THERMODYNAMICS I. TERMS AND DEFINITIONS A. Review of Definitions 1. Thermodynamics = Study of the exchange of heat, energy and work between a system and its surroundings. a. System = That part of universe of interest. (reaction vessel, etc.) b. Surroundings = Rest of universe. c. Some types of systems. 1) Open systems = Mass can be exchanged in addition to heat, energy and work. These are important to engineers in flow systems 2) Closed systems = No mass can be exchanged, only heat, energy and work. These are usually encountered in the laboratory 3) Isolated system = No mass, heat or work can be exchanged. The universe is an example of an isolated system. d. Exchange takes place at boundary between system and surroundings during a change in state of system. 1) State = That condition in which all variables are fixed and unvarying. 2) When one or more of these variables are changed, the system changes state. 3) Examples: Temperature Changes Gas at V 1, P 1, T 1 -----------> Gas at V 2,P 2, T 2 ( PV = nrt is the equation of state for an ideal gas ) Phase changes ( solid ----> liquid; liquid ---> gas: etc. ) Reactants ----------> Products 2. Can control some variables or conditions during a change. a. Isothermal Change = One at constant temperature (supply or take away just enough heat so that the temperature remains constant). b. Adiabatic Change = No heat is exchanged during change, that is, the system is insulated from its surroundings. c. Isobaric Change = One at constant pressure. 1
d. Normal laboratory conditions are isothermal and isobaric ones. e. Reversible change = an idealized change carried out very slowly by allowing one restraining variable to be infinitesimally different from it s equilibrium value; the system is essentially at equilibrium at all times during the change. B. State Functions ( or State Variables ). 1. Variables (properties) whose value depend only on the state of the system. a. Define the state. b. Value does not depend on the past history of the system. c. Examples: State functions -------- T, P, V, Energy. Not State functions --- work(w), heat change(q) 2. Thermodynamics is concerned with how the state variables change during a change of state. It views these changes in the light of three laws. Recall that in science a law is simply a summary of experience. C. Work ( W ) 1. Sense of W. W is positive when work is done on the system by the surroundings. When the system does work, W is negative; the system has to expend energy. 2. Some types of work. a. Mechanical work - exert a force through a distance. W = ( force )x( distance ) or W = f(δx) If force is not constant, W = f dx b. Work of expansion of a gas under constant pressure. W = P V 3. Dimension conversion. a. Because of the different formulas for calculating work, a number of different units for work can be obtained. They should be converted to the usual SI unit of joules (kg m 2 /s 2 ). This can be easily done using the Gas Constant, R, which has a general dimension of Energy mol K. 2
Example. A gas is expanded from a volume of 5.00 L to 9.00 L under a constant pressure of 2.00 atm. Calculate the work done on expansion in joules. W = P(V 2 - V 1 ) P = 2.00 atm V 2 = 9.00 L V 1 = 5.00 L W = (2.00 atm )( 9.00 L - 5.00 L ) = 8.00 L atm What is the work in joules? 8.314 J/molK W = ( 8.00 L atm ) 0.08206 L atm/mol K = 811 J D. Heat (q). 1. Sense of q. a. q is positive if heat energy is absorbed by the system (flows into the system) b. q is negative if heat is evolved by the system (flows out of the system). 2. The value of q can be measured experimentally in calorimetry. II. THE FIRST LAW OF THERMODYNAMICS A. Statement of the first law. 1. Law of Conservation of Energy. The total energy of the universe (or any isolated system ) is constant. Energy can neither be created nor destroyed but can be converted from one form to another. 