THERMODYNAMICS – ENTROPY AND FREE ENERGY Thermodynamics studies the energy of a system, how much work a system could produce, and how to predict the spontaneous changes a system will undergo INTERNAL ENERGY (E) – The total energy of a system Energy is conserved during all processes 3A-1 (of 22)

1847 HERMANN VON HELMHOLTZ Presented a mathematical argument for the Law of Conservation of Energy FIRST LAW OF THERMODYNAMICS – The change in energy of a system is equal to heat that enters the system plus the work done on the system ΔE = q + w STATE FUNCTION – A property of a system that depends only on the current state of the system, not on the way in which the system acquired that state Kinetics Internal Energy is a state function, so ΔE is independent of the pathway of the change Thermodynamics 3A-2 (of 22)

ENTHALPY (H) – The internal energy of a system plus the pressure-volume product of a system H = E + pV ΔH = ΔE + Δ(pV) ΔH = ΔE + (Δp)V + p(ΔV) At constant pressure, Δp = 0: ΔH = ΔE + p(ΔV) From the First Law: ΔE = q + w At constant pressure, w = -p(ΔV): ΔE = q + -p(ΔV) ΔE + p(ΔV) = q At constant pressure the enthalpy change of a system is equal to the heat that enters the system ΔH = qp 3A-3 (of 22)

4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) Calculate the standard enthalpy change for 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) Handout 5 of the class website (data at 298 K) Thermodynamic data is tabulated under specified conditions called the STANDARD STATE, identified with a “º”, such as ΔHº STANDARD STATE – Gaseous matter is at 1 atm, dissolved matter is 1 M, and elemental matter is in its natural state at room temperature and pressure 3A-4 (of 22)

4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) Calculate the standard enthalpy change for 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) Handout 5 of the class website (data at 298 K) ENTHALPY OF FORMATION (ΔHf) – The enthalpy change for a formation reaction FORMATION REACTION – A reaction that forms one mole of a substance from the elements composing it 3A-5 (of 22)

4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) Calculate the standard enthalpy change for 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) ½N2 (g) + 3/2H2 (g) → NH3 (g) ½N2 (g) + O2 (g) → NO2 (g) O2 (g) → O2 (g) H2 (g) + ½O2 (g) → H2O (g) 3A-6 (of 22)

4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) Calculate the standard enthalpy change for 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) ΔHºrxn = Σ(ΔHºf products) - Σ(ΔHºf reactants) 4 mol NO2 (34 kJ/mol NO2) + 6 mol H2O (-242 kJ/mol H2O) - 4 mol NH3 (-46 kJ/mol NH3) - 7 mol O2 (0 kJ/mol O2) = -1132 kJ 3A-7 (of 22)

4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) Calculate the standard enthalpy change for 4NH3 (g) + 7O2 (g) → 4NO2 (g) + 6H2O (g) EXTENSIVE PROPERTY – One that depends on the amount of chemical change -1132 kJ ____________ 4 mol NH3 = -283.0 kJ ____________ mol NH3 3A-8 (of 22)

Why does this work? NH4NO3 → N2O + 2H2O ΔHº = ? N2 + 2H2 + ³∕2O2 → NH4NO3 ΔHºf NH4NO3 = -365 kJ/mol N2 + ½O2 → N2O ΔHºf N2O = 82 kJ/mol H2 + ½O2 → H2O ΔHºf H2O = -242 kJ/mol NH4NO3 → N2 + 2H2 + ³∕2O2 ΔHº = 365 kJ N2 + ½O2 → N2O ΔHº = 82 kJ 2H2 + O2 → 2H2O ΔHº = -484 kJ NH4NO3 → N2O + 2H2O ΔHº = -37 kJ 3A-9 (of 22)

Calculate the standard enthalpy change for N2 + 2H2 → N2H4 From Chem 1A Handout 8: N ≡ N : H – H → H H H – N – N – H . . + 1 mol (941 kJ/mol) = 941 kJ = 864 kJ = -160 kJ = -1564 kJ + 2 mol (432 kJ/mol) - 1 mol (160 kJ/mol) - 4 mol (391 kJ/mol) 81 kJ 3A-10 (of 22)

