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Chapter 16 Spontaneous processes –Occur without assistance –Fast or slow –Thermodynamics predicts if reactions will occur but only focuses on the initial.

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Presentation on theme: "Chapter 16 Spontaneous processes –Occur without assistance –Fast or slow –Thermodynamics predicts if reactions will occur but only focuses on the initial."— Presentation transcript:

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) 1 st law of thermodynamics –Energy of the universe is conserved 2 nd law of thermodynamics –Entropy of the universe increases for any spontaneous process –  S univ =  S sys +  S surr –  S univ = + spont. –  S univ = - spont. In opposite direction

3 Effect of temperature –H 2 O (l)  H 2 O (g) more entropy as gas –  S sys = +  S surr = - –  S surr depends on the  H and T –  S surr = -  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 =  S surr +  S =  S univ

5 –So  G has to be neg. to give a pos.  S univ –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. –Na 2 CO 3 (s)  Na 2 O(s) + CO 2 (g) +  S –2H 2 (g) + O 2 (g)  2H 2 O(g) -  S 3 rd law of thermodynamics –Entropy of a perfect crystal at 0K is zero

7 We can calculate  S by using the following equation  S o =  n p S o prod -  n r S o react Unlike  H all chemicals will have a value Pg 800

8 Free energy and chemical reactions –Standard Free Energy Change  G o  G o =  H o - T  S o 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?  G o = J – (300K)(250J) = J

9 –“Hess’s Law” method -- add reactions together to get an overall reaction and add the  G o to get an overall  G o –Calculate the  G o for the following reaction –C diamond + O 2 (g)  CO 2 (g)  G o = -397 kJ –CO 2 (g)  C graphite + O 2 (g)  G o = 394 kJ –C diamond  C graphite  G o = -397 kJ kJ = -3 kJ

10 –Standard free energy of formation  G o f –  G o f for elements is zero  G o =  n p  G o f prod -  n r  G o f react 2CH 3 OH (g) + 3O 2 (g)  2CO 2 (g) + 4H 2 O (g)  G o = ?  G o = 2n(-394kJ/n)+4n(-229kJ/n)-3n(0)-2n(-163kJ/n)  G o = kJ

11 Dependence on pressure –Since pressure effects entropy it also effects free energy –G = G o + RT ln(P) R = J/nK –  G =  G o + RT ln(Q) Q = (P p a / P r b )

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 =  G o + RT ln(Q) –At equilibrium  G = 0 and Q = K –0 =  G o + RT ln(K) –  G o = - RT ln(K) –  G o = 0 then K = 1 and no shift will occur –  G o = - then K > 1 and the reaction shifts right –  G o = + then K < 1 and the reaction shifts left

14 Free energy and work w max =  G


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