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Transcript
Chemistry 102 Chapter 17 THERMODYNAMICS Thermodynamics is concerned with the energy changes that accompany chemical and physical processes. Two conditions must be fulfilled in order to observe a chemical or physical change: 1. The change must be possible (determined by Thermodynamics) 2. The change must occur at a reasonable speed (determined by Kinetics) Important Terms in Thermodynamics: 1. System and Surroundings: System : The part of the universe we happen to be studying Surroundings: Everything outside the system Three types of systems can be present: Open: exchange of energy and matter Closed: exchange of energy but not matter Isolated: No exchange of energy or matter The State of the system can be evaluated in terms of the fixed properties of the: System Temperature Pressure Components of the system Number of moles Physical State of each system 1 Chemistry 102 Chapter 17 2. Adiabatic and Isothermal Changes: Adiabatic Change Isothermal Change Takes place in a system that is insulated from the surroundings. Takes place in a system that is not insulated from the surroundings. Heat cannot flow between system and surroundings. The temperature of the system changes: Heat can flow between system and surroundings and as such the temperature can be maintained constant. A change that occurs at constant temperature. Temp. increases if rxn is exothermic. Temp. decreases if rxn is endothermic 3. State Function A quantity whose value depends only on the current state of the system, and does not depend on the prior history of the system. Example: Temperature, Internal Energy 2 Chemistry 102 Chapter 17 4. Internal Energy Internal Energy (U) is a state function. It is a property of a system that depends only on its present sate, determined by variables such as temperature and pressure Internal Energy, U = Kinetic Energy of the particles making the system Potential Energy of the particles making the system results from the energy of motion of electrons atoms results from molecules chemical bonding between atoms 3 attractions molecules Chemistry 102 Chapter 17 FIRST LAW OF THERMODYNAMICS The 1st Law of Thermodynamics is the “Law of Conservation of Energy” applied to thermodynamic systems. When a system changes from one state to another (ex: warming), its internal energy changes from one definite energy to another. Change in Internal Energy : U = Uf Ui where: Uf = final value of the internal energy Ui = initial value of the internal energy The Changes in Internal Energy, U : are more important in thermodynamics than the absolute values of Internal Energy are measured by noting the exchanges of energy between the system and the surroundings Heat = q Work = w The energy that moves into or out of the system because of a temperature difference between the system and its surroundings If heat is lost U drops The energy exchange that results when a force F moves an object through a distance, d Work = Force x Distance If work is done by the system U drops q is negative If heat is absorbed U rises w is negative If work is done on the system U rises q is positive w is positive Examples: Determine the sign of the work for each of the changes shown below (carried out at constant P): a) 2 SO2 (g) + O2 (g) 2 SO3 (g) w= b) CaCO3 (s) CaO (s) + CO2 (g) w= c) CO2 (g) + H2O (l) + CaCO3 (s) Ca2+ (aq) + 2 HCO3– (aq) w= 4 Chemistry 102 Chapter 17 HEAT (q) AND WORK (w) RELATIONSHIP IN A PHYSICAL SYSTEM A gas is enclosed in a vessel equipped with a movable piston, momentarily fixed in position Heat passes from the the surroundings to the vessel and the temperature of the gas is increased Assume q = + 165 J The gas pressure increases and causes the gas to expand. This will cause to lift the piston and the weight on top of it In lifting the weight, the system does work = the energy gained by piston and weight Assume w = - 92 J The system gains Internal Energy, U from the heat absorbed : q = + 165 J The system loses internal energy via the work done: The change in Internal Energy = U = (+165 J) + (- 92 J) = + 73 J of the System 5 w = - 92 J Chemistry 102 Chapter 17 FIRST LAW OF THERMODYNAMICS The Change in Internal Energy of a system, U, equals q + w Consider a system where work is done on the system and heat is lost by the system: A gas in enclosed in a vessel equipped with a movable piston, momentarily fixed in position. Work is done on the system by pushing down the piston and thus compressing the gas. U = q + w The gas pressure increases and causes the gas to warm up. This will cause heat to flow out of the system. Assume w = + 92 J Assume q = - 165 J The system gains Internal Energy, U from the work done on it : w = + 92 J The system loses internal energy via the heat lost: q = - 165 J The change in Internal Energy of the System = U = q + w U = (-165 J) + (+ 92 J) = - 73 J Heat (q) and Work (w) relationship in a chemical system Consider the following exothermic chemical reaction carried out in a beaker open to the atmosphere: Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) qp = - 152.4 kJ/mol Zn Zn (s) + 2 H+ (aq) → ZnCl2 (aq) + H2 (g) qp = - 152.4 kJ/mol Zn Note: qp = indicates that the process occurs at constant pressure (atmospheric pressure) The H2 gas that is produced: increases the volume of the system must push against the atmosphere in order to evolve It follows that work must be done by the system. Can the work done by the system be measured and/or calculated? 