1865 RUDOLF CLAUSIUS Proposed that there is always some internal energy that cannot be converted into work He called this ENTROPY ( ΔT ) 3A-11 (of 22)

A large amount of the internal energy of the blocks can be used for work  a low amount cannot be used for work  low entropy A small amount of internal energy of the blocks can be used for work  a high amount cannot be used for work  high entropy What is the spontaneous process? 3A-12 (of 22)

SECOND LAW OF THERMODYNAMICS – In any spontaneous process there is always an increase in total entropy 3A-13 (of 22)

1890 LUDWIG BOLTZMANN Proposed a statistical meaning for entropy ENTROPY (S) – A measure of the number of arrangements available to a system in a given state Boltzmann calculated the entropy of a system by S = k log W S = entropy k = Boltzmann Constant (R/NA, or 1.38 x 10-23 J/moleculeK) W = number of ways particles can be arranged in a given state while keeping the total energy constant 3A-14 (of 22)

Boltzmann calculated the entropy of a system by S = k log W S = entropy k = Boltzmann Constant (R/NA, or 1.38 x 10-23 J/moleculeK) W = number of ways particles can be arranged in a given state while keeping the total energy constant 3A-15 (of 22)

Calculate the entropies of the following samples of CO2 (g) and CO2 (s) W = 20 x 19 = 380 W = 2 x 1 = 2 (1.38 x 10-23 J/K) log 380 (1.38 x 10-23 J/K) log 2 = 3.56 x 10-23 J/K = 4.15 x 10-24 J/K For a particular type of matter, the gaseous state has a higher entropy than the solid state 3A-16 (of 22)

FACTORS IN PREDICTING S FOR CHEMICAL SUBSTANCES 1) State of Matter Sgas > Sliquid > Ssolid More possible arrangements, higher entropy 2) Mass Slarge mass > Slow mass Ei = h2i2 i = 1,2,…  ______ 8mX2    A larger mass means closer spacing of energy levels, so more possible arrangements 3A-17 (of 22)

FACTORS IN PREDICTING S FOR CHEMICAL SUBSTANCES 3) molecular complexity Sless elements < Smore elements Molecules with different elements allows for more possible arrangements Fa Fb Ha Hb Fa – Fb and Ha – Hb Ha – Fa and Hb – Fb or Ha – Fb and Hb – Fa Explain each difference in absolute entropy: Cl2 (g) > Br2 (l) Sgas is higher than Sliquid Cl2 (g) > F2 (g) S is higher for larger masses C3H8 (g) > C2H6 (g) S is higher for larger masses CO (g) > N2 (g) S is higher for more internal complexity 3A-18 (of 22)

THIRD LAW OF THERMODYNAMICS – The entropy of a perfect crystal at 0 K is 0 (1.38 x 10-23 J/K) log 1 (1.38 x 10-23 J/K) log 16 = 0 J/K = 1.66 x 10-23 J/K Because of the 3rd Law, all substances > 0 K have entropy values > 0 3A-19 (of 22)

CALCULATING ΔS FOR CHEMICAL CHANGES Third Law Entropies of 1 mole of many substances in their standard states at 298 K (298Sº or just Sº) are found on Handout 5 Be (s) + O2 (g) + H2 (g) → Be(OH)2 (s) 3A-20 (of 22)

CALCULATING ΔS FOR CHEMICAL CHANGES Third Law Entropies of 1 mole of many substances in their standard states at 298 K (298Sº or just Sº) are found on Handout 5 Sº values are not ΔSºf values, Sº values are the absolute entropy (calculated from the Boltzmann equation) of 1 mole of each substance All Sº values at temperatures > 0 K are positive For a chemical reaction, a 298ΔSº can be calculated from Handout 5 298ΔSºreaction = Σ298Sºproducts - Σ298Sºreactants Be (s) + O2 (g) + H2 (g) → Be(OH)2 (s) NO!!! 3A-21 (of 22)

2NiS (s) + 3O2 (g) → 2SO2 (g) + 2NiO (s) Calculate 298ΔSº, the standard entropy change at 298 K, for 2NiS (s) + 3O2 (g) → 2SO2 (g) + 2NiO (s) 2 mol SO2 (248 J/molK SO2) + 2 mol NiO (38 J/molK NiO) - 2 mol NiS (53 J/molK NiS) - 3 mol O2 (205 J/molK O2) = -149 J/K ΔS predicts whether a system is becoming more ordered or disordered This says the system is becoming more ordered 3A-22 (of 22)

