[madscientist]

Hi all,

Im having a bit of difficulty integrating a rate equation for a bimolecular reaction which is 2nd order overall:

For a unimolecular reaction or a bimolecular reaction with equal concentrations of reactants its relatively easy, its just:

A + A ---> product  'or'  A --> product

Rate = k[A] ²

Upon integrating becomes:

(1/[A]_t) - (1/[A]₀) = kt


But for a bimolecular reaction with diifferent concentrations of reactants it becomes a little more tricky:

aA + bB ---> products

Rate = k[A]*[ B]

This is what i have come up with:

(1/(b*[ A] ₀ - a*[ B] ₀)) * ((ln([ B]₀/[ A]₀) - (ln([ B]_t/[ A]_t)) = kt

Is this right?

My next problem is how to write this equation as a straight line equation in order to obtain the rate constant or the [A]t or [ B]t values (if given the value for k).

Ive read that you plot ln([ B]t/[A]t) v's time, but am unsure as to what the value of k would be? is it the slope of this plot?

I understand that the straight line equation is:

y = mx + b 

y = ln([ B]t/[A]t), m = slope, x = time, b = y intercept (at x = 0)

k = ?

Any help, hints or suggestions would be greatly appreciated,

Mad

[enahs]

Second order rate equation assuming the reaction is first order in both A and B

A+Β -k-> P


      1        *    ln ( [Β]/[Β]_o  )   =  kt
[Β]_o-[A]_o           ( [A]/[A]_o  )   



How I read about what you say about how you want to plot/graph, what you are talking about plotting is sometimes referred as Pseudo First Order.

If you hold either [A] or [Β] constant, it can be considered a first order reaction because it only depends on the concentration of one reactant.

If we hold [Β] constant.
r= k[A][Β] = k'[A]
Where k' = k*[Β]_o

The easiest way to do this is to keep [Β] in large excess, ([Β]>>[A]), so as the reaction proceeds only a small amount of Β is consumed and the concentration of [Β] can be approximated as being constant.

By obtaining k' for many reactions in which [Β]>> [A] in excess but varying amounts, a plot of k' .vs. [Β] gives k as the slope.


I am not aware of any other way to really make it a straight line. That is kinda the definition of a second order rate equation, it is not linear!

[sdekivit]

if the reaction aA + bB --> C with rate equation

     1      *   ln ( [Β]/[Β]o  )   =  kt
[Β]o-[A]o      ( [A]/[A]o  )   

you can create a linear curve by plotting plotting left side of the equation against the time with slope = k.

[enahs]

Plotting inverse of concentration as the dependent variable and t as the independent variable is only valid for second order rate equations of the form R = k[A]², I thought ? Maybe I am just mistaken.

[sdekivit]

you're confused with the equation 1/[A] - 1/[A]₀ = kt

when s = k * [A] * [ B] then

    1      *   ln ( [Β]/[Β]o  )   
[Β]o-[A]o      ( [A]/[A]o  )   

must be completely plotted against t and so your y-axis is both dependent on [A] and [ B] since both can be measured at time t.