Thermodynamics Spontaneity. 150/151 Thermochemistry Review. Spontaneity. Ch. 16: Thermodynamics 12/14/2017
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1 Ch. 16: Thermodynamics Geysers are a dramatic display of thermodynamic principles in nature. As water inside the earth heats up, it rises to the surface through small channels. Pressure builds up until the water turns to steam, and steam is expelled forcefully through a hole at the surface. (credit: modification of work by Yellowstone National Park) 1 Chemistry: OpenStax Thermodynamics Recall thermochemistry (study of heat changes) from CHM 150/151. Chapter 5 sections: 1 (Conservation of energy), 2 (Energy change, system, surroundings) 3 (Enthalpies of Physical and Chemical Changes), 3 (Calculating H o rxn using H o f Standard Enthalpy of Formation). We ll introduce the concept of Entropy ( S) this semester Spontaneity Thermodynamics allows us to determine IF a reaction will occur (is spontaneous) as well as telling us the direction in which (and extent to which) a reaction will occur (like Q vs K). A spontaneous process is one that proceeds on its own without any external influence. A spontaneous process always moves a reaction mixture toward equilibrium Spontaneity Other spontaneous processes include: Ice melting at room temperature Solid sodium reacting violently with water A ball rolling downhill An egg boiling in hot water Heat flowing from a hot object to a cold one Figure 16.4: An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a second flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly distributed between the flasks. 3 Figure 16.5: When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object. 4 Spontaneity Spontaneity is determined by enthalpy and entropy of a reaction. A system tends to move toward a lower state of energy. The enthalpy ( H in kj/mol) of many spontaneous processes are exothermic (e.g., combustion reactions), but some spontaneous processes can be endothermic under the right conditions (e.g., melting of ice above 0 o C). A system tends to become more disordered. The entropy ( S in J/mol K) of a spontaneous system tends to increase. There is an increase in the disorder or randomness of a system (i.e., your room or office naturally becoming messier over time) /151 Thermochemistry Review 1 st Law of Thermodynamics: Energy is conserved, it cannot be created or destroyed If a system / reaction gives off heat, the surroundings / universe must absorb it (and vice versa). Heat (enthalpy, H) is defined with reference to the system: System absorbs heat, H > 0 Associated with breaking bonds or Intermolecular forces. Breaking tends to be positive: heat energy goes into a system to break the bonds or pull atoms/molecules apart. System gives off heat, H < 0 Associated with making bonds or Intermolecular forces Making tends to be negative: energy is released as heat, light, sound, etc. when bonds are formed 6 1
2 Predicting Sign of H o rxn Predict signs of enthalpy for the following processes: Decomposition Melting/fusion Combination Boiling/vaporization Acid-Base Neut. Freezing Combustion Condensation Sublimation Deposition In Chemistry reference manuals, only H o fusion and H o vaporization are listed. The opposite processes have the same values but are opposite in sign. H o sublimation = H o fusion + H o vaporization 7 Defining H o f; Calculating H o rxn Heat of formation, H o f, is the enthalpy change when 1 mole of a substance is made from its elements in their standard state (defined in section 5.3) The o symbol refers to the standard state, 1 atm pressure, 25.0 o C, 1 M for solutions (see section 16.3) H o rxn = Σ n H o f (products) - Σ n H o f (reactants) n is moles (coefficient in balanced equation) H o f values, Appendix G (Standard Thermodynmic Properties), reported in kj/mol H o f for elements is 0 (no energy needed!) 