1
A solution to Problems（物理化学 I）Problem 3
1)
273
273
273
1 ⎤
⎡
ΔH = ∫ CP dT = ∫ (62.76 + 0.0084T )dT = ⎢62.76 + 0.0084 × T 2 ⎥ =17446-510023
2 ⎦773
⎣
773
773
= - 492577Ε − 493 (kJ/mol)
ΔU = ΔH − PΔV =ΔH- 1×0.004×1.0132×102= ΔH- 0.41 = - 492577 – 0.41= - 492577.41
=492.577 (kJ/mol)
1dm3 atm=1.0132×102
2)
等温変化に際して理想気体の内部エネルギーは変化せず一定である
⎛ ∂U ⎞
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜ ⎟ dV + ⎜ ⎟ dT より dT = 0, ⎜ ⎟ = 0 for ideal gas. 故に dU = 0
⎝ ∂V ⎠ T
⎝ ∂T ⎠ V
⎝ ∂T ⎠ V
dU = dq + dw より、dq＝—dw= pdv
2
2
1
1
2
nRT
dV ,
V
1
∫ dq = −∫ dw = ∫
Q = - W = nRTIn(V2/V1) = (50/32) ×8.314×300×2.303log(45/1.5)
=13.256(kJ/mol)
ΔH = ΔU+ Δ(PV) = ΔU + Δ(nRT) = ΔU + nRΔT = 0
3)
⎛ ∂U ⎞
⎛ ∂U ⎞
For ideal gas: dU = dq + dw = ⎜ ⎟ dV + ⎜ ⎟ dT
⎝ ∂V ⎠ T
⎝ ∂T ⎠ V
⎛ ∂U ⎞
dT = 0, ⎜ ⎟ = 0
⎝ ∂T ⎠ V
=0
dU = dq + dw より、dq＝—dw= pdv
Q = - W = nRTIn(V2/V1) = 8.314×300×2.303log(1/10)
= - 5.744(kJ/mol)
For van der Waals Gas: dU = TdS – PdV より
⎛ nRT
nR
n 2a ⎞ n 2a
⎛ ∂U ⎞
⎛ ∂S ⎞
⎛ ∂P ⎞
−⎜
−
⎜ ⎟ = T⎜ ⎟ − P = T⎜ ⎟ − P = T
⎟ = 2
⎝ ∂V ⎠ T
⎝ ∂V ⎠ T
⎝ ∂T ⎠ V
V
V − nb ⎝ V − nb V 2 ⎠
where
⎛ ∂P ⎞
⎛ ∂S ⎞
dA = -PdV- SdT より， ⎜ ⎟ = ⎜ ⎟ ,
⎝ ∂T ⎠ V ⎝ ∂V ⎠ T
van der Waals Gas equation :
⎛
n 2a ⎞
⎜ P + 2 ⎟ (V − nb ) = nRT,
V ⎠
⎝
nRT
n 2a
P=
−
,
V − nb V 2
nR
⎛ ∂P ⎞
⎜ ⎟ =
⎝ ∂T ⎠ V V − nb
For 1mol (at constant temperature for 1mol):
a
⎛ ∂U ⎞
⎛ ∂U ⎞
dU = ⎜ ⎟ dV + ⎜ ⎟ dT =
dV , ΔU =
⎝ ∂V ⎠ T
⎝ ∂T ⎠ V
V2
2
⎛1 1⎞
a
⎡ a ⎤
∫ V 2 dV = ⎢⎣− V 2 ⎥⎦V = a ⎜⎝ V1 − V2 ⎟⎠
V1
1
V2
V
1
2
⎛1 1⎞
a⎞
V −b
⎛ RT
W = − ∫ PdV = − ∫ ⎜
− 2 ⎟ = RTIn 1
+ a⎜ − ⎟
⎝V−b V ⎠
V2 − b
⎝ V1 V2 ⎠
V1
V1
V2
V2
⎛ 1 1 ⎞ ⎧⎪
⎛1 1 ⎞⎫
V −b
Q = ΔU − W = a ⎜ − ⎟ − ⎨ RTIn 1
+ a⎜ − ⎟ ⎬
V2 − b
⎝ V1 V2 ⎠ ⎪⎩
⎝ V1 V2 ⎠ ⎭
10 − 0.115
= −8.3145 × 300 × 2.303log
1 − 0.115
=- 6.020 (kJ/mol)
1
4)
P2 ⎛ P1 ⎞ γ
⎛ P2 ⎞
RT
PV
P1/P2=(V2/V1)γより 2 = 2 2 =
=
⎜ ⎟
⎜ P⎟
RT1
P1V1
P1 ⎝ P2 ⎠
⎝ 1⎠
＝29.288,
CP = R＋Cv
T2
⎛1⎞
=⎜ ⎟
400 ⎝10⎠
5)
1−
1
γ
⎛1⎞
=⎜ ⎟
⎝10⎠
1−
1
γ
CP/ Cv＝γ
,
ideal gasCP - Cv = R より
∴ γ = (29.288/20.92)=1.4
0.2857
∴ Τ2 = 207.2 Κ
不可逆断熱膨張の場合： Q = 0
∴ΔU =W = n CV(T2 – T1) = 0.35 × (3/2) ×0.08206×(T2 – 400)