2. First Law in symbols. For any change in the state of a system ΔE = q + W a. E = Internal Energy = energy intrinsically possessed by a system (due to mass, structure, temperature, etc.) 1) E is State Function. In a change from some Initial State Final State ΔE = change in E = E final - E initial In general, the sense of Δ is final - initial 2) If ΔE is +, then E final > E initial, the internal energy increases. If ΔE is -, then E final < E initial, the internal energy decreases. 3) Do not know the absolute values of E but are only interested in its changes. b. W is the work done on the system and q is the heat absorbed by the system. Neither W nor q are state functions. 3
3. Changes at constant volume and only PV work possible. a. At constant volume, ΔV = 0, W = 0 b. ΔE = q v (subscript v means constant volume) A measurement of q v gives ΔE for the change. This can be measured in a constant volume (bomb) calorimeter 4. Changes at constant pressure and only PV work. a. At constant pressure, q P = H, H = Enthalpy = E + PV b. If ΔH is +, H final > H initial, enthalpy increases; heat is absorbed. If ΔH is -, H final < H initial, enthalpy decreases; heat is evolved. c. Exothermic change = one in which ΔH is, heat is liberated. Endothermic change = one in which ΔH is +, heat is absorbed. d. Chemical Reactions. 1) For a reaction, ΔH = Σ H products - Σ H reactants 2) The absolute values of H's are not known, however, we can develop a set of Enthalpies of Formation to use in calculating ΔH's for reactions 5. Heats of Reactions. a. Standard Enthalpies of Formation, ΔH f. ΔH f = ΔH when 1 mole of a compound is prepared from its elements under standard conditions. b. Standard Conditions. 1) All gases at partial pressures of 1 atm. 2) All solutes at an effective concentration of 1 Molar. 3) All liquids and solids in their most stable form. 4) Note that no temperature is stipulated for standard conditions. Most data are reported at 25 C. 5) A superscript zero ( ) indicates standard conditions. c. Use of ΔH f Calculate ΔΗ 's for reactions. ΔΗ = ΔΗ f 's (products) - ΔH f 's (reactants) ΔH f s have been measured for most common substances and can be found in Tables. They are usually listed for 25 C. 4
III. SPONTANEOUS CHANGES A. Entropy and the Second Law. 1. Second Law can be stated in several ways. One statement is that it is impossible for an isolated system to spontaneously order itself. a. This does not depend on energy considerations. The First Law deals only with the conservation of energy not on the likelihood of the energy change actually occurring in the direction contemplated. b. For example we know from experience that when a 500g block of Cu is dropped, the spontaneous energy changes are: Potential Energy Kinetic Energy Thermal Energy. As far as the First Law is concerned, the process could just as well work backwards. That is, the Cu block could have absorbed energy, jumped up its original height. We know from experience that this will not spontaneously happen. c. When a cylinder of compressed gas is opened in a room, the compressed gas escapes from the cylinder into the air. Why wouldn't the air in the room escape" into the cylinder? In other words, why do gases spontaneously go from regions of high pressure to regions of low pressure? d. In both cases, the nonspontaneous direction requires that the molecules involved would, at the same instant in time, all decide to go in one particular direction. Although this direction could be energetically feasible, the probability (or chance) of this happening is so small that we can be assured that it will not occur. e. In general, the more ways an isolated system can distribute its energy, the less likely it is that the system will change out of that state to one that is more energy restrictive. The Second Law requires that any spontaneous change be in the direction of increased energy distribution possibilities. 