# bài tập nhiệt động lực học giải chi tiết

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2.1. Calculate the work needed for a 65 kg person to climb through 4.0 m on the surface of (a) the Earth and (b) the Moon (g= 1.60 ms2).Solution: on earth, 2.6 x 103 J, on the moon, 4.2 x 102 J.2.2. A chemical reaction takes place in a container of crosssectional area 100 cm2. As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 atm. Calculate the work done by the system.Solution: 1.0 x 102 J The First Law Exercises Assume all gases are perfect unless stated otherwise Unless otherwise stated, thermochemical data are for 298.15K 2.1 Calculate the work needed for a 65 kg person to climb through 4.0 m on the surface of (a) the Earth and (b) the Moon (g= 1.60 ms-2) Solution: on earth, 2.6 x 103 J, on the moon, 4.2 x 102 J 2.2 A chemical reaction takes place in a container of cross-sectional area 100 cm As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 atm Calculate the work done by the system Solution: -1.0 x 102 J 2.3 (a) A sample consisting of 1.00 mol Ar is expanded isothermally at 0°C from 22.4 dm3 to 44.8 dm3 (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely (against zero external pressure) For the three processes calculate q, w, ∆U, and ∆H Solution: (a) ∆U = ∆H = 0, w = -1.57 kJ, q = + 1.57 kJ; (b) ∆U = ∆H = 0, w = -1.13 kJ, q = + 1.13 kJ; (c) ∆U = ∆H = 0, w = 0, q = 2.4 A sample consisting of 1.00 mol of perfect gas atoms, for which C v,m= R, initially at p1 = 1.00 atm and T1 = 300 K, is heated reversibly to 400 K at constant volume Calculate the final pressure, ∆U, q, and w p2= 1.33 atm, ∆U = + 1.25 kJ, w =0, q = + 1.25 kJ 2.5 A sample of 4.50 g of methane occupies 12.7 dm at 310 K (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increased by 3.3 dm (b) Calculate the work that would be done if the same expansion occurred reversibly Solution: (a) -88 J; (b) -167 J, 2.6 A sample of 1.00 mol H2O(g) is condensed isothermally and reversibly to liquid water at 100°C The standard enthalpy of vaporization of water at 100°C is 40.656 kJ.mol-1 Find w, q, ∆U, and ∆H for this process Solution: ∆H= -40.656 kJ, q = -40.656 kJ, w = +3.10 kJ, ∆U= -37.55 kJ 2.7 A strip of magnesium of mass 15 g is dropped into a beaker of dilute hydrochloric acid Calculate the work done by the system as a result of the reaction The atmospheric pressure is 1.0 atm and the temperature 25°C Solution: -1.5 kJ 2.8 The constant-pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression Cp (J K -1) = 20.17 + 0.3665(T/K) Calculate q, w, ∆U, and ∆H when the temperature is raised from 25°C to 200°C (a) at constant pressure, (b) at constant volume Solution: (a) q=∆H=+28.3kJ,w=-1.45kJ, ∆U=+26.8kJ; (b) ∆H=+28.3 kJ, ∆U=+26.8 kJ, w= 