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AP Chemistry 7: Thermodynamics A. Enthalpy (H): Bond Energy (5.3 to 5.5, 8.8) 1. chemical reactions typically involve breaking bonds between reactant atoms and forming new bonds 2. breaking bonds takes energy chemical system gains bond energy; surroundings lose energy (heat, etc.) 3. forming bonds releases energy chemical system loses energy, surroundings gain energy 4. change in energy called “change in enthalpy”—H a. when energy required to break bonds > energy released to form new bonds, +H (endothermic) 1. products at a higher energy state than reactants (weaker bonds) 2. surroundings lose energy (cool down) b. when energy required to break bonds < energy released to form new bonds, –H (exothermic) 1. products at a lower energy state than reactants (stronger bonds) 2. surroundings gain energy (heat up) 5. thermochemical equation a. chemical equation with H 1. listed to the right of equation 2. included as reactant (endothermic) or product (exothermic) b. H can be used in dimensional analysis process 6. H from calorimetry a. reactants are put in an insulated container filled with water, where heat is exchanged between reactants and water, but no heat is lost b. by conservation of energy: Hreaction = –Qwater 1. Q = mcT for simple coffee cup calorimeter— aqueous reactions a. m = mass of water b. c = specific heat of water (4.18 J/g•K) c. T = change in temperature (Tf – Ti) temperature can stay in oC, since 1 oC = 1 K (don't add 273 to ToC!) 2. Q = (C + mc)T for “bomb" calorimeter a. C = “bomb constant” accounts for all non-water components that change temperature b. all other letters are the same as the simple calorimeter 7. H using bond energy (B.E.) data Bond Energies in (kJ/mol) Single Multiple H C N O S F Cl Br I C=C 614 H 436 413 391 463 339 567 431 366 299 C=N 615 C 348 293 358 259 485 328 276 240 C=O 799 N 163 201 272 200 243 N=N 418 O 146 190 203 243 N=O 607 S 266 327 253 218 O=O 495 F 155 253 237 O=S 523 Cl 242 218 208 CC 839 Br 193 175 CN 891 151 CO 1072 I NN 941 S=S 418 a. energy needed to break a bond (i.e. C–H) in a diatomic, gaseous molecule, which contains the bond type 1. is approximately the same for any molecule 2. affected by molecular bonding only works for gaseous species 3. positive value (+ B.E.) for breaking bonds b. forming bonds (– B.E.) c. H = B.E.reactants – B.E.products Name __________________________ B. C. D. Entropy (S): Disorder (19.2) 1. atoms/molecules have inherent disorder depending on a. number of atoms—more internal motion = disorder b. spacing of molecules—farther apart = disorder c. speed of molecules—faster = disorder 2. predict increase in disorder for physical changes (+S) a. spread out: evaporation, diffusion and effusion (solution: spread out solute and solvent (+S), but bond solute-solvent (-S) ?, but usually +S) b. motion: melting and boiling 3. predict increase in disorder for chemical changes (+S): moles gaseous products > moles gaseous reactants Thermodynamic Data (5.6 to 5.7, 19.4) So (kJ/mol•K) Species Hfo (kJ/mol) Al 0.0 +0.0283 Al3+ -531.0 -0.3217 Al2O3 -1675.7 +0.0509 1. standard heat of formation (Hfo) data a. Ho for the formation of one mole of compound from its elements at standard temperature (25oC) Al: Al(s) Al(s) no reaction Al3+: Al(s) Al3+ + 3 eAl2O3: 2 Al(s) + 3/2 O2(g) Al2O3(s) b. Hfo for elements in natural state = 0.0 c. more negative = more stable (harder to decompose) 2. standard entropy (So) data a. amount of disorder compared to H+ (simplest form of matter), which is zero by definition b. listed in J/mol•K on AP exam, so you will have to convert to kJ/mol•K for most calculations 3. calculations using the thermodynamic data chart a. altering Hfo 1. opposite sign for the reverse reaction C + 2 Cl2 CCl4 = –139.4 kJ CCl4 C + 2 Cl2 = +139.4 kJ 2. multiply by number of moles (coefficient) 1 mole CCl4= –139.4 kJ 2 mole CCl4 = –278.8 kJ b. calculate H for a reaction using Hfo 1. H Ho = Hfoproducts – Hforeactants 2. Hess’s Law: H for a multi-step reaction equals the sum of H for each step -(-74.8) CH4(g) C + 2 H2 -393.5 C + O2 CO2(g) 2(-241.8) + 2 H2 + O2 2 H2O(g) CH4(g) + 2 O2 CO2(g) + 2 H2O(g) -802.3 c. calculate S for a reaction using So S So = Soproducts – Soreactants Gibbs Free Energy (G): Overall Energy State (19.5 to 19.6) 1. combination of enthalpy and entropy: G = H + TS 2. for a chemical or physical change: Go = Ho – ToSo a. To = 298 K b. where T 298 K:G Ho – TSo 3. determining if a process is spontaneous (G < 0) a. lower potential energy (-H)—chemical reactions b. greater disorder (+S)—physical changes c. depends on temperature 1. threshold temperature (Tthreshold) 2. occurs when G = 0 Tthreshold = Ho/So d. summary chart H S Spontaneous Process (G <0) for temperatures above Tthreshold + + + – at no temperatures – + at all temperatures for temperatures below Tthreshold – – 1. 2. Enthalpy 4 Fe(s) + 3 O2(g) 2 Fe2O3(s) H = -1640 kJ 8. a. How much heat is released to produce 10.0 g Fe2O3? 10.0 g Fe2O3 x 1 mol x -1640 kJ = -51.3 kJ 160 g 2 mol b. How many grams of iron are needed to generate 1.00 x 104 kJ of heat? -1.00 x 104 kJ x 4 mol x 55.8 g = 1360 g -1640 kJ 1mol 9. CaSO4(s) + CO2(g) CaCO3(s) + SO3(g) H = 224 kJ a. How much heat is absorbed when 10.0 g CaSO4 react. 10.0 g CaSO4 x 1 mol x 224 kJ = 16.5 kJ 136 g 1 mol b. How much CaCO3 is produced when 500. kJ is absorbed? Qwater = mcT = (100 g)(4.18 J/g•K)(-1.00 K) = -418 J H = ½H1 + -5H2 = ½(-2512 kJ) + -5(104 kJ) = -1776 kJ the number of moles of NH4Cl dissolved. 1.51 g NH4Cl x 1 mol/53.5 g = 0.0282 mol c. H (in kJ) for the dissolving of one mole of NH4Cl. 0.418 kJ/0.0282 mol = 14.8 kJ/mol 12.8 g of MgSO4 is dissolved in 250. g of H2O in a coffee cup calorimeter. The temperature of the solution increases from 23.8oC to 33.1oC. Determine a. Qwater for the solution process. Qwater = (250 g)(4.18 J/g•K)(33.1 – 23.8)K = 9720 J b. the number of moles of MgSO4 dissolved. 12.8 g MgSO4 x 1 mol/120. g = 0.107 mol c. H (in kJ) for the dissolving of one mole of MgSO4. -9.72 kJ/0.107 mol = -90.8 kJ 5. 6. 