Unformatted text preview: Lecture 3:Entropy & Reac2on Spontaneity • Review of the last class Entropy: sta5s5cal approach vs Heat change • Calculate change of entropy of a reac5on Entropy change in the system Entropy change in the surrounding • Entropy change and the equilibrium state Entropy & Probability S = k ln W Entropy & Heat Change ΔSsys= qrev/T Entropy Changes in the System ΔSorxn
the entropy change that occurs when all reactants and products are in their standard states. ΔSorxn = ΣmΔSoproducts
ΣnΔSoreactants Sample Problem PROBLEM: Calcula5ng the Standard Entropy of Reac5on, ΔSorxn Calculate ΔSorxn for the combus2on of 1 mol of propane at 25oC: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) So CO2=213.7 J/mol•K So H2O=69.9 J/mol•K So C3H8=269.9 J/mol•K) So O2=205.0 J/mol•K Sample Problem Balance the following equa5on, predict the sign of ΔSorxn and calculate its value at 25 0C. NaOH(s) + CO2(g) Na2CO3(s) + H2O(l) Sample Problem: Does the oxida5on of FeO (s) to Fe2O3 occur spontaneously? PLAN: Write the equa2on of the oxida2on; calculate ΔSorxn, calculate ΔSorxn and ﬁnally calculate ΔSosurr SOLUTION: ΔSsys = ΔHsys= ΔHsurr= ΔHuniv= From Appendix B, So values of FeO, O2 and Fe2O3 are 60.75, 205.0, 87.4 J/mol K, respec2vely Entropy Changes in the Surrounding The change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to the temperature at which the heat is transferred. ΔSsurroundings =
ΔHsystem T ΔSsys = ΔH0f of FeO, Fe3O2 and O2 are
272.0,
825.5 and 0 kJ/mol, respec2vely ΔHsys= ΔHsurr= ΔHuniv= Sample Problem PROBLEM: Determining Reac5on Spontaneity At 298 K, the forma2on of ammonia has a nega2ve ΔSosys: N2(g) + 3H2(g) 2NH3(g) ΔSosys =
197 J/K Calculate ΔSouniv, and state whether the reac2on occurs spontaneously at this temperature. PLAN: ΔSouniverse must be > 0 in order for this reac2on to be spontaneous, so ΔSosurroundings must be > 197 J/K. To ﬁnd ΔSosurr, ﬁrst ﬁnd ΔHsys; ΔHsys = ΔHrxn which can be calculated using ΔHof values from tables. Sample Problem Determining Reac5on Spontaneity ΔHorxn = [(2 mol NH3)(
45.9 kJ/mol)]
[(3 mol H2)(0 kJ/mol) + (1 mol N2)(0 kJ/mol)] ΔHorxn =
91.8 kJ ΔSosurr =
ΔHosys/T =
91.8x103J ΔSouniverse = ΔSosys + ΔSosurr 298 K = 308 J/K =
197 J/K + 308 J/K = 111 J/K ΔSouniverse > 0 so the reac2on is spontaneous at 298 K. Entropy Change and Equilibrium State Αt equilibrium: ΔSuniverse = ΔSsurr + ΔSsys= 0 For a process reaching equilibrium ΔSuniverse > 0, but when the process reaches equilibrium ΔSuniverse = 0 Components of ΔSuniverse for spontaneous reac5ons exothermic exothermic endothermic system becomes more disordered system becomes more ordered system becomes more disordered ...
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 Spring '08
 HOEGER

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