2
$\begingroup$

Consider a trimolecular system where the following chemical reactions occur: $$\ce{A + B ->[k_1] A + C}\tag1$$ $$\ce{A ->[k_2] B}\tag2$$ $$\ce{C ->[k_3] C}\tag3$$ Write the differential equations that model the evolution of the three component molar concentrations and determine the stationary solutions (there are two). Finally, if initially $[\ce{A}]=0.75$, $[\ce{B}]=0.25$ and $[\ce{C}]=0$, calculate the time when the mole fraction of $\ce{A}$ is half of its initial value.

I think that the differential equations, following the expression for the reaction velocity, are $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=-k_2[\ce{A}]$$ $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]$$ $$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t}=k_1[\ce{A}][\ce{B}]$$ However, I don't know how to proceed with the stationary solutions. I guess that one of them would be obtained doing: $$-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]=0$$ but I'm not sure.

Finally, I got the relations $$[\ce{B}]=\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+\frac{k_2}{k_1}$$ $$[\ce{C}]=-\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+1-\frac{k_2}{k_1}-[\ce{A}]$$ but I have no idea how to obtain the time. Any help would be appreciated.

$\endgroup$
  • $\begingroup$ What is the meaning of $C-> C$ with rate constant $k_3$ ?? $\endgroup$ – Soumik Das Mar 28 '18 at 18:50
  • $\begingroup$ That's in the problem statement, and it doesn't give more information about it... I'm physicist and I got no idea about chemistry, so I cannot give you the answer $\endgroup$ – user326159 Mar 28 '18 at 19:15
  • 2
    $\begingroup$ @SoumikDas No, because A is neither spent nor produced in the first reaction. $\endgroup$ – Ivan Neretin Mar 28 '18 at 19:21
  • 1
    $\begingroup$ The third equation still makes no sense, even if it is a part of the problem statement. Also, what is a stationary solution and why do you think you need one? $\endgroup$ – Ivan Neretin Mar 28 '18 at 19:27
  • 3
    $\begingroup$ Well, then you don't need many equations to do that. A goes to B and never goes back, B goes to C and never goes back, so in the end there will be only C (and maybe some B which can't transform due to the lack of A). $\endgroup$ – Ivan Neretin Mar 28 '18 at 19:50
1
$\begingroup$

Your differential equations are correct.

From $\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_1[\ce{A}][\ce{B}]+k_2[\ce{A}] = 0$ follow the stationary solutions $[\ce{A}] = 0$ or $[\ce{B}] = \frac{k_2}{k_1}$.

Reaction (2) obeys a first order decay law. Its half-life is $t_{\frac{1}{2}} = \mathrm{ln}(2) / k_2$.

The image below shows how $[\ce{B}]$ reaches the stationary state soon after the reaction has started ($k_1 = 0.3, k_2 = 0.2$).

k1=0.3, k2=0.2

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.