7. H = H1 + 4H2 = -126 kJ + 4(-179 kJ) = -842 kJ When 1.51 g of NH4Cl are dissolved in 100. g of water the temperature drops 1.00oC. Determine a. Qwater for the solution process. b. 4. >0 0 < 0 x Melting ice at 0oC x CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) x CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) x Distilling alcohol-water mixture Thermodynamic Data 2 Na2O2(s) + 4 HCl(g) 4 NaCl(s) + 2 H2O(l) + O2(g) Determine H from the thermochemical reactions below. 2 Na2O2(s) + 2 H2O(l) s) + O2(g) H1 = -126 kJ NaOH(s) + HCl(g) s) + H2O(l) H2 = -179 kJ C2H2(g) + 5 N2O(g) 2 CO2(g) + H2O(g) + 5 N2(g) Determine H from the thermochemical equations below. 2 C2H2(g) + 5 O2(g) 2(g) + 2 H2O(g) H1 = -2512 kJ N2(g) + ½ O2(g) H2 = 104 kJ 2O(g) 500 kJ x 1 mol/224 kJ x 100 g/1 mol = 223 g 3. Entropy Predict whether S > 0, S < 0 or S 0. H2(g) + F2(g) 2 HF(g) Estimate H for the reaction using the bond energy values. H = (H–H) + (F–F) – 2(F–H) H = 436 + 155 – 2(567) = -543 kJ C2H2(g) + 2 H2(g) C2H6(g) Estimate H for the reaction using the bond energy values. H = 2(C–H) + 1(CC) + 2(H–H) – 6(C–H) – 1(C–C) H = 2(413) + 839 + 2(436) – 6(413) – 348 = -289 kJ A bomb calorimeter with a constant of 921 J/oC contains 1,000 g of water. The combustion of 1.00 g of ethene (C2H4) increases the temperature 9.3oC. Determine a. Qwater for the combustion process. Q = (C + mc)T Q = [921 + (1000 g)(4.18 J/g•K)](9.3 K) = 4.8 x 104 J b. the number of moles of ethene reacted. 1.00 g C2H4 x 1 mol/28.0 g = 0.0357 mol c. H (in kJ) for the combustion of one mole of C2H4. -48 kJ/0.0357 mol = - 1300 kJ d. Write the equation for the combustion of ethene (C2H4). C2H4(g) + 3 O2(g) 2 CO2(g) + 2 H2O(g) e. Calculate H using bond energies. H = 4(C–H) + 1(C=C) + 3(O=O) – 4(C=O) – 4(O–H) H = 4(413) + 614 + 3(495) – 4(799) – 4(463) = -1297 kJ/mol 10. NO(g) + O(g) NO2(g) Determine H for the above reaction using the following thermochemical equations. NO(g) + O3(g) NO2(g) + O2(g) H1 = -198.9 kJ O3(g) 3/2 O2(g) H2 = -142.3 kJ O2(g) 2 O(g) H3 = 495.0 kJ H = H1 + -H2 + -½H3 H = -198.9 kJ + 142.3 kJ + -½(495.0 kJ) = -304.1 kJ 12. a. Write the equation for the combustion of methanol, CH3OH(l). (other reactants and products are gaseous). 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g) b. Calculate H using Hfo values. H = 2 HfoCO2 + 4 HfoH2O – 2 HfoCH3OH – 3 HfoO2 H = 2(-393.5) + 4(-241.8) – 2(-238.7) – 3(0.0) = -1276.8 kJ 1.00 g of methanol is burned in a bomb calorimeter that contains 1200 g of water. The temperature increases 3.4 K. c. Calculate the heat generated by the combustion reaction. 1.00 g CH3OH x 1 mol x 1276.8 kJ = 20.0 kJ 32.0 g 2 mol d. Calculate the calorimeter constant of the bomb. Q = (C + mc)T 20000 J = [C + (1200 g)(4.18 J/g•K)](3.4 K) C = 866 J/K 13. Ca(s) + SO3(g) + 2 H2O(l) CaSO3•2 H2O(s) H = -795 kJ and S = -0.2535 kJ/K for the reaction. a. Calculate Hfo for CaSO3•2 H2O. H = HfoCaSO3•2 H2O – HfoCa – HfoSO3 – 2 HfoH2O -795 = HfoCaSO3•2 H2O – 0 – (-395.7) – 2(-285.8) HfoCaSO3•2 H2O = -1762 kJ b. Calculate So for CaSO3•2 H2O. So = SoCaSO3•2 H2O – SoCa – SoSO3 – 2SoH2O -0.2535 = SoCaSO3•2 H2O – 0.0414 – 0.2567 – 2(0.0699) SoCaSO3•2 H2O = 0.1844 kJ/K Gibbs Free Energy 14. Consider the reaction at 25oC: Cu(s) + 4 H+(aq) + 2 NO3-(aq) Cu2+(aq) + 2 NO2(g) + 2 H2O(l). a. Calculate Ho using Hfo values. H = HfoCu2+ + 2 HfoNO2 + 2 HfoH2O – HfoCu – 4 HfoH+ – 2 HfoNO3 H = 64.8 + 2(33.2) + 2(-285.8) – 0 – 4(0) – 2(-205.0) = -30.4 kJ b. Calculate So using So values. So = SoCu2+ + 2 SoNO2 + 2 SoH2O – SoCu – 4 SoH+ – 2 SoNO3So = -.0996 + 2(.2400) + 2(.0699) – .0332 – 2(.1464) = 0.1942 kJ/K c. Calculate Go. Go = H –TS Go = -30.4 kJ – (298 K)(0.1942 kJ/K) = -88.3 kJ 11. 