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Chem 1045 Lab hess-s_law
 

Chem 1045 Lab hess-s_law

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    Chem 1045 Lab hess-s_law Chem 1045 Lab hess-s_law Presentation Transcript

    • Energy Relationshipsin Chemical Reactions
    • Energy is the capacity to do work• Thermal energy is the energy associated with the random motion of atoms and molecules• Chemical energy is the energy stored within the bonds of chemical substances• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom• Electrical energy is the energy associated with the flow of electrons• Potential energy is the energy available by virtue of an object’s position
    • Thermochemical DefinitionsSystem : That part of the Universe whose change we are going to measure.Surroundings : Every thing else that is relevant to the change is defined as the “surroundings”.Internal Energy : The sum of the kinetic and potential energy of all the particles in a system.Heat (q) : Is the energy transferred between a system and it’s surroundings as result in the differences in their temperatures only!Work (w) : The energy transferred when an object is moved by a force. Therefore: E=q+w
    • Energy Changes in Chemical ReactionsHeat is the transfer of thermal energy between two bodies thatare at different temperatures.Temperature is a measure of the thermal energy. Temperature = Thermal Energy 900C 400C greater thermal energy 6.2
    • Change in Enthalpy = H Enthalpy is defined as the system’s internal energy plus the product of its pressure and volume. H = E + PV For a change in enthalpy: H = E+ PV Exothermic and Endothermic Reactions: H = H final - H initial = H products - H reactantsExothermic : H final H initial H 0Endothermic : H final H initial H 0
    • For processes occurring at constantpressure the enthalpy change equalsthe heat gained or lost. H = qpenthalpy of reaction or heat ofreaction. (Energy change + smallcorrection factor.)
    • Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = ms t qcal = Ccal t Reaction at Constant P H = qrxnNo heat enters or leaves! 6.4
    • The specific heat (s) of a substance is the amount of heat (q)required to raise the temperature of one gram of thesubstance by one degree Celsius.The heat capacity (C) of a substance is the amount of heat(q) required to raise the temperature of a given quantity (m)of the substance by one degree Celsius. C = ms Heat (q) absorbed or released: q = ms t q=C t t = tfinal - tinitial 6.4
    • How much heat is given off when an 869 g iron bar coolsfrom 940C to 50C?s of Fe = 0.444 J/g • 0C t = tfinal – tinitial = 50C – 940C = -890Cq = ms t = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
    • 0The standard enthalpy of reaction ( Hrxn ) is the enthalpy ofa reaction carried out at 1 atm. aA + bB cC + dD Ho = [ c Hof (C) + d Hof (D) ] - [ a Ho (A) + b Ho (B) ] rxn f f Ho = rxn n Hof (products) - m Hfo (reactants)Hess’s Law: When reactants are converted to products, thechange in enthalpy is the same whether the reaction takesplace in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
    • Hess’s Law of Heat Summation The enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps.Example: Problem: Calculate the energy involved in the oxidation of elemental sulfur to sulfur trioxide from reactions: 1) S (s) + O2 (g) SO2 (g) H1 = -296.0 kJ 2) 2 SO2 (g) + O2 (g) 2 SO3 (g) H2 = -198.2 kJ 3) S (s) + 3/2 O2 (g) SO3 (g) H3 = ?
    • Hess’s Law of Heat Summation The enthalpy change of an overall process is the sum of the enthalpy changes of its individual steps.Example: Problem: Calculate the energy involved in the oxidation of elemental sulfur to sulfur trioxide from reactions:2 X 1) S (s) + O2 (g) SO2 (g) 2H1 = -296.0 kJ X2+ 2) 2 SO2 (g) + O2 (g) 2 SO3 (g) H2 = -198.2 kJ 3) S (s) + 3/2 O2 (g) SO3 (g) H3 = ? H3 = 2H1 + H2