Download Chapter 6

Chemical Thermodynamics
6.1
FIRST LAW OF THERMODYNAMICS
Section A
Level-I
1. Extensive : The variables dependent on mass.
Intensive : The variables independent of mass.
Extensive : Enthalpy, Gibbs free energy
Intensive : Density, gas constant, molarity
2. State Function : The variables defining condition of existence of system. e.g. pressure, temperature,
volume. The functions which depends only on the initial and final state of the system.
Path dependent function : The variable whose value changes with route or path when system moves
from state I to state II. e.g. work, heat
3. Internal energy is the intrinsic form of energy. The two components of internal energy are kinetic
energy and potential energy.
4. As per first law of thermodynamics
q = DU + PDV when V is constant DV = 0
\
q = DU & when P is constant
qp = DU + PDV
and
DU = U2 – U1
PDV = PV2 – PV1
\ (U2 – U1) + (PV2 – PV1) Þ (U2 + PV2) – (U1 + PV1)
qp = H2 – H1 = DH
\
3
U + PV = H
5.
(i) 0
(ii) 1
(iii) –2
6. DH = DU + Dng RT = -87.42 + (–2 ´ 8.314 ´ 10–3 ´ 298) KJ
(iv) –2
= -92.37 KJ
7. C6H6 (l) +
15
O2 (g) ® 6CO2 (g) + 3H2O (l)
2
0.532 g gives DU = 22.3 KJ
22.3
´ 78 = 3269.5 KJ
0.532
15
Dng = 6 –
= -1.5 mol
2
DH = DU + Dng RT
\ 78 g (1 mole) gives DU =
= [-3269.5 + (-1.5) ´ 8.314 ´ 10–3 ´ 353]
= -3274 KJ
8. q = 10000 J
DV = (20 – 10) = 10 L
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w = -PDV
= -1 ´ 10 = -10 L atm
= -10 ´ 101 = -1010 J
DU = q + w
= (10000 – 1010)J
DU = 8990 J
9. 2H2O2 (l) ® 2H2O (l) + O2 (g)
DH°r = 2 DH°f H2O (l) – 2 DH°f H2O2(l)
= 2 (-286) - 2(-188)
= -572 + 376
= -196 KJ
The reaction is exothermic.
10. OF2 (g) + H2O (g) ® O2 (g) + 2HF (g)
DHo = [2 DH°f (HF)] – [ DH°f (OF2) + DH°f (H2O)]
= [2 (-268.6)] – [23 + (-241.8)]
DH = -318.4 KJ
o
Dng = 1 mol
\
DUo = DHo – Dng RT
= -318.4 KJ – (1) ´ 8.314 ´ 10–3 ´ 298 KJ
= -318.4 – 2.5
= - 320.9 KJ
11.
1
1
H2 (g) + Cl2 (g) ® HCl (g)
2
2
DH°f = [
1
1
BE (H2) +
BE (Cl2)] – BE(HCl)
2
2
1
1
´ 436 +
´ 242] – (431)
2
2
= -92 KJ/mol
=[
H
H
12. H — C º N (g) + 2H2 (g) ® H — C — N — H (g)
H
DH = [BE (C – H) + BE (C º N) + 2BE (H – H)] – [3BE (C – H) + BE (C – B) + 2BE (N – H)]
-150 = [416 + BE (C º N) + 2 ´ 436] – [3 ´ 416 + 293 + 2 ´ 369]
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-150 = 416 + BE ( C º N) + 872 – 2279
-150 = BE (C º N) – 991
BE (C º N) = -150 + 991 = 841 kJ/mol
Cl
13. H — C — H (g) ® C (g) + 2H (g) + 2Cl (g)
Cl
DH = 2BE (C - Cl) + 2BE (C - H)
