Chapter 17 Spontaneity, Entropy, and Free Energy The goal of this chapter is to answer a basic question: will a given reaction occur “by itself” at a particular.

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Chapter 17 Spontaneity, Entropy, and Free Energy The goal of this chapter is to answer a basic question: will a given reaction occur “by itself” at a particular temperature and pressure, without the exertion of any outside force?

Thermodynamics The science that deals with heat and energy effects. First Law of thermodynamics / The Law of Conservation of Energy: –Energy can be neither created nor destroyed by a process, it is only transformed from one form to another. Any energy lost by a system must be gained by the surroundings and vice versa –CH 4 +2O 2(g)  CO 2(g) + 2H 2 O (g) + energy Here potential energy has been converted to thermal energy.

Figure 16.1 Methane and Oxygen React

Spontaneous Processes A process is said to be spontaneous if it occurs without outside intervention. Thermodynamics lets us predict whether a process will occur or not. Spontaneous processes may be fast or slow. Thermodynamics can tell the direction in which a process will occur but can say nothing about the speed of the process.

Figure 16.2 Rate of Reaction

Entropy What common characteristic cause the processes to be spontaneous? –The driving force for the spontaneous process is an increase in the Entropy (denoted by s) of the universe What is Entropy? –Entropy is a measure of randomness or disorder –State property depends upon the state of a system –Thermodynamic function that describe the number of arrangements  S = S final - S initial

Microstate Each configuration that gives a particular arrangement is called a microstate. Which arrangement is most likely to occur? –One with greatest number of microstate. Positional probability (microstates) which depends on the number of configurations in space. –Positional probability increases (entropy increases) in going from solid to liquid to gas. S solid < S liquid << S gas

Figure 16.3 The Expansion of an Ideal Gas into an Evacuated Bulb

Figure 16.4 Three Possible Arrangements (states) of Four Molecules in a Two-Bulbed Flask

Examples Choose the compound with the greatest positional entropy in each case. a. 1 mol H 2 (at STP) or 1 mol H 2 (at 100 o C, 0.5 atm) H 2 at 100 o C and 0.5 atm; higher temperature and lower pressure means greater volume and hence, greater positional entropy. b. 1 mol N 2 (at STP) or 1 mol H 2 (at 100 K, 2.0 atm) N 2 at STP has the greater volume. c. 1 mol H 2 O (s) (at 0 o C) or 1 mol H 2 O (l) (at 20 o C) H 2 O (l) is more disordered than H 2 O (s)

Second Law of Thermodynamics In any spontaneous process there is always an increase in the entropy of the universe. –Universe System: Portion of the universe in which we are interested Surroundings: everything else in the universe besides the system.

Change in entropy of the universe  S univ =  S sys +  S sur Where  S sys and  S surr represent the changes in entropy  S univ - positive the process is spontaneous in the direction written.  S univ - negative the process is spontaneous in the opposite direction.  S univ - zero the system is at equilibrium.

The Effect of Temperature on Spontaneity Consider, H 2 O (l)  H 2 O (g) 1 mole, 18 grams, 18 mL  1 mole, 18 grams, 31 litters (1 atm, 100 o C) Positional probability increases – entropy of the system increases,  S sys  Positive –The sign of  S surr depends on the direction of the heat flow –The magnitude of  S surr depends on the temperature –  S surr depends directly on the quantity of heat transferred and inversely on temperature.

Continued.... Exothermic process:  S surr = + quantity of heat (J) / temperature (K) Endothermic process:  S surr = - quantity of heat (J) / temperature (K) Heat flow (constant P) = Change in enthalpy =  H Endothermic  H positive and exothermic  H negative  S surr = -  H/T (at constant temp. and pressure)

Example Calculate  S surr for the following reactions at 25 o C and 1 atm. C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (l)  H o = -2221kJ  S surr = -  H/T = -(-2221 kJ) / (25+273)K = 7.45 kJ/K = 7.45 X 10 3 J/K 2NO 2(g)  2NO (g) + O 2(g)  H = 112 kJ  S surr = -  H/T = -112 kJ/298 K = -0.376 kJ/K = -376 J/K

Free Energy  Free energy is a thermodynamic function related to spontaneity and is useful in dealing with temperature dependence of spontaneity, defined by the relationship: G = H – TS where, G is free energy, H is enthalpy, T is Kelvin temperature, S is entropy For a process that occurs at constant temperature, the changes in free energy (  G) is given by the equation,  G =  H – T  S

Lets see how this equation relates to spontaneity,  -  G/T = -  H/T – T  S/-(T) [Divide both side by –T] -  G/T = -  H/T +  S =  S surr +  S =  S univ [Recall,  S surr = -  H/T]   S univ = -  G/T at constant T and P   G negative – process is spontaneous at constant T and P

Example Example: The boiling point of chloroform (CHCl 3 ) is 61.7 o C. The enthalpy of vaporization is 31.4 kJ/mol. Calculate the entropy of vaporization.  G =  H - T  S At the boiling point,  G = 0, so T  S =  H  S =  H/ T = 31.4 kJ/mol ÷ (273.2+61.7)K =9.38 X 10 -2 kJ/K.mol = 93.8 J/K.mol

Example: For mercury, the enthalpy of vaporization is 58.51 kJ/mole and the entropy of vaporization is 92.92 J/K.mole. What is the normal boiling point of mercury? At the boiling point,  G = 0, so  H = T  S T=  H/  S=58.51X10 3 J/mol  92.92 J/K.mol = 629.7 K

