Page 187 - Essentials of physical chemistry

Basic Chemical Kinetics 149

This is really an amazing mathematical trick because we should see that the two terms in N B would
k b t
be part of a derivative of a product if only there was a factor of e , so why not multiply the whole
equation by that factor? Now we can integrate both sides of the equation:
ð ð (k b k a )t
e
k b t
(k b k a )t
k b t
d(e N B ) ¼ k a N 0 e dt ) e N B ¼ (k a N 0 ) þ C but N B ¼ 0at t ¼ 0, so we
(k b k a )
k a N 0 e (k a k b )t
k a N 0 k a N 0
k b t
(k b k a ) (k b k a ) (k b k a )
have C ¼ and e N B ¼ . Then collecting common factors we
k a N 0 e (k a k b )t
k b t k a N 0 k a N 0 (k b k a )t
have e N B ¼ ¼ e 1 and after we divide both sides
(k b k a ) (k b k a ) (k b k a )
k b t
by e ,we finally reach
k a N 0 k a t k b t
e e :
N B ¼
(k b k a )
That gives us a formula for N B at any time t > 0. When will the contents of tank B reach a

dN B dN B
maximum? We need to set ¼ 0 and solve for t max N B . Thus, ¼
dt dt

k a N 0 k a t k b t k a t k b t
k a e þ k b e ¼ 0. Therefore k a e ¼ k b e ; now take the natural log of
(k b k a )

k a
the whole equation, so ln k a k a t ¼ ln k b k b t or ln ¼ (k a k b )t. Thus, t max N B ¼
k b

ln k a
k b
using the numerical values,
. Now find t max N B
(k a k b )
0:693
0 1
3 days
B C
ln B C
k a @ 0:693 A 8
ln ln
8 days 3
k b
t max N B ¼ ¼ ¼ ¼ 6:7936 days:
(k a k b ) 0:693 0:693 8 3
0:693

3 days 8 days 24 days
With this information we can find out how much water=solution is in tank B at that time:

0:693
(1000 gal)
k a N 0 k a t k b t 3 days h 0:693 0:693 i
e e e ð 3 Þ(6:7936) e ð 8 Þ(6:7936) ,
(k b k a ) 0:693 0:693
N B ¼ ¼
8 days 3 days
and so N B (max) ¼ ( 1600 gal) ( 0:346975) ¼ 555:16 gal. We might as well calculate the con-
tents of tank A and tank C at this time, assuming the outlet valve of tank C is closed.
0:693
N A ¼ (1000 gal)e ð 3 days Þ(6:7936 days) ¼ 208:19 gal, so assuming conservation of the initial 1000 gal
(neglecting evaporation) we can calculate the volume in tank C as whatever is not in tank A or
tank B. The volume in tank C ¼ (1000 208.19 555.16) ¼ 236.65 gal at t ¼ 6.7936 days. We also
realize that eventually all the solution will end up in tank C with tanks A and B completely empty.
If this was a problem in nuclear decay, we might have so few individual atoms starting from a
number like 1000 atoms that we would have to round the values of N A , N B , and N C to integer values