2. Entropy, S. a. Measures the dispersal of energy as a function of temperature. The dimensions are in J/K or J/mol K (ds = dq rev /T). The more ways a system has the distribute its energy, the higher will be its entropy. For example in going from solid ----> liquid ----> gas entropy increases. 5
b. A perfectly ordered crystal of a substance at 0 K, is most restrictive in the ways its energy can be distributed and would therefore have zero entropy (S = 0). This is a statement of the Third Law of Thermodynamics and allows absolute entropies to be determined. The entropies of substances depend on their molecular complexity, their phase and the temperature. The molar entropies for most common substances have been measured under standard conditions and are found in standard thermodynamic tables. Some values are listed below for 25 C. Substance S (J/mol K) H 2 O (s) 47.91 H 2 O (l) 69.91 H 2 O (g) 188.83 CH 4(g) 186.26 C 2 H 6(g) 229.60 C 3 H 8(g) 237.4 C 4 H 10(g) 310.1 1) S increases as you go from solid ----> liquid ---> gas 2) S increases as the complexity of the molecule increases. 3) S increases as temperature increases. For H 2 O (l) at 50 C S = 75.3 J / mol Κ, and at 100 C, S = 86.8 J / mol K. 2. Standard Entropy Changes. a. Entropy is a state function, that is, for any change in state, the change in entropy, ΔS = S final S initial b. For phase changes, ΔS can be obtained from the S 's of the pure phases. Example. For water ΔS fus = S (liquid) S (solid) = 69.6 J / mol K 47.6 J / mol K = 22.0 J / mol K ΔS vap = S (gas) S (liquid) = 188.8 J / mol K 69.6 J / mol K = 118.9 J / mol K These are the values at 25 C. The values at the normal melting and boiling points will be slightly different. c. For chemical reactions, ΔS = Σ S (products) Σ S (reactants). B. Spontaneous Reactions. 1. For spontaneous chemical reactions the total entropy must increase. This means that 6
ΔS tot = ΔS sys + ΔS surr > 0 a. ΔS sys can be determined from the S 's of the reactants and products. However, ΔS surr can not be easily obtained. b. If heat is absorbed by the surroundings, its entropy should increase. Therefore, at constant temperature, ΔS surr = q/t. At constant pressure, q p = ΔH. c. Therefore, at constant temperature and pressure ΔS surr = "H Surr T But H surr = H sys ΔS surr =!H sys T ΔS tot = ΔS sys!h sys T > 0. TΔS tot = TΔS sys ΔH sys > 0 We now have a set of parameters we can use to predict spontaneity at constant T and P. 2. Free Energy ( Gibbs Free Energy or Gibbs Energy ), G a. G = H TS Since H, S, and T are state functions, G is also a state function. b. At constant temperature and pressure, ΔG = ΔH TΔS Note TΔS tot = TΔS sys Δ Η sys = ΔG sys at constant T and P ΔG sys is negative for all spontaneous changes ΔG sys = 0 at equilibrium. c. Note if ΔG sys > 0, the change is not spontaneous. The reverse direction is spontaneous direction. 3. Consider the phase change H 2 O(s) ----> H 2 O(l ) at 1 atm pressure. Given that ΔH (fus) = 6010 J / mol and ΔS (fus) = 22.0 J / mol K. Assuming that ΔH and ΔS are independent of temperature, calculate ΔG (fus) at -10.0 C, 0.0 C, and 10.0 C t, C T, K ΔΗ (fus) TΔS (fus) ΔG (fus)= H (fus)-t S (fus) -10.0 263.2 6010 5790 +220 Nonspontaneous 0.0 273.2 6010 6010 0 Equilibrium 10.0 283.2 6010 6230 220 Spontaneous 7
