bài tập nhiệt động lực học giải chi tiết

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 2. 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 = 0.
3
2

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 3 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 3. (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.
1


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 3 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 2 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
0
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.
2


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 2 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.
7 kJ.mol-1, ∆H of (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 = -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 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.
3


Theo 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,
V2

44,8

dw = -pdV = -nRTln V = -1.8,31.273.ln 22,4 = -1,572 J = -1,57 kJ,
1

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

V2
P
400
= const → P2 = P1.
= 1.
= 1,33 atm
V1
T
300

∆U = n. Cv,m(T2 – T1) =

3
3
R.100 = .8,31.100 = 1245 J = 1,25 kJ,
2
2

Vì V = const → w =0, q = ∆U = 1,25 kJ.
2.5. a) nCH =
4

4,5
= 0, 28125mol Mà 1 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
1

= -167J

2.6. a) ΔH= -40,656 kJ = q Mà 1 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

4

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 2 / 2 I ≈ 28,3kJ

ΔH = q
ΔU = q + W = 28,3 − 1, 45 = 26,8kJ

b) W = -P ΔV = 0
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
3

= 131K

5/3−1

Đơn nguyên tử: γ = 5/3
2.10. W = ΔU = nCv∆T =
γ −1

T1V1

Với T2 = V
2

γ −1

=

2, 45
.28, 796. ( 179,8 − 300 ) ≈ −193 J
44

300.0,5
3

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
2

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, 4. ( 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 đây là quá trình giãn đoạn nhiệt thuận nghịch nên q = 0.
Công chống áp xuất ngoài là:
ω = −p ng .∆V = −

600
atm. ( 3.20 − 20 ) dm 3 .101,3 = −3, 2 ( kJ )
760

∆U = q + ω = -3,2 kJ
ω = vCv∆T = n(Cp – R)∆T

5

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 quá trình giãn đoạn nhiệt thuận nghịch:
1γ

P 
P T = P T vì vậy V2 = V1  1 ÷
 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 1

γ
2 2

1

 3,25  1,40
=> V2 = 7,83. 
÷ = 9,44 (l)
 2,50 
P2 V2 2,5.9,44
=
= 288 K
Vì vậy T2 =
nR 1.0,082
Công đoạn nhiệt thuận nghịch là
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
0
ΔH vap = 26,0 kJmol-1
0
Đối với 0,5 mol chất thì ∆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
0
Theo đề ta có: ΔHsn (C8 H10 ) = -12,5 kJmol-1
Phản ứng đốt cháy etylbenzen:
C8H10 +

21
O2 → 8CO2 + 5H2O
2

0
0
Với: ΔH CO = -393,51 kJmol-1; ΔH H O = -285,83 kJmol-1
Áp dụng định luật Hess ta có:
2

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
2



(

)

Entanpy tiêu chuẩn của quá 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:
6


C3H6 +

9
O2 → 3CO2 + 3H2O
2

0
0
Với: ΔH CO = -393,51 kJmol-1; ΔH H O = -285,83 kJmol-1
Áp dụng định luật Hess ta có:
2

2

9


∆H pu = ( ∑ ν i ∆H i0 ) − ( ∑ ν i ∆H i0 ) = 3∆H CO2 + 3∆H H 2O −  ∆H ( CH2 ) + ∆H O2 ÷
sp
tg
3
2


⇒ ∆H( CH 2 ) = 3∆H CO2 + 3∆H H2 O − ∆H pu

(

(

3

)

)

= 3. (-393,51) + 3.(-285,83) - (-2091) = 53 kJmol-1
Hình thành propen từ xyclopropan ( CH 2 ) 3 → CH 3 = CH − CH 3
∆H pu = ∆H propen − ∆H xyclopropane = 20, 42 − ( +53) = −32, 6 ( 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 ra khi đốt cháy C10H8 là
qp = n ΔH = (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/ ΔT → C = 4,828/ 3,05 = 1,58 kj/k
Trong trường hợp củng dùng nhiệt kế đó để đốt phenol ta có
C6H5OH + 7O2

→ 3H2O + 6CO2 ΔH°2 = ?

ΔH°2 = 3ΔH°H2O + 6ΔH°CO2 - ΔH°C6H5OH - 7ΔH°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
qp = n ΔH°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 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

7


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 - 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 2 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.
0
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 0 = ΔHN2O40 - 2∆HNO20 = 9,16 – 2.33,18 = -57,2 kJmol-1
∆Hr tại 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.

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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)= 0

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 của phản ứng Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s) từ các thông tin trong bảng 2.7
trong phần dữ liệu. -218.66 kJmol-1.
Giải:

ΔHpư = ∆H298, Zn2+ + ∆H298, Cu - ∆H298, Zn - ∆H298, Cu2+ = -153,89 + 0 – 64,77 – 0 = - 218,66 (kJ/mol)

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