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Topic: Integrated Rate Laws: 2nd order reaction & half-life  (Read 9283 times)

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Offline Winga

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Integrated Rate Laws: 2nd order reaction & half-life
« on: February 22, 2005, 12:10:58 PM »
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.

Demotivator

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Re:Integrated Rate Laws: 2nd order reaction & half-life
« Reply #1 on: February 22, 2005, 02:42:23 PM »
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.





« Last Edit: February 22, 2005, 10:03:56 PM by Demotivator »

Offline Winga

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Re:Integrated Rate Laws: 2nd order reaction & half-life
« Reply #2 on: February 23, 2005, 09:44:56 AM »
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?

Demotivator

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Re:Integrated Rate Laws: 2nd order reaction & half-life
« Reply #3 on: February 23, 2005, 10:23:16 AM »
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

Offline Winga

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Re:Integrated Rate Laws: 2nd order reaction & half-life
« Reply #4 on: February 23, 2005, 10:46:56 AM »
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?

Demotivator

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Re:Integrated Rate Laws: 2nd order reaction & half-life
« Reply #5 on: February 23, 2005, 11:52:21 AM »
Obviously, the boundaries of integration have to be the same variable as the differential elements dx and dt over which you are integrating.

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