Note that T fp =!H o fus o In the same way T!S bp = fus C. G s for chemical Reactions. Two ways to obtain. 1. From H s and S s. Given the following data at 25 C Substance ΔΗ f (kj / mol) S ( J / mol K ) SO 3 (g) 395.72 256.76 SO 2 (g) 296.83 248.22 NO(g) 90.25 210.76 N 2 O(g) 82.05 219.85 o!h vap o!s vap For the reaction N 2 O(g) + SO 3 (g) ----> 2NO(g) + SO 2 (g) Calculate H, S and G at 25 C. a. ΔH. ΔH = ΔH f (SO 2 ) + 2ΔH f (NO) ΔH f (N 2 O) ΔH f (SO 3 ) ΔH = 296.83 kj + 2( 90.25 kj) ( 82.05 kj) ( 395.72 kj) ΔH = 197.34 kj/mol b. ΔS. ΔS = S (SO 2 ) + 2S (NO) - S (N 2 O) - S (SO 3 ) ΔS = 248.22 J / K + 2(210.76 J / K) 219.85 J / K 256.76 J / K = 193.13 J /mol K c. ΔG at 25 C. ΔG = ΔΗ TΔS Νote ΔH is in kj/mol and ΔS is in J / mol K ΔG = 197,340 J/mol (298 K )(193.13 J /mol K) = 139,787 J/mol = 139.77 kj/mol The dimension kj/mol (or J/mol) means per mol of the balanced equation. Note that the reaction is nonspontaneous under standard conditions. That is, It is nonspontaneous at 25 C when PN2O = PNO = PSO 3 = PSO2 = 1.0 atm. 2. From standard Free Energies of Formation ( ΔG f s) a. ΔG f = ΔG when one mole of a compound is prepared from its elements. c. ΔG f 's have been calculated at 25 C for many compounds and are listed with ΔH f 's and S s. As with ΔH f 's, ΔG f 's of elements = 0 d. Given the following ΔG f 's at 25 C, calculate ΔG for the reaction 8
N 2 O (g) + SO 3(g) -----> 2NO (g) + SO 2(g). Substance ΔG f ( kj / mol ) SO 3 (g) 371.06 SO 2 (g) 300.19 NO(g) 86.55 N 2 O(g) 104.20 Δ G = 2ΔG f (NO) + ΔG f (SO 2 ) ΔG f (SO 3 ) ΔG f (N 2 O) = 2(86.55 kj / mol) + ( 300.19 kj / mol ) ( 371.06 kj / mol ) 104.20 kj / mol = 139.77 kj / mol which is the same answer obtained earlier from ΔH f 's and S 's. D. ΔG Under Nonstandard Conditions. 1. Chemical Potential ( Free Energy per mole) a. The Chemical Potential of a component, i, of a mixture, µ i = Change in G i with moles of i G i = µ i n i µ i in J / mol. b. The value of µ depends on the nature of the substance, its concentration, and temperature. In general µ i = µ i + RT ln a i where a i is the activity of the component = effective concentration. Will assume: 1) For gases, activity is the partial pressure in atm. a i = P i 2) For solutes, activity is the Molarity a i = M i 3) For pure liquids and solids, a i = 1 2. Consider the gas phase reaction cc + bb ---> dd + ee ΔG = G(products) - G(reactants) = dµ D + eµ E cµ C bµ B Since µ C = µ C + RT ln a C, etc. ΔG = dµ D + eµ E cµ C bµ B + drtlna D + ertlna E crtlna C brtlna B but ΔG = dµ D + eµ E cµ C bµ B ΔG = ΔG + RTlna d D + RTlna e E RTlna b B RTlna c C 9
# "G = "G 0 + RTln a d e Da E & % a b c ( $ B a C ' If all reactants and products are gases, such as, cc(g) + bb(g) ---> dd(g) + ee(g), and using partial pressures (in atm) for activities, you get "G = "G +RTln P d e DP E P c b C P B a. G for is usually written in the form # "G = "G 0 + RTln a d e Da E & # % a b c ( = "G +RTlnQ where Q = a d e Da E & % $ B a C ' a b c ( $ B a C ' Q is called the Reaction Quotient (or the Mass Action Expression). For gaseous reactions where the partial pressures are used for activities, a subscript P is attached to Q (Q P ). The subscript P indicates that partial pressures are used to express the activities of gases. b. For the reaction Zn(s) + 2 H + ---> Zn 2+ + H 2 (g), "!G =!G 0 + RTln a a Zn 2 + H % $ 2 ' 2 # $ a Zn(s) a H + &' =!G " 0 + RTln [Zn2+ ]P H % 2 # $ [H + ] 2 &' c. Q measures the extent of the reaction. At the start of the reaction when no products are present Q = 0 and ΔG = -, that is, a large negative number and any reaction would be spontaneous. As the reaction proceeds, Q increases and ΔG becomes less negative. Will reach a point where ΔG = 0. At this point equilibrium is established between reactants and products and the reaction is complete. d. At equilibrium Q reaches a constant value. This value is called the Equilibrium