17. 18. 19. Heat of Reaction Lab 20. Use calorimetry to determine H for a series of reactions, compare the results with thermodynamic data, and combine the results to verify Hess' law. Heat about 75 mL of water to about 70oC. Place a Styrofoam cup in a 250-mL beaker. Add 50.0 mL cold tap water to the cup. Record the temperature, TC. Measure out 50.0 mL of the hot water and place in a second Styrofoam cup. Record the temperature, TH. Pour the hot water into the cold water, cover the cup, insert the thermometer in the hole, and mix gently. Record the temperature every 20 seconds for 3 minutes. Discard the water. a. (1) Record the temperatures. TC TH time (s) 20 40 60 80 100 120 140 160 180 ToC (2) Graph the temperature vs. time data. Draw a best fit straight line (use a ruler). Temperature (oC) 16. NH4NO3(s) NH4+(aq) + NO3-(aq) Determine the following for the above reaction. a. Is the reaction exothermic or endothermic? H = HfoNH4+ + HfoNO3- – HfoNH4NO3 H = (-132.5) + (-205.0) – (-365.6) = 28.1 kJ endo b. Is there an increase or decrease in entropy? S = SoNH4+ + SoNO3- – SoNH4NO3 S = 0.1134 + 0.1464 – 0.1511 = 0.1087 kJ/K increase c. Is the reaction spontaneous at 25oC? Go = H –TS = 28.1 kJ – (298 K)( 0.1087kJ/K) Go = -4.3 kJ spontaneous 2 SO2(g) + O2(g) 2 SO3(g) a. Calculate Ho. Ho = 2 HfoSO3 – 2 HfoSO2 – HfoO2 Ho = 2(-395.7) – 2(-296.8) – 0 = -197.8 kJ b. Calculate So. So = 2 SoSO3 – 2 SoSO2 – SoO2 So = 2(0.2567) – 2(0.2481) – 0.2050 = -0.1878 kJ/K c. Calculate G at 400 K. G = Ho –TSo G = -197.8 – (400)(-0.1878) = -122.7 kJ d. Determine the temperature range where the reaction is spontaneous. Tthreshold = Ho/So = (-197.8)/(-0.1878) = 1053 K All temperatures below 1053 K C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(l) a. Calculate Ho. Ho = 2 HfoCO2 + 3 HfoH2O – HfoC2H5OH – 3 HfoO2 Ho = 2(-393.5) + 3(-285.8) – (-277.6) – 3(0) = -1366.8 kJ b. Calculate So. So = 2 SoCO2 + 3 SoH2O – SoC2H5OH – 3 SoO2 So = 2(.2136) + 3(.0699) – (.1607) – 3(.2050) = -.1388 kJ/K c. Calculate G at 20oC. G = Ho –TSo = -1366.8 – (293)(-0.1388) G = -1326.1 kJ d. At which temperature (if any) will the reaction be spontaneous? T = Ho/So = -1366.8/-0.1388 = 9847 K All temperatures below 9847 K When H2SO4(l) is dissolved in water, the temperature of the mixture increases. Predict the sign of H, S and G for this process (justify your answer). +/– Justification – Temperature increases H S +/– Can't predict about dissolving – dissolving occured G C2H5OH(l) C2H5OH(g) Calculate the boiling point (threshold temperature) given the information: H = 37.95 kJ and S = 0.1078 kJ/K. T = Hvap/Svap = 37.95/0.1078 = 352 K (79oC) 20 60 100 140 180 Time (s) (3) Use the y-intercept to determine Tmix. Tmix (y-intercept) b. Calculate the following. (1) Average of the hot and cold temperatures. Tav = (TH – TC)/2 (2) Heat lost from the water. QL = mc(Tav – Tmix) (3) C. C = QL/(Tmix – TC) Place a Styrofoam cup in a 250 mL beaker. Add 50.0 mL of 3.00 M NaOH. Record the temperature, To. Pour 50.0 mL of 3.00 M HCl into the NaOH, cover, insert the thermometer, and mix gently. Record the temperature every 20 seconds for 3 minutes. Discard the mixture. c. (1) Record the temperatures. To time (s) 20 40 60 80 100 120 140 160 180 ToC (2) Graph the temperature vs. time data. Draw a best fit straight line (use a ruler). 20 60 Temperature (oC) 15. 100 140 180 Time (s) (3) Use the y-intercept to determine Tmix. Tmix (y-intercept) d. Calculate Hreaction per mole of reactant based on the calorimetry data. T (K) Qwater (kJ) Hreaction/mole e. H % Calculate Hreaction per mole of reactant based on Hfo. OH-(aq) + H+(aq) H2O(l) Place a Styrofoam cup in a 250 mL beaker. Add 50.0 mL of 3.00 M NH4Cl. Record the temperature, To. Pour 50.0 mL of 3.00 M NaOH into the NH4Cl, cover, insert the thermometer, and mix gently. Record the temperature every 20 seconds for 3 minutes. Discard the mixture. f. (1) Record the temperatures. To time (s) 20 40 60 80 100 120 140 160 180 ToC (2) Graph the temperature vs. time data. Draw a best fit straight line (use a ruler). Calculate Hreaction per mole of reactant based on the calorimetry data. T (K) Qwater (kJ) Hreaction/mole k. Calculate Hreaction per mole of reactant based on Hfo. NH3(aq) + H+(aq) NH4+(aq) H Temperature (oC) % Show that the chemical equations and H from part (d) – part (g) = part (j). OH-(aq) + H+(aq) H2O(l) -55.8 kJ NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) 3.3 kJ NH3(aq) + H+(aq) NH4+(aq) -52.5 kJ l. 20 100 140 180 Time (s) (3) Use the y-intercept to determine Tmix. Tmix (y-intercept) g. Calculate Hreaction per mole of reactant based on the calorimetry data. 60 T (K) Qwater (kJ) Hreaction/mole h. j. Calculate Hreaction per mole of reactant based on Hfo. NH4+(aq) + OH-(aq) NH3(aq) + H2O(l) H % Temperature (oC) Place a Styrofoam cup in a 250 mL beaker. Add 50.0 mL of 3.00 M NH3. Record the temperature, To. Pour 50.0 mL of 3.00 M HCl into the NH3, cover, insert the thermometer, and mix gently. Record the temperature every 20 seconds for 3 minutes. Discard the mixture. i. (1) Record the temperatures. To time (s) 20 40 60 80 100 120 140 160 180 ToC (2) Graph the temperature vs. time data. Draw a best fit straight line (use a ruler). 20 60 100 140 Time (s) (3) Use the y-intercept to determine Tmix. Tmix (y-intercept) 180 Practice Quiz Multiple Choice (no calculator) Briefly explain why the answer is correct in the space provided. 