定圧膨張の場合：W = - PΔV =0.5 × (5 - 1) = - 2.0(dm3 atm)
∴0.35 × (3/2) ×0.08206×(T2 – 400),
T2=353.8 K
ΔH = ΔU+ Δ(PV) = - 2.0 + 0.5(5-1) = 0
6)
Q = 0,
∴ W = ΔU = n CV(T2 – T1) = (10/32) × 25.1 ×(333– 298) = 274.5 (J/mol)
∴ W = -P2(V2 –V1) = -P2{(nRT2/P2) - (nRT1/P1)}
= (-10/32) ×8.3145×333+ P2{(-10/32) ×8.3145×298}=274.5 (J/mol)
∴ P2=1.47
From P2V2=nRT2
7)
a)
P2=1.47atm
V2=nRT2 / P2 =5.8 (dm3)
2CO(g) + O2
2CO2(g)
ΔH =2×(- 395.51) - 2×(- 110.53) = - 569.96 (kJ/mol)
ΔH = ΔU+ Δ(PV) = ΔU + Δn(RT) よりΔU = ΔH - Δn(RT) = - 569.96 – (2 -3)
×8.3145×(273+25) ×10-3= - 567.5 (kJ/mol)
b)
4HCl(g) + O2
2H2O (g) + 2Cl2(g)
ΔH =2×(- 241.82) + 0 - 4×(- 92.31) = - 114.4 (kJ/mol)
ΔU = ΔH - Δn(RT) = = - 114.4 – (4 - 5) ×8.3145×(273+25) ×10-3
= - 111.9(kJ/mol)
c)
SO3(g) + H2O(l)
H2SO4(l)
ΔU = ΔH - Δn(RT) = = - 132.44 – (1 - 2) ×8.3145×(273+25) ×10-3
= - 129.9 (kJ/mol)
d) AgBr(s) + 1/2Cl2
AgCl(s) + 1/2 Br2(g)
ΔU = ΔH - Δn(RT) = = - 11.25–{(1 + 1/2) – (1+1/2)} ×8.3145×(273+25) ×10-3
2
3
= - 11.25(kJ/mol)
e)
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O
ΔU = ΔH - Δn(RT) = = - 3127.66–{(4 + 6) – (2+7)} ×8.3145×(273+25) ×10-3
= - 3130(kJ/mol)
8)
C6H5COOH(s) + (15/2) O2 (g)
7CO2 (g) +3H2O (l)
Benzoic acid M.W.=122
ΔU = - 26,359(J) ×122= - 3215.8 (kJ/mol)
ΔH = ΔU + ΔνRT よ り ΔH = - 3215.8
+ {7−(15/2)}×8.3145×(273+25) ×10-3
=-3217.039
Heat of formation of C6H5COOH=x
∴ x = - 408.9 (kJ/mol)
7×(-395.5) + 3 × (-285.8) – x = -3217
9)
H20(g); Cp H20 = 34.39+6.28×10-4T+5.6 ×10-8 T2 とおく
1000
ΔH =
1000
∫ C dT = ∫ (34.39 + 6.28 ×10
P
298
10)
−4
T + 5.6 × 10−8 )dT =24.45 (kJ/mol)
298
For NO: N(g) + O (g)
ΔH1
1/2 N2(g) + 1/2 O2(g)
ΔH2
NO(g)
ΔH
ΔH1 = 0 – (472.7+249.17)= - 721.87 (kJ/mol)
ΔH2 = (90.25) - 0= 90.25 (kJ/mol)
∴ ΔH1= ΔH1 + ΔH1 = - 721.87 + 90.25= -631.62 (kJ/mol) : Bond energy for NO
For HCl: H(g) + Cl (g)
1/2 H2(g) + 1/2 Cl2(g)
ΔH1
ΔH2
HCl(g)
ΔH
ΔH1 = 0 – (217.97+121.68)= - 339.65 (kJ/mol)
ΔH2 = (-92.31) - 0= - 92.31 (kJ/mol)
∴ ΔH= ΔH1 + ΔH2 = - 339.65 - 92.31 = -432.0 (kJ/mol) : Bond energy for HCl
11)
For C2H6: 2C(g) + 6H (g)
ΔH1
2 C(s) + 3 H2(g)
ΔH2
C2H6(g)
ΔH
