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Chapter 18 Thermodynamics Week 2 2009

# Chapter 18 Thermodynamics Week 2 2009 - Another Consider...

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Another example of calculating whether a reaction is spontaneous : Consider: CO(g) + 2 H 2 (g) CH 3 OH(l) Question : is the reaction spontaneous left-to-right at 298 K? Answer : calculate sys (= rxn ), surr and thus univ (i) Calculate sys from the Table of S° values (Appendix 3) sys =(1mol)(126.8J/K.mol)-{(1mol)(197.9J/K.mol)+(2mol)(131.0J/K.mol) S ° sys = -331.1 J/K ( Note : sign agrees with prediction) (ii) Calculate sys from Table of f values (Appendix 3) sys = (1mol)(-238.7kJ/mol) - {(1mol)(-110.5 kJ/mol) + (2mol)(0 kJ/mol)} H ° sys = -128.2 kJ Now, surr = - sys /T surr = -(-128.2 kJ/298K)(1000 J/kJ) = +430.2 J/K (iii) Calculate univ = sys + surr = -331.1 J/K + 430.2 J/K univ = + 99.1 J/K univ > 0 the reaction is spontaneous Section 18.5. Gibbs Free Energy (G) To avoid having to consider sys and surr all the time, a new thermodynamic quantity was introduced in the 19 th century called the Gibbs Free Energy (G). Since S univ = S sys + S surr and S surr = T H - sys S univ = S sys - T H sys T S univ = T S sys - H sys and -T S univ = H sys - T S sys = G sys 1

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G sys = H sys - T S sys or usually just as G = H - T S Since G sys = -T S univ and S univ > 0 for a spontaneous reaction: G < 0 (i.e., negative) for a spontaneous rxn. G > 0 (i.e., positive) for a non-spontaneous rxn. G = 0 for a process at equilibrium This is the most convenient form of the 2 nd Law of Thermodynamics i.e. saying: G sys < 0 for a spontaneous reaction is the same as saying S univ > 0 Free Energy is a state function and G° = standard free energy change (kJ/mol). Note : since H values not available, G values also are not, and we can only calculate G values. Appendix 3 lists form values for this. Calculating G ° rxn (= G ° sys ) for reactions. Example Calculate rxn for: 4 KClO 3 (s) 3 KClO 4 (s) + KCl (s) Method 1 : Calculate rxn and rxn , use G° = H° - T rxn = [3mol(-433.5 kJ/mol) + 1mol(-435.9 kJ/mol)-4mol(-391.2kJ/mol)] rxn = -171.6 kJ (exothermic) rxn = [3mol(151.0 J/mol.K) + 1mol(82.7 J/mol.K) – 4mol(143.0 J/mol.K)] rxn = -36.3 J/K = -36.3 x 10 -3 kJ/K rxn = rxn - T rxn = -171.6 kJ –[(298K)(-36.3 x 10 -3 kJ/K)] rxn = -161 kJ (spontaneous) 2
Method 2 : Use f values of products and reactants f = standard free energy of formation = free energy change to make substance from its component elements Like for f , the f = 0 for an element in its standard state.

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