7.6 They are both equal to 0.05 M.7.7 [B]/[C] =10 and [B] +[C] =0.1 M
10[C] +[C] =11[C] =0.1 M →[C] =0.009 M and [B] =0.097.8 (a)(b) =−(2.0 M−^1 s−^1 )(0.1 M)(0.1 M) =−0.02 Ms−^1=−(2.0 M−^1 s−^1 )(0.1 M)(0.1 M) =−0.02 Ms−^1=(2.0 M−^1 s−^1 )(0.1 M)(0.1 M) −(1.0 s−^1 )(0.1 M)
+(5.0 s−^1 )(0.1 M) =0.42 Ms−^1=(1.0 s−^1 )(0.1 M) −(5.0 s−^1 )(0.1 M) =−0.40 Ms−^1(c) At equilibrium the concentrations of A and B are zero since the
reactions are irreversible. The sum of the concentrations must
be equal to 0.4 M and the ratio of the amounts of C and D is
given by the ratio of the two rate constants:[C] +[D] =[C] +0.2[C] =0.4 M →[C] =1/3 M
[D] =0.2[C] =1/15 M
7.9
7.10 (a)(b)(c) Since the reaction is irreversible, the final concentration of A is
zero and that of B is 0.2 M.dB
dAs M Ms[]
[] ( )(. ).
t=+kf =101 01−−^11 =dA
dAsM Ms[]
[] ( )(. ).
t=−kf =− 101 −−^11 =− 01dA
dA
dB
dA
[]
[]
[]
[]
tk
t=−ff=+kAAekt. ek s k
() ln.
=→= 0 −− 05 10 =−05
10
Mor
s=0 0693. s−^1K
k
kf
b[]
[]
===→=.[].[]
D
C
(^2) D C and
2
02 02 [[] [ ] .CD+= 04 M
dD
d[]
tdC
d[]
tdB
d[]
tdA
d[]
tdC
dAB C D
dD
d[]
[][] [] []
[]
tkkk
ffbt
=−+ 122 ==−kkfb 22 []CD[ ]dA
dAB
dB
dAB
[]
[][]
[]
[][]
tk
t=−ff 11 =−k450 ANSWERS TO PROBLEMS
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