CH301 Worksheet 11 (Answer Key)1. What is the second law of thermodynamics? How does this apply to someone exploding ahydrogen balloon? 2 H2(g) + O2(g) → 2 H20(g)The second law of thermodynamics states that the entropy of the universe is alwaysincreasing. This reaction appears to violate this law because the entropy in 2 moles ofgaseous water is less than the entropy of 3 moles of gaseous hydrogen and oxygen.However, the reaction also releases heat, and this extra heat can contribute to entropy inthe surroundings. In other words, the entropy of the system decreases, but the entropy ofthe surroundings increases, and the entropy of the universe increases.2. Let's talk about signs. What does a positive or negative value mean for change inenthalpy (ΔH), work (w), and change in Gibb's free energy (ΔG)? Remember, be thesystem! And try to do this one from memory. Don't just copy it directly from the notes.A positive value for ΔH means that the system is absorbing heat, and the reaction isendothermic.A negative value for ΔH means that the system is losing heat, and the reaction isexothermic.A positive value for w means that work is being done on the system (the surroundings aredoing work on the system).A negative value for w means that work is being done by the system (the system is doingwork on the surroundings). After pressure is removed, the gas inside a balloon can do workon the surroundings to increase the volume.Try explaining this one to your peers a couple times. It will help you remember thesignificance of the sign.3. A bomb calorimeter is filled with 2 L of water. After a reaction, the temperature of thewater raises from 25.0 °C to 28.3 °C. The density and heat capacity of water are 1 g/mLand 4.184 J/(g∙K), respectively. The heat capacity of the calorimeter is 85 J per K.Determine ΔH of the reaction.m = 2000 mL∙1 g/mL = 2000 gCw= 4.184 J/(g∙K)ΔT = 301.3 K - 298 K = 3.3 KCcal= 85 J/KΔH = m∙Cliq∙ΔT + Ccal∙ΔT = 2000 g∙4.184 J/(g∙K)∙3.3 K + 85 J/k∙3.3 K = 27894.9 J4. The same bomb calorimeter is filled with 2 L of a liquid that has a density of 1.7 gramsper mL. A reaction releases 250 kJ of heat, and the temperature of the liquid increases from25 °C to 27 °C. What is the heat capacity of the liquid?ΔH = 250 kJ = 250,000 Jd = 1.7 g/mLV = 2 Lm = 2000 mL∙1.7 g/mL = 3400 gΔT = 300 K - 298 K = 2 KCcal= 85 J/KΔH = m∙Cliq∙ΔT + Ccal∙ΔTCliq= [ΔH - Ccal∙ΔT] / [m∙ΔT] = [250000 J - (85 J/K∙2 K)] / [3400 g∙2K] = 36.7 J/(g∙K)5. The liquid is allowed to cool down to 25 °C. The calorimeter is equipped with anotherreaction that releases 400 kJ of heat. What is the final temperature of the liquid after thereaction is complete?Ti= 298 KΔH = 400,000 Jm = 3400 gCcal= 85 J/KCliq= 36.7 J/(g∙K)ΔH = m∙Cliq∙ΔT + Ccal∙ΔTΔT = ΔH / [m∙Cliq+ Ccal] = 400,000 J / [3400 g∙36.7 J/(g∙K) + 85 J/K] = 3.2 KΔT = Tf- TiTf= ΔT + Ti= 3.2 K + 298 K = 301.2 K = 28.2 °C6. Balance the following reaction of hydrazine with methanol. Calculate the work done.N2H2(l) + CH3OH(l) → CH2O(g) + N2(g) + H2(g)N2H2(l) + CH3OH(l) → CH2O(g) + N2(g) + 2 H2(g)Δngas= 4 mol - 0 mol = 4 molw = -Δngas∙R∙T = -4 mol∙8.314 J/(K∙mol)∙298 K = -9.9 kJ7. If the work done on a system is 5.7 kJ, and