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Energetics

Energetics. Topic 5. Energy & Chemistry. ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature. System and Surroundings. SYSTEM The object under study SURROUNDINGS

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Energetics

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  1. Energetics Topic 5

  2. Energy & Chemistry ENERGY is the capacity to do work or transfer heat. HEAT is the form of energy that flows between 2 objects because of their difference in temperature.

  3. System and Surroundings • SYSTEM • The object under study • SURROUNDINGS • Everything outside the system

  4. Directionality of Energy Transfer • Energy transfer as heat is always from a hotter object to a cooler one. • EXOthermic: energy transfers from SYSTEM to SURROUNDINGS.

  5. Directionality of Energy Transfer • Energy transfer at heat is always from a hotter object to a cooler one. • ENDOthermic: heat transfers from SURROUNDINGSto theSYSTEM.

  6. James Joule 1818-1889 UNITS OF ENERGY 1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC. 1000 cal = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) But we use the unit called the JOULE 1 cal = exactly 4.184 joules

  7. Which has the larger heat capacity? HEAT CAPACITY The heat required to raise an object’s T by 1 ˚C.

  8. Specific Heat Capacity How much energy is transferred due to T difference? The heat (q) “lost” or “gained” is related to a) sample mass b) change in T and • specific heat capacity (is the amount of heat needed to raise the temperature of 1 gram of a substance 1°C.) q C = ------------ m x Δ T

  9. heat gain/lose = q = (specific heat)(mass)(∆T) Specific Heat Capacityexample: If 25.0 g of Al cools from 310 oC to 37 oC, what amount of energy (J) has been transferred by the Al? where ∆T = Tfinal - Tinitial q = (0.897 J/g•K)(25.0 g)(37 - 310)K q = - 6120 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

  10. ENTHALPY Most chemical reactions occur at constant Pressure, Heat transferred at constant P = qp qp = ∆Hwhere H = enthalpy ∆H = change in heat contentof the system ∆H = Hfinal - Hinitial

  11. ENTHALPY ∆H = Hfinal - Hinitial If Hfinal > Hinitial then ∆H is positive Process is ENDOTHERMIC If Hfinal < Hinitial then ∆H is negative Process is EXOTHERMIC

  12. Thermochemical Equations • Energy amount can be written as a reactant or product: CaO + H2O  Ca(OH)2 + 65.2 kJ or CaO + H2O  Ca(OH)2 ∆H = -65.2kJ ____________________________________ 2NaHCO3 + 85kJ  Na2CO3 + H2O +CO2 or 2NaHCO3  Na2CO3 + H2O + CO2ΔH = 85kJ

  13. Example: 2NaHCO3 + 85kJ  Na2CO3 + H2O +CO2 Calculate the amount of heat required to decompose 2.24 mol NaHCO3 85 kJ ΔH = 2.24 mol NaHCO3 x ---------------------- 2 mol NaHCO3 ΔH = 95 kJ

  14. Standard Enthalpy Values Most ∆H values are labeled ∆Ho which is enthalpy measured under standard conditions P = 100 kPa Concentration = 1 mol/L T = usually 25 oC with all species in standard states (STP) Examples: C = graphite and O2 = gas Standard enthalpy of formation of a substance is the enthalpy change when 1 mole of a substance is formed from its elements in their standard states under standard conditions. ΔH° = ΔH° (products) - ΔH° (reactants)

  15. Example Making liquid H2O from H2 + O2 involves twoexothermic steps. H2 + O2 gas H2O vapor Liquid H2O

  16. Example Consider the formation of water H2(g) + 1/2 O2(g)  H2O(g) + 241.8 kJ Exothermic reaction — energy is a “product” and∆H = – 241.8 kJ

  17. Example Making H2O from H2 involves two steps. H2(g) + 1/2 O2(g)  H2O(g) + 242 kJ H2O(g)  H2O(l) + 44 kJ -------------------------------------------------------- H2(g) + 1/2 O2(g)  H2O(liq) + 286 kJ Example of HESS’S LAW— If a rxn is the sum of 2 or more others, the net ∆H is the sum of the ∆H’s of the other rxns.

  18. Hess’s Law & Energy Level Diagrams Forming H2O can occur in a single step or in a two steps. ∆rHtotal is the same no matter which path is followed. NOTE: ∆rH stands for the enthalpy change for a reaction, r

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