6 Chemistry 102 Chapter 17 PRESSURE-VOLUME WORK F Assume that the pressure exerted by the atmosphere is replaced by a piston and weights, whose downward force of gravity, F, creates a pressure on the gas equivalent to that of the atmosphere F = force of gravity= atmospheric pressure V = increase in volume of the system due to the production of H2 gas V = (cross sectional area) x( height) height = h = V A The work done by the system in expanding = (Force of Gravity) x (Distance the piston moves) Atmospheric pressure w = -F x h = - F x V F =x A A V Negative sign is given because work “w” is done by the system and represents energy lost by it. It follows: w = - P V 7 Chemistry 102 Chapter 17 Examples: 1. In the reaction shown below, 1.00mol of Zn reacts with excess HCl to produce 1.00 mol of H2 gas. At 25C and 1.00 atm, this amount of H2 occupies 24.5 L. Calculate the change in Internal Energy (∆U) for this reaction. (1 Latm = 0.101 kJ) Zn (s) + 2H3O+ (aq) Zn2+ (aq) + 2 H2O (l) + H2 (g) q = –152.4 kJ/mol ∆U = q + w q = –152.4 kJ w=? w done by system (to push back atmosphere) = –PV = –(1.00 atm)(24.5 L)(0.101 kJ/1 Latm) = –2.47 kJ U = qp + w = ( –152.4 kJ ) + (–2.47 kJ) = –154.9 kJ 2. What is U when 1.00 mol of liquid water vaporizes at 100C? (H0 vap = 40.66 kJ/mol at 100 C) 8 Chemistry 102 Chapter 17 Examples: 3. Consider the combustion (burning) of methane, CH4, in oxygen: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) The heat of reaction at 25 C and 1.00 atm is –890.2 kJ. a) What is the change in volume, V when 1.00 mol CH4 reacts with 2.00 mol O2? (Ignore the volume of liquid water formed in the reaction) b) What is w for this change? c) Calculate U for the change indicated by the chemical equation. a) Find the change in volume (V) Volume occupied by 1 mole of any gas at STP = 22.4 L Volume occupied by 1 mole of any of the gases in the equation at 25 C: V = 22.4 L x CH4(g) 298 K = 24.45 L 273 K + 2 O2(g) CO2(g) + 3 moles of gas Gaseous Volume decreases: 2 H2O (l) H = –890.2 kJ 1 mole of gas 3 moles of gas 1 mole of gas V = Volume occupied by 3 moles – Volume occupied by 1 mole V = (3 x 24.45 L) – 24.45 L = 48.9 L b) Calculate the work (w) done on the system w = + P V = (1.00 atm)(48.9 L)(0.101 kJ/1 Latm) = 4.94 kJ c) Calculate change in internal energy, U = + 4.94 kJ qp = –890.2 kJ U = qp + w = (–890.2 kJ) + (+ 4.94 kJ) = –885.3 kJ Conclusion: Energy leaves the system 9 Chemistry 102 Chapter 17 4. What is U for the following reaction at 25 C? N2 (g) + 3 H2 (g) 2 NH3 (g) 4 moles of gas 2 moles of gas Volume occupied by 1 mole of any of the gases in the equation at 25 C: V = 22.4 L x H = – 91.8 kJ 298 K = 24.45 L 273 K The decrease in Volume: V == 2 x (24.45 L) = 48.9 L Since the Volume decreases, work is done on the chemical system by the atmosphere: w = + P V = (1.00 atm) x (48.9 L)(0.101 kJ/1 Latm) = + 4.94 kJ U = qp + w = (–91.8 kJ) + (+ 4.94 kJ) = – 86.9 kJ Conclusion: Energy leaves the system 5. Calculate the amount of work done in each of the following reactions. In each case, is the work done by or on the system? a) oxidation of HCl at 200 C 4 HCl (g) + O2 (g) 2 Cl2 (g) + 2 H2O (g) b) The decomposition of NO2 at 300 C 2 NO (g) N2(g) + O2 (g) 10 Chemistry 102 Chapter 17 ENTHALPY AND ENTHALPY CHANGE Recall: Enthalpy = The Heat of Reaction, H, at constant pressure (qp) Actually: Enthalpy = H = U + PV U (internal Energy), P (pressure), V (volume) are state functions. If follows: H (Enthalpy) is also a state function Consequences: 1. For a given temperature and pressure, a given amount of a substance has a definite enthalpy. 2. If the enthalpies of substances are known, the change of enthalpy for a reaction, H, can be easily calculated. H = Final Enthalpy – Initial Enthalpy = Hf – Hi At constant pressure, P : Hf = Uf + PVf and Hi = Ui + PVi H = (Uf + PVf) – (Ui + PVI) = (Uf – Ui) + (PVf – PVi ) = U + PV H = U + PV and as found earlier: U = qp – P V It follows: H = (qp – P V) + P V = qp – P V + P V = qp Conclusion: The Enthalpy or Reaction equals the Heat of Reaction at constant pressureApplications: H = qp (Chapter 6) Heat absorbed (endothermic reaction) or given off (exothermic reaction) by a chemical reaction H0 = Standard Enthalpy Change H0 = Standard Enthalpy Change for a Reaction can be calculated if the Standard Enthalpies of formation, H0f for both reactants and products are known. H0f = Standard Enthalpy of Formation of a Substance = Enthalpy change for the formation of one mole of substance in its standard state from its elements in their reference form ( most stable form at 250C and 1.00 atm) and in their standard states. (Given in Table 6.2 and Appendix C) 11 Chemistry 102 Chapter 17 Examples: 1. Calculate the heat absorbed or evolved in the reaction between NH3 gas and CO2 gas that yields aqueous urea (NH2CONH2) and liquid water. 