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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.

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  • $\begingroup$ What is the meaning of $C-> C$ with rate constant $k_3$ ?? $\endgroup$
    – Soumik Das
    Commented Mar 28, 2018 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
    Commented Mar 28, 2018 at 19:15
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    $\begingroup$ @SoumikDas No, because A is neither spent nor produced in the first reaction. $\endgroup$
    – Ivan Neretin
    Commented Mar 28, 2018 at 19:21
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    $\begingroup$ 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
    Commented Mar 28, 2018 at 19:27
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    $\begingroup$ 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
    Commented Mar 28, 2018 at 19:50

1 Answer 1

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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

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