8 Defining H o f, Calculating H o rxn Write the equation for the heat of formation of water. H 2 (g) + ½ O 2 (g) H 2 O (l) kj/mol This is how the values in Appendix G (Standard Thermodynamic Properties) are obtained. We then use these values ( H o f) to calculate the enthalpy of a reaction: H o rxn = Σ n H o f (products) - Σ n H o f (reactants) Watch physical states when looking up values (l) versus (g) has a different value! 9 Calculate H o rxn Calculate H o rxn for the following reaction: C 2 H 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Predict sign of H o rxn first. Find H o f values in Appendix G. C 2 H 4 (g) = 52.5 kj/mol O 2 (g) = 0 kj/mol CO 2 (g) = kj/mol H 2 O(g) = kj/mol Entropy To predict spontaneity, both the enthalpy and the entropy of a process must be known. Entropy (S) is a measure of disorder. It tells us how spread out, or dispersed the system s molecules are or how random the motion of atoms/molecules is. S is the change in entropy of reaction (from reactants to products) and is measured in J/mol K. Entropy of Substances What factors would affect the entropy of atoms/molecules? Physical state S (entropy change for a system) = S final - S initial S > 0, increase in disorder S < 0, decrease in disorder (or increase in order)
3 Entropy of Substances What factors would affect the entropy of atoms/molecules? Relative size: more atoms and bonds = greater entropy Entropy of Substances What factors would affect the entropy of atoms/molecules? Temperature Entropy is measured by motion: vibration of bonds, rotation, and translation. Greater molecular motion = greater the number of possible microstates = higher the entropy. Image from Principles of General Chemistry v1.0/s19-chemical-equilibrium.html, CC-BY-NC-SA 3.0 license Figure 16.11: As temperature increases, entropy increases because of increased kinetic energy and molecular motion. 14 Group work: Which has greater entropy? Determine which example in each pair will have more entropy. Explain why. O 2 (g) C 4 H 10 (g) Fe(NO 3 ) 2 (s) He(g) C(s) at 25 o C 1 mol O 2 (g) in 1 L O 2 (l) C 3 H 6 (g) Fe(s) Ne(g) C(s) at 35 o C 1 mol O 2 (g) in 2 L 15 Entropy of Reactions What factors would affect the entropy of a reaction? Change in physical state: phase change; create a solid or gas 16 Entropy of Reactions What factors would affect the entropy of a reaction? Change in number of moles of gas (more moles in products is increase in entropy) Entropy of Reactions What factors would affect the entropy of a reaction? Reaction type Decomposition: H 2 O 2 (g) H 2 (g) + O 2 (g) Combination: 2 H 2 (g) + O 2 (g) 2 H 2 O(g) Dissolution of solid in water: Depends on the type of solid (molecular vs ionic)
4 Entropy of Dissolution Dissolving a substance can lead to an increase or a decrease in entropy, depending on the nature of the solute. Molecular solutes (i.e. sugar): entropy increases Ionic compounds: entropy can decrease or increase NaCl (s): S soln = J/K AlCl 3 (s): S soln = J/K Reactions where S > 0 Entropy usually increases for the following: Melting, vaporization, or sublimation Decomposition reactions Reaction resulting in a greater number of gas molecules (or greater molar mass of solid) Image from Principles of General Chemistry CC-BY-NC-SA license 20 Does entropy increase ( S > 0) or decrease ( S < 0) for the following? Why? 2SO 2 (g) + O 2 (g) 2SO 3 (g) 2O 3 (g) 3O 2 (g) CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + H 2 O(l) + CO 2 (g) I 2 (s) I 2 (g) C 2 H 4 (g) + 3O 2 (g) 2CO 2 (g) + 2H 