the external pressure is equal to 3.5 atm. Isthe volume of the system increasing or decreasing?Because work is being done on the system, the sign of the work is positive. Also, w = -PΔV.Because P is positive, ΔV must be negative for the work to be positive. So, the volume mustbe decreasing.8. In your own words, what is Hess's Law?Hess's law states that the change in energy of one reaction is equal to the sum of thechanges in energy required for all the smaller steps that can complete the reaction. This lawis why we can use the enthalpies of formation for reactants and products to calculate theoverall change in enthalpy.9. Use the following data to calculate the change in enthalpy.Overall reaction: H2S (g) + 2 O2(g) → S03(g) + H2O(l)1. H2SO4(l) → H2S(g) + 2 O2(g) ΔH = 78.5 kJ2. H2SO4(l) → SO3(g) + H2O(g) ΔH = 20.5 kJ3. H2O(g) → H2O(l) ΔH = -11 kJThe overall reaction is equal to the reverse of reaction 1 plus the forwards reactions ofreactions 2 and 3.ΔHtotal= -ΔH1+ ΔH2+ ΔH3= -78.5 kJ + 20.5 kJ - 11 kJ = -69 kJ10. Use the following data to calculate the change in enthalpy.Overall reaction: N2H4(l) + H2(g) → 2NH3(g)1. N2H4(l) + CH4O(l) → CH2O(g) + N2(g) + 3H2(g) ΔH=-185KJ2. N2(g) + 3H2(g) → 2NH3(g) ΔH=-230KJ3. CH4O(l) → CH2O(g) + H2(g) ΔH=-325KJThe overall reaction is equal to the sum of the forward reactions 1 and 2 and the reversereaction of 3.ΔHtotal= ΔH1+ ΔH2- ΔH3= -185 kJ - 230kJ + 325 kJ = -90 kJ11. Determine the ΔHrxnfor the reaction using the provided bond energies:CH4(g) + I2(g) → CH3I(g) + HI (g)Bond energies:C-H : 416 kJ/mol H-I: 299 kJ/molI-I: 151 kJ/mol C-I: 213 kJ/molIs the reaction endothermic or exothermic?ΔHrxn= sum of bond energies of reactants - sum of bond energies of productsΔHrxn= [4(C-H) + 1(I-I)] - [3(C-H) + 1(C-I) + 1(H-I)]ΔHrxn= [4(416 kJ/mol) + (151 kJ/mol)] - [3(416 kJ/mol0 + (213 kJ/mol) + (299 kJ/mol)]ΔHrxn= +55 kJ/molThe reaction is endothermic because ΔHrxn is positive.12. Determine the boiling point for iron. ΔHvap= 349.6 kJ/mol and ΔSvap= 111.55J/(mol∙K)Phase changes are equilibrium processes, hence ΔG = 0 and ΔH = TΔST = ΔH/ΔS = (349.6 kJ/mol) / (.11155 kJ/(mol∙K) = 3134 KNote: make sure you put entropy and enthalpy both in either kJ or J.13. Calculate the amount of heat given off when 11 grams of manganese (Mn) is oxidized toMn2O3(s) at standard state conditions. ΔHf,Mn2O3(s) = -962.3 kJ/mol2Mn + (3/2)O2→ Mn2O3ΔHf,Mn= 0 kJ/molΔHf,O2(g) = 0 kJ/molΔHrxn= ΣΔHf,products- ΣΔHf,reactantsΔHrxn= ΔHf,Mn2O3q = (-962.3 kJ / mol Mn2O3)(1 mol Mn2O3/ 2 mol Mn)(1 mol Mn / 55g Mn)(11 g Mn)q = -96.4 kJ96.4 kJ of heat released.Note: be careful with the sign. The question asks how much heat was released, the negativeis therefore implied.14. Calculate the work done on the systems with only one mole of reactant:a. 2NO2 (g) → N2O4 (g) at 30 °Cb. 2NO (g) → N2 (g) + O2 (g) at 300 °Cw = -PΔV = -ΔnRTa. NO2 (g) → (1/2)N2O4 (g)T = 30 °C + 273 = 303KΔn = .5 mol - 1 mol = -.5 mol gasw = -(-.5 mol gas)(8.3145 J / K mol) (303 K) (1 kJ / 1000J) = 1.25 kJb. NO (g) → (1/2)N2 (g) + (1/2)O2 (g)Δn = 1 mol - 1 mol = 0 molw = 0 kJ15. Calculate ΔSuniverseafter the completion of the
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