0, q =+26.8 kJ 2.9 Calculate the final temperature of a sample of argon of mass 12.0 g that is expanded reversibly and adiabatically from 1.0 dm3 at 273.15 K to 3.0 dm3 Solution: 131 K 2.10 A sample of carbon dioxide of mass 2.45 g at 27.0°C is allowed to expand reversibly and adiabatically from 500 cm to 3.00 dm3 What is the work done by the gas? Solution: -194J 2.11 Calculate the final pressure of a sample of carbon dioxide that expands reversibly and adiabatically from 57.4 kPa and 1.0 dm3 to a final volume of 2.0 dm3, Take γ= 1.4 Solution: 22 kPa 2.12 When 229 J of energy is supplied as heat to 3.0 mol Ar(g), the temperature of the sample increases by 2.55 K Calculate the molar heat capacities at constant volume and constant pressure of the gas Solution: Cp,m = 30 J K-1mol-1, Cv,m = 22 J K-1mol-1 2.13 When 3.0 mol O2 is heated at a constant pressure of 3.25 atm, its temperature increases from 260 K to 285 K Given that the molar heat capacity of O at constant pressure is 29.4 J K-1mol-1, calculate q, ∆H, and ∆U Solution: qp= +2.2 kJ, ∆H= +2.2 kJ, ∆U = + 1.6 kJ 2.14 A sample of 4.0 mol O2 is originally confined in 20 dm3 at 270 K and then undergoes adiabatic expansion against a constant pressure of 600 Torr until the volume has increased by a factor of 3.0 Calculate q, w, ∆T, ∆U, and ∆H (The final pressure of the gas is not necessarily 600 Torr.) Solution: q = 0, w=-3.2 kJ, ∆U=-3.2 kJ, ∆T=-38 K, ∆H= -4.5 kJ 2.15 A sample consisting of 1.0 mol of perfect gas molecules with C v = 20.8 J K-1 is initially at 3.25 atm and 310 K It undergoes reversible adiabatic expansion until its pressure reaches 2.50 atm Calculate the final volume and temperature and the work done Solution: Vf = 0.0113 m3, Tf = 344 K, w = 7.1x102 J 2.16 A certain liquid has ∆H vap = 26.0 kJmol-1 Calculate q, w, ∆H, and ∆U when 0.50 mol is vaporized at 250 K and 750 Torr Solution: q = 13.0 kJ, w = -1.0 kJ, ∆U = 12.0 kJ 2.17 The standard enthalpy of formation of ethylbenzene is -12.5 kJmol -1 Calculate its standard enthalpy of combustion Solution: -4564.7 kJ mol-1 2.18 The standard enthalpy of combustion of cyclopropane is -2091 kJ.mol -1 at 25°C From this information and enthalpy of formation data for CO 2(g) and H2O(g), calculate the enthalpy of formation of cyclopropane The enthalpy of formation of propene is +20.42 kJmol-1 Calculate the enthalpy of isomerization of cyclopropane to propene Solution: ∆Hf[(CH2)3,g] =+53 kJmol-1, ∆H = -33 kJmol-1 2.19 When 120 mg of naphthalene, C10H8(s), was burned in a bomb calorimeter the temperature rose by 3.05 K Calculate the calorimeter constant By how much will the temperature rise when 10 mg of phenol, C 6H5OH(s), is burned in the calorimeter under the same conditions? Solution: ∆U 0c = -5152 kJ rnol-1, C = 1.58 kJ K-l, ∆T= +0.205 K 2.20 Calculate the standard enthalpy of solution of AgCl(s) in water from the enthalpies of formation of the solid and the aqueous ions Solution: 65.49 kJ.mol-1 2.21 The standard enthalpy of decomposition of the yellow complex H 3NSO2 into NH3 and SO2 is +40 kJmol-1, Calculate the standard enthalpy of formation of H3NSO2 Solution: -383 kJmol-1 2.22 Given the reactions (1) and (2) below, determine (a) ∆H 0r and ∆U 0r for reaction (3), (b) ∆H 0f for both HCl(g) and H2O(g) all at 298 K (1) H2(g) + Cl2(g) → 2HCl(g) ∆H 0r = -184.62 kJmol-1 (2) 2H2(g) + O2(g) → 2H2O(g) ∆H 0r = -483.64 kJmol-1 (3) 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) Solution: (a) ∆H 0r = -114.40 kJ.mol-1, ∆Ur = -111.92 kJ.mol-1, (b) ∆H 0f (HCl,g) = -92.31 kJ.mol-1, ∆H 0f (H2O,g) = -241.82 kJ.mol-1 2.23 For the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g), ∆U 0r = -1373 kJmol-1 at 298 K Calculate ∆H 0r Solution: -1368 kJ.mol-1 2.24 Calculate the standard enthalpies of formation of (a) KClO 3(s) from the enthalpy of formation of KCl, (b) NaHCO3(s) from the enthalpies of formation of CO and NaOH together with the following information: 2KClO3(s) → 2KCl(s) + 3O2(g) ∆H 0r = -89.4 kJmol-1 NaOH(s) + CO2(g) → NaHCO3(s) ∆H 0r = -127.5 kJmol-1 Solution: (a) -392.1 kJ mol-1; (b) -946.6 kJmol-1 2.25 Use the information in Table 2.5 to predict the standard reaction enthalpy of 2NO2(g) → N2O4(g) at l00°C from its value at 25°C Solution: -56.98 kJmol-1 2.26 From the data in Table 2.5, calculate ∆H 0r and ∆U 0r at (a) 298 K, (b) 378 K for the reaction C(graphite) + H2O(g) → CO(g) + H2(g) Assume all heat capacities to be constant over the temperature range of interest (a) ∆H 0r (298 K) =+131.29 kJmol-1, ∆U 0r (298 K) =+128.81 kJmol-1, (b) ∆H 0r (378 K) =+132.56 kJmol-1, ∆U 0r (378 K) =+129.42 kJmol-1 2.27 Calculate ∆H 0r for the reaction Zn(s) + CuSO 4(aq) → ZnSO4(aq) + Cu(s) from the information in Table 2.7 in the Data section -218.66 kJmol-1 kJ.mol-1, ∆H of (metallocene, 583 K) = +116.0 kJ.mol-1 2.1 Công mà người thực là: A=F.s Với: F= m.a Trên mặt đất gia tốc 10: A= 65x10x4=2600 (j) Trên mặt trăng: A= 65x4x1.6= 416 (j) Lời giải: Wtđ = mgtđh = 65 10 = 2,6 103 J Wtr = mgtrh = 65 1,6 = 416 J 2.2 W = -P ΔV = -1 100 10 = -1 103 (ml.atm) = -1 103 10-6 105 = -1,0 102 J 2.3 (a): Quá trình thuận ngịch: Vì trình đẳng nhiệt nên ∆U = ∆H = A= -pdV= - (nRT/V)dV = nRTln(V1/V2) = 1x273x8.314xln(22.4/44.8) = 1572.9 (j) Vì ∆U = 0, Q= -A = -1572.9 (j) (b): (Áp suất đầu áp suất cuối, không hiểu, làm không giống kết quả) (c): Vì trình đẳng nhiệt nên ∆U = ∆H = Theo giả thiết với áp suất nên : P= 0, suy A=-Q=0 Lời giải: (a) Vì trình giản nở đẳng nhiệt T = const, nên dT = dU = CvdT = 0, dH = CpdT = → ∆U = ∆H = 0, V2 44,8 dw = -pdV = -nRTln V = -1.8,31.273.ln 22,4 = -1,572 J = -1,57 kJ, q = -w = + 1.57 kJ; (b) Vì trình giản nở đẳng nhiệt T = const, nên dT = dU = CvdT = 0, dH = CpdT = → ∆U = ∆H = 0, Ta có P1 = atm (vì n = mol, V1 = 22,4 lít, T = 273K Vì T = const nên PV = const nên V2 = 2V1 P2 = 0,5 P1 = 0,5 atm w = -P2(V2 – V1) = 0,5.22,4 = 11,2 l.atm = 11,2.101 = -1131 J -1.13 kJ, q = -w = + 1.13 kJ; (c) Vì trình giản nở đẳng nhiệt T = const, nên dT = dU = CvdT = 0, dH = CpdT = → ∆U = ∆H = 0, Vì P = nên w = P(V2 – V1) = 0, q = w = 2.4 -Vì trình đẳng tích nên A = - Ta có : P2xT1=P1xT2, suy ra: p2= 1x400/300 = 1.33 (at) - Với trình đẳng tích ∆U= Q = Cv(T2-T1) = 3/2x8.314x(400 – 300) = 1247.1 (j) Lời giải: Vì V = const nên V2 P 400 = const → P2 = P1 = = 1,33 atm V1 T 300 ∆U = n Cv,m(T2 – T1) = 3 R.100 = 8,31.100 = 1245 J = 1,25 kJ, 2 Vì V = const → w =0, q = ∆U = 1,25 kJ 2.5 a) nCH = 4,5 = 0, 28125mol Mà Torr = 133,32 N/m2 16 W = -P ΔV = -200 133,32 3,3.10-3 = -88 J V2 3,3+12,7 b) W = -nRTln V = -2,8125 8,314 310.ln 12,7 = -167J 2.6 a) ΔH= -40,656 kJ = q Mà atm = 101,33 kPa W = -P ΔV = -1 101,33 (-30.10-3) = 3,1 kJ ΔH= q + W= -40,656+3,1= -37,556 kJ 15 8,314 (25+273) = -1,5 kJ 24 2.7 W = -nRT = - 2.8 a) ΔH 0473 = ΔH 0298 + 473 ∫ ∆CpdT 298 473 ΔH 0473 = ∫ ( 20,17 + 0,3665T ) dT = 20,17T 298 473 473 298 298 I + 0,3665T / I ≈ 28,3kJ ΔH = q ΔU = q + W = 28,3 − 1, 45 = 26,8kJ b) W = -P ΔV = ΔU = q + W = q = 26,8kJ γ −1 γ −1 2.9 TVγ −1 = const → T1V1 = T2V2 γ −1 → T2 = 5/3−1 T1V1 = γ −1 V2 273,15.1 = 131K 5/3−1 Đơn nguyên tử: γ = 5/3 2.10 W = ΔU = nCv∆T = γ −1 T1V1 Với T2 = V γ −1 = 2, 45 28, 796 ( 179,8 − 300 ) ≈ −193 J 44 300.0,5 9/7−1 9/7−1 = 179,8 K γ γ γ 2.11 PV = const  → P1V1 = P2 V2 γ → P2 = P1V1 γ V2 = 57,4.11,4 1,4 = 22kPa 2.12 Lời giải: Cp,m = ∆H 229 = = 30 J / k mol n∆T 3.2,55 Cv , m = C p ,m − R = 30 − 8,314 = 21, 686 J / k mol 2.13 ΔU p = ΔH p = nCp∆T = 3.29, ( 285 − 260 ) ≈ 2, 2kJ ΔU = q + W → q = ΔU − W -3 Với W = -P ΔV = -3,25 101325 1,89.10 = -622J ⇒ q = 2,2 + 0,622 = 2,8 kJ 2.14 Solution: q = 0, w = -3,2 kJ, ΔU = -3,2 kJ, ΔT = -38 K, ΔH = -4,5 kJ Vì trình giãn đoạn nhiệt thuận nghịch nên q = Công chống áp xuất là: ω = −p ng ∆V = − 600 atm ( 3.20 − 20 ) dm 101,3 = −3, ( kJ ) 760 ∆U = q + ω = -3,2 kJ ω = vCv∆T = n(Cp – R)∆T Ta có: Đối với O2: Cp,m=29.355 J K-1 mol-1 ω -3,2.103 = = -38 kJ => ΔT = n(Cp - R) 4(29,355-8,314) ∆H = ∆U + nR∆T = -3200 + 4.8,314.(-38) = - 4500J = - 4,5 kJ 2.15 Solution: Vf = 0,0094 m3, Tf = 288 K, w = -4,66x102 J Đối với trình giãn đoạn nhiệt thuận nghịch: 1γ P  P T = P T V2 = V1  ÷  P2  Cp R + Cv R 8, 314 = =1 + =1 + =1, 40 Với γ = Cv Cv Cv 20,8 nRT1 1.0,082.310 V1 = = = 7,83 (l) P1 3,25 γ 1 γ 2  3,25  1,40 => V2 = 7,83  ÷ = 9,44 (l)  2,50  P2 V2 2,5.9,44 = = 288 K Vì T2 = nR 1.0,082 Công đoạn nhiệt thuận nghịch ω = nCv∆T = 1.20,8.