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Chapter 6 The Nature of Energy

Chapter 6 The Nature of Energy. What is Energy? The capacity to do work or to produce heat The Law of Conservation Of Energy- energy is neither created or destroyed but converted from one form to another. Types of Energy. 1. Potential Energy-energy due to position

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Chapter 6 The Nature of Energy

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  1. Chapter 6The Nature of Energy • What is Energy? • The capacity to do work or to produce heat • The Law of Conservation Of Energy- energy is neither created or destroyed but converted from one form to another

  2. Types of Energy • 1. Potential Energy-energy due to position • 2. Kinetic Energy- energy due to motion • KE = ½ mv2

  3. More Terms • Heat- the transfer of energy between two objects due to temperature differences • Temperature- a property which determines the direction heat will flow when two objects are brought into contact

  4. Con. • Work- a force acting over a distance • State Function- a property that depends only on its present state • Energy is a state function • System- part of the universe which we are focusing on( usually the reactants and products)

  5. More Terms!!! • Surroundings- everything else in the universe(examples could be the reaction container, the room, and anything else other than the reactants and products) • Exothermic-energy flows out of the system( heat is produced) • Endothermic-energy flows into the system( heat is absorbed

  6. Internal Energy(E) E = q + w E is the change in the system’s internal energy q represents heat w represents work q= +x (endothermic, heat flows in) q= -x (exothermic, heat flows out) w is neg. if energy flows out(system does work on the surroundings) w is pos. if energy flows in( surroundings do work on the system

  7. Example • Calculate E for a system where 1.40 kJ of work is done on the system undergoing an endothermic process in which 15.6 kJ of heat flows into the system. E = q + w = 15.6kJ + 1.4kJ= 17.0kJ

  8. PV Work w = -PV Example: Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

  9. Examplep.247 6.3 • A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from 4.00x106 L to 4.50x106 L by the addition of 1.3x108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process. ( To convert between L•atm and J, use 1 L•atm=101.3 J.

  10. Enthalpy and Calorimetry Enthalpy (H) = E + PV Remember that E = qp + w so E = qp - P V so qp = E + PV Now if H = E + (PV) so H = E + PV H = qp

  11. Example • When one mole of methane is burned at a constant pressure, 890 kJ of energy is released as heat. Calculate H for a process in which a 5.8 g sample of methane is burned at constant pressure.

  12. Calorimetry • - a device use to determine the heat associated with a chemical reaction • Heat Capacity(C)= heat absorbed increase in temp. Specific heat capacity- the energy required to raise the temperature of one gram of a substance by one degree Celsius Molar heat capacity- the energy required to raise the temperature of one mole of a substance by one degree Celsius

  13. Constant Pressure Calorimetry • Is used in determining the changes in enthalpy(heats of reactions) for reactions occurring in solution • Energy released= s x m x T where • S= specific heat capacity • M= mass of solution • T= increase in temperature

  14. Example p. 252, 6.5 • When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25.0°C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18J/°C•g, and that the density of the final solution is 1.0 g/ml, calculate the enthalpy change per mole of BaSO4 formed.

  15. Hess’s Law • In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

  16. Characteristics of Enthalpy Changes • If a reaction is reversed, the sign of H is also reversed. • If the coefficients in a balanced reaction are multiplied by an integer, the value of H is multiplied by the same integer.

  17. Example • Two forms of carbon are graphite and diamond. Using the enthalpies of combustion for graphite and diamond, calculate H for the conversion of graphite to carbon: Cgraphite(s)  Cdiamond(s) The combustion reactions are Cgraphite(s) +O2(g)  CO2(g) H =-394kJ Cdiamond(s) + O2(g)  CO2(g) H =-396kJ

  18. Example Calculate H for the synthesis of diborane from its elements according to the following: 2B(s) + 3H2(g)  B2H6(g) Use the following reactions: 2B(s) +3/2O2(g) B2O3(s) H =-1273kJ B2H6(g) +3O2(g)  B2O3(s)+3H2O(g) H =-2035 kJ H2(g) + ½O2(g)  H2O(l) H= -286 kJ H2O(l)  H2O(g) H = 44kJ

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