Sample quiz and test questions – Chapter 2.I. Terms and short answers1. A system that can exchange neither matter nor energy with its surroundings is called isolated2. A process that releases heat into the surroundings is calledexothermic3. Work needed to raise a mass m through a height h on the surface of the Earth is given byequationw = mgh4. Work of isothermal compression of an ideal gas can be calculated using the equationw = -nRTln(Vf/Vi)5. A reversible process is defined asA process that can be reversed by an infinitesimal change in a parameter6. The internal energy isThe sum of all kinetic, potential, rotational, vibrational, and electronic energy of allthe atoms, ions, and molecules in the systemII. Sketch the enthalpy change diagram for the following processes and reactions:1. CO2(g) à CO2(s) ∆depHo = - 33.56 kJ2. C(gr) + 2S(s) à CS2(l) ∆fHo = 88 kJIII. Calculations of ∆H, ∆U, q, and w.Calculate the quantities ∆H, ∆U, q, and w for the following processes. Be sure to give signs!Gases may be considered ideal, and the volumes of liquids and solids (but not gases) may beignored. Thermodynamic data are in Appendix A1.2.initial state final state(1) 0.3 N2 (g), 300 K, 1 bar → 0.3 N2 (g), 300 K, 0.1 barif this expansion occurs against a constant external pressure of 0.05 bar.w = -pext*∆V Vi = 0.0075 m3Vf = 0.075 m3w = - (.05*10^5) * 0.067 = - 340 J q = ∆U - w = +340 J∆U=0 (isothermal, ideal gas; U for an ideal gas depends only on T)∆H = ∆U + ∆(pV); pV = nRT (ideal gas);∆(pV) = nR(∆T) = 0 (isothermal, ideal gas) ∆H = ∆U = 0initial state final state(2) 0.3 N2 (g), 300 K, 1 bar → 0.3 N2 (g), 300 K, 0.1 barif this expansion occurs reversibly, with pext = pgas.w = - nRT ln(Vf/Vi) (see p. 45)CO2(g)CO2(s)∆H∆HC(gr) + 2S(s)CS2(l)∆depH∆fHw = - 1720 J q = ∆U - w = +1720 J; ∆U = 0 (isothermal, ideal gas)∆H = ∆U + ∆(pV) = 0 (isothermal, ideal gas)initial state final state(3) 0.3 N2 (g), 300 K, 1 bar → 0.3 N2 (g), 300 K, 0.0 barexpansion into a vacuum, pext = 0.w=0, q = 0, ∆U=0, ∆H=0initial state final state(4) 1 g Au(s), 273 K, 1 bar → 1 g Au(s), 373 K, 1 barwith pext = 1 bar.w = -pext∆V = 0; (no gases involved, volume changes are negligible)∆H = qp = n*Cp,m*(∆T) = (1/197)*25.42*100 = 12.9 JCp,m = 25.42 J-mol-1-K-1(Appendix A1.2)∆U = ∆H - ∆(pV) = ∆H = 12.9 J (ignore volume changes)initial state final state(5) CaO(s) + CO2(g) → CaCO3(s)with all reactants and products at 298 K, 1 bar.∆H° = ∆fH°(CaCO3(s)) - ∆fH°(CaO(s)) - ∆fH°(CO2(g))∆H° = (-1206.9) - (-635.1) - (-393.5)= -178.3 kJ ∆H°=qp = -178.3 kJw = -pext∆V = - (1.00*105 Pa) * (-n*R*T/pgas) = +2480 J (neg ∆V gives pos w)∆U = q + w = -178.3 + 2.5 = -175.8 kJinitial state final state(6) 10 g H2O, solid, 250 K, 1 bar → 10 g H2O, liquid, 350 K, 1 bar∆H1 ∆H2∆H33 steps: (solid at 250K → solid at 273 K→ liquid at 273 K → liquid at 350K)∆H1 = n * Cp,m(H20(s)) * ∆T = (10/18) * 37 * (273-250) = +473 J (Table 2.1)∆H2 = n * Cp,m(H20(s)) * ∆T = (10/18) * 6010 = +3340 J (Table 2.2)∆H3 = n * Cp,m(H20(l)) * ∆T = (10/18) * 75.3 * (350-273) = +3220 J (Table 2.1)∆H = q = +7030 kJ ∆U = ∆H (no gases involved) w = 0 (no gases involved)IV. Problems1. 