Constant and is given the symbol K. For gaseous reactions, Q is Q P and K is K P. The subscript p indicates that gas concentrations are expressed in partials pressures in atm. At equilibrium: ΔG = 0 Q = K G = 0 = G + RT ln K or ln K =!G RT K = e!g RT 10
the values of the equilibrium constants for reactions can be calculated from tabulated thermodynamic data. 4. Q vs. K Since G = RTlnK, the G equation could be written as 3. Example. G = RTlnK + RTlnQ = RTln Q K If Q < K, then G is and the reaction is spontaneous as written If Q > K, then G is + and the reaction is not spontaneous as written, the reverse reaction is spontaneous. Given the following data at 25 C Substance ΔH f (kj/mol) S (J.mol K) NO(g) 90.25 210.76 N 2 O(g) 82.05 219.85 NO 2 (g) 33.18 240.06 For the reaction. N 2 O(g) + NO 2 (g) ----> 3NO(g), a. calculate H, S and G at 25 C. H = 3 H f (NO) - H f (N 2 O) H f (NO 2 ) = 3(90.25) 82.05 33.18 = 155.52 kj/mol S = 3S (NO) S (N 2 O) S (NO 2 ) = 3(210.76) 219.85 240.06 = 172.37 J/mol K or 0.17237 kj/mol K G = H T S = 155.52 kj/mol (298 K)(0.17237 kj/mol K) = 104.15 kj/mol b. calculate G when P NO = 1.0x10 3 atm, P N 2 O = 500 atm and P NO 2 = 3.0x10 4 atm. Q P = 3 P NO P P = N2 O NO 2 ( 1.0x10 3 ) 3 500 ( ) = 6.7x10 17 ( ) 3.0x10 4 G = G + RTlnQ P = 104.15 kj/mol + (8.314x10 3 kj/mol K)(298 K)ln(6.7x10 17 ) G = 11.9 kj/mol The reaction is still not spontaneous c. calculate K P at 25 C. 11
lnk P =!G RT = K P = e 42.039 = 5.5x10 19 104.15 kj/mol 8.314x10 3 kj/mol K ( ) 298 K ( ) = 42.039 Q P must be less than 5.5x10-19 for the reaction to be spontaneous at 25 C. E. Variation of ΔG with Temperature. Temperature Dependence of K. 1. For small temperature changes, ΔH and ΔS do not change much and can be considered as being temperature independent. However, ΔG changes with temperature. a. The temperature dependence of ΔG can be used to calculate the variation of K p with temperature. ln K =!G RT (!H T!S ) = RT = "H RT + "S R = # "H R $ 1 ' & ) + "S % T( R b. This means that a plot of ln K p vs. 1/T will be a straight line whose slope = ΔH /R. Note: If ΔH is negative (the reaction is exothermic), then K p decreases as T increases. If ΔH is positive (the reaction is endothermic), then K p increases as T increases. 2. Plots. C ln K P B A 1/T Of the three reactions, A is exothermic, and B and C are endothermic. Of the two endothermic reactions, reaction B has the higher ΔH. 12
3. For calculations Let K P1 = the equilibrium constant at temperature T 1 K P2 = the equilibrium constant at temperature T 2 Then!H!S lnk P2 = - + RT2 R :!H!S lnk P2 = - + RT1 R Subtract the two equations to eliminate ΔS /R ln K p2 K p1!h = - R 1 1 - T T1 2 This equation can be used to estimate ΔH if K p is known at two temperatures or to estimate K p at some temperature when ΔH is known and a K p at some other temperature. 4. Example. For the reaction N 2 O (g) + NO 2(g) -----> 3NO (g) ΔH = 155.52 kj / mol, ΔS = 173.37 J / mol K, and K p at 25 C = 6.24x10-19. Calculate K p at 500 C. Two ways to work this. a. Calculate ΔG at 500 C. Note K = 500 + 273 = 773K ΔG 773 = 155,520 J / mol (773 K)(173.37 J / mol K) = 21,505 J / mol ln K p =!Go RT ln K p = e 3.46 = 3.1x10 2 = 21,505 J / mol) / (8.314 J / mol K)(773K) = 3.46! b. Use ln K p # 2 "# K p1 $ & %& = 'H R o! 1 1 $ # " T 2 T & 1 % K p1 = 5.5x10-19 T 1 = 298K; K p2 =? T 2 = 773K K p2 155,520 J/mol ln =! 1 5.5x10 19 8.314 J/mol K 773 K 1 # " 298 K$ 13
