Chapter 6 THERMOCHEMISTRY ppt

Chapter 6 THERMOCHEMISTRY ppt - THERMOCHEMISTR Y Both...

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Unformatted text preview: THERMOCHEMISTR Y Both chemical and physical changes involve the gain or loss of energy Energy: the capacity to do work or supply heat Types: Kinetic ..energy of motion Potential..stored energy, related to position SI Unit of energy is a Joule: kg m2 s2 calorie: the amount of energy required to raise the temperature of 1 gram of water by 1 C 1 calorie = 4.184 joule 1 kilocalorie = 4.184 kilojoule Food Calorie = 1 kilocalorie Temperature: a measure of the average kinetic energy of some part of the universe Heat: Energy transferred from one substance to another due to a difference in temperature or average kinetic energy. Chemical Energy: Potential energy stored in bonds First Law of Thermodynamics: The energy of the Universe is constant Then if we isolate a portion of the universe, the heat (q) flows from the system to its surroundings Exothermic event : q < 0 the heat flows from the surroundings to the system Endothermic event: q > 0 Euniverse = 0 and for this to be true qsystem + qsurroundings = 0 qsystem = qsurroundings work (w) done on the system is > 0 work (w) done by the system is < 0 Change in internal energy : E = Efinal Einitial E is a state function or path independent E = q + w where w is P V pressure volume work at constant Volume, there will be no PV work and E = qv; Bomb calorimetry at constant Pressure, qp = E + P V; coffee cup calorimetry Enthapy or heat content H E + PV; H = E + P V p = H q H is also a state function: H = Hfinal Hinitial; H = Hprod H react Types of H: enthalpy of reaction enthalpy of solution enthalpy of vaporization enthalpy of fusion enthalpy of crystallization enthalpy of condensation enthalpy of solidification enthalpy of combustion enthalpy of formation For reversible changes: H for the forward change = H for the reverse change or Hsolution = H crystallization Enthalpy of Formation: the energy required or given off when one mole of a substance is formed from its constituent elements at standard conditions: 25C, 1 atm for gases and 1 M solutions. Elements are assigned Hf 0 as is the hydrogen ion, H+. Hformation Hf Thermochemical Equation: mass and energy balance C(s) + 2 H2(g) CH4(g) + 74.8 kJ or C(s) + 2 H2(g) CH4(g) H = 74.8 kJ What this means is that 74.8 kJ of energy are given off when the reaction between C and H2 occurs and one mole of methane is formed at 25C and 1 atm pressure. Using thermodynamic data tables, you calculate the Hreaction . Hrxn = Hf products Hfreactants Energy Balance: Hf products= Hfreactants + Hrxn E N T H A L P Y Enthalpy of Reactants H Enthalpy of Products Remember! that the states are important when you look up the values of Hf that elements and H+ may not appear in the tables since they are defined at having Hf= 0. that the data will vary a little from table to table so not to worry unless it is quite variant that the sign and magnitude of Hf give information about stability. A large positive Hf means that the compound is unstable. A large negative Hf means that the compound is stable. The larger the magnitude, the greater the effect. that the units of Hf are kJ/mole of substance that the equation may be balanced with nonintegral stoichiometric coefficients Specific Heat Capacity and Molar Heat Capacity Specific Heat ( c or SH) q = Quantity of Heat c (or SH) is specific heat units are J g K Molar Heat Capacity (C) molar heat capacity (C) = q moles x T J mol K Calculating the Amount of Heat from the Specific Heat Capacity Problem: Calculate the quantity of heat required to heat a gold ring with a mass of 21.63 g from 22.68oC to 255.05oC Plan: We know the mass of gold, and can look up the specific heat capacity, the change in temperature is the difference in the temperatures. Solution: The specific heat capacity (c) of gold is 0.129 J/g K. T = Tfinal-Tinitial = 255.05oC- 22.68oC = 232.37oC = 232.37 K q = mass x c x T = 0.129 J x 21.63 g x 232.37 K = 648.37504 J gK q = 648 J = 0.648 kJ A Coffee-Cup Calorimeter Fig. 6.11 Coffee Cup Calorimetry EPS cups isolate the system and its surroundings from the rest of the universe. qsystem + qsurroundings = 0 and E is conserved So we measure the temperature change of the surroundings (water) and calculate qwater qwater = mass x c x T if q water >o then the event is exothermic qreaction= qwater and also qreaction= Hreaction These q's are constant pressure(open system)..qp Hess's Law of Heat Summation - I 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 = -296.8 kJ 2) 2 SO2 (g) + O2 (g) 3) S (s) + 3/2 O2 (g) 2 SO3 (g) SO3 (g) H2 = -198.4 kJ H3 = ? Hess's Law of Heat Summation - II To get equation #3 we have to add the first reaction to 1/2 times reaction #2, and sum them: S (s) + O2 (g) + 1/2 x [ SO2 (g) +1/2 O2 (g) S (s) + 3/2 O2 (g) + SO2 (g) SO2 (g) H1 = -296.8 kJ SO3 (g)] 1/2 H2 = -99.2 kJ SO2 (g) + SO3 (g) H3 = -296.8 +(-99.2) = -396.0 kJ S (s) + 3/2 O2 (g) SO3 (g) Hf (SO3) = -396.0 kJ/mol Applying Hess's Law: Formation of WC - I Problem: What is the enthalpy of reaction, H, for the formation of tungsten carbide, WC, from the elements? Given the equations: 1) 2 W(s) + 3 O2 (g) 2 WO3 (s) H = -1680.6 kJ 2) C(graphite) + O2 (g) CO2 (g) H = -393.5 kJ 3) 2 WC(s) + 5 O2 (g) 2 WO3 (s) + 2 CO2 (g) H = - 2391.6 kJ Plan: We need to rearange the three equations to be able to add them, and get the reaction for tungsten carbide's formation from the elements. Solution: The equation for the production of tungsten carbide from the elements: W(s) + C(graphite) WC(s) H=? A Bomb Calorimeter Fig. 6.12 qv= E internal energy change qreaction + qcalorimeter or bomb + qwater = 0 Bomb calorimeters are stainless steel and so the qcalorimeter = Ccalorimeter x T (C is mass x c ) constant for a calorimeter ( T is the same for water and the calorimeter) IF the amount of water used is held constant the energy balance is: qreaction + qcalorimeter= 0 READ THE PROBLEM CAREFULLY Constant Volume Calorimetry Bomb Calorimetry H from Bond Energies H=Bond Energies(Reactants)Bond energies (Products) Bond Energies > 0 ENDOTHERMIC Breaking chemical bonds is endothermic. Making chemical bonds is exothermic So to use BE's, you break all the bonds in reactants and make all the bonds in products ...
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This note was uploaded on 03/31/2008 for the course CHEM 101 taught by Professor Farahh during the Spring '02 term at UNC.

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