 Example: For ammonia (NH 3 ), the enthalpy of fusion is 5.65 kJ/mol and the entropy of fusion is 28.9 J/K.mol a. Will NH 3(s) spontaneously melt at 200 K? b. What is the approximate melting point of ammonia? a. NH 3(s)  NH 3(l)  G =  H - T  S = 5650 J/mol – 200 K.(28.9 J/K.mol) = 5650 J/mol - 5780 J/mol = - 130 J/mol Yes, NH 3 will melt since  G < 0 (negative) at this temperature b. At the melting point  G = 0  H = T  S T =  H /  S = 5650 J/mol  28.9 J/K.mol = 196 K

Entropy Changes in Chemical Reactions The entropy changes in the system (the reactants and the products of the reaction) are determined by positional probability. eg. In the ammonia synthesis reaction N 2(g) + 3H 2(g)  2NH 3(g) –Four reactant molecules becomes two product molecules. –Fewer molecules mean fewer possible configuration.

Entropy Changes in Chemical Reactions Does positional entropy increases or decreases for the following reaction 4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) 9 gaseous molecule  10 gaseous molecule  Positional entropy increases If the number of molecules of the gaseous products is greater than the number of molecules of the gaseous reactants, positional entropy increases, and  S will be positive for the reaction.

Figure 16.5 Entropy

Predicting the Sign of  S o  Predict the sign of  S o for each of the following changes a. AgCl (s)  Ag + (aq) + Cl - (aq) Increase in disorder;  S o (+) b. 2H 2(g) + O 2(g)  2H 2 O (l) Decrease in disorder;  n < 0;  S o (-) c. H 2 O (l)  H 2 O (g) Increase in disorder;  n > 0;  S o (+) d. Na (s) + 1/2Cl 2(g)  NaCl (s) Decrease in disorder;  n < 0;  S o (-) e. 2SO 2(g) + O 2(g)  2SO 3(g) Decrease in disorder;  n < 0;  S o (-)

Third Law of Thermodynamics The entropy of a perfect crystal at 0 K is zero Entropy is a state function of the system Entropy change for a given chemical reaction can be calculated by taking the difference between the standard entropy values of products and those of the reactants:  S o reaction =  n p S o products -  n r S o reactants Entropy is an extensive property (depends on amount)

Example Calculate  S o at 25 o C for the reaction 2NiS (s) + 3O 2(g)  2SO 2(g) + 2NiO (s)  S o reaction =  n p S o products -  n r S o reactants = 2S o SO2(g) + 2S o NiO(s) – (2S o NiS(s) + 3S o O2(g) ) = 2 mol(248 J/K.mol) + 2mol(38 J/K.mol) - 2mol(53 J/K.mol) - 3mol(205 J/K.mol) = 496 J/K + 76 J/K - 106 J/K - 615 J/K = -149 J/K We would expect  S o to be negative because the number of gaseous molecules decreases.

Free Energy and chemical Reactions  G o, the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. N 2(g) + 3H 2(g)  2NH 3(g)  G o = -33.3 KJ  G o =  H o - T  S o

Example: C (s) + O 2(g)  CO 2(g) The values of  H o and  S o are -393.5 KJ and 3.05 J/K, calculate  G o at 298 K.  G o =  H - T  S o = -3.935 x 10 5 J – (298)(3.05 J/K) = -3.944 x 10 5 J = -394.4 KJ (per mol of CO 2 )

Free Energy Change and Chemical Reactions  G o = standard free energy change that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states.  G o =  n p  G o f (products) –  n r  G o f (reactants) The standard free energy of formation of an element in its standard state is zero.

Free Energy and Pressure G = G o + RTln(P) where, G o = the free energy of the gas at a pressure of 1 atm G = the free energy of the gas at a pressure of P atm R = the universal gas constant T = the Kelvin temperature  G =  n p G products –  n r G reactants Where, G products = G o products + RTln(P products ) G reactants = G o reactants + RTln(P reactants )  G =  G o + RTln(Q) where, Q = reaction quotient from the law of mass action

Free Energy and Equilibrium  G =  G o + RTln(Q) At equilibrium,  G = 0 (G products = G reactants ) and Q = K (equilibrium constant) So,  G = 0 =  G o + RTln(K)  G o = -RTln(K)

Figure 16.8 The Dependence of Free Energy on Partial Pressure

Continued…. Case 1:  G o = 0, the system is at equilibrium when the pressures of all reactants and products are 1 atm, which means that K = 1 Case 2:  G o < 0, in this case G o products < G o reactants. The system will adjust to the right to reach equilibrium, K will be greater than 1, since the pressure of the products at equilibrium will be greater than 1 atm and the pressure of the reactants will be less than 1 atm. Case 3: G o > O, in this case G o reactants < G o products. The system will adjust to the left to reach equilibrium. The value of K will be less than 1.

Temperature Dependence of K  G o = -RTln(K) =  H o - T  S o ln(K) = -  H o /RT +  S o /R ln(K) = -[  H o /R][1/T] +  S o /R This is a linear equation of the form y = mx + b where, y = ln(K), m = -  H o /R = slope, x = 1/T, and b =  S o /R = intercept (  H o and S o independent of temperature over a small temperature range)

Summary First law of thermodynamics Spontaneous process Entropy  S = S final – S initial Microstate Second law of thermodynamics Effect of temperature System Surroundings

Summary  S univ =  S sys +  S surr Free energy  G =  H – T  S Entropy change in chemical reactions  S o reaction =  n p S o products -  n r S o reactants  G o =  n p  G o f(products) =  n r  G o f(reactants)  G o = G o + RTln(p) = G o + RTln(Q)  G o = -RTln(K) ln(K) = -[  H o /R][1/T] +  S o /R

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