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Topic: rate equation  (Read 2493 times)

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Offline davon806

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rate equation
« on: June 08, 2014, 01:05:16 AM »
In a reaction aA + bB ---> cC + dD
the rate velocity = (-1/a) (d[A]/dt) = (-1/b)(d[ B]/dt)
rate equation,say: v1 = k[A]^m [ B]^n
Now if we double the concentration of A,
the new rate velocity v2 = (-2/a)(d[A]/dt)
so imagine we are performing a experment in order to find the value of m,
using the initial rate method,the concentration of B is kept constant.

v2/v1 = [2A/A]^m
      2 = 2^m

Here v2/v1 is the ratio in rate velocity
so m = 1 and the reaction is first order wrt A
However,if the above is true,we would never get reaction of order >1 wrt to a particular species.
It is because the increase in concentration of a reactant will also increase its rate velocity.When you cancel all other terms in v2/v1 = [tA/A]^m ,where t is the factor of increment in conc. of A,you simply reduce the equation into t = [t]^m which means m must be 1.
What's wrong with the above ? Thx :)
« Last Edit: June 08, 2014, 04:04:29 AM by Borek »

Offline mjc123

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Re: rate equation
« Reply #1 on: June 09, 2014, 04:43:44 AM »
Why do you say v2=-(2/a)d[A]/dt? The rate law remains the same:  v = -(1/a)d[A]/dt. But d[A]/dt will increase if [A] is increased.
Given your rate equation and keeping B constant
v2/v1 = 2^m
Measure v2/v1 and equate it to 2^m. Thus you get m.
v2/v1 is not equal to 2 unless m = 1. If m = 2, v2/v1 = 4.

Offline davon806

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Re: rate equation
« Reply #2 on: June 09, 2014, 03:44:52 PM »
But v = (-1/a)d[A]/dt (a is to eliminate the stoichometric coefficient)
If we double the concentration,[A] --> [2A], so we take the 2 out can get v= (-2/a)(d[A]/dt),which is the twice of the initial rate?

Offline mjc123

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Re: rate equation
« Reply #3 on: June 10, 2014, 04:52:16 AM »
This is bad and confusing notation. [A] means the concentration of A, whatever that may be. The velocity is always -(1/a)d[A]/dt, i.e. -1/a times the rate of change of [A] then and there. If you want to refer to specific values of [A] you should use indices, e.g. [A]1,0 for the initial concentration of A in experiment 1. This is a constant, not a variable like [A]. Your problem is that you are confusing a variable with particular values of that variable.
Then you can say [A]2,0 = 2*[A]1,0
v1,0 = -(1/a)(d[A]/dt)1,0     {Note, not d[A]1,0/dt because [A]1,0 is a constant}
v2,0 = -(1/a)(d[A]/dt)2,0
But, in the general case, (d[A]/dt)2,0 ≠ 2*(d[A]/dt)1,0. Only if m = 1. The ratio is 2^m, as d[A]/dt is proportional to [A]m.

{Incidentally, in another thread someone asked about comparisons between chemistry and computer programming. Here is an important difference. In computer programming you can reassign the value of a variable by a statement like x = 2*x (i.e the variable x is assigned a value twice its previous value). Here, saying [A]  :rarrow: [2A] causes confusion because, as I said, it is confusing a variable with particular values of that variable.}

Offline davon806

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Re: rate equation
« Reply #4 on: June 10, 2014, 07:02:09 AM »
Thx  ;D.Feeling ashamed to be fooled by these notations   :P

Offline mjc123

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Re: rate equation
« Reply #5 on: June 10, 2014, 08:29:08 AM »
Don't feel bad. We've all made mistakes. Important thing is you learn from them.

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