= 2 ´ 326 + 2 ´ 416
= 1484 KJ/mol
Dng = 4
\
DU = DH - Dng RT
= 1484 KJ - 4 ´ 8.314 ´ 10-3 ´ 298 KJ
= 1474 KJ
14. m = 52 g
Dt = t2 – t1 = 275 – 25 = 250oC
s = 0.444 J/g oC
\
q = ms Dt = 52 ´ 0.444 ´ 250 = 5772 J = 5.77 kJ
15. First we calculate the heat changes for the water and the bomb calorimeter.
q = ms D t
qwater = 1200 ´ 4.184 ´ (23.05 – 21.22)
= 9188 J
qbomb = 1726 ´ (23.05 – 21.22)
= 3159 J
But
qreaction = qbomb + qwater
= 3159 + 9188 = 12347 J
The molar mass of octane is 114g, so the heat of combustion of 1 mol of octane is
12347
´ 114
0.257
= 5476879 J/mol
= 5476.9 KJ/mol
16. q = ms D t
qwater = 150 ´ 4.184 ´ (35.3 – 24.5)
= 6778 J
qbomb = 125 ´ (35.3 – 24.5) = 1350 J
\
Chapter-06 OLC.p65
qreaction = 6778 + 1350 = 8128 J
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6.4
Physical Chemistry for IIT-JEE
The molar mass of magnesium is 24 g, so the heat of combustion of 1 mole of magnesium is
8128
´ 24 = 45899 J/mol
.0425
= 459 KJ/mol
Level-II
1. C3H8 (g) + 5O2 (g) ® 3CO2 (g) + 4H2O (l) DH = -2220 kJ
2220 kJ of heat is supplied by 44 g C3H8
\ 350 kJ of heat is supplied by
44 ´ 350
= 6.94 g of C3H8
2220
2. 2 C (gr) + 2H2 (g) ® C2H4 (g) DH = ?
Given:
… (1)
C(gr) + O2 (g) ® CO2 (g)
… (b)
DH = -394 kJ/mol
1
O2 (g) ® H2O (l)
2
… (c)
DH = -286 kJ/mol
H2(g) +
C2H6(g) +
7
O2 (g) ® 2CO2 (g) + 3H2O (l)
2
… (d)
C2H4(g) + H2 (g) ® C2H6 (g)
… (5)
DH = -1560 kJ/mol
DH = -138 kJ/mol
In order to find DH for reaction (1), we have
DH = eq.(b) ´ 2 + eq.(3) ´ 3 – eq.(4) – eq.(5)
= -394 ´ 2 + (-286) ´ 3 - (-1560) - (-138)
= 52 kJ/mol
3. As per Born-Haber cycle
DHf (RbCl) = DHatom (Rb) + IE (Rb) +
1
BE(Cl2) + EA of Cl + URbCl
2
Putting the data
1
´ 242 + EA of Cl + (-675)
2
EA of Cl = -371 kJ/mol
-431 = 86 + 408 +
\
4. (i) Cyclohexene has one double bond and naphthalene has five double bonds. So, heat of hydrogenation of naphthalene
= 5 ´ (-120) = -600 kJ/mol
+ 5H 2
(ii)
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Chemical Thermodynamics
Using BE data
DHhyd = 5 ´ BE (C = C) + 5BE (H – H) – 5BE (C – C) – 10 BE (C – H)
= (5 ´ 610) + (5 ´ 436) – (5 ´ 346) – (10 ´ 413)
= -630 kJ/mol
(iii) The C = C energy in Cyclohexene is different from C = C bond energy in naphthalene which
leads to difference in values.
