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Thermochemistry

Thermochemistry . Chapter 6. Section 1: The Nature of Energy. Energy is the capacity to do work or produce heat. Law of Conservation of Energy. Total energy of the universe is constant. Energy lost = Energy gained by something else. Potential Energy = energy due to position= mgh

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Thermochemistry

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  1. Thermochemistry Chapter 6

  2. Section 1: The Nature of Energy • Energy is the capacity to do work or produce heat. Law of Conservation of Energy Total energy of the universe is constant. Energy lost = Energy gained by something else. Potential Energy = energy due to position= mgh Kinetic Energy = energy of motion = ½ mv2

  3. Heat is a transfer of energy • System vs Surroundings • State Function – a property that depends only on the present state of the system…not on the changes it has or will experience. • Internal energy • Pressure • Volume • Enthalpy System Surroundings

  4. Thermodynamic Quantities Consist of two parts 1) the number – indicates how much 2) the sign- indicates direction of flow HEAT q WORK w ENTHALPY H Internal Energy E Negative Values = flow out of system Positive Values = Flow into system

  5. Heat Lost & Heat Gained Draw… …a graph of energy vs. reaction time for a reaction that gives energy to the surroundings AND a second one for a reaction that absorbs energy from the surroundings

  6. Exothermic Reaction Feels hot Heat transferred to surroundings (lost by system) Negative enthalpy and heat values Endothermic Reaction Feels cold Heat transferred to system (gained by system) Positive enthalpy and heat values Exothermic and Endothermic

  7. PV Work Work (w) = Force*displacement W = F * d = F * Dh W = P * A * Dh P = Force/Area When pressure of system does not change W = -PDV VOLUME!

  8. Try Me Problem • A balloon is inflated to its maximum capacity by heating. If the volume changes from 4.0 x 106L to 4.5 x 106L by addition of 1.3 x 108J energy as heat. Assuming that the balloon expands against constant 1.0 atm pressure. Calculate Internal Energy. • (1 L*atm = 101.3 J)

  9. How is enthalpy different? • Enthalpy, H, is the amount of energy capable of doing work in a system. • The amount of energy contained within the bonds of chemicals involved in the system. • H = E + PV Use the equation above and the total internal energy equation to determine an alternate equation for enthalpy when pressure is constant Answer Now

  10. Section 4: Enthalpies of Formation The enthalpy of formation (Hfo) for an element in its standard state is ZERO. Standard State is at 1 atm and 25oC The more negative the value of Hfo, the more stable the compound. DHrxn = S[np(DHfoprod)]-S[nr(DHforct)]

  11. Try Me 1 Find the enthalpy for the reaction: 4NH3(g) + HCl(l) 4NH4Cl(s) Try Me 2 Find the enthalpy for the reaction: 2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s) Sample Problems

  12. Section 2: Enthalpy & Calorimetry • Specific/Molar Heat of Combustion q = mcDT q = ncDT -The heat needed to raise the temperature of 1 g (or 1 mol) of substance 1 degree K.

  13. A Third Way to find DHrxn • DHrxn = DHproducts – DHreactants • How do you measure this stuff? • CALORIMETRY!

  14. Section 3: Hess’s Law • Enthalpy is a state function • The value will be the same regardless of how many steps are needed to complete the reaction. Hess’s Law States: The enthalpies of individual steps in a reaction mechanism can be added together to calculate the enthalpy of the overall reaction.

  15. Fundamentals for Applying Hess’s Law • Reverse the reaction, reverse the sign on enthalpy. • Multiply the reaction by a coefficient, multiply the enthalpy by the same coefficient. • Add the reactions together, add the enthalpies together. why?

  16. Try Me! Overall: N2O4(g)=>N2(g) + 2O2(g) Reaction Mechanism: NO2(g)  ½ N2(g) + O2(g) H=-84.75 kJ 2NO2(g)  N2O4(g) H=-145.5 kJ This is fun! Let’s do some more!

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