2 NH3 (g) + CO2 (g) From Appendix C: H0f (NH3) H0f (CO2) H0f [CO(NH2)2] H0f (H2O) 2 NH3 (g) + 2 x (– 45.9 kJ/mol) H0 CO(NH2)2 (aq) + gas gas aqueous liquid CO2 (g) (– 393.5 kJ/mol) H2O (l) = – 45.9 kJ/mol = – 393.5 kJ/mol = – 319.2 kJ/mol = – 285.8 kJ/mol CO(NH2)2 (aq) + (– 319.2 kJ/mol) H2O (l) (– 285.8 kJ/mol) = n H0f (products) – n H0f (reactants) H0 = [(– 319.2) + (– 285.8 kJ)] – [2 x (– 45.9 ) + (– 393.5)] = – 119.7 kJ Since H0 is negative, we conclude that the reaction is exothermic. 2. Use data in appendix C to find H0 for the reactions shown below: a) NH4NO3 (s) NO2 (g) + 2 H2O (l) b) SiO2 (s) + 3 C (graphite) SiC (s) + 2 CO (g) 12 H0 = ? Chemistry 102 Chapter 17 THE SECOND LAW OF THERMODYNAMICS Spontaneous Processes: Physical or chemical changes that occurs by themselves, without outside assistance Spontaneous Physical Change: A ball rolling from the top of a hill High Potential Energy Low Potential Energy Spontaneous Chemical Change: The reaction of solid K(potassium) with liquid water spontaneous 2 K(s) + 2 H2O (l) 2 KOH(aq) + H2(g) In order for spontaneous changes to go in the opposite direction work would have to be expended, and the changes would be called non-spontaneous. Non-spontaneous Physical Change: Rolling a ball uphill Non-spontaneous Chemical Change Change: Obtaining K(s) from KOH(aq) non-spontaneous 2 KOH (aq) + H2 (g) 2 K(s) + 2 H2O(l) This process requires several chemical reactions. NOTE: The Second Law of Thermodynamics determines if a reaction is spontaneous or not. Exothermic Reactions (H 0) are not necessarily spontaneous and Endothermic Reactions (H 0) are not necessarily non-spontaneous 13 Chemistry 102 Chapter 17 ENTROPY (S) Entropy is the thermodynamic quantity that is a measure of the randomness or disorder in a system. Entropy is a state function: - the quantity of entropy in a substance depends only on variables that determine the state of the substance (temperature and pressure) Entropy is measured in J/K (SI unit) Entropy increases when the disorder in a sample of the substance increases: Example: Consider the melting of ice at O0C and 1 atm. MELTING ICE H2O molecules occupy regular fixed positions (crystal lattice) Ordered crystalline structure LIQUID WATER H2O molecules move about freely, giving a disordered structure Less-ordered structure Lower Entropy Higher Entropy Si = Initial Entropy = Entropy of Ice (41 J/K) Sif= Final Entropy = Entropy of liquid water (63 J/K) melting at 00C and 1 atm H2O(s) Si = 41 J/K H2O(l) Sf = 63 J/K S = 22 J/K In any natural process the degree of disorder (Entropy) increases. Natural processes are changes that occur by themselves (spontaneous changes). Second Law of Thermodynamics in reference to the system and its surroundings The Total Entropy of a System and its Surroundings always increases for a Spontaneous Process. ENTROPY is created during a spontaneous change 14 ENERGY can be neither created or destroyed during a spontaneous change Chemistry 102 Chapter 17 SECOND LAW OF THERMODYNAMICS IN REFERENCE TO THE SYSTEM ONLY Consider a spontaneous change and the changes that occur in the system at a given temperature: 1. Entropy is created 2. Heat (q) flows into or out of the system Entropy accompanies the heat flow Entropy Change q associated with heat flow (q) T Entropy created S = during the spontaneous process + Change in entropy associated with heat flow S + Entropy Flow S = = Entropy Created Entropy Created q T + 15 Chemistry 102 Chapter 17 S = Entropy Created + q T For a Spontaneous change: S 0 q T Entropy Created 0 may be positive or negative q It follows: S T For a Spontaneous Process at a Given Temperature, the Change in Entropy of the System is greater than the Heat divided by the Absolute Temperature 16 Chemistry 102 Chapter 17 ENTROPY CHANGE FOR A PHASE TRANSITION The entropy of a sample of matter increases as it changes state from a solid to a liquid or from a liquid to a gas. This is due to the increase in disorder of molecules in liquid compared to solid and gas compared to liquid. We can therefore predict the sign of ∆S for processes involving changes of state (or phase). In general, entropy increases (∆S > 0) for each of the following: the phase transition from a solid to a liquid the phase transition from a solid to a gas the phase transition from a liquid to a gas an increase in the number of moles of a gas during a chemical