2 O(g) 4 Al(s) + 3 O 2 (g) 2Al 2 O 3 (s) Example 16.3 Predict the sign of S nd and 3 rd Laws of Thermo 3 rd Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero. This allows us to calculate entropy (S o : (J/mol K)) and changes in entropy ( S o ); unlike enthalpy - where we can only measure change! Use Standard Molar Entropies, S o, (Table 16.2 and Appendix G) - entropy of 1 mole of a pure substance at 1 atm and 25 o C S o rxn = Σ n S o (products) - Σ n S o (reactants) Standard Entropy Values (S o ) Which phase has the lowest entropy? Highest? Substance S o (J mol -1 K -1 ) Substance S o (J mol -1 K -1 ) C (s, diamond) 2.38 H (g) C (s, graphite) H 2 (g) Al (s) 28.3 CH 4 (g) Al 2 O 3 (s) HCl (g) BaSO 4 (s) H 2 O (g) CO (g) H 2 O (l) 70.0 CO 2 (g) Hg (l) 75.9 C 2 H 4 (g) H 2 O 2 (l) C 2 H 6 (g) Br 2 (l) CH 3 OH (g) CCl 4 (l) CCl 4 (g) Note: The standard entropy for a pure element in its most stable form is NOT zero (like for H o 23 f). Calculate Entropy of Reaction Calculate S o rxn for: C 2 H 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Predict sign of S o rxn first. Find S o values in Appendix G. C 2 H 4 (g) = J/mol K, O 2 (g) = J/mol K, CO 2 (g) = J/mol K H 2 O(g) = J/mol K Examples 16.5,
5 Entropy and the 2 nd Law The second law of thermodynamics states that in any spontaneous process, the combined entropy of a system and its surrounding always increases. S universe (or S total ) = S sys + S surr > 0 S sys (or S rxn ) (that we just calculated) S surr (or S H2O ) is entropy change for surroundings Entropy Changes in Surroundings S surr is related to the enthalpy of the reaction; if heat enters surroundings, molecules have more energy and entropy increases. S total > 0; reaction is spontaneous S total < 0; reaction is nonspontaneous S total = 0; system is at equilibrium Heat Entropy Heat Entropy Entropy Changes in Surroundings Temperature is inversely related to S surr If surroundings have a low temperature, adding heat results in a substantial change in randomness. (Like throwing a huge rock into a calm lake.) Conversely, if surroundings have a high temperature, adding heat makes a marginal difference in change in randomness. (Like throwing a pebble into the ocean during a hurricane.) Putting it all together we get: Total Entropy Changes S total = S rxn + S surr S total = S rxn H rxn /T Multiply both sides by T -T S total = G rxn (note the opposite signs!) G rxn = H rxn T S rxn G rxn is Gibbs Free Energy; G < 0 indicates spontaneous reaction Free Energy & Spontaneity Gibbs free energy (G) or free energy can be used to express spontaneity from the perspective of the system. G (kj/mol) is the maximum amount of energy available to do work on the surroundings and takes into account both enthalpy and entropy. In a spontaneous process at constant temperature and pressure, the free energy of the system always decreases. The actual amount of work (w max ) obtained is always less than the maximum available because of energy lost in carrying out a process (given off as heat, light, sound, etc. energy). G sys = H sys - T S sys G < 0 The reaction is spontaneous (still moving forward). G > 0 The reaction is nonspontaneous (moves in reverse direction). G = 0 The reaction is at equilibrium Figure Predicting Spontaneity G sys = H sys - T S sys
