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Understand Spontaneous Processes in Thermodynamics

Learn about spontaneous processes, entropy, laws of thermodynamics, free energy, and predicting reactions in this comprehensive guide.

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Understand Spontaneous Processes in Thermodynamics

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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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