reaction Examples: 1. Predict the sign of ∆S for each of the following processes: a) the boiling of water ∆S = b) I2 (g) → I2 (s) ∆S = c) CaCO3 (s) → CaO (s) + CO2 (g) ∆S = 17 Chemistry 102 Chapter 17 ENTROPY CHANGE FOR A PHASE TRANSITION The criterion for spontaneity is an increase in the entropy of the universe. However, there are several spontaneous processes that include a decrease in entropy. For example, when water freezes below 0C, the entropy of the water decreases, yet the process is spontaneous. The 2nd Law of Thermodynamics states that for a spontaneous process, the entropy of universe increases (∆Suniv > 0). In the example above, even though the entropy of water decreases, the entropy of the universe must increase in order for the process to be spontaneous. Similar to the distinction of a system and surroundings in thermodynamics, we can distinguish between the entropy of the system and the surroundings. In the example of the freezing water, ∆Ssys is the entropy change of water, and the ∆Ssurr is the entropy change of the surroundings. The entropy change of the universe is then the sum of the entropy changes of the system and the surrounding. ∆S univ = ∆Ssys + ∆Ssurr If a spontaneous process includes a decrease in ∆Ssys , then it would follow that there must be a greater increase in ∆Ssurr in order for the ∆Suniv to increase. Since freezing of water is an exothermic process, it would follow that the heat given off by the process increases the entropy of the surroundings by a greater amount than the decrease of entropy of the system, making it a spontaneous process. To summarize: An exothermic process increase the entropy of the surroundings. An endothermic process decreases the entropy of the surroundings. 18 Chemistry 102 Chapter 17 QUANTIFYING ENTROPY CHANGE OF SURROUNDINGS & SYSTEM We know that when a system exchanges heat with surroundings, it changes the entropy of the surroundings. At constant pressure, qsys can be used to quantify the entropy change of the surroundings. In general, A process that emits heat into the surroundings (qsys = negative) increases the entropy of the surroundings (positive ∆Ssurr) A process that absorbs heat from the surroundings (qsys = positive) decreases the entropy of the surroundings (negative ∆Ssurr) The magnitude of the change in entropy of the surroundings is proportional to the magnitude of the qsys ∆Ssurr α – qsys It can also be shown that for a given amount of heat exchanged with the surroundings, the magnitude of ∆Ssurr is inversely proportional to the temperature ∆Ssurr α 1/T - qsys - Hsys Combining the above relationships: Since the sum of ∆Ssys and ∆Ssurr must be positive for a spontaneous process, it would then follow that for a spontaneous process Ssurr = Ssys > T = T (constant P, T) qsys T In an equilibrium process, such as a phase change, no significant amount of entropy is created by molecular disorder, and almost all the change in entropy is due to the absorption of heat. Therefore, Ssys = qsys (equilibrium process) T 19 Chemistry 102 Chapter 17 Examples: 1. Calculate S for the melting of ice (Heat absorbed = Heat of Fusion = Hfus = 6.0 kJ/mol) Ssurr = Hfus 6.0x103 J = = 22 J/K T 273 K 2. Liquid ethanol, C2H6O (l), at 250C has an entropy of 161 J/(mol x K). If the heat of vaporization, Hvap at 25 C is 42.3 kJ/mol, what is the entropy of the vapor in equilibrium with the liquid at 25 C? When the liquid evaporates, it absorbs heat: Hvap = 42.3 kJ/mol at 250C q Hvap 42.3 kJ/mol Entropy Change = S = = = = 141.9 J/K x mol T (25 + 273) K 298 K The entropy of 1 mole of vapor is calculated using the entropy of 1 mole of liquid (161 J/K) to which the entropy change resulting from the heat absorption (141.9 J/K) is added: Entropy of Vapor = 161 J/K + 141.9 J/K = 303 J/K 3. Consider the combustion of propane gas: C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) ∆Hsys = – 2044 kJ a) Calculate the ∆Ssurr associated with this reaction occurring at 25C. b) Determine the sign of ∆Ssys c) Determine the sign of ∆Suniv. Will this reaction be spontaneous? 20 Chemistry 102 Chapter 17 2ND LAW OF THERMODYNAMICS AND SPONTANEITY OF REACTIONS Consider the preparation of urea at constant temperature and pressure: 2 NH3 (g) + CO2(g) CO(NH2)2 (aq) + H2O (l) Is this a spontaneous reaction? (Does it go from left to right as written?) For a spontaneous reaction: (constant T and P) For a spontaneous reaction: (constant T and P) qp H S = T T H S T H – S 0 T smaller larger Multiplying each term by the positive quantity T yields: For a spontaneous reaction: H – T S 0 (constant T and P) If S is available and H is calculated (Table 6.2), one can predict if the reaction is spontaneous or not: H – T S 0 The reaction is spontaneous H – T S 0 The reaction is non-spontaneous ( ) The reaction is spontaneous ( ) The reaction is at equilibrium ( ) H – T S = 0 For a spontaneous reaction: H – T S 0 21 ( (constant T and P) ) Chemistry 102 Chapter 17 INTERPRETING SPONTANEITY Two factors determine the spontaneity of a chemical change: 1. Nature’s tendency for the potential energy to be at a minimum. Any change that tends to lower the potential energy tends to occur spontaneously The quantity that is related to the lowering of the potential energy of changes that take place at constant pressure is H For exothermic reactions H 0 Exothermic reactions “tend” to occur spontaneously 2. Nature’s tendency toward disorder Any change that tends to increase the disorder of a system tends to occur spontaneously The quantity that is related to the increase in the degree of disorder for changes that take place at constant pressure is S The larger the entropy of a system, the greater its statistical probability Reactions that go in the direction of greater entropy “tend” to occur spontaneously. S = Sfinal – Sinitial When S 0 Spontaneity is favored The entropy (S) of a system depends on several factors: (a) Physical state of the components of the system S (solid) S (liquid) S (gas) (b) Temperature of the system Entropy (S) increases with temperature (c) Mixing of components If the volume of the mixture is larger than the volumes of the components the Entropy (S) increases If two liquids are mixed to form a solution, S increases If two gases are mixed, S increases only if the total volume increases 22 Chemistry 102 Chapter 17 THIRD LAW OF THERMODYNAMICS A substance that is perfectly crystalline at 0 K has an entropy of zero This law helps determine the Entropy of different substances at different temperatures, if one considers the following facts: When the temperature of a substance is raised from 0 K, the substance absorbs heat, and the substance becomes more disordered The entropy change that occurs when heat is absorbed, S= q T Examples: 1. What is S for a substance if it is heated from near 0.0 K to 2.0 K and the heat absorbed is 0.19 J? 0.19 J 0.19 J S = = = 0.19 J/K (at 1.0 K) T(average) 1.0 K 2. What is S for the same substance if it is heated from near 2.0 K to 4.0 K and the heat absorbed is 0.88 J? 0.88 J 0.88 J S = = = 0.29 J/K (at 3.0 K) T(average) 3.0 K Proceeding this way, the entropy at any temperature can be determined. Standard entropy of methyl chloride, CH3Cl, at various temperatures Notes: 1. The entropy increases gradually as the temperature increases. 2. The entropy increases sharply when a phase change occurs. 23 Chemistry 102 Chapter 17 Standard Entropy of a Substance or Ion (Absolute Entropy), S0 S0 is the entropy value for the standard state of the species. For a substance: Standard State is the pure substance at 1 atm For a species in solution: Standard state is the 1 M solution S0 is temperature dependent; it is usually given for 25 C Entropy Changes (S) for a Reaction A. S can sometimes be predicted The entropy usually increases in the following situations: (a) A reaction in which a molecule is broken into two or more smaller molecules. Example: fermentation of glucose (grape sugar) to alcohol: C6H12O6 (s) 2 C2H5OH (l) + 2 CO2 (g) 1 large molecule 4 smaller molecules (b) A reaction in which there is an increase in moles of gas. This may result from a molecules breaking up, in which case Rule (a) and Rule (b) are related. Example: the burning of acetylene in oxygen: 2 C2H2 (g) + 3 O2(g) 4 CO2 (g) + 2 H2O (g) 5 moles of gas 6 moles of gas n gas = 6 – 5 = +1 (c) A process in which a solid changes to a liquid or gas or a liquid changes to a gas melting Ice at 0 C Liquid Liquid Water at 0 C vaporization Ethyl alcohol vapor at 25 C Ethyl alcohol at 25 C sublimation Solid CO2 at 25 C CO2 gas at 25 C 24 Chemistry 102 Chapter 17 B. S can be calculated if the values of standard entropies are available Examples: 1. Calculate the change in entropy, S0 for the fermentation of glucose. The standard entropy of glucose is C6H12O6(s) is: 212 J/mol x K The standard entropy for C2H5OH (l) is: (ethyl alcohol) 161 J/mol x K The standard entropy for CO2(g) is: 213.7 J/mol x K C6H12O6 (s) S : 1 x (212) 2 C2H5OH (l) 2 x (161) S0 = - 0 nS0(products) S0 = [(2 x 161) + 2 x 213.7)] + 2 CO2 (g) 2 x (213.7) nS0(reactants) - [1 x 212] = 537 J/K 2. Predict the sign of S0 for each reaction shown below. If you cannot predict the sign for any reaction, state why. a) C2H2 (g) + 2 H2 (g) C2H6 (g) S0 = b) N2 (g) + O2 (g) 2 NO (g) S0 = c) 2 C (s) + O2 (g) 2 CO (g) S0 = 3. Using S0 values in your textbook, calculate S0 for the synthesis of ammonia. N2 (g) + 3 H2 (g) 2 NH3 (g) 25 Chemistry 102 Chapter 17 PREDICTING SPONTANEITY Recall: For a spontaneous reaction: H – T S 0 (constant T and P) The value of H – T S can be used as a criterion for spontaneity: If: H – T S 0 H – T S 0 H – T S = 0 The reaction is spontaneous from left to right ( The reaction is non-spontaneous from right to left ( ) The reaction is non-spontaneous from left to right ( ) The reaction