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We are given the following two equation,

The first reaction is an equilibrium reaction. $$A_2 \rightarrow A+A \quad (fast)$$ $$A+B\rightarrow P \quad (slow)$$

Where A is the intermediate. Now I have to find the rate law using the steady state approximation.

I have come to this following equation for the net change of [A],

$$\frac{d[A]}{dt}=k_1[A_2]-k_2[A]^2-k_3[A][B]=0$$

Where I have said that $k_1$ is for the forwards reaction, $k_2$ is the for the backward reaction in the first one. $k_3$ is the final reaction's rate constant. Now what I dont understand is this - How can I solve the above equation and find an expression for $[A]$.

Is there some trick that I can use, or something? I haven't been able to solve it, since it's a quadratic formula and I cannot really solve a quadratic formula with just this given information. I even tried using the quadratic formula, but that was a dead end after some lines of working out.

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  • $\begingroup$ Why are you trying to find $\ce{[A]}$? The third term is the rate of reaction, almost certainly what you actually want. Combine the first two terms into the equilibrium expression for the fast reaction. $\endgroup$ – Zhe Nov 21 '16 at 22:52
  • $\begingroup$ Well but [A] is the intermediate, so that is why i can not use the third term. I need to find an expression in terms of only $[A_2]$ and [B]. $\endgroup$ – MathCurious314 Nov 21 '16 at 22:58
  • $\begingroup$ No, the second step is fast. There will not be a $\ce{B}$ term in the rate. But the third term is the rate of product formation. I'll try to write this out for you when I get back to my desk. Typing LaTeX on a phone is hard... $\endgroup$ – Zhe Nov 22 '16 at 1:16
  • $\begingroup$ Also, I can't read, so the second step isn't fast... :( $\endgroup$ – Zhe Nov 22 '16 at 1:53
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I apologize for my comments; I shouldn't be answering Chem.SE questions during my commute without writing stuff out...

I renamed some of your rate constants. In addition, notice that as written, the reverse of the first reaction does not lead directly to a component of $\frac{d\ce{[A]}}{dt}$ because two $\ce{A}$'s are being consumed.

$$\ce{A2 \underset{k_{-1}}{\overset{k_{1}}{<=>}} 2A}$$ $$\ce{A + B \overset{k_{2}}{->} P}$$ $$\mathrm{rate} = \frac{d\ce{[P]}}{dt} = k_{2}\ce{[A][B]}$$

Your steady state equation is almost correct:

$$\frac{d\ce{[A]}}{dt}=2k_{1}\ce{[A2]}−\frac{k_{-1}}{2}\ce{[A]}^{2}−k_{2}\ce{[A][B]}=0$$

I actually ended up solving the quadratic equation. It wasn't that bad:

$$\ce{[A]} = \frac{\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}}{k_{-1}}$$

so:

$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}$$

Sanity check the two limits:

  1. $k_{2}\ce{[B]} \gg \sqrt{4k_{1}k_{-1}\ce{[A_{2}]}}$
  2. $k_{2}\ce{[B]} \ll \sqrt{4k_{1}k_{-1}\ce{[A_{2}]}}$

Case 1 means that the second reaction is very fast. Case 2 means that the second reaction is very slow.

Case 1: $$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}=\frac{k_{2}\left(k_{2}\ce{[B]}\sqrt{1 + \frac{4k_{1}k_{-1}\ce{[A2]}}{k_{2}^{2}\ce{[B]}^{2}}}-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}$$

Taylor expand the root: $\sqrt{1+x^{2}} \approx 1+\frac{x^{2}}{2}$

$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\left(k_{2}\ce{[B]}\left(1 + \frac{2k_{1}k_{-1}\ce{[A2]}}{k_{2}^{2}\ce{[B]}^{2}}\right)-k_{2}\ce{[B]}\right)\ce{[B]}}{k_{-1}}=2k_{1}\ce{[A2]}$$

In other words, the intermediate is consumed in the forward direction at the rate that we can fragment $\ce{A2}$. This seems right.

Case 2: $$$$ $\sqrt{k_{2}^{2}\ce{[B]}^{2}+4k_{1}k_{-1}\ce{[A2]}}-k_{2}\ce{[B]}$ reduces to $\sqrt{4k_{1}k_{-1}\ce{[A2]}}$.

$$\frac{d\ce{[P]}}{dt} = \frac{k_{2}\sqrt{4k_{1}k_{-1}\ce{[A2]}}\ce{[B]}}{k_{-1}}$$

The thing you want is somewhere between the complicated first rate I wrote and the limit in case 2. This seems reasonable in that it is first order in $\ce{B}$ as you might expect from a transition state that involves a single $\ce{B}$.

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    $\begingroup$ The square root term in your last equation is in effect the equilibrium constant for the $\ce{A2 = A + A}$ reaction as would be expected as this is the fastest reaction $\endgroup$ – porphyrin Nov 23 '16 at 9:47
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If the first reaction is fast, then the third term can be neglected in comparison to the other two terms in the equation $$\frac{d[A]}{dt}=k_1[A_2]-k_2[A]^2-k_3[A][B]=0$$This immediately gives:$$[A]=\sqrt{\frac{k_1[A_2]}{k_2}}$$Therefore, $$\frac{d[P]}{dt}=k_3\sqrt{\frac{k_1}{k_2}}[B]\sqrt{[A_2]}$$

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