5. The mass of carbon in 10 kg coal is 8 kg
60% of 8 kg gives CO = 4800 g carbon
1
O2 ® CO DH = -110 kJ/mol
2
12 g carbon gives 110 kJ of head
C+
110 ´ 4800
= 44000 kJ
12
40 % of 8 kg = 3200 g carbon gives CO2
\
4800 g carbon will produce
C + O2 ® CO2 DH = -393 kJ/mol
12 g carbon gives 393 kg of heat
393
´ 3200 = 104800 kJ
12
Total heat evolved by 10 kg coal is 104800 + 44000 = 148800 kJ
In case the efficient oven is used, then
\
3200 g carbon will produce
7200 g C burn to give
393
´ 7200 kJ = 235800 kJ of heat
12
110
´ 800 kJ = 7333.33 kJ of heat
12
Total heat evolved by 10 kg coal is 7333.33 + 235800 = 243133.33 kJ
6. We calculate the heat changes for the water and the bomb calorimeter.
and
800 g C burn to give
q = ms Dt
qwater = (2000) ´ (4.184) ´ (25.84 – 20.17) = 4.74 ´ 104 J
qbomb calorimeter = (1.80 ´ 103) (25.84 – 20.17) = 1.02 ´ 104 J
Now, we write qreaction = -(qwater + qbomb) = -(4.74 ´ 104 + 1.02 ´ 104)
= -5.76 ´ 104 J
The molar mass of naphthalene is 128 g, so the heat of combustion of 1 mol of naphthalene is
- 5.76 ´ 104
´ 128 = -5.14 ´ 106 J/mol = -5.14 ´ 103 kJ/mol
1.435
7. If we assume that no heat is lost to the surroundings, we write
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Physical Chemistry for IIT-JEE
qsystem = qsoln. + qcalorimeter + qreaction = 0
qreaction = -(qsoln. + qcalorimeter)
\
Mass of a 100 mL solution is 100 g (because d = 1.0 g/mL), then
qsolution = (1.0 ´ 102 + 1.0 ´ 102) ´ 4.184 ´ (24.90 – 22.50) = 2.01 ´ 103 J
qcalorimeter = 335 ´ (24.90 – 22.50) = 804 J
We write, qreaction = (-2.01 ´ 103 + 804) = -2.81 ´ 103 J = -2.81 kJ
This is the heat released by reacting 0.05 mol NaOH and 0.05 mol HCl, Therefore, molar heat of
neutralisation is
- 2.81
= -56.2 kJ/mol
0.05
8. qcalorimeter = Cv ´ DT = -20.7 kJ/K ´ (298.89 – 298)K
= -18.4 kJ (-ve sign indicates the exothermic reaction)
DH for the combustion of the 0.562 g of carbon is -18.4 kJ
(DH = DU as Dng = 0)
- 18.4 ´ 12
= -392.88 kJ/mol
0.562
Thus, the enthalpy of combustion of carbon graphite is -329.88kJ/mol
9. The given reaction can be obtained as follows:
DH = -36 kJ mol-1
B2H6 (g) ® O2 (g) + 3H2 (g)
For 1 mol carbon combustion, DH =
3
O2 (g) ® B2O3 (s)
2
3[H2O(l) ® H2O (g)]
2B (s) +
3[H2 (g) +
1
O2(g) ® H2O (l)]
2
Add
B2H6 (g) + 3O2 (g) ® B2O3 (s) + 3H2O (g)
10. The given cycle process is shown in figure. It is given that
DH = -1273 kJ mol-1
DH = 3 ´ 44 kJ mol-1
DH = -3 ´ 286 kJ mol-1
DH = -2035 kJ mol-1
VA
V
= 2 and D = 4
VB
VB
TA = 27°C
(a) The process A ® B in which the plot of V versus T is linear occurs at constant pressure condition. Hence, we will have
VA
V
= B
TA
TB
æV ö
TB = ç B ÷ TA = (b) (300) (K) = 600 K
è VA ø
or
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Chemical Thermodynamics
(b) The process A ® B occurs at constant pressure, Hence
qA ® B = nCp,m (TB - TA)
æ5 ö
= (2 mol) ç R ÷ (6000 K - 300 K)
è2 ø
= (1500 mol K) R
The process B ® C occurs at constant temperature. From first law of thermodynamics,
dU = dq - dw
Since the internal energy of an ideal gas depends only on temperature, will have
dU = 0 and dq = dW
q B ® C = WB ® C =
= nRTB ln
ò PdV = nRTB ò
dV
V
VC
V
= nRTB ln D
VB
VB
(as VC = VD)
æV V ö
= nRTB ln ç D A ÷
è VA VB ø
= (2 mol) R (600 K) ln (4/2)
= (12 mol K) R In 2
The process C ® D occurs at constant volume. Hence
qC ® D = nCv,m (TA - TB)
æ3 ö
= (2 mol) ç R ÷ (300 K - 600 K)
è2 ø
= -(900 mol K) R
The process D ® A occurs at constant temperature. Hence
qD ® A = wD ® A = nRTA In
VA
VD
= (2 mol) R (300 K) In (1/4)
= - (1200 mol K) R In 2
(c) Since the process ABCDA is a cyclic process, we will have
D =0
W=q
where
q = qA ® B + qB ® C + qC ® D + qD ® A
= (1500 mol K) R + (1200 mol K) R In 2 - (900 mol K) R - (1200 mol K) R In 2
= (600 mol K) R
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11. From the first law of thermodynamics, we have
dU = dq + dw
Since, the processes occur adiabatically, dq = 0. This gives
dU = dw
For the first process
liquid (2 bar, 100 mL) ® liquid (100 bar, 100 mL)
dV = 0. Hence dw = -PdV = 0. For this process dU = 0 or DU = 0
For the second process
liquid (100 bar, 100 mL) ® liquid (100 bar, 98 mL)
w = -PDV = -(100 bar) (98 mL - 100 mL) = 200 bar mL
Hence, for both the processes
DU = 0 + 200 bar mL = 200 bar mL = (200) (105 N m-2) (10-2 m)3
= 20 J
DH = DU + D(PV) = 200 bar mL + [(100 bar) (98 mL) - (2 bar) (100 mL)]
= 100 bar ml + 9600 bar mL
= 9700 bar mL = 9700 (105 N m-2) (10-2 m)3
= 970 J
SECOND LAW OF THERMODYNAMICS
Section A
Level-I
1. DHfus = 6.025 ´ 103 J mol-1
for 1 g ice into H2O (l)
DH =
\
DSfus =
6.025 ´ 103 J
18
6.025 ´ 103 J
= 1.23 J K-1 g-1
18 ´ 273
2. DG will be negative and the reaction will occur spontaneously when TDS >> DH, and this is possible
only at high temperature.
