1 Second Law of Thermodynamics As the reaction goes to products our system becomes more disordered and the entropy of our system increases. One driving.

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1 Second Law of Thermodynamics As the reaction goes to products our system becomes more disordered and the entropy of our system increases. One driving forces for a chemical reaction is one of the products is a gas.

2 19.3 GIBBS FREE ENERGY 19.3 GIBBS FREE ENERGY ΔS universe = ΔS surroundings + ΔS system = - ΔH system / T + ΔS system -T ΔS universe = ΔH system - T ΔS system = ΔG system ΔG is the change in free energy for the system. ΔG o system = ΔH o system - T ΔS o system (standard state) ΔG o rxn from ΔG o f and from ΔH o rxn and ΔS o rxn ΔG o rxn = S ΔG f o (products) - S ΔG f o (reactants ΔG o rxn = ΔH o rxn - T ΔS o rxn (Don't forget to change the entropy term from J to kJ)

3 Gibbs Free Energy, G  S univ =  S surr +  S sys  S univ =  H sys T +  S Multiply through by -T -T ΔS univ = ΔH sys - T ΔS sys -T ΔS univ = change in Gibbs free energy for the system = ΔG system Under standard conditions — Under standard conditions — Δ G o = Δ H o - T Δ S o Δ G o = Δ H o - T Δ S o

4 Gibbs Free Energy, G ΔG o = ΔH o - T ΔS o Gibbs free energy change = total energy change for system - energy lost in disordering the system total energy change for system - energy lost in disordering the system If reaction is exothermic (ΔH o negative) and entropy increases (ΔS o is +), then ΔG o must be negative and reaction product-favored in the standard state. If reaction is endothermic (ΔH o is +), and entropy decreases (ΔS o is -), then ΔG o must be + and reaction is reactant-favored in the standard state.

5 Product-Favored or Reactant-Favored? The reaction is product-favored if ΔG is negative.The reaction is product-favored if ΔG is negative. We can see that this is always true if ΔH is negative and ΔS is positive.We can see that this is always true if ΔH is negative and ΔS is positive. If ΔH is positive and ΔS is negative, ΔG is always positive.If ΔH is positive and ΔS is negative, ΔG is always positive.

6 Product-Favored or Reactant-Favored? The other two cases are temperature dependent with a positive ΔS favoring spontaneity at high temperature, thus overcoming the positive ΔH, and a negative ΔH favoring spontaneity at a low temperature when ΔS is negative.The other two cases are temperature dependent with a positive ΔS favoring spontaneity at high temperature, thus overcoming the positive ΔH, and a negative ΔH favoring spontaneity at a low temperature when ΔS is negative. Table next slide and Figure 20.8Table next slide and Figure 20.8

7 Gibbs Free Energy, G ΔG o = ΔH o - T ΔS o ΔG o = ΔH o - T ΔS o ΔH o ΔS o ΔG o Reaction exo(-)increase(+)-Prod-favored endo(+)decrease(-)+React-favored exo(-)decrease(-)?T dependent endo(+)increase(+)?T dependent

8 Figure 20.8

9 Gibbs Free Energy, G ΔG o = ΔH o - T ΔS o ΔG o = ΔH o - T ΔS o Two methods of calculating ΔG o a)Determine Δ H o rxn and Δ S o rxn and use Gibbs equation. b)Use tabulated values of free energies of formation, Δ G f o.  G o rxn =   G f o (products) -   G f o (reactants)

10 Calculating  G o rxn Combustion of acetylene C 2 H 2 (g) + 5/2 O 2 (g) --> 2 CO 2 (g) + H 2 O(g) Use enthalpies of formation to calculate ΔH o rxn = -1238 kJ ΔH o rxn = -1238 kJ Use standard molar entropies to calculate ΔS o rxn = -97.4 J/K or -0.0974 kJ/K ΔS o rxn = -97.4 J/K or -0.0974 kJ/K ΔG o rxn = -1238 kJ - (298 K)(-0.0974 kJ/K) = -1209 kJ = -1209 kJ Reaction is product-favored (in the standard state) in spite of negative ΔS o rxn. Reaction is “enthalpy driven”.

11 Calculating  G o rxn Is the dissolution of ammonium nitrate product- favored? If so, is it enthalpy- or entropy-driven? NH 4 NO 3 (s) + heat NH 4 NO 3 (aq)

12 Calculating  G o rxn From tables of thermodynamic data: ΔH o rxn = +25.7 kJ ΔH o rxn = +25.7 kJ ΔS o rxn = +108.7 J/K or +0.1087 kJ/K ΔS o rxn = +108.7 J/K or +0.1087 kJ/K ΔG o rxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) ΔG o rxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = -6.7 kJ = -6.7 kJ Reaction is product-favored (in the standard state) in spite of negative ΔH o rxn. Reaction is product-favored (in the standard state) in spite of negative ΔH o rxn. Reaction is “entropy driven”. Reaction is “entropy driven”. NH 4 NO 3 (s) + heat NH 4 NO 3 (aq)