6 Gibbs Free Energy and Temperature G = H T S: Notice that G is temperature dependent! Temperature plays a role in determining spontaneity. Also note the absence of standard conditions notation ( o ). Endothermic reactions will be spontaneous at higher temperatures (where T S > H) Exothermic reactions will be spontaneous at lower temperatures (where H > T S) Higher and lower temps are reaction-specific. Every reaction has a cut-off temperature depending on values of H and S. G = 0: 1) turning point between spontaneous and nonspontaneous; 2) phase changes (occur at equilibrium) G = 0 0 = H - T S Temperature Dependence Solve for T T = H / S This temperature tells us the temperature cut-off for high and low temps. We can also use this to determine the temperature for a phase change (system at equilibrium). Note: this G is the non-standard value! 32 Gibbs Free Energy G = Gibbs Free Energy (or free energy), kj/mol: Uses both enthalpy and entropy to determine if a reaction will be spontaneous. Calculate G when H = -227 kj, S = J/K, T = 1450 K. Are you calculating Standard or Non-Standard G? Is this process spontaneous? At what temperature will it become spontaneous? Is it spontaneous above or below this temperature? Temperature of Phase Change 1) Calculate the melting point of ice? H 2 O(s) H 2 O(l) H fusion = 6.01 kj/mol; S fusion = 22.0 J/K mol 263K (-10 o C): G = kj/mol (non-spontaneous) 283K (10 o C): G = kj/mol (spontaneous) Practice: Calculate the boiling point (in o C) of ethanol if the entropy of vaporization is J/K and the enthalpy of vaporization is 39.3 kj T = H / S = 39.3 kj / kj/k = K = 78.2 o C Examples 16.9, Standard Free Energy Changes We can calculate G o rxn = H o rxn T S o rxn for the reaction: H o rxn = kj S o rxn = J/K T = K (because we re calculating G o, T = 25 o C) Predicting Spontaneity - Practice Formation of rust: 4 Fe (s) + 3 O 2 (g) 2Fe 2 O 3 (s) Predict signs of H o rxn, S o rxn, and G o rxn. Is the reaction spontaneous or nonspontaneous at 500 o C? At what temperature does it change spontaneity? H o f (kj/mol) S o (J/K mol) Fe(s) O 2 (g) Fe 2 O 3 (s) Example Example
7 Predicting Spontaneity Practice H o rxn = [2 mol * kj/mol] [0] = kj S o rxn = [2 mol * J/K mol] [4 mol * 27.3 J/K mol + 3 mol * J/K mol] = J/K = kj/k G = H o T S o = kj ( K * kj/k) = kj, spontaneous At what temperature does the reaction change spontaneity? Above or below this temperature? T = H / S = 2997 K (FYI, surface of the sun is 5,778 K!); below this is spontaneous Standard Free Energy Changes The standard free energy of reaction ( G o rxn) is freeenergy change for a reaction when it occurs under standard-state conditions. Standard-state conditions are: Solids, liquids, and gases in their pure form 1 M solutions, 1 atm gases, 25 o C (298 K) G (non-standard) predicts spontaneity G o (standard at 25 o C) predicts reactant- or productfavored at equilibrium; relates to K 38 Standard Free Energy of Formation We can calculate G o rxn using the equation: G o rxn = Σ n G o f (products) - Σ n G o f (reactants) G o f is the free energy change for the formation of 1 mole of a substance from elements in their standard states. G o f for any element in its most stable form at 1 atm is defined as 0 (just like H o f values). Values obtained from Appendix G (just like H o f and S o ) Practice Calculation of G o rxn Calculate G o rxn for the following reaction: 1 C 2 H 4 (g) + 3 O 2 (g) 2 CO 2 (g) + 2H 2 O(g) Predict the sign of G o rxn first. Find G o f values in Appendix G. C 2 H 4 (g) = 68.4 kj/mol CO 2 (g) = kj/mol H 2 O(g) = kj/mol Answer: kj (we calculated kj using G o = H o T S o ) Should they be the same or different? Example Standard vs Non-Standard G Sign of G determines spontaneity (whether a reaction is still moving forward to reach equilibrium). Sign of G o determines whether a reaction is reactantor product-favored at equilibrium. The relationship between G and G o is: G o = Products - Reactants G is determined by the slope of the tangent at any point on the graph. R is the gas constant (8.314 J/K mol). T is temperature in Kelvin. Q is the reaction quotient (from Ch. 13). 41 Figure
