CHM1045 Lecture 17Outline of Current LectureI. EnergyII. 2 Types of Processes:III. Thermodynamics: IV. State Functions:V. First law of thermodynamics VI. Work Done By the System On the SurroundingsVII. Enthalpy and the First Law of ThermodynamicsVIII. Enthalpy:IX. Thermochemical EquationsX. A Comparison of ∧H and ∧UCurrent LectureEnergy- Energy is the capacity to do worko Work: energy change- Radiant Energy- ex. sun- Thermal Energy- random motion of atoms- Chemical Energy- Energy stored with in bonds- Nuclear- Stored within collectionHeat- transfer between energy of 2 bodies at different temperaturesTemperature: measure of thermal energyThermochemistry: Study of heat change in chemical/reactionsSystem: Specific part of universe we’re interested in2 Types of Processes:1.) Exothermic: given off heat- transfers thermal energya. 2H2 + O2 2H2O2.) Endothermic- heat is absorbed/applieda. Energy + 2HgO2Hg+O2b. Energy+H20H20Thermodynamics: Def. Study of inter conversion of heat and other kind of energyState Functions:Properties determined by state of system regardless how you got their getting from point A to BFirst law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.DUsystem + DUsurroundings = 0DUsystem = -DUsurroundingsC3H8 + 5O2 3CO2 + 4H2O = Exothermic chemical reactionWork Done By the System On the Surroundingsw = -P DV DV > 0-PDV < 0w < 0Example 1:A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against a vacuum(b) against a constant pressure of 1.2 atma.) P=0W= -p ∧ VW= -0*4W=0b.) p=1.2 atmw= -(1.2)(4)w=4.8L*atm-4.8*101.3J/1L*atm= -4.9*102 JExample 2:The work done when a gas is compressed in a cylinder like that shown in Figure 6.5 is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the energy change for this process ∧ U=q+ww= +462q= -128∧ U= -128+462= 334JEnthalpy and the First Law of Thermodynamics∧U = q + wAt constant pressure: q = ∧H and w = -P∧V∧U = ∧H - P ∧ V ∧H = ∧U + P∧V Enthalpy:Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.DH = H (products) – H (reactants)Thermochemical Equations• The stoichiometric coefficients always refer to the number of moles of a substanceH2O (s) H2O (l) ∧H = 6.01 kJ/mol- If you reverse a reaction, the sign of ∧H changesH2O (l) H2O (s) ∧H = -6.01 kJ/mol• If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.2H2O (s) 2H2O (l) ∧H = 2 x 6.01 = 12.0 kJ• The physical states of all reactants and products must be specified in thermochemical equations.H2O (s) H2O (l) ∧H = 6.01 kJ/molH2O (l) H2O (g) ∧H = 44.0 kJ/molExample3:Given the thermochemical equations2SO2 + O2 2SO2 ∧H-198.2kj/molCalculate the heat evolved when 87.9g of SO2 is converted to SO3-87.9gSO2*1molSO2/64.07gSO2=1.372 mol SO21.372gSO2 * 198.2kj/2 mol SO2= -136 kj/molA Comparison of ∧ H and ∧ U2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) DH = -367.5 kJ/molDU = -367.5 kJ/mol – 2.5 kJ/mol = -370.0
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