13 Calculating  G o rxn Combustion of carbon C(graphite) + O 2 (g) --> CO 2 (g) ΔG o rxn = ΔG f o (CO 2 ) - [ΔG f o (graph) + ΔG f o (O 2 )] ΔG o rxn = -394.4 kJ - [ 0 + 0 ] Note that free energy of formation of an element in its standard state is 0. Note that free energy of formation of an element in its standard state is 0. ΔG o rxn = -394.4 kJ Reaction is product-favored as expected.  G o rxn =   G f o (products) -   G f o (reactants)

14 If we assume that Δ H o and Δ S o are relatively independent of temperature, we can calculate Δ G o at any particular temperature we choose.If we assume that Δ H o and Δ S o are relatively independent of temperature, we can calculate Δ G o at any particular temperature we choose. ΔG T o rxn = ΔH o rxn - T ΔS o rxn ΔG T o rxn = ΔH o rxn - T ΔS o rxn Free energy and Temperature

15 Free Energy and Temperature 2 Fe 2 O 3 (s) + 3 C(s) ---> 4 Fe(s) + 3 CO 2 (g) ΔH o rxn = +467.9 kJ ΔS o rxn = +560.3 J/K ΔG o rxn = +300.8 kJ Reaction is reactant-favored at 298 K At what T does ΔG o rxn just change from being (+) to being (-)? When ΔG o rxn = 0 = ΔH o rxn - T ΔS o rxn. T=  H rxn  S = 467.9 kJ 0.5603 kJ/K = 835.1 K

16 For the reaction below: Calculate ΔG o at 298.15 K two ways, explain the sign of ΔS o, determine if the reaction in the standard state is product-favored at 298.15 K, determine which term or terms favor spontaneity, and calculate the temperature at which the reaction would first become reactant favored in the standard state.For the reaction below: Calculate ΔG o at 298.15 K two ways, explain the sign of ΔS o, determine if the reaction in the standard state is product-favored at 298.15 K, determine which term or terms favor spontaneity, and calculate the temperature at which the reaction would first become reactant favored in the standard state. 3 H 2(g) + CO (g) ====> CH 4(g) + H 2 O (g) Free Energy and Temperature

17 20.4 THERMODYNAMICS AND THE EQUILIBRIUM CONSTANT At equilibrium, ΔG T = 0, and ΔG T o = -RT ln K T Figure 20.11, shows the relationship between Q and K, which comes from the concentration dependence of Free Energy.Figure 20.11, shows the relationship between Q and K, which comes from the concentration dependence of Free Energy.Figure 20.11Figure 20.11 One cannot calculate a new K by simply changing the T in the equation since ΔG T o is a function of temperature.One cannot calculate a new K by simply changing the T in the equation since ΔG T o is a function of temperature. Δ G T o rxn = Δ H o rxn - T Δ S o rxn Δ G T o rxn = Δ H o rxn - T Δ S o rxn A useful combined form of these equations is:A useful combined form of these equations is: Δ G T o = Δ H o - T Δ S o = -RT ln K T.

18 Figure 20.11

19 K eq is related to reaction favorability. When ΔG o rxn < 0, reaction moves energetically “downhill” ΔG o rxn is the change in free energy as reactants convert completely to products. But systems often reach a state of equilibrium in which reactants have not converted completely to products. In this case ΔG rxn is < ΔG o rxn, so state with both reactants and products present is more stable than complete conversion. Thermodynamics and K eq

20 Product-favored reaction. 2 NO 2 ---> N 2 O 4 ΔG o rxn = -4.8 kJ Here ΔG rxn is less than ΔG o rxn, so the state with both reactants and products present is more stable than complete conversion. Thermodynamics and K eq

21 Reactant-favored reaction. N 2 O 4 --->2 NO 2 ΔG o rxn = +4.8 kJ Here ΔG o rxn is greater than ΔG rxn, so the state with both reactants and products present is more stable than complete conversion. Thermodynamics and K eq

22 K eq is related to reaction favorability and so to ΔG o rxn. The larger the value of ΔG o rxn the larger the value of K. ΔG o rxn = - RT lnK ΔG o rxn = - RT lnK where R = 8.31 J/Kmol Thermodynamics and K eq

23 Calculate K for the reaction N 2 O 4 --->2 NO 2 Δ G o rxn = +4.8 kJ Δ G o rxn = +4800 J = - (8.31 J/K)(298 K) ln K G o rxn = - RT lnK  G o rxn = - RT lnK lnK = - 4800 J (8.31 J/K)(298K) = - 1.94 Thermodynamics and K eq K = 0.14 When ΔG o rxn > 0, then K 0, then K < 1

