7.11 (a)
(b)
(c)
7.12 (a)
(b)
(c) Since the reactions are irreversible, the concentrations of A and
B are zero and that of C is 0.3 M.
7.13
=17.62 kJ mol−^1
7.14
=107.8 kJ mol−^1
7.15
=− −
⎛
⎝
−
.
5
0 008314
1
277
1
295
kJ mol^1
kJ/(Kmol) K K
⎜⎜⎜
⎞
⎠
⎟⎟→=k 2 878 s−^1
lnkkln ln
E
RT T
21 A k
21
2
11
−=− −
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟→−− 691.
E
Rk k
A TT
(ln ln )
//
.
=−
−
−
(^21) =
(^1121)
0 008314 kJJ/(Kmol)
KK
(. .)
(/ ) (/
−+
−
10 41 11 87
1 293 1 303 ))
lnkkln
E
RT T
21 A EA
21
11
−=− −
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟→=−−
−
−
Rk k
TT
(ln ln )
//
21
(^1121)
E
Rk k
A TT
(ln ln )
//
.
=−
−
−
(^21) =
(^1121)
0 008314 kJ/((Kmol)
KK
(ln ln )
(/ ) (/ )
20 10
1 330 1 298
−
−
lnkkln
E
RT T
21 A EA
21
11
−=− −
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟→=−−
−
−
Rk k
TT
(ln ln )
//
21
(^1121)
dC
d
BsMMs
[]
[] (. )(. ).
t
=−kf 2 = 01 −−^1101 = 01
dB
d
AB sM
[]
[] [] (. )(. )
t
=+kkff 12 − =+ 01 −^101 −=(. 01 sM−^1 )(. 01 ) 0
dA
d
AsMMs
[]
[] (. )(. ).
t
=−kf 1 =− 01 −^101 =− 01 −−^1
dA
d
A
dB
d
AB
[] d
[]
[]
[] []
[
t
k
t
=−ff 11 =+kk−f 2
CC
d
B
]
[]
t
=−kf 2
[]
[]
[] [].
A
B
s
s
==so A ==B M
−
−
1
1
101
1
1
dB
d
sMsM
[]
()(.)()(.)
t
=+101 101 0−−^11 − =
dA
d
sMsM
[]
()(.)()(.)
t
=−101 101 0−−^11 + =
dA
d
AB
dB
d
A
[]
[] []
[]
[]
t
kk
t
=−fb+ =+kkfb− [[]B
ANSWERS TO PROBLEMS 451
9781405124362_5_end.qxd 4/29/08 9:17 Page 451