8 Key features/information we can determine from graph: 1) G o = Products Reactants (constant for a reaction) This determines whether the reaction is product- or reactantfavored at equilibrium Can determine relative value of K from this (> 1, < 1, or 1) 2) Equilibrium: the minimum on the graph ( G = 0) This confirms whether the reaction is product- or reactantfavored at equilibrium 3) Spontaneity at a given point: Slope of the tangent at that point tells us the sign of G. G < 0: spontaneous (left of equilibrium) G > 0: non-spontaneous (right of equilibrium) What is the sign of G o? Is the reaction reactantor product-favored at equilibrium? What is the value of K? Where is: G < 0? G > 0? G = 0? 44 What would the graph look like for a reaction where Κ = 1? What is G o? Standard vs Non-Standard G G is the actual free energy change under non-standard conditions. It changes as a reaction proceeds (as concentrations, pressures, etc. change). G o is standard and does NOT change. It is a constant for a reaction (like H o and S o ). G o < 0 is spontaneous and product-favored. G changes as a reaction proceeds (initially decreases). This value determines the spontaneity of a reaction at a given point in time At equilibrium, G = 0 and Q = K: G = G o + RT ln K becomes: G o = - RT ln K This equation can be used for any system and any type of K value (K c, K p, K a, K b, K w, K sp, etc.) G o K Comment G o < 0 K > 1 Equilibrium mixture contains mostly products. G o > 0 K < 1 Equilibrium mixture contains mostly reactants. G o = 0 K = 1 Equilibrium mixture contains appreciable amount of reactants and products at equilibrium 47 G o rxn for the dissociation of ammonia is 27.1 kj/mol. NH 3 (aq) + H 2 O(aq) NH 4+ (aq) + OH - (aq) Calculate the equilibrium constant, K b, at 25 o C. Example
9 G o rxn for the dissociation of ammonia is 27.1 kj/mol. NH 3 (aq) + H 2 O(aq) NH 4+ (aq) + OH - (aq) Calculate G at 95 o C when [NH 4+ ] = M, [OH - ] = M, and [NH 3 ] = 1.15 M. Is the reaction spontaneous at this point? 1) Calculate G o rxn for the following reaction at 25 o C: P 4 (g) + 6 Cl 2 (g) 4 PCl 3 (g) 2) H o fusion for the melting of gold is kj/mol. S o fusion for the melting of gold is 9.38 J/mol K. What is the melting point (in o C) of gold? What is G o for at 25 o C? Is this reactant- or product-favored? 3) Calculate G for S(s) + O 2 (g) SO 2 (g) when [O 2 ] = M and [SO 2 ] = 1.24 M at 25 o C. In which direction will the reaction proceed? Example ) If K for the reaction CuO(s) Cu(s) + O 2 (g) at 25 o C is 2.66 x 10-46, what is G o? 50 1) Calculate G o rxn for the following reaction at 25 o C: P 4 (g) + 6 Cl 2 (g) 4 PCl 3 (g) G o f (P 4 (g)) = 24.4 kj/mol G o f (Cl 2 (g)) = 0 kj/mol G o f (PCl 3 (g)) = kj/mol Answer: G o rxn = [4 mol( kj/mol] [1 mol(24.4 kj/mol) + 0] = kj 51 2) H o fusion for the melting of gold is kj/mol. S o fusion for the melting of gold is 9.38 J/mol K. What is the melting point (in o C) of gold? Answer: T = H / S = kj/mol / kj/molk T = K 273 = 1060 o C What is G o for at 25 o C? Is this reactant- or productfavored? Answer: G o = H o T S o = kj/mol (298K)( kj/molk) = 9.75 kj/mol; reactantfavored 52 3) Calculate G for S(s) + O 2 (g) SO 2 (g) when [O 2 ] = M and [SO 2 ] = 1.24 M at 25 o C. In which direction will the reaction proceed? Need to calculate G o and Q. G o f (SO 2 (g)) = kj/mol Answer: G o rxn = kj Q = 1.24 M / M = G = kj; spontaneous in the forward direction. 53 4) If K for the reaction CuO(s) Cu(s) + O 2 (g) at 25 o C is 2.06 x 10-46, what is G o? Answer: G o = -RT ln K = J/K mol K (ln 2.06x10-46 ) = J/mol = 261 kj/mol (reactant-favored) 54 9
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