XXV
3. 3HC º CH (g) WX
X C6H6 (g)
DG ° = DG° f (C6H6) -3DG° f (CH º CH)
= 1.24 ´ 105 -3 ´ 2.09 ´ 105
= -5.03 ´ 105 J
æ - Ä G° ö
K = antilog ç
è 2.303 RT ÷ø
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Chemical Thermodynamics
æ
ö
5.03 ´ 105
= antilog ç
÷
è 2.303 ´ 8.314 ´ 298 ø
K = 1.43 ´ 1088
As K > 1, the process can be recommended for making benzene.
4. H2O (g) + C (gr) ® CO (g) + H2 (g)
DH ° = DH°f (CO) -DH° f H2O (g)
= -110.5 -2 (-242)
= + 131.5 KJ/mol
DS ° = (197.6 + 130.6) - (189 + 5.69)
= 133.51 J mol-1 K-1
Temperature at which the DG° = 0 is
T=
Ä H° 131.5 ´ 103
=
ÄS°
133.51
T = 985 K
At T > 985 K, the reaction would become spontaneous.
XXV
5. 2H2O (l) WX
X 2H2 (g) + O2 (g)
\
H2 (g)
K = 6.8 ´ 10-84
1
XXV
O2 (g) WX
X H2O (l)
2
1
æ
ö
K¢ = ç
è 6.8 ´ 10 -84 ÷ø
1/2
K¢ = 3.83 ´ 1041
DG° f (H2O) = -2.303 RT log K¢
= -2.303 ´ 8.314 ´ 298 log 3.83 ´ 1041
= -237271 J/mol
= -237.271 kJ/mol
6. The temperature at which the DG is zero is equal to
T=
T=
ÄH
ÄS
30.56 ×103 J mol -1
66 J K -1 mol -1
T = 463.03 K
Above this temperature, the sign of DG is negative.
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Physical Chemistry for IIT-JEE
7. The reactions given are:
C + O2 ® CO2
(1)
DG = -380 kJ/mol
XXV
2C + O2 WX
X 2CO
4Al + 3O2 ® 2Al2O3
2Pb + O2 ® 2PbO
2Al2O3 ® 4Al + 3O2
(b)
(c)
(d)
(5)
DG
DG
DG
DG
Multiply eq. (2) by
= -500 kJ/mol
= -22500 kJ/mol
= -120 kJ/mol
= 22500 kJ/mol
3
2
3
XXV
O2 WX
(6)
DG = -750 kJ/mol
X 3CO
2
Add eq. (5) and eq. (6) (representing reduction) (Divide eq. (5) by 2)
3C +
22500
-750 = 10500 kJ/mol
2
since, DG is positive, so reduction of Al2O3 not possible by C.
Similarly, if we couple eq. (1) with eq. (5), DG is positive i.e. non spontaneous, and when same
principle is employed for PbO reduction, DG comes out to be negative i.e. reduction of PbO with C
forming CO or CO2 is spontaneous.
8. Work to be done by person = mgh = 60 ´ 9.81 ´ 100 = 58860 J
This is 40% of energy required by combustion of glucose.
Al2O3 + 3C ® 2Al + 3CO
DG =
58860
= 147150 J
0.4
180 g glucose gives = 2808 ´ 1000 J.
2808 ´ 103 J of energy required 180 g glucose.
