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Chapter 16

Chapter 16. Spontaneous processes Occur without assistance Fast or slow Thermodynamics predicts if reactions will occur but only focuses on the initial and final states Entropy (S) Randomness of a system. S(gas) >> S(liq.) > S(solid) 1 st law of thermodynamics

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Chapter 16

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  1. Chapter 16 • Spontaneous processes • Occur without assistance • Fast or slow • Thermodynamics predicts if reactions will occur but only focuses on the initial and final states • Entropy (S) • Randomness of a system

  2. S(gas) >> S(liq.) > S(solid) • 1st law of thermodynamics • Energy of the universe is conserved • 2nd law of thermodynamics • Entropy of the universe increases for any spontaneous process • Suniv = Ssys + Ssurr • Suniv = + spont. • Suniv = - spont. In opposite direction

  3. Effect of temperature • H2O (l)  H2O (g) more entropy as gas • Ssys = + Ssurr = - • Ssurr depends on the H and T • Ssurr = - H / T T must be in Kelvin

  4. Free Energy (G) G = H - TS all are system G = - spont. = + rev. spont. = 0 equil. • G / T = - H / T + S - G / T = Ssurr + S = Suniv

  5. So G has to be neg. to give a pos. Suniv • From the equation we can predict when reactions will be spontaneous given H and S G = H - TS H S G + + - at high T + - + - - - at low T - + -

  6. Entropy in chemical reactions • We can predict entropy for reactions based on the states of the substances and/or the amount of gas particles present. • Na2CO3(s)  Na2O(s) + CO2(g) + S • 2H2(g) + O2(g)  2H2O(g) - S • 3rd law of thermodynamics • Entropy of a perfect crystal at 0K is zero

  7. We can calculate S by using the following equation So =npSoprod - nrSoreact Unlike H all chemicals will have a value Pg 800

  8. Free energy and chemical reactions • Standard Free Energy Change Go Go = Ho - TSo What’s the free energy for an exothermic reaction that releases 45 kJ of energy and has an increase in entropy of 250 J at 300K? Go = -45000 J – (300K)(250J) = -120000 J

  9. “Hess’s Law” method -- add reactions together to get an overall reaction and add the Go to get an overall Go • Calculate the Go for the following reaction • C diamond + O2 (g)  CO2 (g) Go = -397 kJ • CO2 (g)  C graphite + O2 (g) Go = 394 kJ • C diamond  C graphite Go = -397 kJ + 394 kJ = -3 kJ

  10. Standard free energy of formation Gof • Gof for elements is zero Go= npGof prod - nrGofreact 2CH3OH (g) + 3O2 (g)  2CO2 (g) + 4H2O (g) Go= ? Go = 2n(-394kJ/n)+4n(-229kJ/n)-3n(0)-2n(-163kJ/n) Go= -1378 kJ

  11. Dependence on pressure • Since pressure effects entropy it also effects free energy • G = Go + RT ln(P) R = 8.3145 J/nK • G = Go + RT ln(Q) Q = (Ppa / Prb)

  12. Meaning of Free Energy • G tells us which side of a reaction is favored but it doesn’t tell us that the reaction will go to completion. • The lowest G is at the equilibrium point

  13. Free energy and Equilibrium • G = Go + RT ln(Q) • At equilibrium G = 0 and Q = K • 0 = Go + RT ln(K) • Go = - RT ln(K) • Go = 0 then K = 1 and no shift will occur • Go = - then K > 1 and the reaction shifts right • Go = + then K < 1 and the reaction shifts left

  14. Free energy and work wmax = G

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