PREDICTING ΔS FOR CHEMICAL CHANGES ΔS is + for reactions that have 1) more gaseous products than reactants 2) solid reactants dissolving CaCO3 (s) → CaO (s) + CO2 (g) Ag+ (aq) + Cl- (aq) → AgCl (s) 0 gas molecules → 1 gas molecule dissolved ions precipitate ΔS = + ΔS = – 2SO2 (g) + O2 (g) → 2SO3 (g) H+ (aq) + HCO3- (aq) → H2O (l) + CO2 (g) 3 gas molecules → 2 gas molecules 0 gas molecules → 1 gas molecule ΔS = – ΔS = + 3A-22 (of 22)

SPONTANEITY SPONTANEOUS PROCESS – One that proceeds without outside intervention SECOND LAW OF THERMODYNAMICS – In any spontaneous process there is always an increase in total entropy For a spontaneous process: ΔSuniv > 0 ΔSsys + ΔSsurr = ΔSuniv HO 5 -ΔHsys _________ T 3B-1 (of 19)

1873 JOSIAH WILLARD GIBBS Proposed a way to determine the maximum amount of internal energy that could be converted into work GIBBS FREE ENERGY (G) – The maximum amount of internal energy of a system that can be converted into work 3B-2 (of 19)

G = H - TS ΔGsys = ΔHsys - TΔSsys -ΔGsys = -ΔHsys + ΔSsys ________ ________ T T ΔSsurr -ΔGsys = ΔSsurr + ΔSsys ________ T ΔSuniv 3B-3 (of 19)

-ΔGsys = ΔSuniv ________ T ΔSuniv > 0 for a spontaneous process  ΔGsys < 0 for a spontaneous process The change in Gibbs Free Energy (ΔG) indicates if a process is spontaneous or nonspontaneous 3B-4 (of 19)

ΔG = ΔH - TΔS Spontaneous processes are favored by a decrease in enthalpy (processes that are exothermic – ΔH negative) an increase in entropy (processes that increase disorder – ΔS positive) 3B-5 (of 19)

CONTRIBUTIONS OF ΔH AND ΔS TO SPONTANEITY ΔG = ΔH - TΔS = (-) - (+) = (+) - (+) = (+) - (-) = (-) - (-) = (+) = (?) = (?) = (-) ΔH - + exothermic endothermic ΔS + - more disorder more order ΔG - + ? spontaneous nonspontaneous spontaneous at high T spontaneous at low T 3B-6 (of 19)

Calculate the ΔGº at 298 K for the reaction NH4Cl (s) ⇆ NH3 (g) + HCl (g) 298ΔGº 1 mol NH3 (-17 kJ/mol NH3) + 1 mol HCl (-95 kJ/mol HCl) - 1 mol NH4Cl (-203 kJ/mol NH4Cl) = 91 kJ When starting with solid NH4Cl, 1 atm NH3, and 1 atm HCl, the reaction is nonspontaneous  reverse reaction is spontaneous 3B-7 (of 19)

Calculate the ΔGº at 298 K for the reaction NH4Cl (s) ⇆ NH3 (g) + HCl (g) 298 ΔHº 1 mol NH3 (-46 kJ/mol NH3) + 1 mol HCl (-92 kJ/mol HCl) - 1 mol NH4Cl (-314 kJ/mol NH4Cl) = 176 kJ 298 ΔSº 1 mol NH3 (193 J/molK NH3) + 1 mol HCl (187 J/molK HCl) - 1 mol NH4Cl (96 J/molK NH4Cl) = 284 J/K 298 ΔGº = 298 ΔHº - T 298 ΔSº = 176 kJ - (298 K)(0.284 kJ/K) = 91 kJ 3B-8 (of 19)