ΔH1 = 0 – (716.68×2+217.97×6)= - 2741.28 (kJ/mol)
ΔH2 = (-84.68) - 0= - 84.68 (kJ/mol)
∴ ΔH= ΔH1 + ΔH2 = - 2741.28 - 84.68 = - 2825.96 (kJ/mol)
Text P44, Tab3.4 より C-H Bond energy=414 (kJ/mol)
(H-CH3)
Therefore C-C bond energy for C2H6=2825.86 - 6×414=342 (kJ/mol)
12)
Text P44, Tab3.4 より
C2H6 :- (414 ×6+347)= - 2831(kJ/mol) ,C2H5OH :-(414×5+351+460+347)=- 322(kJ/mol)
C2H5SH :-(414×5+255+368+347)=- 3040(kJ/mol),これらのエネルギーは成分原子か
らの生成熱であるから次の様なサイクルを考える．
1: 2C(g) + 6H
C2H6
ΔH=- 2834 (kJ/mol)
3
4
2: 2C(g) + 6H +O
C2H5OH
ΔH= -3228(kJ/mol)
3: 2C(g) + S +6H
C2H5SH
ΔH=- 3040 (kJ/mol)
4: C(graphite)
C
ΔH= 716.68 (kJ/mol)
5: H2
2H
ΔH=217.97×2=435.94(kJ/mol)
6: O2
2O
ΔH=249.17×2=498.34 (kJ/mol)
7: S(crystal)
S(g)
ΔH=278.81(kJ/mol)
2C(g)+6H(g)
C2H6(g)
For C2H6: 2C(s) + 3H2(g)
ΔH1
ΔH2
ΔH
ΔH1 = 716.68×2+435.94×5= - 2741.18 (kJ/mol)
ΔH2 = -1 ×(C-C bond energy) -6×(C-H bond energy) = -347- 414×6 = - 2834 (kJ/mol)
∴ ΔH= ΔH1 + ΔH2 = - 2741.18 - 2834 = - 89.8 (kJ/mol)
同様に For C2H5OH: ∴ ΔH= ΔH1 + ΔH2 = 2990 - 3224 = - 234 (kJ/mol)
For C2H5SH: ∴ ΔH= ΔH1 + ΔH2 = 3020 - 3040 = - 20 (kJ/mol)
13)
ΔH1673
CO(g) + 1/2 O2(g)
ΔCp=CP,CO2 - ( CP,CO+1/2 CP,O2)
CO2(g)
⎡
⎤
−3
−7 2
⎢
⎥
T
−148.3
×10
T
)
(25.99
+
43.50
×10
1673
1673
⎥
⎢
= ΔH291 + ∫ ΔCP dT = −282880 + ∫ ⎢−(26.861 + 6.966 ×10−3 T − 8.201 ×10−7 T 2 ) ⎥
291
291 ⎢
⎥
1
−3
−7 2
⎢− (25.723 + 12.98 ×10 T − 38.62 ×10 T )⎥
⎦
⎣ 2
= - 279.839(kJ/mol)
where
16732=2798929, 2912=84681, 16733=4682608217, 2913=24642171, T=(1673- 291)=1382,
T2=[2798929-84681]=2714248, T3=4657966046
14)
⎡(25.99 + 43.50 ×10−3 T −148.3 ×10−7 T 2 )
⎤
⎢
⎥
1673
T
+2 × (30.359 + 9.615 ×10−3 T + 11.84 ×10−7 T 2 )⎥
⎢
ΔHT = ΔH0 + ∫ ΔCP dT = ΔH0 + ∫
−3
−7 2
⎢
⎥
291 −(14.146 + 75.496 ×10 T −179.9 ×10 T )
0
⎢
⎥
−3
−7 2
⎢⎣−2 × (25.723 + 12.98 ×10 T − 38.62 ×10 T ) ⎥⎦
= ΔH0+21.116T-19.363×10-3T2+44.173×10-7T3
where: ΔCp=(CP,CO2(g)+2 CP,H2O(g)) - ( CP,CH4(g)+2 CP,O2(g))
CP,CO2(g);25.99+43.50×10-3T – 148.3×10-7T2,
CP,H2O;30.359+9.615×10-3T + 11.84×10-7T2,
CP,CH4(g);14.146+75.496×10-3T – 179.9×10-7T2,
CP,O2;25.723+12.98×10-3T – 38.62×10-7T2,
4