(288 – 310) = -466 J 2.16 Solution: q = 13,0 kJ, w = -1,0 kJ, ΔU = 12,0 kJ ΔH vap = 26,0 kJmol-1 Đối với 0,5 mol chất ∆H = q = 0,5 ΔH vap = 0,5.26 = 13 kJ Mà ∆H = ∆U + nRT => ∆U = ∆H – nRT = 13000 – 0,5.8,314.250 = 12000 J = 12 kJ ∆U = q + ω => ω = ∆U – q = 12 – 13 = -1kJ 2.17 Solution: -4564,7 kJ mol-1 Theo đề ta có: ΔHsn (C8 H10 ) = -12,5 kJmol-1 Phản ứng đốt cháy etylbenzen: C8H10 + 21 O2 → 8CO2 + 5H2O 0 Với: ΔH CO = -393,51 kJmol-1; ΔH H O = -285,83 kJmol-1 Áp dụng định luật Hess ta có: 2 21   ∆H pu = ( ∑ ν i ∆H 0r ) − ( ∑ ν i ∆H 0r ) = 8∆H CO2 + 5∆H H 2O −  ∆H C8H10 + ∆H O2 ÷ sp tg   ( ) Entanpy tiêu chuẩn trình đốt cháy C8H10: ⇒ ∆H p.u = 8∆H CO2 + 5∆H H 2O − ∆H C8H10 ( ) = 8.(-393,51) + 5.(-285,83) - (-12,5) = -4564,73 (kJmol-1) 2.18 Solution: ΔHf [(CH2)3, g] = + 53 kJmol-1, ΔH = -33 kJmol-1 Quá trình đốt cháy cyclopropan: C3H6 + O2 → 3CO2 + 3H2O 0 Với: ΔH CO = -393,51 kJmol-1; ΔH H O = -285,83 kJmol-1 Áp dụng định luật Hess ta có: 2   ∆H pu = ( ∑ ν i ∆H i0 ) − ( ∑ ν i ∆H i0 ) = 3∆H CO2 + 3∆H H 2O −  ∆H ( CH2 ) + ∆H O2 ÷ sp tg   ⇒ ∆H( CH ) = 3∆H CO2 + 3∆H H2 O − ∆H pu ( ( ) ) = (-393,51) + 3.(-285,83) - (-2091) = 53 kJmol-1 Hình thành propen từ xyclopropan ( CH ) → CH = CH − CH ∆H pu = ∆H propen − ∆H xyclopropane = 20, 42 − ( +53) = −32, ( kJmol −1 ) 2.19: C10H8 + 12O2 → 4H2O + 10CO2 ΔH° = ? ΔH°C10H8 = 4ΔH°H2O + 10ΔH°CO2 - ΔH°C10H8 - 12ΔH°O2 = 4(-285,83 ) + 10 ( 393,51) – 78,53 – 12x0 = - 5156,95 kj/mol Nhiệt lượng tỏa đốt cháy C10H8 qp = n ΔH = (120.10-3/128) x ( - 5156,95)= - 4,828 kj Nhiệt kế hấp thu lượng nhiệt q = 4,828 kj Hằng số nhiệt độ nhiệt kế C = q/ ΔT → C = 4,828/ 3,05 = 1,58 kj/k Trong trường hợp củng dùng nhiệt kế để đốt phenol ta có C6H5OH + 7O2 → 3H2O + 6CO2 ΔH°2 = ? ΔH°2 = 3ΔH°H2O + 6ΔH°CO2 - ΔH°C6H5OH - 7ΔH°O2 = ( -285,83) + ( -393,51) – ( -165) – x0 = 3053,55 kj/mol nhiệt lượng sinh trình đốt cháy qp = n ΔH°2 = (10.10-3/94)x (-3053.55) = - 0,25 kj Nhiệt kế hấp thu lượng nhiệt q = 0,25 kj Mà q = C x ΔT → ΔT = q/C = 0,25/1,58 = 0,205 K 2.20: AgClr → Ag+aq + Cl-aq ΔH° = ? ΔH° = ΔH° Ag+aq + ΔH° Cl-aq - ΔH° AgClr = 105,9 - 167,44 + 127,03 = 65,49 kj/mol 2.21: Ta có : H3NSO2 → NH3 + SO2 ΔH°1 = 40 kj/mol Ta có 3/2 H2 (k) + 1/2N2(k) + S(r) + O2(k) ΔH°NH3 NH3 ΔH° ΔH°SO2 ΔH° H3NSO2 ΔH°1 SO2 = ΔH°NH3 + ΔH°SO2 - ΔH1 = -46,11 – 296,83 – 40 = - 382,94 kj/mol 2.22: a) (1) H2(g) + Cl2(g) → 2HCl(g) ΔH°1 = -184,62 kj/mol → 2H2O(g) ΔH°2 = -483,64 kj/mol (2) 2H2(g) + O2(g) (3) 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) ΔH°3 = ? Ta có ΔH°3 = ΔH°2 - ΔH°1 = - 483,64 – 2( -184,62) = - 114,4 kj/mol ADCT: ΔU = ΔH -P ΔV = ΔH – ΔnRT (Δn = n(sau) - n(trước)) = - 114,4 –(-1)x8,34x298.10-3 = 111,92 kj/mol b) Tính ΔH°(s)HCl g = ½ ΔH°1 = -92,31 kj/mol ΔH°(s)H2O g = ½ ΔH°2 = -241,82 kj/mol 2.23 For the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g), ∆U 0r = -1373 kJmol-1 at 298 K Calculate ∆H 0r Solution: -1368 kJ.mol-1 ADCT: ΔU = ΔH - ΔnRT  ΔH= ΔU + ΔnRT = -1373 + (2+3-3).8,314.298/1000 = -1368 kJ.mol-1 2.24 Calculate the standard enthalpies of formation of (a) KClO 3(s) from the enthalpy of formation of KCl, (b) NaHCO3(s) from the enthalpies of formation of CO and NaOH together with the following information: 2KClO3(s) → 2KCl(s) + 3O2(g) (a) ∆H 0r = -89.4 kJmol-1 NaOH(s) + CO2(g) → NaHCO3(s) (b) ∆H 0r = -127.5 kJmol-1 Solution: (a) -392.1 kJ mol-1; (b) -946.6 kJmol-1 r KClO3(s) → ∆H KCl(s) + 3/2O2(g) (a) ∆H 0r = -89.4/2kJmol-1 ΔH1 ΔH2 K + 1/2Cl2 + 3/2 O2 ∆H1 = - ∆H 0r + ΔH2 = -ΔHr0 + ΔHKCl0 + ΔHO20 = 89.4/2 - 436,75 = -392,05 kJ mol-1 NaOH(s) + CO2(g) → NaHCO3(s) (b) ∆H 0r = -127.5 kJmol-1 ΔH1 ΔH2 Na + 1/2C + 3/2 O2 + 1/2H2 =ΔH2 = ∆H 0r + ΔH1 = ΔHr0 + ΔHNaOH0 + ΔHCO20 = -127,5 -425,61 -393,51 = -946.62 kJmol-1 2.25 Use the information in Table 2.5 to predict the standard reaction enthalpy of 2NO 2(g) → N2O4(g) at l00°C from its value at 25°C Solution: -56.98 kJmol-1 ∆H NaHCO3(s) ∆Hr = ΔHN2O40 - 2∆HNO20 = 9,16 – 2.33,18 = -57,2 kJmol-1 ∆Hr 100oC : Định luật Kirchhoff: ∆H0r(T2)= ∆Hor(T1) +∆Cp∆T (∆Cp = ∆CpN2O4 - 2∆CpNO2) =-57,2 + (77,28- 2.37,2).(373-298) = -56.98 kJmol-1 2.26 From the data in Table 2.5, calculate ∆H 0r and ∆U 0r at (a) 298 K, (b) 378 K for the reaction C(graphite) + H2O(g) → CO(g) + H2(g) Assume all heat capacities to be constant over the temperature range of interest (a) ∆H 0r (298 K) =+131.29 kJmol-1, ∆U 0r (298 K) =+128.81 kJmol-1, (b) ∆H 0r (378 K) =+132.56 kJmol-1, ∆U 0r (378 K) =+129.42 kJmol-1 ΔHf°C(graphite)=0 ΔHf°H2O(g)= -241.82 kJ mol-1 ΔHf°CO(g)= -110.53 kJ mol-1 ΔHf°H2 (g)= C°pC(graphite)= 8.527 J K-1mol-1 C°pCOg= 29.14 J K-1mol-1 C°p H2O(g)= 33.58 J K-1mol-1 C°pH2 (g)= 28.824 J K-1mol-1 (a) Tại 298 K ∆H°r= ΔHf°COg - ΔHf°H2Og => ∆H°r = -110.53 - (-241.82) ∆H°r =131.29 kJ mol-1 => ∆U°r=∆H°r - ∆nRT => ∆U°r = 131.29 - (8.314×10-3) (298 K) => ∆Ur°= 131.29 - 2.48 => ∆U°r = 128.81 kJ mol-1 (b) Tại 378 K ∆H°r 378 K = ∆H°r298 K+ (378 K-298 K)∆Cr°p ∆Cr°p = C°p COg + C°p H2 g- C°pCgraphite+C°pH2Og ∆Cr°p = (29.14 +28.824) - ( 8.527 + 33.58 ) => ∆C°rp = 15.857 J K-1mol-1 = 1.5857 ×10-2 kJ K-1mol-1 => ∆Hr°378 K =131.29 + 80 1.5857 ×10-2 => ∆Hr°378 K = 131.29 + 1.27 = =+132.56 kJmol-1 => ∆U°r378K = ∆H°r - ∆nRT => ∆U°r = 132.56 - (8.314×10-3) (378 K) = =+129.42 kJmol-1 2.27 Tính ΔHpư phản ứng Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s) từ thông tin bảng 2.7 phần liệu -218.66 kJmol-1 Giải: ΔHpư = ∆H298, Zn2+ + ∆H298, Cu - ∆H298, Zn - ∆H298, Cu2+ = -153,89 + – 64,77 – = - 218,66 (kJ/mol)
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