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I'm blocked in a step of getting to the equation $(7)$, more precisely in the step between $(5)$ and $(6)$. I show my way to solve it down, but I don't get the same as $(6)$, so I wanted to know if I was wrong and an explanation of the correct aproach. $$\ce{A ->[k_\mathrm{1}]I->[k_\mathrm{2}]P}$$

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_\mathrm{1}[\ce{A}] \tag{1}$$ $$\frac{\mathrm d[\ce{I}]}{\mathrm dt} = k_\mathrm{1}[\ce{A}] -k_\mathrm{2}[\ce{I}] \tag{2}$$ $$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = k_\mathrm{2}[\ce{I}] \tag{3}$$

$$[\ce{A}]_\mathrm{0}\neq0 ; [\ce{I}]_\mathrm{0}=0; [\ce{P}]_\mathrm{0}=0$$ $$[\ce{A}]=[\ce{A}]_\mathrm{0}e^{-k_\mathrm{1}t} \tag{4}$$

$$\frac{\mathrm d[\ce{I}]}{\mathrm dt} = k_\mathrm{1}[\ce{A}]_\mathrm{0}e^{-k_\mathrm{1}t} -k_\mathrm{2}[\ce{I}] \tag{5}$$

$$[\ce{I}]=\frac{k_\mathrm{1}}{k_\mathrm{2}-k_\mathrm{1}}(e^{-k_\mathrm{1}t}-e^{-k_\mathrm{2}t})[\ce{A}]_\mathrm{0} \tag{6}$$

$$[\ce{A}]_\mathrm{0}=[\ce{A}]-[\ce{I}]-[\ce{P}]$$ $$[\ce{P}]=[\ce{A}]-[\ce{A}]_\mathrm{0}-[\ce{I}]$$

$$[\ce{P}]=(\frac{k_\mathrm{1}e^{-k_\mathrm{2}t}-k_\mathrm{2}e^{-k_\mathrm{1}t}}{k_\mathrm{2}-k_\mathrm{1}}+1)[\ce{A}]_\mathrm{0} \tag{7}$$

And here is what I've done:

$$\int_{0}^{[\ce{I}]} \, \mathrm{d}[\ce{I}]=k_\mathrm{1}[\ce{A}]_\mathrm{0}\int_{0}^{t}e^{-k_\mathrm{1}t} \, \mathrm{d}t-k_\mathrm{2}[\ce{I}]\int_{0}^{t} \, \mathrm{d}t \tag{A}$$

$${[\ce{I}]}=-e^{-k_\mathrm{1}t}[\ce{A}]_\mathrm{0}-k_\mathrm{2}t[\ce{I}] \tag{B}$$

$${[\ce{I}]}=-[\ce{A}]_\mathrm{0}(\frac{e^{-k_\mathrm{1}t}}{k_\mathrm{2}t+1}) \tag{C}$$

Anyone could help me? Thanks :)

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  • $\begingroup$ This symbolic integrator to resolve both infinite and finite integrals may help you: integral-calculator.com // It is very powerful. $\endgroup$
    – Poutnik
    Aug 16, 2022 at 9:21
  • $\begingroup$ Thanks a lot, but I actually know how to solve thos integrals and I've just checked them. The question was how to solve the differential equation to reach equation (6). $\endgroup$ Aug 16, 2022 at 9:27
  • $\begingroup$ It looks to me you need to separate terms to integral equation for I and t on respective sides. $\endgroup$
    – Poutnik
    Aug 16, 2022 at 13:43
  • $\begingroup$ I end up with the same because I can't pass "dt" to the right without affecting "k2[I]". $\endgroup$ Aug 16, 2022 at 13:51
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    $\begingroup$ @JorgeBonifaz I claimed it was related. I did not claim it was a duplicate. $\endgroup$ Aug 18, 2022 at 9:00

1 Answer 1

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This is a form of a linear first-order differential equation, and you can solve it like this (or just plug it into wolfram alpha).

We have

$$\frac{d[I]}{dt} = k_1[A]_0e^{-k_1t} - k_2[I]$$ $$\frac{d[I]}{dt} + k_2[I] = k_1[A]_0e^{-k_1t}$$

If we let $P(t) = k_2$, $Q(t) = k_1[A]_0e^{-k_1t}$, and $I(t) = e^{\int P(t)dt} = e^{k_2t}$ the solution to this sort of differential equation is

$$[I] = \frac{1}{I(t)} \left( \int I(t)Q(t)dt + C \right)$$

where C is a constant. Plugging in the values set for this problem, we have

$$[I] = \frac{1}{e^{k_2t}} \left( \int e^{k_2t}\cdot k_1[A]_0e^{-k_1t} dt + C\right)$$

$$[I] = Ce^{-k_2t} + k_1[A]_0\cdot\frac{1}{k_2-k_1}e^{(k_2-k_1)t}\cdot e^{-k_2t}$$

$$[I] = Ce^{-k_2t} + k_1[A]_0\cdot \frac{1}{k_2-k_1}e^{-k_1t}$$

Now plugging in $t = 0$, we know $[I]_0 = 0$. So:

$$C = -\frac{k_1[A]_0}{k_2-k_1}$$

Reinputting C to the previous equation gives:

$$[I] = -\frac{k_1[A]_0}{k_2-k_1} e^{-k_2t} + \frac{k_1[A]_0}{k_2-k_1} e^{-k_1t}$$

$$[I] = \frac{k_1}{k_2-k_1}(e^{-k_1t} - e^{k_2t})[A]_0$$

We have to do this because [I] depends on t as well, so you can't just take it out of the integral like you did.

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    $\begingroup$ Thanks a lot. I'm not in university yet, so I didn't know how to solve differential equations, but now I understand it. $\endgroup$ Aug 17, 2022 at 13:16

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