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
Trang 1The 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 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
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
(b) ∆U = ∆H = 0, w = -1.13 kJ, q = + 1.13 kJ;
(c) ∆U = ∆H = 0, w = 0, q = 0
2.4 A sample consisting of 1.00 mol of perfect gas atoms, for which Cv,m=23 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
p 2 = 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 dm3 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 dm3 (b) Calculate the work that would be done if the same expansion occurred reversibly
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
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
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
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
Trang 22.10 A sample of carbon dioxide of mass 2.45 g at 27.0°C is allowed to expand reversibly and adiabatically from 500 cm3 to 3.00 dm3 What is the work done by the gas?
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
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
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 O2 at constant pressure is 29.4 J K-1mol-1, calculate q, ∆H, and ∆U
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.)
2.15 A sample consisting of 1.0 mol of perfect gas molecules with Cv = 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: V f = 0.0113 m 3 , T f = 344 K, w = 7.1x10 2 J
2.16 A certain liquid has ∆H0
vap = 26.0 kJmol-1 Calculate q, w, ∆H, and ∆U when 0.50 mol is vaporized at 250 K and 750 Torr
2.17 The standard enthalpy of formation of ethylbenzene is -12.5 kJmol-1 Calculate its standard enthalpy of combustion
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 CO2(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
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, C6H5OH(s), is burned in the calorimeter under the same conditions?
c= -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
2.21 The standard enthalpy of decomposition of the yellow complex H3NSO2 into NH3 and SO2 is +40 kJmol-1, Calculate the standard enthalpy of formation of H3NSO2
Trang 3Solution: -383 kJmol -1
2.22 Given the reactions (1) and (2) below, determine (a) ∆H 0
r and ∆U0
r for reaction (3), (b) ∆H0
f for both HCl(g) and H2O(g) all at 298 K
(1) H2(g) + Cl2(g) → 2HCl(g) ∆H0
r = -184.62 kJmol-1 (2) 2H2(g) + O2(g) → 2H2O(g) ∆H0
r = -483.64 kJmol-1 (3) 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)
r = -114.40 kJ.mol -1 , ∆U r = -111.92 kJ.mol -1 ,
f (HCl,g) = -92.31 kJ.mol -1 , ∆H0
f (H 2 O,g) = -241.82 kJ.mol -1
2.23 For the reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g),
∆U0
r = -1373 kJmol-1 at 298 K Calculate ∆H0
r
2.24 Calculate the standard enthalpies of formation of (a) KClO3(s) from the enthalpy
of formation of KCl, (b) NaHCO3(s) from the enthalpies of formation of CO2 and NaOH together with the following information:
2KClO3(s) → 2KCl(s) + 3O2(g) ∆H0
r = -89.4 kJmol-1 NaOH(s) + CO2(g) → NaHCO3(s) ∆H0
r = -127.5 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
2.26 From the data in Table 2.5, calculate ∆H0
r and ∆U0
r 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) ∆H0
r (298 K) =+131.29 kJmol -1 , ∆U0
r (298 K) =+128.81 kJmol -1 ,
r (378 K) =+132.56 kJmol -1 , ∆U0
r (378 K) =+129.42 kJmol -1
2.27 Calculate ∆H0
r for the reaction Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) from the information in Table 2.7 in the Data section
-218.66 kJmol -1
7 kJ.mol-1, ∆Ho
f (metallocene, 583 K) = +116.0 kJ.mol-1
2.1 Công mà người đó thực hiện là: A=F.s
Với: F= m.a
Trên mặt đất gia tốc bằng 10: A= 65x10x4=2600 (j)
Trên mặt trăng: A= 65x4x1.6= 416 (j)
Lời giải:
Wtđ = mgtđh = 65 10 4 = 2,6 103 J
Wtr = mgtrh = 65 1,6 4 = 416 J
2.2 W = -PΔVV = -1 100 10 = -1 103 (ml.atm) = -1 103 10-6 105 = -1,0 102 J
2.3 (a): Quá trình thuận ngịch:
Vì quá trình đẳng nhiệt nên ∆U = ∆H = 0
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 bằng áp suất cuối, không hiểu, làm không giống kết quả)
(c): Vì quá trình đẳng nhiệt nên ∆U = ∆H = 0
Trang 4Theo giả thiết với áp suất bằng 0 nên : P= 0, suy ra A=-Q=0.