0.850 mol of an ideal gas initially at a pressure of 15 atm and 300 K is allowed to expandisothermally until its final pressure is 2 atm. Calculate the wotk done if the expansion iscarried out against a constant external pressure of 1 atm.n = 0.85 p1 = 15 atm T = const = 300 Kp2 = 2 atm pext = 1 atm = 101325 Pap1V1 = nRT à V1 = nRT/p1 = 0.85x0.082x300/15 L = 1.39 Lat const temp. p1V1 = p2V2 à V2 = p1V1/p2 = 15x1.39/2 L = 10 46 Lw = -pext∆V = -101325 Pa (10.46 - 1.39)x10-3m3 = -919 J2. What is the specific heat capacity of a compound that rose in temperature by 2.54oC whenits sample of a mass 5 g was supplied with 346 J of heat.C = q/(m∆T) = 346J/(5g x 2.54K) = 27.24 J/(g K)3. When 0.9862 g of benzoic acid was burned in a bomb calorimeter, the temperature rosefrom 21.84oC to 25.67oC. What is the heat capacity of the calorimeter?In a separate experiment a quantity of 0.4654 g of glucose was burned in the same calorimeterand the temperature rose from 21.22 oC to 22.28oC. Calculate the enthalpy of combustion ofglucose.Information from the tables:Benzoic acid: C6H5COOH, Mw = 122.13 g/mol, ∆combHo = -3227 kJ/molGlucose: C6H12O6, Mw = 180.16 g/moln(benzoic acid) = 0.9862g/122.13 g/mol = 0.0081 molheat supplied to the calorimeter q = 0.0081mol x 3227kJ/mol = 26.06 kJC = q/∆T = 26.06J/(25.67 - 21.84)K = 6.8 kJ/Kn(glucose) = 0.4654g/180.16g/mol = 0.0026molheat evolved by cobustion of glucose sample q = C∆T = 6.8 kJ/K x (22.28 - 21.22)K =7.21 kJ∆combHo(glucose) = -q/n = -7.21kJ/0.0026mol = -2772 kJ/mol4. From the following data, calculate the enthalpies of transformations from the more stableform to the less stable form:1. C(graphite) + O2(g) à CO2(g) ∆rxnHo1 = -393.51 kJ2. C(diamond) + O2(g) à CO2(g) ∆rxnHo2 = -395.41 kJ3 = 2 - 1 : C(d) à C(gr)∆rxnH3 = ∆rxnH2 - ∆rxnH1 = -395.41kJ -(-393.51)kJ = -1.9 kJ5. The enthalpies of hydrogenation of ethylene and benzene have been determined at 298 K:C2H4(g) + H2(g) à C2H6(g) ∆rxnHo = -132 kJC6H6(g) + 3H2(g) à C6H12(g) ∆rxnHo = -246 kJWhat would be the enthalpy of hydrogenation of benzene if it contained three isolated,unconjugated double bonds? How would you account for the difference between the calculatedvalue based on this assumption and the measured value?The enthalpy of hydrogenation would be -396 kJ. The difference between the measuredand calculated value can be explained as follows: in hydrogenation of a double bond theresultant heat is the difference between the energy required to break C=C bond and formtwo C-H bonds. This difference will be greater negative value for 3 separate double bondsthan for benzene due to the delocalization energy.6. Metabolic activity in the human body releases about 1x104 kJ of heat per day. Assumingthat the body is 50 kg of water, what temperature rise would the body experience if it were anisolated system? How much water must the body eliminate as perspiration to maintain thenormal body temperature (98.6oF). The heat of vaporization of water may be taken as 2.41 kJ/gand the specific heat capacity as 4.18 J/gK.∆H = 1x104kJ = 1x107J = mC∆T à ∆T = 1x107J/(5x104g x 4.18J/gK) =47.85K∆H = m x ∆vapH à m =
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