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−^17.14
=107.8 kJ mol−^17.15
=− −
⎛
⎝
−
.5
0 008314
1
277
1
295
kJ mol^1
kJ/(Kmol) K K⎜⎜⎜
⎞
⎠
⎟⎟→=k 2 878 s−^1lnkkln lnE
RT T
21 A k
21211
−=− −
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟→−− 691.
E
Rk k
A TT(ln ln )
//.
=−
−
−
(^21) =
(^1121)
0 008314 kJJ/(Kmol)
KK
(. .)
(/ ) (/
−+
−
10 41 11 87
1 293 1 303 ))
lnkklnE
RT T
21 A EA
2111
−=− −
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟→=−−
−
−
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
−
−
lnkklnE
RT T
21 A EA
2111
−=− −
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟→=−−
−
−
Rk k
TT(ln ln )
//21(^1121)
dC
d
BsMMs
[]
[] (. )(. ).
t=−kf 2 = 01 −−^1101 = 01dB
dAB sM[]
[] [] (. )(. )
t=+kkff 12 − =+ 01 −^101 −=(. 01 sM−^1 )(. 01 ) 0dA
dAsMMs[]
[] (. )(. ).
t=−kf 1 =− 01 −^101 =− 01 −−^1dA
dA
dB
dAB
[] d
[][]
[] []
[
tk
t=−ff 11 =+kk−f 2CC
dB
]
[]
t=−kf 2[]
[]
[] [].
A
B
s
s==so A ==B M−
−1
1
101
1
1dB
dsMsM[]
()(.)()(.)
t=+101 101 0−−^11 − =
dA
dsMsM[]
()(.)()(.)
t=−101 101 0−−^11 + =
dA
dAB
dB
dA
[]
[] []
[]
[]
tkk
t=−fb+ =+kkfb− [[]BANSWERS TO PROBLEMS 451
9781405124362_5_end.qxd 4/29/08 9:17 Page 451