So total energy required =
\
147150 J of energy requires =
180 ´ 147150
2808 ´ 103
= 9.43 g
9. For the given reaction, we have
CO (g) +
1
O2 (g) ® CO2 (g)
2
Dr S° = -0.094 kJ K-1 mol-1
D f G ° of CO (g) = -137.2 kJ mol-1; Df Go of CO2 (g) -394.4 kJ mol-1
The free-energy change of the reaction is
Dr G ° = Df G° (CO2, g) -Df G° (CO, g)
= (-394.4 + 137.2) kJ mol-1
= -257.2 kJ mol-1
Since Dr G° is negative, the reaction is spontaneous. The enthalpy change of the reaction is
Dr H ° = Dr G° + T Dr S °
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Chemical Thermodynamics
= [-257.2 + (300) (-0.094)] kJ mol-1
= -285.4 kJ mol-1
Since, Dr H° is negative, the reaction is exothermic.
10. We have
1- pentyne
1.3 %
(A)
2 - pentyne
95.2 %
(B)
1, 2- pentadiene
(3.5 %) (C)
XXV
For the equilibrium B WX
X A, we get
Keql =
DG01
[A]eq
[B]eq
=
1.3
= 1.366 ´ 10-2
95.2
= - RT In Keql = - (8.314 J K-1 mol-1) (448 K) In (3.666 ´ 10-2)
= 15991 J mol-1
XXV
For the equilibrium B WX
X C, we get
Keq2 =
[C]eq
[B]eq
=
3.5
= 3.676 ´ 10-2
95.2
DG02 = - RT In Keq2 = - (8.314 J K-1 mol-1) (448 K) In (3.676 ´ 10-2)
= 12304 J mol-1
XXV
For the equilibrium B WX
X A, we get
DG01 = Df G° A - Df G° B = 15991 J mol-1
Df G° A = Df G° B + 15991 J mol-1
or
XXV
similarly for B WX
X C we have
Df G° C = Df G°B + 12304 J mol-1
Diagrammatically, we have
G
C(g) + H2(g)
A
C
B
15991 J mol–1
12304 J mol–1
Hence, the order of stability of isomers is
B>C>A
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Physical Chemistry for IIT-JEE
11. (i) The p-V diagram of the given processes a, b and c are shown in figure below.
(a) Reversible isobaric expansion
(1 atm, 20 L) ® (1 atm, 40 L)
(b) Reversible isochoric expansion
(1 atm, 40 L) ® (0.5 atm, 40 L)
(c) Reversible isothermal compression
(0.5 atm, 40 L) ® (1 atm, 20 L)
1.0
(a)
(1 atm, 20 L)
(1 atm, 40 L)
0.8
p/atm
(b)
(c)
0.6
0.4
(0.5 atm, 40 L)
0.2
10
20
30
40
V/L
(ii) Work involved in the given processes are as follows:
w1 = p(DV) = (1 atm) (40 L - 20 L) = 20 L atm
Process a
= 20 ´ 101.325 J = 2026.5 J
Process b
w2 = 0, since the process occurs at constant volume
Process c
w3 = nRT In
V2
V
= p1 V1 In 2
V1
V1
= (1 atm) (20 L) In (20 L/40 L) = -13.86 L atm
= -13.86 ´ 101.325 J = -1404.4 J
The total work done is
w = w1 + w2 + w3 = 2026.5 J + 0 - 1404.4 J
= 622.1 J
Since, the given processes constitute a cyclic process, DU = 0, and from the first law of thermodynamics, we get
q = - w = - 622.1 J
(iii) Since the overall process is a cyclic process.
DU = 0 DH = 0
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DS = 0
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6.13
Chemical Thermodynamics
12. Initially, we have
æ 5 mol ö
(20 bar) = 10 bar
p(NO2) = x NO2 p = ç
è 10 mol ÷ø
æ 5 mol ö
p(N2O4) = x N2O4p = ç
(20 bar) = 10 bar
è 10 mol ÷ø
Qp =
(p NO2 ) 2
p N 2O 2
=
(10 bar)2
= 10 bar
(10 bar)
For the given reaction,
Dr G ° = 2Df G° (NO2) - Df G° (N2O4)
= 2 (50 kJ mol-1) - 100 kJ mol-1
=0
From the given reaction, DrG = Dr G° + RT ln Qp
we get Dr G = 0 + (8.314 J K-1 mol-1) (298 K) In (10)
= (8.314 ´ 298 ´ 2.303) J mol-1
= 5705.85 J mol–1
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