Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Winga on February 22, 2005, 12:10:58 PM
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Q.1
A + A --> B
d[A]/dt = -k[A]2
(Why d[A]/dt no need to time 1/2?)
Q.2
A + B --> C
d[A]/dt = -k[A]
Let x be the extent of reaction,
[A] = [A]o - x
= o - x
d[A]/dt = -k([A]o - x)(o - x)
since d[A]/dt = -dx/dt
dx/dt = k([A]o - x)(o - x)
Integrate dx/([A]o - x)(o - x) = Integrate kdt
(How to integrate the left-hand side?)
Sorry, this question is about calculus.
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1) half life:
t(1/2) = 1/k[A]o
2. Need to use method of partial fractions, then integrate:
1/(A-x)(B-X) = C/A-X + D/B-X
where C and D are constants to be evaluated. Continuing,
C(B-X) + D(A-X) =1
CB - CX +DA - DX = 1
Therefore
-CX - DX = 0, D = -C
and
CB + DA = 1
substituting -C for D above,
CB - CA =1
C = 1/B-A
hence, D= -1/B-A
Therefore
dx/([A]o - x)(o - x) =
dx C/([A]o - x) + dx D/(o -x) =
(1/(o - [A]o))dx /([A]o-x) - (1/(o - [A]o))dx/(o - x) =
(1/(o - [A]o)) (dx /([A]o-x) - dx/(o - x))
now, the separated terms can be integrated via natural log.
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About Q.1, I mean...
e.g.
A + 2B --> 3C
-d[A]/dt = -d/dt x 1/2 = -d[C]/dt x 1/3
A + A --> B
2A --> B
Why the equation is only d[A]/dt = -k[A]2 but not d[A]/dt x 1/2 = -k[A]2?
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You are confusing the two issues, between expressing the rate law as k[A]2 and expressing it in terms of another species rate of formation.
In terms of dB/dt , its -dA/dt = 2dB/dt
or (-1/2)dA/dt = dB/dt
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I have solved it.
By the way,
Integrate from x to 0 dx/([A]o - x)(o - x) = Integrate from t to 0 kdt
Why integrate from x to 0?
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Obviously, the boundaries of integration have to be the same variable as the differential elements dx and dt over which you are integrating.