is spontaneous from right to left ) The reaction is at equilibrium ( ( ) Examples: 1. Consider the reaction by which urea is prepared from NH3 and CO2 , at 25 C 2 NH3 (g) + CO2 (g) CO(NH2)2 (aq) + H2O (l) H0 = – 119.7 kJ Determine if the reaction is spontaneous from left to right as written. The value of H – T S must be determined: H0 = – 119.7 kJ T = 25 + 273 = 298 K S = ? From Table 18.1: S0(NH3) gas S0 (CO2) gas 2 NH3 (g) 2 x 193 S0 = 193 J/mol x K = 214 J/mol x K + S0 [CO(NH2)2] aqueous = 174 J/mol x K S0(H2O) liquid = 70. J/mol x K CO2 (g) 214 CO(NH2)2 (aq) 174 + S0= n S0(products) – n S0(reactants) = [(174 + 70] – [2 x (193 ) + (214)] = H2O (l) 70 – 0.356 kJ/K H – T S = (119.7 kJ) (298 K) ( 0.356 kJ/K) = 13.6 kJ ) The reaction is spontaneous under standard conditions from left to right, as written. 26 Chemistry 102 Chapter 17 GIBBS FREE ENERGY Willard Gibbs (American Physicist) brought together the H (Enthalpy Change) and S (Entropy Change) into a single thermodynamic quantity G: G = Gibbs Free Energy = H TS As a reaction proceeds at a given temperature and pressure: REACTANTS PRODUCTS n Hf (reactants) m Hf (products) n S (reactants) H= n Hf (products) - m Hf(reactants) mS (products) S = nS (products) - mS(reactants) Result: For a change at constant T and P : G = H TS For any spontaneous change that takes place in a system: Free Energy G must be lowered (Gproducts Greactants) G must be negative: G = n G (products) - m G(reactants) 0 – 0 + G < 0 G > 0 Reaction is spontaneous Product-favored reaction Forward reaction is favored Reaction is non-spontaneous Reactant-favored reaction Reverse reaction is favored Conclusion: G gives a composite of the two factors that contribute to spontaneity (H and S) 27 Chemistry 102 Chapter 17 SPONTANEITY & FREE ENERGY G = H TS Consider the following situations: G = (–) – T(+) = negative 1. H is negative and S is positive (exothermic) (increase in entropy) G = (+) – T(–) = positive 2. H is positive and S is negative (endothermic) G = (+) – T(+) = ? (increase in entropy) The temperature plays the determining role in controlling spontaneity: (a) When T is large: TS H G = (+) – Tlarge(+) = negative spontaneous (b) When T is small: TS H G = (+) – Tsmall(+) = positive non-spontaneous G = (–) – T(–) = ? 4. H is negative and S is negative (exothermic) (decrease in entropy) The temperature plays the determining role in controlling spontaneity: TS > H G = (–) – Tlarge(–) = positive (b) When T is small: TS < H G = (–) – Tsmall(–) = negative (a) When T is large: non-spontaneous (decrease in entropy) 3. H is positive and S is positive (endothermic) spontaneous non-spontaneous spontaneous The effects of the algebraic signs of H and S and the effect of temperature on spontaneity can be summarized as: H S G = H – TS Outcome (–) (+) negative Spontaneous at all temp. (+) (–) positive Non-spontaneous regardless of temp. (+) (+) negative only at high T Spontaneous only at high temp. (–) (–) negative only at low T Spontaneous only at low temp. 28 Chemistry 102 Chapter 17 FREE ENERGY AT STANDARD STATE For interpreting thermodynamic data, standard states are chosen. Standard States: Standard states are indicated by a superscript degree sign on the symbol of the quantity: G0 = standard free-energy change H0f = standard enthalpy change = n H0f (products) - m H0f (reactants) S0 = standard enthropy change = nS0 (products) - mS0(reactants) Standard states are: Pressure 1 atm pressure (for pure liquids and solids) 1 atm partial pressure (for gases) Concentration Temperature 1 M (for solutions 25 ºC (298 K) Examples: 1. Calculate the standard free-energy change (G0) for the following reaction at 25 C, and determine whether it is spontaneous or not. CH4(g) + H0f : − 74.9 S0 : 186.1 2 O2 CO2(g) + − 393.5 213.7 2x0 (2) x (205.0) 2 H2O(g) (2) x ( − 241.8) (2) x (188.7) H0 = [- 395.5 + (2) x (-241.8)] kJ – [- 74.9 + 0] kJ = S0 = [213.7 + (2) x (188.7)] J/K – [186.1 + (2) x (205.0)] = − 802.2 kJ − 5.0 J/K G0 = H0 - TS0 = - 802.2 kJ – (298 K) (- 5.0 x 103 kJ/K) = – 800.7 kJ (spontaneous) 2. Calculate the standard free-energy change (G0) for the following reaction at 25 C, and determine whether it is spontaneous or not. N2 (g) H0f (kJ/mol) S0 (J/mol.K) 0 191.5 + O2 (g) 0 205.0 2 NO (g) 90.25 210.7 29 Chemistry 102 Chapter 17 STANDARD FREE ENERGY OF FORMATION (G0f) G0f is the free-energy change that occurs when 1 mol of substance is formed from its elements in their most stable states at 1 atm and 25 C G0f has units of kJ/mol of substance G0f is a constant, with values available in textbooks. NOTE: Standard free energies of formation of elements in their most stable form = 0 G0 can be calculated directly for any reaction by using: G0 = nG0f (products) - mG0f (products) Examples: 1. Calculate the standard free energy of the following reaction at 25 C, using standard free energies of formation given below: Na2CO3 (s) + G0f : – 1048.1 H+ (aq) 2 Na+ (aq) + 0 (2) x (– 261.9) HCO3(aq) – 587.1 G0f = [(2) x (–261.9) + (–587.1)] – [(–1048.1) + 0] kJ = – 62.8 kJ 3. Calculate the standard free-energy change (G0) for the reaction shown below, and determine whether it is spontaneous or not. (Use G0f values in your textbook). 