1 2 3 4 5 6 7 8 9 10 11 12 B D B D D D C B A C D B 1. I2(g) + 3 Cl2(g) 2 ICl3(g) According to the data in the table below, what is the value of Ho , in kJ, for the reaction represented above? Bond CI–CI I–I I–Cl Bond Energy (kJ/mole) 150 240 210 (A) - 870 (B) - 390 (C) +180 (D) + 450 H = (I-I) + 3(Cl-Cl) – 6(I-Cl) H = (150) + 3(240) – 6(210) = -390 kJ 2. C2H4(g) + 3 O2(g) 2 CO2(g) + 2 H2O(g) For the reaction, H is -1,300 kJ. What is the value of H, in kJ, if the combustion produced liquid water rather than water vapor? (H for H2O(l) H2O(g) is 45 kJ/mol) (A) -1,300 (B) -1,210 (C) -1,345 (D) -1,390 H = Hcomb – 2HH2O(l) H = -1300 kJ – 2(45 kJ) = -1390 kJ 3. CH4 (g) + 2 O2(g) CO2(g) + 2 H2O(l) Ho = -900 kJ What is the standard heat of formation of CH4, in kJ/mol, as calculated from the data below? (HfoH2O = -300 kJ/mol, HfoCO2 = -400 kJ/mol) (A) -200 (B) -100 (C) 100 (D) 200 H = HfoCO2 +2 HfoH2O – HfoCH4 -900 kJ = -400 kJ + 2(-300 kJ) – HfoCH4HfoCH4 = -100 kJ 4. H2(g) + ½ O2(g) H2O(l) Ho = x 2 Na(s) + ½ O2(g) Na2O(s) Ho = y Na(s) + ½ O2(g) + ½ H2(g) NaOH(s) Ho = z What is H for the reaction below? Na2O(s) + H2O(l) 2 NaOH(s) (A) x + y + z (B) x + y – z (C) x + y - 2z (D) 2z - x - y the first and second reactions are reversed and the third reaction is doubled H = -x + -y + 2z 5. Which is true when ice melts at its normal melting point? (A) H < 0, S > 0, G = 0 (B) H < 0, S < 0, G > 0 (C) H > 0, S < 0, G < 0 (D) H > 0, S > 0, G = 0 Bonds are broken +H, liquids more disordered than solids +S, threshold temperature G = 0 6. Which of the following reactions has the largest positive value of S per mole of Cl2? (A) H2(g) + Cl2(g) 2 HCl(g) (B) Cl2(g) + O2(g) Cl2O(g) (C) Mg(s) + Cl2(g) MgCl2(s) (D) 2 NH4Cl(s) 4 H2(g) + Cl2(g) Large positive S for reactions that produce many more moles of gas compared to reactants. 7. Ice is added to hot water in an insulated container, which is then sealed. What has happened to the total energy and the total entropy when the system reaches equilibrium? (A) Energy and entropy remain constant (B) Energy remains constant, entropy decreases (C) Energy remains constant, entropy increases (D) Energy decreases, entropy increases The total energy constant because insulation. Entropy increases because melted ice has more disorder. 8. N2(g) + 3 H2(g) 2 NH3(g) The above reaction is thermodynamically spontaneous at 298 K, but becomes nonspontaneous at higher temperatures. Which of the following is true at 298 K? (A) G, H, and S are all positive. (B) G, H, and S are all negative. (C) G and H are negative, but S is positive. (D) G and S are negative, but H is positive. Nonspontaneous at high temperature then H and S are negative. Spontaneous at 298 K G is negative. 9. 3 C2H2(g) C6H6(g) What is the standard enthalpy change, Ho, for the reaction represented above? (HfoC2H2 is 230 kJ•mol-1; HfoC6H6 is 80 kJ•mol-1) (A) -610 kJ (B) 150 kJ (C) -770 kJ (D) 610 kJ H = HfoC6H6 – 3HfoC2H2 H = (80 kJ) – 3(230 kJ) = -610 kJ 10. When solutions of NH4SCN and Ba(OH)2 are mixed in a closed container, the temperature drops and a gas is produced. Which of the following indicates the correct signs for G, H, and S for the process? (A) –G –H –S (B) –G +H –S (C) –G +H +S (D) +G –H +S Temperature drop means +H. The gas product means +S. The reaction occurs –G. 11. X(s) X(l) Which of the following is true for any substance undergoing the process represented above at its normal melting point? (A) S < 0 (B) H = 0 (C) H = TG (D) H = TS Threshold temperature, G = 0. G = H – TS = 0 H = TS. 12. For a reaction, Ho = -150 kg/mol and So = -50 J/mol•K. Which statement is true about this reaction? (A) It is spontaneous at high temperature only. (B) It is spontaneous at low temperature only. (C) It is spontaneous at all temperatures. (D) It is non-spontaneous at all temperatures. Ho and So are negative range of spontaneous temperatures are all temperatures below the threshold. Free Response (calculator) 1. Consider the combustion of butanoic acid at 25oC: HC4H7CO2(l) + 5 O2(g) 4 CO2(g) + 4 H2O(l Ho= -2,183.5 kJ o So (kJ/mol•K) Substance Hf (kJ/mol) CO2(g) -393.5 0.2136 H2O(l) -285.8 0.0699 O2(g) 0.0 0.2050 C3H7COOH(l) ? 0.2263 2. a. Calculate Hfo, for butanoic acid. Ho = 4HfoCO2 + 4HfoH2O – HfoC3H7COOH – 5HfoO2 -2183.5 = 4(-393.5) + 4(-285.8) – HfoC3H7COOH – 5(0) HfoC3H7COOH = -533.7 kJ/mol b. Calculate So for the combustion reaction at 25oC. So = 4SoH2O + 4SoCO2 – SoC3H7COOH -5SoO2 So = 4(0.0699) + 4(0.2136) – (0.2263) – 5(0.2050) So = -0.1173 kJ/K c. Calculate Go for the combustion reaction at 25oC. Go = Ho –TSo Go = -2183.5 kJ – (298 K)(-0.1173 kJ/K) Go = -2148.5 kJ d. What is the spontaneous temperature range? T = Ho/So = -2183.5 kJ/-0.1173 kJ/K = 19,000 K < 19,000 K Consider the synthesis reaction: N2(g) + 3 F2(g) 2 NF3(g) (Ho298 = -264 kJ mol-1, So298 = -278 J K-1 mol-1) a. Calculate Go298 for the reaction. Go = Ho –TSo = -264 kJ – (298 K)(-0.278 kJ/K) Go = -181 kJ b. For what temperature range is the reaction spontaneous? T = Ho/So = -264 kJ/-0.278 kJ/K = 950 K and below c. Calculate the heat released when 0.256 mol of NF3(g) is formed from N2(g) and F2(g) at 1.00 atm and 298 K. 0.256 mol NF3 x -264 kJ/2 mol NF3 = -33.8 kJ d. 3. 4. Calculate the F–F bond energy using the information above and the bond energies (NN = 946 kJ/mol, N–F = 272 kJ/mol). Ho = (NN) + 3(F–F) – 6(N–F) -264 kJ = (946 kJ) + 3(F–F) – 6(272 kJ) (F–F) = 141 kJ/mol The combustion of carbon monoxide is represented by the equation: CO(g) + ½O2(g) 2(g) a. Determine Ho for the reaction above using the values. C(s) + ½ O2(g) CO(g) Ho298 = -110.5 kJ•mol-1 C(s) + O2(g) CO2(g) Ho298 = -393.5 kJ•mol-1 CO(g) ½ O2(g) + C(s) Ho = 110.5 kJ C(s) + O2(g) CO2(g) Ho = -393.5 kJ CO(g) + ½ O2(g) CO2(g) Ho = -283.0 kJ o b. Determine S for the reaction above using the table CO(g) CO2(g) O2(g) Substance So (J/mol•K) 197.7 213.7 205.1 So = SoCO2 – SoCO - ½ SoO2 So = 213.7– 197.7– ½(205.1) = -86.5 J/K c. Determine Go for the above reaction at 298 K. Go = Ho - TSo Go = (-283 kJ) - (298 K)(-.0865 kJ/K) = -257.2 kJ The dissolving of AgNO3(s) in water is represented by the equation: AgNO3(s) Ag+(aq) + NO3-(aq) a. Is G positive, negative, or zero? Justify your answer. G is negative because AgNO3 is soluble in water, the solution process is spontaneous and G < 0. b. The solution cools when AgNO3(s) is dissolved. Is H positive, negative or zero? Justify your answer. H is positive because heat is absorbed during dissolving. c. Is S positive, negative, or zero? Justify your answer. S is positive because G = H – TS and G < 0 and H > 0 S must be > 0.