A standard table of DHfo for common compounds is available • Hess’s Law can be written as an equation • DHo = SDHfo(products) –SDHfo (reactants)

  27. Lets apply this to the Mg lab we did • 1st react Mg(s) + 2HCl  MgCl2 + H2 DH1 • 2nd react MgO(s) + 2HCl  MgCl2 + H2O DH2 • Hess’s Law states: DH3 = DH1 + DH2 • So: DH1 + DH2 = DHfo (MgO) + DHfo (H2) - DHfo (Mg) - DHfo (H2O) • By definition: DHfo (H2) and DHfo (Mg) are 0 because elements are at std.state • DHfo(MgO) = DH1 + DH2 - DHfo (H2O)

  28. To see how well you did check the values you deduced for the DHfo(MgCl2) in exp. 1 and 2 • You need the following: • DHfo(MgO)(s) = -601.7kJ • DHfo(H2O)(l) = -285.9kJ • DHfo(HCl)(aq) = -167.4kJ

  29. FOODS • Carbohydrates: fast breakdown in body for energy • C6H12O6 + 6O2 6CO2 + 6H2O DHo=-2816kJ • Carbohydrates have an average fuel value of • 17kJ/g or 4kcal/g (4Cal/g) • Remember conv. of 1cal = 4.184J • Proteins provide approx. the same energy as carbs • Fats also break down into CO2 and H2O and release energy • Avg. fuel value of fat is 38kJ/g or 9kcal/g • Food labels show fats,carbs and proteins in avg. sitting

  30. How much energy in food? • Red beans: 62%carb, 22%protein, 1.5%fat • Food value? • Apple: 13% carb, 0.4%protein, 0.5%fat • Peanuts: 22%carb, 26%protein, 39%fat

  31. FUELS • When combusted: C converted into CO2 and H converted into H2O • Both have large DH = neg. value • Fossil fuels: • Coal- solid, hydrocarbons, high molecular wt., S,O,N cmpds. • Petroleum- liquid, hydrocarbons(100’s), organic of S,O,N • Natural gas- gaseous, hydrocarbons, mostly CH4, C2H6, C3H8, C4H10

  32. For environmental purposes coal is being researched to form syngas • Coal + steam -------- complex mixture------- CH4 + H2 + CO conversion purification • Process removes most S and N compounds which allows clean burning • Is easily transported due to a gas

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