2 NOCl (g) 2 NO (g) 30 + Cl2 (g) Chemistry 102 Chapter 17 G0 AS A CRITERION FOR SPONTANEITY G < –10 kJ Large negative number Reaction is spontaneous Reactants Products Reactants transform entirely to products –10 kJ <G < +10 kJ Small negative or positive number Reaction gives an equilibrium mixture Reactants Products Significant amount of both reactants and products present G > +10 kJ Large positive number Forward reaction is non-spontaneous Reverse reaction is spontaneous Reactants Products Mostly reactants present Examples: 1. Which of the following reactions are spontaneous in the direction written? (See table in your textbook for G0f values. (A) C(graphite) 0 Gf : 0 + 2 H2 (g) 0 G0 = [(–50.80 – (0)] kJ = –50.8 kJ Ag+ (aq) (B) Gf 0 : 77.1 + CH4 (g) – 50.8 (spontaneous reaction) I(aq) –51.7 AgI (s) –66.3 G0 = 2. Consider the reaction of nitrogen, N2 and oxygen, O2 to form nitric oxide, NO: N2 (g) + O2 (g) 2 NO (g) The standard free energy of formation of NO is + 86.60 kJ/mol. Do you expect the reaction to be spontaneous as written? Recall: Standard free energies of formation of elements in their stablest form = 0 G0 = 2 (+ 86.60 kJ) = + 173.2 kJ 31 (non-spontaneous reaction) Chemistry 102 Chapter 17 FREE ENERGY and USEFUL WORK In a spontaneous reaction: the “free” energy is lowered as reactants change to products, the change in “free” energy is released as free-energy change( G) this “G” can be harnessed to perform useful work A spontaneous reaction can be used to obtain useful work Examples: 1. The combustion of gasoline is used to move a car 2. The reaction in a battery generates electricity that can drive a motor The free – energy change (G) is the maximum energy available (“free”) to do useful work Maximum useful work = In an ideal situation: wmax = G G completely Useful work converted into In real situations: G converted Some Useful Work + Some Entropy into Conclusions: 1. In theory, all of the free-energy decrease liberated during a spontaneous chemical change can be used to do work (this would be wmax) 2. In practice, less work is obtained and the difference appears as an increase in entropy Example: Living systems are able to convert only about 40% of the free energy available in the oxidation of glucose to other forms of stored chemical energy (for example , ATP). The rest (60%) appears as heat, which ensures the body’s effective temperature control system. 32 Chemistry 102 Chapter 17 FREE-ENERGY CHANGE DURING A SPONTANEOUS REACTION (combustion of gasoline) Gasoline + O2 CO2(g) + H2O(g) Free energy decreases as the reaction proceeds (G 0) At equilibrium, free energy is at a minimum (equilibrium mixture is mostly products) FREE-ENERGY CHANGE DURING A NON-SPONTANEOUS REACTION (synthesis of NO from elements) N2(g) + O2(g) 2 NO(g) Free energy increases as the reaction proceeds (G 0) There is a small decrease in free energy as the system goes to equilibrium (some reaction occurs to give the equilibrium mixture which consists mostly of reactants) N2(g) + O2(g) NO(g) 33 Chemistry 102 Chapter 17 G0 & THE EQUILIBRIUM CONSTANT G determines the maximum amount of energy that is available to perform useful work as a system passes from one state to another (G = Gproducts – Greactants) As a reaction proceeds: the system’s capacity to perform useful work decreases (G diminishes), the system eventually reaches equilibrium and G 0 (Free Energy ceases to change) The value of G determines where the system stands with respect to equilibrium: When G < 0 The reaction is spontaneous and proceeds in the forward direction towards equilibrium (G decreases) When G = 0 When G > 0 The system is in a state of dynamic equilibrium The reaction is not spontaneous in the forward direction. The reaction is spontaneous in the reverse direction. Qualitative Relationship between G0 and the Position of Equilibrium The direction in which a reaction proceeds toward equilibrium is determined by where the system lies with respect to the free-energy minimum. G0 G Direction of reaction Reactants Products Position of equilibrium NOTE: The reaction proceeds spontaneously only in a direction that gives rise to a decrease in free energy (G is negative) 34 Chemistry 102 Chapter 17 RELATIONSHIP BETWEEN G0 AND EQUILIBRIUM CONSTANT G = G0 + RT lnQ Q = Thermodynamic Reaction Quotient for reactions involving gases Q is obtained from partial pressures. for reactions in solutions, Q is obtained from molar concentrations R = Gas Law Constant in units of energy = 8.314 J/mol K G0 = Standard Free-Energy Change G = Free-Energy Change when Reactants in nonstandard states are changed to