Calculate the ΔGº at 298 K for the reaction NH4Cl (s) ⇆ NH3 (g) + HCl (g) 298 ΔHº = 176 kJ Works against spontaneity , and predominates 298 ΔSº = 284 J/K Works for spontaneity 298 ΔGº = 298 ΔHº - T 298 ΔSº = 176 kJ - (298 K)(0.284 kJ/K) = 91 kJ 3B-9 (of 19)

Calculate the ΔHº, ΔGº, and ΔSº at 298 K for the reaction 2C2H6 (g) + O2 (g) ⇆ 2C2H5OH (l) 298 ΔHº 2 mol C2H5OH (-278 kJ/mol C2H5OH) - 2 mol C2H6 (-84.7 kJ/mol C2H6) - 1 mol O2 (0 kJ/mol O2) = -386.6 kJ 298 ΔGº 2 mol C2H5OH (-175 kJ/mol C2H5OH) - 2 mol C2H6 (-32.9 kJ/mol C2H6) - 1 mol O2 (0 kJ/mol O2) = -284.2 kJ 3B-10 (of 19)

Calculate the ΔHº, ΔGº, and ΔSº at 298 K for the reaction 2C2H6 (g) + O2 (g) ⇆ 2C2H5OH (l) 298 ΔSº 2 mol C2H5OH (161 J/molK C2H5OH) - 2 mol C2H6 (229.5 J/molK C2H6) - 1 mol O2 (0 J/molK O2) = -137 kJ ΔGº = ΔHº - TΔSº ΔGº - ΔHº = ΔSº ____________ -T = -284.2 kJ – (-386.6 kJ) ____________________________ -298 K = -0.34 kJ/K 3B-11 (of 19)

Calculate the ΔHº, ΔGº, and ΔSº at 298 K for the reaction 2C2H6 (g) + O2 (g) ⇆ 2C2H5OH (l) ΔH and ΔS are temperature independent, so Handout 5 can be used to calculate a ΔHº and ΔSº at any temperature However, ΔG depends on temperature, so the ΔGºf’s on Handout 5 can ONLY be used to calculate ΔGº at 298 K  To calculate ΔGº at any temperature other than 298 K, use ΔHº - TΔSº 3B-12 (of 19)

Calculate ΔHº, ΔSº, and ΔGº at 308 K for 2SO2 (g) + O2 (g) → 2SO3 (g) 298ΔHº = 308ΔHº (constant at all temperatures) 2 mol SO3 (-396 kJ/mol SO3) - 2 mol SO2 (-297 kJ/mol SO2) - 1 mol O2 (0 kJ/mol O2) = -198 kJ ∴ reaction is exothermic 3B-13 (of 19)

Calculate ΔHº, ΔSº, and ΔGº at 308 K for 2SO2 (g) + O2 (g) → 2SO3 (g) 298ΔSº = 308ΔSº (constant at all temperatures) 2 mol SO3 (257 J/molK SO3) - 2 mol SO2 (248 J/molK SO2) - 1 mol O2 (205 J/molK O2) = -187 J/K ∴ reaction becomes more ordered 3B-14 (of 19)

Calculate ΔHº, ΔSº, and ΔGº at 308 K for 2SO2 (g) + O2 (g) → 2SO3 (g) 298ΔGº ≠ 308ΔGº (not constant at all temperatures) 308 ΔGº = ΔHº - TΔSº = -198 kJ - (308 K)(-0.187 kJ/K) = -140. kJ ∴ when all reactants and products are present and start in their standard states, the forward reaction is spontaneous at 308 K 3B-15 (of 19)

For the reaction: H2O (s) ⇆ H2O (l) Find ΔGº at 20.0ºC Calculated from Handout 5: 298 ΔHºfus = 6,038 J 298 ΔSºfus = 22.1 J/K 293.2ΔGº = 293.2 ΔHº - T 293.2 ΔSº = 6,038 J - (293.2 K)(22.1 J/K) = -442 J reaction is spontaneous  forward reaction is spontaneous H2O (s) melts at 20.0ºC 3B-16 (of 19)

For the reaction: H2O (s) ⇆ H2O (l) Find ΔGº at -20.0ºC Calculated from Handout 5: 298 ΔHºfus = 6,038 J 298 ΔSºfus = 22.1 J/K 253.2ΔGº = 253.2 ΔHº - T 253.2 ΔSº = 6,038 J - (253.2 K)(22.1 J/K) = 442 J reaction is nonspontaneous  reverse reaction is spontaneous H2O (l) freezes at -20.0ºC Notice ΔGº is different at the two different temperatures: -442 J and +442 J 3B-17 (of 19)

