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Kinetics - effective 1st order rate constant

For the question attached, can someone just clarify that my answers to the marked question are correct?

bi) 2CH3 > 2CH3* Activation
2CH3* > 2CH3 De-activation
2CH3* + M > C2H6 + M Reaction

ci) Is this as simple as plotting ln(k) against 1/T, getting the gradient and re-arranging -Ea/R to get Ea? There is no trick hidden in it? I remember something about when you have a series of bi-molecular steps, it changes what you plot? It was something to do with using the Boltzman distribution where K=[A*]/

(edited 10 years ago)

Reply 1

Big-Daddy

Original post by DonnieBrasco

ci) Is this as simple as plotting ln(k) against 1/T, getting the gradient and re-arranging -Ea/R to get Ea? There is no trick hidden in it? I remember something about when you have a series of bi-molecular steps, it changes what you plot? It was something to do with using the Boltzman distribution where K=[A*]/[A] = exp(-deltaE/RT)? Or am I over complicating things?

cii) Because this is an effective 1st order rate constant doesnt that mean that we are treating a second order reaction as pseudo first? So given that it is a second order reaction would the half life = 1/k[A]₀?

Thanks! :smile:

What do we do when we handle a "pseudo first-order reaction" which leads to an "effective" rate constant? How do these come about? Here's the answer:

Let's take a general case r=k₁[A]^a*[ B]^b*[C]^c. Now let's say I want to just model this reaction as first-order in [C]. Can I do it? No. But I can model it as cth-order in C and zero-order in [A] and [ B], and integrating when the only variable is [C] is easier than when you have three ([A],[ B],[C]). This model is performed by setting the concentrations of [A] and [ B] so high that they are in "large excess": the values [A] and [ B] are taken as constant with time. So now, why don't we subsume them into the rate constant: r=k_1mod*[C]^c, where k_1mod=k₁[A]^a*[ B]^b, and these latter are all constants so this works. (k_1mod is the effective rate constant)

Now half-life should be clear to you. If c=1, then the reaction is pseudo-first order. You can treat it precisely the same as you treat a normal first-order reaction (except that you cannot work with concentrations of A and B changing with time; that's the price for making the {[A]>>[C], [ B]>>C} approximations), with your new constant, k_1mod, acting as the first-order rate constant usually does. So you can write the half-life equation in the same way you do for first-order reactions: t_1/2=log_e(2)/k_1mod.

As for the Arrhenius equation, I don't know what process you would be using to find E_A if you had to treat the rate law as multireactant but if your reaction is pseudo-cth order, by design, you will always be using the same formula for calculations with this reaction as for a usual cth-order reaction, using k_1mod in the place of k. (Unless obviously you cannot call A and B concentrations constant, perhaps because you have to model them with time as well - in which case, you either engineer the reaction to be pseudo ath-order in them, or you just solve the differential equation, what's the big deal anyway? :tongue:

)

(edited 10 years ago)

Reply 2

DonnieBrasco

Original post by Big-Daddy

What do we do when we handle a "pseudo first-order reaction" which leads to an "effective" rate constant? How do these come about? Here's the answer

Very good explanation thank you! So can I just clarify, we would use the half life for a 1st order reaction (t=ln(2)/k but then use our pseudo 1st order rate constant? sorry I cant rep you, PRSOM and all.

Thanks

Reply 3

Big-Daddy

Original post by DonnieBrasco

Very good explanation thank you! So can I just clarify, we would use the half life for a 1st order reaction (t=ln(2)/k but then use our pseudo 1st order rate constant?

No worries, happy to help. It helps me make sure I understand too!

Yup, t_1/2=ln(2)/k_1mod for a pseudo-first order reaction. And for a pseudo-cth order reaction, you just use the modified rate constant in the place of where you would use the normal rate constant for a normal c-order reaction.

The thing to bear in mind is that this isn't the best approximation unless [A] and [ B] are vastly greater than [C] (because obviously, if their concentrations change, your effective rate "constant" is not constant!) Other than that, just remember to calculate the effective constant properly as I showed in my above post.

Hope this helps! :smile:

(edited 10 years ago)

Reply 4

DonnieBrasco

Original post by Big-Daddy

No worries, happy to help. It helps me make sure I understand too!

Yup, t[sub[1/2

=ln(2)/k_1mod for a pseudo-first order reaction. And for a pseudo-cth order reaction, you just use the modified rate constant in the place of where you would use the normal rate constant for a normal c-order reaction.

The thing to bear in mind is that this isn't the best approximation unless [A] and [ B] are vastly greater than [C] (because obviously, if their concentrations change, your effective rate "constant" is not constant!) Other than that, just remember to calculate the effective constant properly as I showed in my above post.

Hope this helps! :smile:

Yep got it :smile:

Thanks again