[SinkingTako]

(see attached ) for this reaction, the reaction of A proceeds competitively towards product B and C. where B is the kinetic product and C is the thermodynamic product.

B    A C

the various rates are given:
k1=1
k-1=0.01
k2=0.1
k-2=0.0005(min-1 )

estimate the product ratio B/C in the 1st 4minutes of the reaction.

I have absolutely no idea. thinking along the lines of steady state (forward rate =backward rate). however reaction is not in equilibrium,  so I'm really confused... Thank-you!

[curiouscat]

Well, show your attempt.

If I were to guess I'd say 10:1

[SinkingTako]

Oh okay. So for reaction A B, rate=k1[A]= [itex]\frac{d[B ]}{dt}[/itex]

Then same for reaction A   C, rate=k2[A]= [itex]\frac{d[C]}{dt}[/itex]


So [itex]\frac{d[B ]}{d[C]}=\frac{B}{C}=\frac{k_1}{k_2}=10[/itex]


However, does this not consider the backward reaction that is always taking place too? So while A B, wouldn't B   A happen at the same time? Albeit the backward rate is quite slow, how to account for this?

Also I don't really get the thing about the 4 minutes. Why 4 minutes??? How will it affect... The second part of the problem states that estimate the product ratio B/C after 4days (when supposedly the reaction is in equilibrium... but how could one know?)

Sorry, I'm really noob... Thank yoU!

[curiouscat]

At large times, assume equilibrium prevails:

[tex]

\frac{[ B]}{[A]}=100 \\
\frac{[C]}{[A]}=200 \\

\therefore \frac{[ B]}{[C]}=0.5 \\
[/tex]

However, does this not consider the backward reaction that is always taking place too? So while A  B, wouldn't B   A happen at the same time? Albeit the backward rate is quite slow, how to account for this?

Yes. By using math.

The full solution is given by simultanously solving the pair of ODE's as an initial value problem:

[tex]
\frac{dB}{dt}=k_1c_A - k_{-1} c_B\\
\frac{dC}{dt}=k_2c_A - k_{-2} c_C\\
[/tex]

At all times
[tex]
\begin{align} [A]_0 = [A] + [ B] + [C]  \end{align}
[/tex]

Sorry, I don't have a CAS package handy and am too lazy to solve this by hand.

[SinkingTako]

Oh. So just assume time is short its initial rate then if its after a few days the reactionwill be in equilibrum...Thanks! I haven't learnt differential equations yet so will try to figure out that later on: P thanks again!

[curiouscat]

Oh. So just assume time is short its initial rate then if its after a few days the reactionwill be in equilibrum...Thanks! I haven't learnt differential equations yet so will try to figure out that later on: P thanks again!

I suppose so. Unless there's some other way to do it.