CH301 Fall 2008 Worksheet 12 Answer key 1. Assume we want to use a bomb calorimeter to determine the specific heat capacity of anunknown liquid. We use 3 L of the unkown liquid and perform a known reaction thatreleases 400 kJ of heat. We measure an initial and final temperature of 25 ºC and 28.7 ºC,respectively. If the heat capacity of the calorimeter is 85 J·K-1, and the density of the liquidis 2.34 g·mL-1, what is the specific heat capacity of the unknown liquid?ΔH = 400 kJm = 3 L * 1000 mL·L-1* 2.34 g·mL-1= 7020 gΔT = Tf- Ti= 28.7 ºC - 25 ºC = 3.7 ºC = 3.7 Kccal= 85 J·K-1* .001 kJ·J-1= 0.085 kJ·K-1ΔH = m·c·ΔT + ccal·ΔTc = (ΔH - ccal·ΔT)/(m·ΔT)= (400 kJ - 0.085 kJ·K-1* 3.7 K)/(7020 g * 3.7 K)= 0.01539 kJ·g-1·K-1= 15.39 J·g-1·K-12. Lets say we filled the calorimeter above with 3 L of water and performed the same knownreaction above. We measured a final temperature of 57.56 ºC, but forgot to measure theinital temperature. Considering the density and specific heat capacity of water are 1 g·mL-1and 4.184 J·g-1·K-1, could we calculate what the initial temperature must have been? If so,what was the initial temperature?Yes, we can determine Ti. We don't know the value of either ΔT or Ti, so we have twounknowns, but we also have two equation, so we can solve:ΔH = m·c·ΔT + ccal·ΔT and ΔT = Tf- TiΔT = ΔH/(m·c + ccal) = 400,000 J/(3000 g * 4.184 J·g-1·K-1+ 85 J·K-1)= 32.34 K = 32.34 ºCTi = Tf - ΔT = 57.56 ºC - 32.34 ºC = 25.22 ºC3. Given the following data:P4(s) + 6Cl2(g) ↔ 4PCl3(g) ΔH = -1,225.6 kJ·mol-1P4(s) + 5O2(g) ↔ P4O10(s) ΔH = -2,967.3 kJ·mol-1PCl3(g) + Cl2(g) ↔ PCl5(g) ΔH = -84.2 kJ·mol-1PCl3(g) + 1/2O2(g) ↔ Cl3PO(g) ΔH = -285.7 kJ·mol-1calculate ΔH for the reactionP4O10(s) + 6PCl5(g) ↔ 10Cl3PO(g)ΔHrxn= (1 * -1,225.6 kJ·mol-1) + (-1 * -2,967.3 kJ·mol-1) + (-6 * -84.2 kJ·mol-1) + (10* -285.7 kJ·mol-1)= -610.1 kJ·mol-14. Given the following data:2O3(g) ↔ 3O2(g) ΔH = -427 kJ·mol-1O2(g) ↔ 2O(g) ΔH = 495 kJ·mol-1NO(g) + O3(g) ↔ NO2(g) + O2(g) ΔH = -199 kJ·mol-1calculate ΔH for the reactionNO(g) + O(g) ↔ NO2(g)ΔHrxn= (-1/2*-427 kJ·mol-1) + (-1/2*495 kJ·mol-1) + (1*-199 kJ·mol-1)= -233 kJ·mol-15. Hyrdoiodic acid (HI) and sodium hydroxide (NaOH) are a strong acid and strong baserespectively. Calculate the change in enthalpy for their neutralization reaction:HI(g) + NaOH(s) → NaI(s) and H2O(l)Consult Appendix 2 in your ebook for standard enthalpy of formation values.ΔHrxn= ΣHf,products- ΣHf,reactants= (-287.78 kJ·mol-1+ -285.83 kJ·mol-1) - (26.48 kJ·mol-1+ -425.61 kJ·mol-1)= -174.48 kJ·mol-16. Calculate the change in enthalpy for the reaction of hydroxylamine and hydrogenperoxide to form ammonia, water and ozone.2NH2OH(s)2 + H2O2(aq) → 2NH3(aq) + H2O(l) + O3(g)Consult Appendix 2 in your ebook for standard enthalpy of formation values.