products in non-standard states At equilibrium: 1. The free energy ceases to change: G = 0 2. The Thermodynamic Reaction Quotient, Q becomes equal to the Thermodynamic Equilibrium Constant (K): G = G0 + RT lnQ By rearrangement: becomes: G0 = – RT ln K Q=K 0 = G0 + RT lnK or G0 = – 2.303 RT log K Basic Equations relating the standard free-energy change to the equilibrium constant. Natural logarithms (ln) and logarithms to the base 10 (log) are related by: ln x = 2.303 log x G0 = - 2.303 RT log K Thermodynamic Equilibrium Constant The equilibrium Constant in which: the concentration of gases are expressed in partial pressures in atmospheres. the concentration of solutes are expressed in molarities It follows that: 1. For reactions involving only gases: K = Kp 2. For reactions involving only solutes in liquid solutions: K = Kc 3. For net ionic equations: K = Ksp 35 Chemistry 102 Chapter 17 RELATIONSHIP OF G0, K, AND SPONTANEITY G0 = –2.303 RT log K When K > 1 When K 1 When K <1 Reactants transform almost entirely to products Significant amounts of both reactants and products are present Mostly reactants present Reactants Reactants Products Products Reactants Products log K > 0 log K 0 log K < 0 G0 < 0 G0 = 10 kJ K = (0.018 – 57) G0 > 0 Reaction is spontaneous Reaction gives an equilibrium mixture Forward reaction is non-spontaneous Reverse reaction is spontaneous Examples: 1. Use the standard free energies of formation given to determine the equilibrium constant (K) for the following reaction at 25 C: N2 (g) G0f (kJ/mol) 0 + 3 H2 (g) 2 NH3 (g) –16.66 0 G0 = [ 2 x G0f (NH3)] – [ G0f (N2) + 3 x G0f (H2)] = 2 x (–16.66) – 0 = –33.32 kJ G = –RT ln K 0 ln K = ΔG 0 ( 33, 320 J) = = 13.45 RT (8.314 J/K)(298 K) G0 RT K=e G 0 /RT K = e13.45 = 6.9 x 105 2. Use the standard free energies in your text to calculate G0 and K for the following reaction at 25 C: 2 H2 (g) + O2 (g) 2 H2O (g) 36 Chemistry 102 Chapter 17 CHANGE OF FREE ENERGY WITH TEMPERATURE How can G0 be found for temperatures other than standard temperatures (25 C)? An approximate method used to calculate G0T is based on the assumption that both H0 and S0 are constant with respect to temperature (only approximately true) Then: NOTE: G0T = H0 - TS0 (a convenient approximation for G0T ) G0T is strongly temperature dependent G0T = Change in free Energy for a substance: at 1 atm of pressure (standard pressure) and at the specified temperature, T (nonstandard temperature) depends on the value of G SPONTANEITY determined by 0 TEMPERATURE T It follows that SPONTANEITY IS TEMPERATURE DEPENDENT! Meaning: Some chemical changes may be non-spontaneous at one temperature but spontaneous at another temperature. Effect of Temperature on the Spontaneity of Reactions G0T = H0 - TS0 For a Spontaneous Reaction: G0T 0 H0 S0 Temperature (–) (+) no effect (+) (–) no effect (–) (+) G0 Negative regardless of temp. Positive regardless of temp. Low Negative High Positive Low Positive High Negative (–) (+) 37 Spontaneity Spontaneous at any temp. Non-spontaneous at any temp. Spontaneous at low temp. Non-spontaneous at high temp. Non-spontaneous at low temp. Spontaneous at high temp. Chemistry 102 Chapter 17 Examples: 1. When glucose ferments, it produces ethyl alcohol and carbon dioxide. The reaction is exothermic. Under what temperature conditions is this reaction spontaneous? C6H12O6 (s) glucose 1 molecule of solid 2 C2H5OH (l) + ethyl alcohol breaks 2 molecules of liquid and 2 molecules of gas into G0T = H0 - TS0 G0T = H0 - TS0 H0 0 2 CO2(g) = (Negative Number) = S0 0 – [(Temperature) x (Positive Number)] – (–) [ (+) ] NOTE: T is absolute temperature in K cannot be negative G0T is negative (regardless of temperature) Reaction is spontaneous at any temperature 2. Sodium Carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3. Estimate the temperature at which NaHCO3 decomposes to products at 1 atm. 2 NaHCO3 (s) Na2CO3 + H2O (g) + CO2 (g) First calculate the H0 and S0 , using the given Hf 0 and S0 values: Hf : S0 : 0 2 NaHCO3 (s) 2 x (–947.7) 2 x 102 Na2CO3 –1130.8 139 + H2O (g) + –241.826 188.72 CO2 (g) –393.5 213.7 H0 = [(–1130.8) + (–241.826) + (–393.5)] – [ (2) x (–947.7)] = 129.274 kJ S0 = [(139) + (188.72) + (213.7)] – [ 92) x (102)]= 337.4 J/K = 0.3374 kJ/K Let G0 = 0 Substitute these values into G0 = H0 – TS0 ; H0 – TS0 = 0 H0 129.274 kJ T = = = 383 K (110 C) S0 0.3374 kJ/K [T 383 K (110 C)] The temperature at which NaHCO3 decomposes to products at 1 atm. should be at least = 383 K (110 C) 38 Chemistry 102 Chapter 17 3. Oxygen was first prepared by heating mercury(II) oxide, HgO. Estimate the temperature at which HgO decomposes to O2 at 1 atm. 2 HgO (s) 2 Hg (g) + O2(g) First calculate the H0 and S0 , using the given Hf 0 and S0 values: 2 HgO(s) Hf : –90.8 S0 : 70.3 0 2 Hg (g) 61.3 174.9 + O2 (g) 0 205.0 H0 = S0 = Substitute these values into G0 = H0 - TS0 ; H0 - TS0 = 0 H0 T = = S0 39 Let G0 = 0