K p2 / 5.5x10-19 = e 38.572 = 5.64x10 16 K p2 = ( 5.5x10-19 )( 5.64x10 16 ) = 3.1x10-2 5. Variation of G with temperature Sign of Change in G (= H T S ) Example H S with Temperature ( H & S ) + + G is + at low temperatures, but H 2 O(s) ----> H 2 O(l) becomes at high temperatures H = 6.01 kj/mol, S = 22.0 J/mol K + G is + at any temperature 3 O 2 (g) ----> 2 O 3 (g) H = 286 kj/mol, S = 137 J/mol K + G is at all temperatures 2 H 2 O 2 (l) ----> 2 H 2 O(l) + O 2 (g) H = 196 kj/mol, S = 125 J/mol K G is at low temperature, but Na + (aq) + Cl (aq) ----> NaCl(s) + at high temperatures H = 3.9 kj/mol, S = 43.2 J/mol K C. F. SUMMARY Using the thermodynamic tables one can: a. Calculate the heat change at constant pressure (ΔH ). b. Determine if a reaction is spontaneous under standard conditions at 25 C. c. Determine if a reaction is spontaneous under any set of concentrations at 25 d. Calculate the equilibrium constant of the reaction at 25 C. e. Calculate G at temperatures other than 25 C. f. Determine if a reaction is spontaneous under standard conditions at any temperature. g. Determine if a reaction is spontaneous under any set of concentrations at any temperature. h. Estimate the equilibrium constant of the reaction at any temperature. 14
THERMODYNAMICS PROBLEM SET CHEM 1304 Use the table of thermodynamic data given at the end of this problem set and in the appendix 3 of your textbook as necessary for working these problems. 1. Without referring to the thermodynamic tables, indicate the sign of S for the following reactions. a. 3O 2 (g) ----> 2O 3 (g) b. 2H 2 O 2 (l) ----> 2H 2 O (l) + O 2 (g) S (H 2 O 2 (l)) = 110.0 J/mol c. 2KCl (s) + 4 O 2 (g) ----> 2KClO 4 (s) d. Pb 2+ (aq) + 2 Cl (aq) ----> PbCl 2 (s) e. Mg(OH) 2 (s) ----> MgO (s) + H 2 O (l) 2. Calculate G for each of the reactionsin question 1 at 25 C. Under standard conditions, which of these reactions will be spontaneous? 3. For which of the reactions in question1, will K increase as the temperature increases? 4.. Calculate S, H, and G for the reaction in which ethane C 2 H 6 is burned in oxygen to form H 2 O (l) and CO 2 (g) at 25 C. 5. For the reaction CO (g) + 2H 2 (g) ----> CH 3 OH (l) calculate the values of G and K at 25 C. 6. Using the thermodynamic data in your text, and the fact that H f (CS 2 (g)) = 115.3 kj/mol and S (CS 2 (g)) = 237.8 J/mol K, calculate the normal boiling point of CS 2 (l). 7. For the reaction PCl 3 (g) + Cl 2 (g) ----> PCl 5 (g) at 25 C, calculate H, S, G, and K. 8. For the reaction N 2 H 4 (l) + 2NO 2 (g) ----> 2N 2 O (g) + 2H 2 O (g) calculate: a. G, H, and S at 25 C. b. G at 25 C when P N2 O = 1.0 x 10-4 atm, P H2 O = 2.0 x 10-2 atm, P NO2 = 1000 atm c. K at 500 C 15
9. Calculate the equilibrium vapor pressure of N 2 H 4 (l) at 25 C and at 50 C. 10. Calculate the vapor pressure of I 2 (s) at 25 C. Thermodynamic Data at 25 C Substance H f (kj/mol) S (J/mol K) PCl 3 (g) -287.0 311.7 PCl 5 (g) -374.9 364.5 I 2 (g) 62.4 260.6 N 2 H 4 (l) 50.6 121.2 N 2 H 4 (g) 95.4 238.4 H 2 O 2 (l) -187.8 110.0 Thermodynamics Problem Set Answers 1. S is : a, c, d. S is +: b, e 2. H (kj/mol) S (J/mol-K) G (kj/mol) a. 3O 2 (g) ----> 2O 3 (g) 284.4-139.8 326.1 b. 2H 2 O 2 (l) ----> 2H 2 O (l) + O 2 (g) 196.4 124.8 233.6 c. 2KCl (s) + 4 O 2 (g) ----> 2KClO 4 (s) 4.8 683.4 208.4 d. Pb 2+ (aq) + 2 Cl (aq) ----> PbCl 2 (s) 26.4 2.1 27.0 e. Mg(OH) 2 (s) ----> MgO (s) + H 2 O (l) 37.1 33.6 27.1 3. a, c and e reactions b and d will be spontaneous 4. 2 C 2 H 6 + 7 O 2 ---> 4 CO 2 + 6 H 2 O(l) S = -621.2 J/K, H = -3119 kj, and G = -2935 kj 5. G = -166.3 ( 137.3) = 29.0 kj ln K = 2.90x10 4 /(8.31)(298) = 11.71 ; K = 1.21x10 5 = 1 P co P H2 2 6. CS 2 (l) --> CS 2 (g) H = 27.4 kj, S = 86.8 J/K, T = 315.7 K or 42.5 C 7. H = -87.9 kj S = -170.2 J/K G = -37.2 kj K = 3.3x10 6 8. a. G, H, and S at 25 C. H = -439.0 kj, S = 215.2 J/K, G = -503.2 kj 16
b. G at 25 C when P N2 O = 1.0 x 10-4 atm, P H2 O = 2.0 x 10-2 atm, P NO2 = 1000 atm G = -602.4 kj c. K at 500 C K = 8.0x10 40 9. 14.3 Torr, 57.9 Torr 10. G = 19.34 kj/mol; P = 0.31 Torr 17