ΔHrxn= ΣHf,products- ΣHf,reactants= (2*-80.29 kJ·mol-1+ -285.83 kJ·mol-1+ 142.7 kJ·mol-1) - (2*-114.2 kJ·mol-1+-191.17 kJ·mol-1)= 115.86 kJ·mol-17. Using bond enthalpies, approximate the change in enthalpy for the reaction in question 6above. Would the reaction be more or less endothermic with every species in the gasphase? Consult tables 6.7 and 6.8 in your ebook for mean bond enthalpiesΔHrxn= ΣBEreactants- ΣBEproducts= (2*N-H + N-O + 3*O-H + O-O) - (3*N-H + 2*O-H + O-O + O=O)= (2*388 kJ·mol-1+ 210 kJ·mol-1+ 3*463 kJ·mol-1+ 157 kJ·mol-1) - (3*388 kJ·mol-1+2*463 kJ·mol-1+ 157 kJ·mol-1+496 kJ·mol-1)= -211 kJ·mol-1The reaction would be less endothermic in the gas phase; it would in fact be exothermic.8. Calculate the change in standard molar entropy for the reaction in question 6 above.Consult Appendix 2 in your ebook for standard molar entropy values. Assume thathydroxyalmine has a standard molar entropy of 0. Also, discuss whether we areoverestimating or underestimating the change in standard molar entropy for the reaction .ΔSrxn= ΣSm,products- ΣSm,reactants= (2*111.3 J·mol-1·K-1+ 69.91 J·mol-1·K-1+ 238.93 J·mol-1·K-1) - (143.9 J·mol-1·K-1)= 387.54 J·mol-1·K-1By assuming that a reactant has no entropy, we are overestimating ΔSrxn.9. Based on the values of ΔHrxnand ΔSrxncalculated in question 6 and question 8, howwould the reactions spontaneity be effected by temperature? Explain your answer.The reaction would become more spontaneous as temperature increases. Based on theMaxwell free energy equation, ΔGrxn= ΔHrxn+ TΔSrxn, and the convention that a negativefree energy corresponds to a spontaneous process, we can state that in spite of the changein enthalpy being positive (non-spontaneous), if the temperature is sufficiently high, thenthe change in entropy (also positive, i.e spontaneous) can drive the reaction and make itspontaneous.10. Based on the values of ΔHrxnand ΔSrxncalculated in question 6 and question 8, at whattemperature would the reaction switch from being non-spontaneous to spontaneous. Inother words, what would T be if ΔGrxn= 0. Considering your answer to number 8, are weoverestimating or underestimating the temperature at which the reaction switches fromnon-spontaneous to spontaneous?ΔGrxn= ΔHrxn- TΔSrxn= 0T = ΔHrxn/ΔSrxn= 115,860 kJ·mol-1/387.54 J·mol-1·K-1= 299 KBecause the equilibrium temperature is inversely proportional to ΔSrxn, overestimatingΔSrxnwould mean underestimating T.11. In the list of elements below, mark (circle, underline, etc.) all of the elements that arenot shown in their standard state.Cdiamond(s) Ca(s) B2(s) Na(s) Fe(s) Hg(s)Br2(l) Mo(s) H(g) He(g) Xe(g) Rb2(s)Cd(l) As(s) N2(l) O2(l) Si60(s) F2(g)12. Write the standard formation reactions for the following chemical speciesNH3(g)1/2N2(g) + 3/2H2(g) → NH3(g)Fe2O3(s)2Fe(s) + 3/2O2(g) → Fe2O3(s)O2(l)O2(g) → O2(l)O3(g)3/2O2(g) → O3(g)NH2OH(s)1/2N2+ 3/2H2+ 1/2O2→ NH2OH(s)13. State in your own words the first law of thermodynamics. What are some of theconsequences of the first law?The energy of the universe is a constant. Some important consequences are that theenergy of a system can be increased or decreased by adding/subtracting heat or doing workon/by the system. Etc...14. What is a state function? List all of the state functions you can.A state function is a parameter or value of a system that depends only on the state of thesystem (whence the name), and not on how the system arrived at that state. Etc...Temperature, pressure, volume, density, enthalpy, entropy, Gibbs free energy, Helmholzfree energy,
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