Lời giải:
(a) Vì quá trình giản nở đẳng nhiệt T = const, nên dT = 0
dU = CvdT = 0, dH = CpdT = 0
→ ∆U = ∆H = 0,
dw = -pdV = -nRTln
1
2
V
V
= -1.8,31.273.ln 2244,,48 = -1,572 J = -1,57 kJ,
q = -w = + 1.57 kJ;
(b) Vì quá trình giản nở đẳng nhiệt T = const, nên dT = 0
dU = CvdT = 0, dH = CpdT = 0
→ ∆U = ∆H = 0,
Ta có P1 = 1 atm (vì n = 1 mol, V1 = 22,4 lít, T = 273K Vì T = const nên
PV = const nên khi V2 = 2V1 thì 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 quá trình giản nở đẳng nhiệt T = const, nên dT = 0
dU = CvdT = 0, dH = CpdT = 0
→ ∆U = ∆H = 0,
Vì P ngoài = 0 nên w = P(V2 – V1) = 0, q = w = 0
2.4
-Vì là quá trình đẳng tích nên A = 0
- Ta có : P2xT1=P1xT2, suy ra: p2= 1x400/300 = 1.33 (at)
- Với quá trình đẳng tích ∆U= Q = Cv(T2-T1) = 3/2x8.314x(400 – 300) = 1247.1 (j)
Lời giải:
Vì V = const nên TP = const → P2 = P1
1
2
V
V
= 1 300400 = 1,33 atm
∆U = n Cv,m(T2 – T1) = 23 R.100 = 23 8,31.100 = 1245 J = 1,25 kJ,
Vì V = const → w =0, q = ∆U = 1,25 kJ
2.5 a) 4 4,5 0, 28125
16
CH
n mol Mà 1 Torr = 133,32 N/m2
W = -PΔVV= -200 133,32 3,3.10-3 = -88 J
b) W = -nRTln 2
1
V
V = -2,8125 8,314 310.ln3,3+12,7= -167J
12,7 2.6 a) ΔVH= -40,656 kJ = qMà 1 atm = 101,33 kPa
W = -PΔVV= -1 101,33 (-30.10-3) = 3,1 kJ
ΔVH= q + W= -40,656+3,1= -37,556 kJ
Trang 52.7 W = -nRT = -15
24 8,314 (25+273) = -1,5 kJ 2.8 a) 0 0 473
473 298
298
298
ΔVU q W 28,3 1, 45 26,8 kJ
b) W = -PΔVV= 0
ΔVU q W q 26,8kJ
1 1 2 2
TV const T V T V
1 5/3 1
1 1 2
2
T V 273,15.1
Đơn nguyên tử: γ = 5/3
2.10 ΔVU 2, 45.28,796 179,8 300 193
44
Với
1 9/7 1
1 9/7 1
1 1 2
2
T V 300.0,5
179,8
2.11 PVγ = const P1 1V P2V2
1,4
1 1 2 2
V 57,4.1
22
P
2.12 Lời giải:
Cp,m = 229 30 /
3.2,55
H
J k mol
n T
2.13
ΔVUp ΔVHp nCp T 3.29, 4 285 260 2, 2kJ
ΔVU q W q ΔVU W
Với W = -PΔVV= -3,25 101325 1,89.10-3 = -622J q = 2,2 + 0,622 = 2,8 kJ
2.14 Solution: q = 0, w = -3,2 kJ, ΔVU = -3,2 kJ, ΔVT = -38 K, ΔVH = -4,5 kJ
Vì đây là quá trình giãn đoạn nhiệt thuận nghịch nên q = 0
Công chống áp xuất ngoài là:
ng
600
p V atm 3.20 20 dm 101,3 3, 2 kJ
760
U = q + = -3,2 kJ
= vCvT = n(Cp – R)T
Trang 6Ta có: Đối với O2: Cp,m=29.355 J K-1 mol-1
=>
3
p
n(C - R) 4(29,355-8,314)
H = U + nRT = -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 quá trình giãn đoạn nhiệt thuận nghịch:
1 1 2 2
P T = P T vì vậy
1 γ 1
2
P
V = V
P
C C C 20,8
1
1
1
nRT 1.0,082.310
=>
1 1,40 2
3,25
