CH302 Worksheet 0 Answer Key: A Little Thermo Review 1. What two processes (a.k.a. path functions) can transfer internal energy between asystem and its surroundings? What symbols are used for these variables?Heat and work. q for heat and w for work.2. Write a good definition for a state function.A state function is a property of a system which is dependent exclusively on the state ofthe system and not the processes leading to that state.3. Name some state functions.Temperature, Volume, Pressure, Number of Moles, Gibb's Free Energy, Entropy,Enthalpy, Internal Energy, etc.4. The first law states that the value of which state function is conserved in an isolatedsystem? What two symbols are used for this state function?Internal energy (E or U).5. What is an isolated system? Name the most obvious example of an isolated system (hint:big).A closed system is one that exchanges neither matter nor energy with its surroundings.The universe is the most obvious example because it has no surroundings.6. What inequality is often associated with the second law of thermodynamics? What doesit mean in plain English?ΔSuniv> 0. This means that the entropy of the universe is always increasing.7. What equality is often associated with the second law of thermodynamics? What does itmean in plain English?ΔSuniv= ΔSsystem+ ΔSsurroundings. This means that the universe's change in entropy isthe sum of the system's and surrounding's change in entropy8. What does the third law of thermodynamics state?It states that the entropy of a perfect crystal will approach zero as its temperatureapproaches zero.9. How many translational, rotational and vibrational modes, respectively, does C2H4have?It has 3 translational, 2 rotational, and 13 vibrational modes.10. What would be the total internal energy associated with the vibrational motion of 1molecule of C2H4? What about 1 mole of C2H4?since E = 0.5kT for each mode, one molecule would have 6.5kT and one mole 6.5RT.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. Assume we want to use a bomb calorimeter to determine the specific heat capacity ofan unknown 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-114. Given the following data:H2SO4(l) ↔ H2S(g) + 2O2(g) ΔH = 793 kJ·mol-1H2O(g) + SO3(g) ↔ H2SO4(l) ΔH = -176 kJ·mol-1H2O(l) + SO3(g) ↔ H2S(g) + 2O2(g) ΔH = 661 kJ·mol-1calculate ΔH for the process: H2O(l) → H2O(g)ΔH = (-1*793 kJ·mol-1) + (-1*-173 kJ·mol-1) + (1*661 kJ·mol-1)= 41 kJ·mol-115. Consider the reaction below and approximate the value of the work function at roomtemperature. (Note: this doesn't require a calculator.)C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)At room temperature (~300K), the product of the RT term in w = -ΔngasRT isapproximately 2.5 kJ. Since Δngasis -2 for this reaction, the work function is approximately5 kJ.16. If a system at -272 °C absorbs 545 J of heat, what is its change in entropy?545 J·K-117. Ammonia (the common name for NH3, the odor of windex and cat urine), has a ΔH°vap= 23.35 kJ·mol-1and a ΔS°vap= 97.43 J·mol-1·K-1. What is the normal boiling point ofammonia expressed in centigrade?Because boiling is an equilibrium process, ΔG°vap= 0 = ΔH°vap- TΔS°vap.And so TΔS°vap= ΔH°vapand T = ΔH°vap/ΔS°vap= 23,350 J·mol-1/97.43 J·mol-1·K-1= 238.7 K = -33.3 °C.18.Consider the reaction below:HNO3(l) + H2(g) ↔ H2O(l) + NO2(g)Using the provided table values, calculate ΔG°rxnif it is performed under standardconditions.ΔH°f(kJ·mol-1)ΔS°m(J·mol-1·K-1)HNO3(l) -174.1 156H2(g) not provided 131H2O(l) -285.8 70NO2(g) 33.2 240ΔHrxn= ΣHf,products- ΣHf,reactants= (-285.8 kJ·mol-1+ 33.2 kJ·mol-1) - (-174.10 kJ·mol-1)= -78.5 kJ·mol-1ΔSrxn= ΣSm,products- ΣSm,reactants= (70 J·mol-1·K-1+ 240 J·mol-1·K-1) - (156 J·mol-1·K-1+ 131 J·mol-1·K-1)= 23 J·mol-1·K-1= 0.023 kJ·mol-1·K-1ΔG°rxn= ΔH°rxn- TΔS°rxn= -78.5 kJ·mol-1- 273 K*0.023 J·mol-1·K-1= -85 kJ·mol-119. Rank the following compounds in terms of decreasing standard molar entropy: CO(s),CO2(g), CO2(l), CO(l).CO2(g) > CO2(l) > CO(l) > CO(s)20. Assuming you have one mole of each of the following, rank them in terms of decreasingvalue of the term W in the Boltzmann equation: HF, NaF, ClF, F2.ClF > HF > NaF >
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