[Big-Daddy]

Is there a standard procedure - using a mathematical tool, for example, such as WolframAlpha - to "integrate both sides" of an equation? For instance, let us say that the rate of loss of A in the reaction A+B->Products is related directly to the concentrations of A and B ([A] and [B ] respectively) by the equation r=k*[A]*[B ]. By the definition of that rate, this means -d[A]/dt=k*[A]*[B ]. I have seen this rearranged into -d[A]/d[A][B ]=k*dt. All fine. The next step is where it gets tricky: somehow by "integrating on both sides" we get the expression directly relating [A]₀, [B ]₀, k, t, [A] and [B ]. How is this done in general? And how can I use a mathematical software to extrapolate this approach to problems with more complex solutions (e.g. -d[A]/dt=k*[A]³*[B ]²). Thanks.

[curiouscat]

Key is [A] and [ B] are not independently variable. Link them through stoichiometry e.g.

[A]0 + [ B]0 = [A] + [ B]

Now eliminate, say, [ B] and your integrals are in [A] and t alone.

The general procedure always works; you can use Wolfram if the integrals are hard. 

[Big-Daddy]

Key is [A] and [ B] are not independently variable. Link them through stoichiometry e.g.

[A]0 + [ B]0 = [A] + [ B]

How does this work? It will only be true at the very beginning of the experiment surely, by definition, since after that both A and B (being reactants) with experience a drop in concentration.

e.g. for the reaction 2A+B -> Products, what is the stoichiometric link that enables me to eliminate B?

And more problematically, how do I reach the integrated state (using WolframAlpha) in the first place? This means "integrate on both sides", but how do I do that on WolframAlpha? (Because for some rate laws it's too much to do in my head) Only once the integrated rate law is found can I use stoichiometry to eliminate one of the reactants (enabling me to track concentration of that reactant with time based purely on k, t, [A]0 and [ B]0). And even so I'm not sure how to do the second step, but I suppose one at a time and my first problem is using WolframAlpha to find the integrated rate law for the more complex rate equations (rate=kAB is complex enough, but others, such as rate=kA²B, are much more complicated).

[curiouscat]

How does this work? It will only be true at the very beginning of the experiment surely, by definition, since after that both A and B (being reactants) with experience a drop in concentration.

e.g. for the reaction 2A+B -> Products, what is the stoichiometric link that enables me to eliminate B?

It doesn't. My bad. Brain fart.

Here's what does work:

[A]0 - [ A] = [B ]0 - [ B]

Makes sense?

e.g. for the reaction 2A+B -> Products, what is the stoichiometric link that enables me to eliminate B?

[A]0 - [A] = 2×([ B ]0 - [ B ])

Only once the integrated rate law is found can I use stoichiometry to eliminate one of the reactants

No. Right from the start.

my first problem is using WolframAlpha to find the integrated rate law for the more complex rate equations (rate=kAB is complex enough, but others, such as rate=kA2B, are much more complicated).

Doesn't matter how complex. Bring dt to one side. Everything else to the other. Integrate.

[Big-Daddy]

[A]0 - [A] = 2×([ B ]0 - [ B ])

Oh I see, thank you! This makes sense now.

No. Right from the start.

Doesn't matter how complex. Bring dt to one side. Everything else to the other. Integrate.

OK, so let's take 2A+B -> Products with a rate law rate=k*A²*B. We thus know that -d[A]/dt=k*A²*B (this should be [A] and [ B], sorry for the lack of formality in representation). This we rearrange to get -d[A]/([A²*B)=k*dt. Could you maybe show me (with a link to the sum) how to integrate this using Wolfram?

[curiouscat]

OK, so let's take 2A+B -> Products with a rate law rate=k*A²*B. We thus know that -d[A]/dt=k*A²*B (this should be [A] and [ B], sorry for the lack of formality in representation). This we rearrange to get -d[A]/([A²*B)=k*dt. Could you maybe show me (with a link to the sum) how to integrate this using Wolfram?

Remove B from your last expression. I already gave you an expression to do that:

[A]0 - [A] = 2×([ B ]0 - [ B ])

[Big-Daddy]

OK, so let's take 2A+B -> Products with a rate law rate=k*A²*B. We thus know that -d[A]/dt=k*A²*B (this should be [A] and [ B], sorry for the lack of formality in representation). This we rearrange to get -d[A]/([A²*B)=k*dt. Could you maybe show me (with a link to the sum) how to integrate this using Wolfram?

Remove B from your last expression. I already gave you an expression to do that:

[A]0 - [A] = 2×([ B ]0 - [ B ])

As far as I can see that doesn't work until I've integrated first. My integration of -d[A]/dt=k*A²*B should lead to an expression in k, t, [A], [ B], [A]0 and [ B]0. I can then remove [ B] from being a variable using the expression above and then solve for [A] if I want to find the concentration of [A] at any given point in time (or remove [A] and then solve for [ B] if I want the concentration of [ B] at any given time). However, if I try and do this now, before integrating, I would be attempting to integrate something with [A]0 and [ B]0 already in it, and I don't understand how this would work (since as far as I know [A]0 and [ B]0 only come in as the constants after the integration). If you could do this example (or some other one with both [A] and [ B] involved - possibly rate=k[A][ B] for A+B -> Products) on Wolfram I might be able to understand how it's done.

[curiouscat]

However, if I try and do this now, before integrating, I would be attempting to integrate something with [A]0 and [ B ]0 already in it,

Yes.  So what?

and I don't understand how this would work (since as far as I know [A]0 and [ B ]0 only come in as the constants after the integration).

Why not? A0 and B0 are just known constants. Not constant of integration.

e.g. When you integrate (P+x)dx isn't P a constant?

Write what you get as integrand if you eliminate B. We can go from there.

[Big-Daddy]

However, if I try and do this now, before integrating, I would be attempting to integrate something with [A]0 and [ B ]0 already in it,

Yes.  So what?

and I don't understand how this would work (since as far as I know [A]0 and [ B ]0 only come in as the constants after the integration).

Why not? A0 and B0 are just known constants. Not constant of integration.

e.g. When you integrate (P+x)dx isn't P a constant?

Write what you get as integrand if you eliminate B. We can go from there.

-dA/dt=k*A²*B
B=B₀-(1/2)*A₀+(1/2)*A
-dA/dt=k*A²*(B₀-(1/2)*A₀+(1/2)*A)

-dA/(A²*(B₀-(1/2)*A₀+(1/2)*A))=k*dt

I frankly have no clue how to integrate that. And nor do I know how to use Wolfram to integrate on both sides. Obviously k*dt will go to k*t as always, but the left hand side is an enigma to me, particularly using Wolfram to do it.

[curiouscat]

http://www.wolframalpha.com/input/?i=integrate+%28-1%2F%28A%5E2*%28%28B0-%281%2F2%29*A0%2B%281%2F2%29*A%29%29%29%29+dA+

You'll have to do the right limit. At t=0 A=A0 and at t=t A=A.

OTOH, this may not be the cleanest way to do this.

PS. I didn't have time to check any of this. Be warned!