V = 7,83 = 9,44 (l)
2,50
Vì vậy 2 2 2
P V 2,5.9,44
nR 1.0,082
Công đoạn nhiệt thuận nghịch là
= nCvT = 1.20,8.(288 – 310) = -466 J
2.16 Solution: q = 13,0 kJ, w = -1,0 kJ, ΔVU = 12,0 kJ
ΔVH0
vap = 26,0 kJmol-1
Đối với 0,5 mol chất thì H = q = 0,5 ΔVH0
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 +
=> = U – q = 12 – 13 = -1kJ
2.17 Solution: -4564,7 kJ mol-1
Theo đề ta có: 0
sn 8 10
Phản ứng đốt cháy etylbenzen:
C8H10 + 21
2 O2 8CO2 + 5H2O Với: 0 2
CO
ΔVH = -393,51 kJmol-1; 02
H O
ΔVH = -285,83 kJmol-1
Áp dụng định luật Hess ta có:
21
2
Entanpy tiêu chuẩn của quá trình đốt cháy C8H10:
= 8.(-393,51) + 5.(-285,83) - (-12,5) = -4564,73 (kJmol-1)
2.18 Solution: ΔVHf [(CH2)3, g] = + 53 kJmol-1, ΔVH = -33 kJmol-1
Quá trình đốt cháy cyclopropan:
Trang 7C3H6 + 9
2O2 3CO2 + 3H2O Với: 0 2
CO
ΔVH = -393,51 kJmol-1; 02
H O
ΔVH = -285,83 kJmol-1
Áp dụng định luật Hess ta có:
9
2
CH 2 3 CO 2 H O 2 pu
= 3 (-393,51) + 3.(-285,83) - (-2091) = 53 kJmol-1
Hình thành propen từ xyclopropanCH 23 CH 3 CH CH 3
pu propen xyclopropane
2.19: C10 H 8 + 12O 2 4H 2 O + 10CO 2 ΔVH = ?
ΔVH C10H8 = 4ΔVH
H2O + 10ΔVH
CO2 - ΔVH
C10H8 - 12ΔVH
O2
= 4(-285,83 ) + 10 ( 393,51) – 78,53 – 12x0 = - 5156,95 kj/mol Nhiệt lượng tỏa ra khi đốt cháy C 10 H 8 là
q p = n ΔVH = (120.10 -3 /128) x ( - 5156,95)= - 4,828 kj
Nhiệt kế đã hấp thu lượng nhiệt là q = 4,828 kj
Hằng số nhiệt độ của nhiệt kế là
C = q/ ΔVT C = 4,828/ 3,05 = 1,58 kj/k
Trong trường hợp củng dùng nhiệt kế đó để đốt phenol ta có
C 6 H 5 OH + 7O 2 3H 2 O + 6CO 2 ΔVH
2 = ? ΔVH 2 = 3ΔVH
H2O + 6ΔVH
CO2 - ΔVH
C6H5OH - 7ΔVH
O2
= 3 ( -285,83) + 6 ( -393,51) – ( -165) – 7 x0 = 3053,55 kj/mol
vậy nhiệt lượng sinh ra trong quá trình đốt cháy là
q p = n ΔVH
2 = (10.10-3/94)x (-3053.55) = - 0,25 kj Nhiệt kế đã hấp thu lượng nhiệt là q = 0,25 kj
Mà q = C x ΔVT ΔVT = q/C = 0,25/1,58 = 0,205 K
2.20: AgClr Ag +
aq + Cl
-aq ΔVH = ? ΔVH = ΔVH Ag +
aq + ΔVH Cl
-aq - ΔVH AgCl r
= 105,9 - 167,44 + 127,03 = 65,49 kj/mol
2.21: Ta có : H3 NSO 2 NH 3 + SO 2 ΔVH
1 = 40 kj/mol
Ta có 3/2 H 2 (k) + 1/2N 2(k) + S (r) + O 2(k) ΔVH H 3 NSO 2
ΔVH NH3 ΔVH
SO2 ΔVH
1
NH 3 SO 2
ΔVH = ΔVH
NH3 + ΔVH
SO2 - ΔVH 1 = -46,11 – 296,83 – 40 = - 382,94 kj/mol
2.22: a) (1) H2(g) + Cl 2(g) 2HCl (g) ΔVH
1 = -184,62 kj/mol
Trang 8(2) 2H 2(g) + O 2(g) 2H 2 O (g) ΔVH
2 = -483,64 kj/mol (3) 4HCl (g) + O 2(g) 2Cl 2(g) + 2H 2 O (g) ΔVH
3 = ?