For the reaction: H2O (s) ⇆ H2O (l) Find ΔGº at 0.0ºC Calculated from Handout 5: 298 ΔHºfus = 6,038 J 298 ΔSºfus = 22.1 J/K 273.2ΔGº = 273.2 ΔHº - T 273.2 ΔSº = 6,038 J - (273.2 K)(22.1 J/K) = 0 J The temperature of a phase change (like melting) is when the two phases are in equilibrium For any process at equilibrium, ΔG = 0 (if it is a standard state reaction like this, then ΔGº = 0) 3B-18 (of 19)

For the reaction: H2O (l) ⇆ H2O (g) Find the normal boiling point of water A phase change temperature is an equilibrium between 2 phases, so ΔG = 0 “Normal” means vapor pressure is 1 atm when boiling ∴ standard state, so ΔGº = 0 Calculated from Handout 5: 298 ΔHºvap = 44,400 J 298 ΔSºvap = 119 J/K TΔGº = ΔHº - TΔSº 0 = ΔHº - TeqΔSº -ΔHº = - TeqΔSº ΔHº = Teq _____ ΔSº = 44,400 J ___________ 119 J/K = 373 K = 100. ºC 3B-19 (of 19)

THERMODYNAMIC FUNCTIONS (1) Enthalpy (H) – The total energy plus pV product of a system Change in Enthalpy (ΔH) – Indicates if a process is exothermic or endothermic at constant pressure (2) Entropy (S) – The energy of a system that cannot be converted to work, related to the number of arrangements available to a system in a given state Change in Entropy (ΔSsys) – Indicates if a process is becoming more ordered or more disordered Change in Entropy (ΔSuniv) – Indicates if a process is spontaneous or nonspontaneous (3) Gibbs Free Energy (G) – The maximum energy of a system that can be converted into work Change in Gibbs Free Energy (ΔG) – Indicates if a process is spontaneous or nonspontaneous 3C-1 (of 13)

PREDICTING ΔG FOR NON-STANDARD STATE CHEMICAL REACTIONS ΔGº predicts spontaneity when reactants and products are all present in the container and in their standard state (1 M or 1 atm) ΔG predicts spontaneity when reactants and products are not in their standard state ΔG = ΔGº + RT ln Q 3C-2 (of 13)

The standard free energy change at 298 K is -514 kJ for the reaction 2CO (g) + O2 (g) → 2CO2 (g) If a reaction vessel contains 0.010 atm CO, 0.020 atm O2, and 3.0 atm CO2, calculate the free energy change for the reaction at 298 K ΔG = ΔGº + RT ln Q 298ΔG = 298ΔGº + RTln Q Q = pCO22 __________ pCO2pO2 = (3.0)2 _________________ (0.010)2(0.020) = 4.5 x 106 298ΔG = -514,000 J + (8.314 J/K)(298 K) ln 4.5 x 106 = -476,000 J 3C-3 (of 13)

HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq) (a) Calculate the standard free energy change at 298 K for the reaction HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq) 298ΔGº = 1 mol H3O+ (-237 kJ/mol H3O+) + 1 mol IO2- (-23 kJ/mol IO2-) - 1 mol HIO2 (-55 kJ/mol HIO2) - 1 mol H2O (-237 kJ/mol H2O) = 32 kJ This means that the reverse reaction is spontaneous in a solution that is 1 M HIO2, 1 M H3O+, and 1 M IO2- 3C-4 (of 13)

HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq) (b) If a solution is 0.500 M HIO2, 1.00 x 10-7 M H3O+, and 0.250 M IO2-, calculate the free energy change at 298 K for the reaction HIO2 (aq) + H2O (l) ⇆ H3O+ (aq) + IO2- (aq) ΔG = ΔGº + RT ln Q 298ΔG = 298ΔGº + RTln Q Q = [H3O+][IO2-] _______________ [HIO2] = (1.00 x 10-7)(0.250) _______________________ (0.500) = 5.00 x 10-8 298ΔG = 32,000 J + (8.314 J/K)(298 K) ln 5.00 x 10-8 = -10,000 J This means that the forward reaction is spontaneous in a solution that is 0.500 M HIO2, 1.00 x 10-7 M H3O+, and 0.250 M IO2- 3C-5 (of 13)