24 Let us consider the derivation of these equations to further our understanding of them and their interrelationships.Let us consider the derivation of these equations to further our understanding of them and their interrelationships. This derivation starts with the definition of G, G = H - TSThis derivation starts with the definition of G, G = H - TS At constant temperature, with H = E + PV, and w ext = 0,and w = -PdV, we end up with, Δ G = nRT Δ P/P.At constant temperature, with H = E + PV, and w ext = 0,and w = -PdV, we end up with, Δ G = nRT Δ P/P. THERMODYNAMICS AND THE EQUILIBRIUM CONSTANT

25 G = G o + nRT ln a, where a, is the activity, unitless concentration. G = G o + nRT ln a, where a, is the activity, unitless concentration. For a reaction, we arrive at:For a reaction, we arrive at: Δ G T = Δ G T o + RT ln Q T Δ G T = Δ G T o + RT ln Q T Where Q = K at equilibrium.Where Q = K at equilibrium.

26 DERIVATIONSDERIVATIONS Derive the relationships between ΔG T o and K T.Derive the relationships between ΔG T o and K T. Derive the relationships between ΔH o, K T, and T.Derive the relationships between ΔH o, K T, and T.

27 THE EQUILIBRIUM CONSTANT The temperature dependence of K can be calculate using the equation:The temperature dependence of K can be calculate using the equation: Δ G T o = Δ H o - T Δ S o = - RT ln K T. This equation can be written at two temperatures, T 1 and T 2, and combined to eliminate Δ S o.This equation can be written at two temperatures, T 1 and T 2, and combined to eliminate Δ S o. This produces the very useful equation:This produces the very useful equation:                   1 2 11 1 2 TT R H K K n T T

28 THE EQUILIBRIUM CONSTANT Δ A plot of ln K vs. 1/T yields a straight line with a slope of - Δ H o /R.

29 20.5 THERMODYNAMICS AND TIME The first and second Laws of Thermodynamics cannot be proven, they are laws of experience and tell us the direction of time in any given "picture".The first and second Laws of Thermodynamics cannot be proven, they are laws of experience and tell us the direction of time in any given "picture".

30 Real World Examples of: Chapter 20: Second Law of Thermodynamics “ The total entropy of the universe is always increasing ”

31 Faster molecules can break intermolecular forces like H-bonding to escape a solution

32 Kinetic Molecular Theory of Gases

33 Experiment: Two Florence flasks with H 2 O, one closed and the other open. This experiment demonstrates: 1.There is sufficient heat in the surroundings to allow liquid water to escape to the atmosphere; 2.Air currents and gas diffusion prevent the gaseous water from making contact with the water surface.

34 The Clausius-Clapeyron equation is a method for obtaining enthalpy of vaporization, at any temperature.

35 In a closed system … Questions 1.Can we change the equilibrium in this system? 2.Is there any reason for wanting to change the equilibrium in this system?

36 Closed system equilibrium Q < K Reaction favors reactants to products Q > K Reaction favors products to reactants Q = K Reaction is at equilibrium

37 Frost-free freezers Air in the freezer is warmed then dried. The vapor pressure of ice is 4.579 torr. Warm, desiccated air can remove water vapor. Q < K favors reactants to products

38 Second Law of Thermodynamics As the reaction goes to products our system becomes more disordered and the entropy of our system increases. One driving forces for a chemical reaction is one of the products is a gas.

39 Gibbs free energy

40  G > 0 reactant favored  G = 0 equilibrium  G < 0 product favored, reaction proceeds to products

41 If Q < K product favored What are the conditions necessary for the spontaneous formation of products? A Q of.0313 allows water vapor to evaporate. The atmosphere has a P H2O of.001%-4% water vapor (3%-100% humidity), in the winter P vap can be as high as 20 torr.

42 Water will Evaporate even though the process takes energy Kinetic Molecular Theory Evidence that is easily manipulated under normal conditions Highly ordered H 2 O(l) has a large  S, this is the main driving force for producing a higher partial pressure of water. Summary

43 As a rule of thumb the rate of an rxn doubles, or triples for every 10 degrees increase in temperature. What factors affect the rate of a reaction? The concentration of the reactants. The more concentrated the faster the rate;  Temperature. Usually reactions speed up with increasing temperature;  Physical state of reactants;  Catalyst (or inhibitor). A catalyst speeds up a reaction, an inhibitor slows it down.

44 Concentrations of the Ester and H 2 O are held constant, only the temperature changes. ExpTemp Kk [L/mol  s] KKK 2 /K 1 1288.0521 2298.101101.93 3308.184101.82 4318.332101.80

45 Arrhenius equation

46 Testing the “Rule of Thumb” Setting K 2 /K 1 = 2, T 1 = 273, T 2 = 283 Solve for E a, E a = 44.5kJmol -1

47 Theoretical Results T1T1 T2T2 k 2 /k 1 2732832.00 3733831.45 4734831.26 5735831.17 6736831.12 7737831.09

48 When can relying on the rule be dangerous Once the rxn is initiated, no further heat is needed.

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