Ta có ΔVH 3 = ΔVH
2 - 2 ΔVH
1 = - 483,64 – 2( -184,62) = - 114,4 kj/mol ADCT: ΔVU = ΔVH -P ΔVV
= ΔVH – ΔVnRT (ΔVn = n(sau) - n(trước))
= - 114,4 –(-1)x8,34x298.10 -3
= 111,92 kj/mol
b) Tính ΔVH (s)HCl g = ½ ΔVH
1 = -92,31 kj/mol ΔVH (s)H2O g = ½ ΔVH
2 = -241,82 kj/mol 2.23 For the reaction C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(g),
∆U 0
r = -1373 kJmol -1 at 298 K Calculate ∆H 0
r
ADCT: ΔVU = ΔVH - ΔVnRT
ΔVH= ΔVU + ΔVnRT = -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) NaHCO 3 (s) from the enthalpies of formation of CO 2 and NaOH together with the following information:
2KClO 3 (s) → 2KCl(s) + 3O 2 (g) (a) ∆H 0
r = -89.4 kJmol -1
NaOH(s) + CO 2 (g) → NaHCO 3 (s) (b) ∆H 0
r = -127.5 kJmol -1
KClO 3 (s) → KCl(s) + 3/2O 2 (g) (a) ∆H 0
r = -89.4/2kJmol -1
K + 1/2Cl2 + 3/2 O 2
∆H 1 = - ∆H 0
r + ΔVH 2 = -ΔVH r0 + ΔVH KCl0 + ΔVH O20 = 89.4/2 - 436,75 = -392,05 kJ mol -1
NaOH(s) + CO 2 (g) → NaHCO 3 (s) (b) ∆H 0
r = -127.5 kJmol -1
Na + 1/2C + 3/2 O 2 + 1/2H 2
∆H NaHCO3(s) =ΔVH 2 = ∆H 0
r + ΔVH 1
= ΔVH r0 + ΔVH NaOH0 + ΔVH CO20
= -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) → N 2 O 4 (g)
at l00°C from its value at 25°C.
∆H r 0 = ΔVH N2O40 - 2∆H NO20 = 9,16 – 2.33,18 = -57,2 kJmol -1
∆H r tại 100 o C :
Định luật Kirchhoff: ∆H 0
r (T 2 )= ∆H o
r (T1) +∆C p∆T (∆C p = ∆C pN2O4 - 2∆C pNO2 )
=-57,2 + (77,28- 2.37,2).(373-298) = -56.98 kJmol -1.
∆H0r
Trang 92.26 From the data in Table 2.5, calculate ∆H 0
r and ∆U 0
r at (a) 298 K, (b) 378 K for the reaction C(graphite) + H 2 O(g) → CO(g) + H 2 (g) Assume all heat capacities to be constant over the temperature range of interest.
(a) ∆H0
r (298 K) =+131.29 kJmol -1 , ∆U0
r (298 K) =+128.81 kJmol -1 ,
r (378 K) =+132.56 kJmol -1 , ∆U0
r (378 K) =+129.42 kJmol -1
ΔHH f ° C(graphite)= 0 ΔHH f ° H2O(g) = -241.82 kJ mol-1 ΔHHf° CO(g) = -110.53 kJ mol-1
ΔHHf°H2 (g)= 0 C° pC(graphite) = 8.527 J K-1mol-1 C° p H2O(g) = 33.58 J K-1mol-1
C°p COg = 29.14 J K-1mol-1 C° pH2 (g) = 28.824 J K-1mol-1
(a) Tại 298 K ∆H° r = ΔHH f ° COg - ΔHH f ° H2Og => ∆H° r = -110.53 - (-241.82)
=> ∆U° r =∆H° r - ∆nRT => ∆U° r = 131.29 - (8.314×10-3) (298 K)
=> ∆U r °= 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)∆C r °p
∆C r °p = C°p COg + C° p H2 g - C°p Cgraphite +C° pH2Og
∆C r °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
=> ∆H r ° 378 K =131.29 + 80 1.5857 ×10 -2
=> ∆H r ° 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
trong phần dữ liệu -218.66 kJmol -1
Giải:
ΔVH pư = H 298, Zn2+ + H 298, Cu - H 298, Zn - H 298, Cu2+ = -153,89 + 0 – 64,77 – 0 = - 218,66 (kJ/mol)