THE MOST IMPORTANT REASON FOR KNOWING ΔGº For a reaction at equilibrium ΔG = 0 ΔG = ΔGº + RT ln Q 0 = ΔGº + RT ln Q 0 = ΔGº + RT ln Keq -RT ln Keq = ΔGº ΔGº = -RT ln Keq Knowing ΔGº is valuable because it is related to a reaction’s equilibrium constant 3C-6 (of 13)

4NO (g) ⇆ 2N2O (g) + O2 (g) The 298ΔGº for the above reaction is -28.2 kJ. Calculate Keq at 298 K. ΔGº = -RT ln Keq ΔGº = ln Keq _____ -RT e-ΔGº/RT = Keq Keq = e-(-28,200 J)/(8.314 J/K)(298 K) = 8.77 x 104 3C-7 (of 13)

HF (aq) + H2O (l) ⇆ H3O+ (aq) + F- (aq) The Keq (Ka) for the above reaction is 7.2 x 10-4 at 298 K. Calculate 298ΔGº. ΔGº = -RT ln Keq 298ΔGº = -(8.314 J/K)(298 K) ln (7.2 x 10-4) 298ΔGº = 18,000 J 3C-8 (of 13)

RELATIONSHIP BETWEEN ΔGº AND Keq The sign of ΔGº indicates the magnitude of the Keq Keq >1 <1 1 ΔGº = -RT ln (>1) ΔGº = -RT ln (<1) ΔGº = -RT ln (1) ln (>1) = + ln (<1) = – ln (1) = 0 ΔGº – + ∴ if ΔGº is – it means Keq > 1 ∴ if ΔGº is + it means Keq < 1 ∴ if ΔGº is 0 it means Keq = 1 Why? ΔGº means all chemicals are 1 atm or 1M ∴ Q = 1 If Q = 1 and Keq > 1 , then the forward reaction must be spontaneous, so ΔGº = – 3C-9 (of 13)

Find 298ΔGº and Keq at 298 K for the reaction N2O4 (g) ⇆ 2NO2 (g) 298ΔGº = 2 mol NO2 (52 kJ/mol NO2) - 1 mol N2O4 (98 kJ/mol N2O4) = 6 kJ e-ΔGº/RT = Keq Keq = e-(6,000 J)/(8.314 J/K)(298 K) = 0.09 3C-10 (of 13)

HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq) Find ΔGº and Ka at 298 K for the reaction HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq) 298ΔGº = 1 mol H3O+ (-237 kJ/mol H3O+) + 1 mol NO2- (-35 kJ/mol NO2-) - 1 mol HNO2 (-54 kJ/mol HNO2) - 1 mol H2O (-237 kJ/mol H2O) = 19 kJ e-298ΔGº/RT = Keq = e-(19,000 J)/(8.314 J/K)(298 K) = 4.7 x 10-4 3C-11 (of 13)

HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq) Find ΔGº and Ka at 398 K for the reaction HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq) 298ΔHº = 1 mol H3O+ (-286 kJ/mol H3O+) + 1 mol NO2- (-174 kJ/mol NO2-) - 1 mol HNO2 (-207 kJ/mol HNO2) - 1 mol H2O (-286 kJ/mol H2O) = 33 kJ 298ΔSº = 1 mol H3O+ (70 J/molK H3O+) + 1 mol NO2- (156 J/molK NO2-) - 1 mol HNO2 (146 J/molK HNO2) - 1 mol H2O (70 J/molK H2O) = 10. J/K 398ΔGº = 33 kJ - (398 K)(0.010 kJ/K) = 29 kJ 3C-12 (of 13)

HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq) Find ΔGº and Ka at 398 K for the reaction HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2- (aq) Keq = e-ΔGº/RT = e-398ΔGº/R(398 K) = e-(29,000 J)/(8.314 J/K)(398 K) = 1.6 x 10-4 3C-13 (of 13)