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We have the following elementary reactions: \begin{align} &\overset{k_1}{\longrightarrow}A\overset{k_2}{\longrightarrow}{B+C}\\ &C \overset{k_4}{\longrightarrow} 2D \overset{k_5}{\longrightarrow}{C}\\ &B \overset{k_3}{\longrightarrow}\\ &D \overset{k_6}{\longrightarrow} \end{align} Let $\alpha(t), \beta(t), \gamma(t), \delta(t)$ denote the concentrations of $A, B, C, D$, respectively. \begin{align} \frac{\mathrm d}{\mathrm dt}\alpha(t) &= k_1 - k_2\alpha(t)\\[3pt] \frac{\mathrm d}{\mathrm dt}\beta(t) &= k_2\alpha(t) - k_3\beta(t)\\[3pt] \frac{\mathrm d}{\mathrm dt}\bigl(2\delta(t)\bigr) = 2\frac{\mathrm d}{\mathrm dt}\delta(t) &= k_4\gamma(t)-k_6\delta(t)^2 \quad \text{or} \quad \frac{\mathrm d}{\mathrm dt}\delta(t) = \frac{k_4}{2}\gamma(t) - \frac{k_6}{2}\delta(t)^2\\[3pt] \frac{\mathrm d}{\mathrm dt}\gamma(t) &= k_5\delta(t)^2 - k_4\gamma(t) + k_2\alpha(t)\\ \end{align} Is my systems of equations correct?

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  • $\begingroup$ No. There must be an instance of $\delta^2$. Also, where did the $\frac{d}{dt}(\beta + \gamma)$ thing come from? $\endgroup$ Commented Dec 17, 2019 at 17:48
  • $\begingroup$ Where is the law of mass action (I'm surprised you are invoking it for a non-equilibrium system)? $\endgroup$
    – Karsten
    Commented Dec 17, 2019 at 18:04
  • $\begingroup$ @IvanNeretin: It came from $A \overset{k_2}{\longrightarrow}B + C$? $\endgroup$ Commented Dec 17, 2019 at 18:06
  • $\begingroup$ @KarstenTheis: Oh no. I must have misunderstood what it meant. Sorry, my background is not in chemistry. I will remove it. Thanks! $\endgroup$ Commented Dec 17, 2019 at 18:10
  • $\begingroup$ @TheLastCipher Well, this reaction does not produce B or C; instead, each time it runs, it produces B and C, so it should be represented by two addends, both in $d\beta\over dt$ and $d\gamma\over dt$. $\endgroup$ Commented Dec 17, 2019 at 18:11

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$\displaystyle\frac{\mathrm d}{\mathrm dt}\alpha(t) = k_1 - k_2\alpha(t)$ is correct.

Just to show you the way, here are the next equations:

$$ \begin{align} \frac{\mathrm d}{\mathrm dt}\beta(t) &= k_2\alpha(t) - k_3\beta(t) \\ \frac{\mathrm d}{\mathrm dt}\gamma(t) &= k_2\alpha(t) - k_4\gamma(t) + k_5\delta(t)^2 \\ \frac{\mathrm d}{\mathrm dt}\delta(t) &= 2k_4\alpha(t) - k_6\delta(t) - k_5\delta(t)^2 \end{align} $$

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  • $\begingroup$ we are left with $2\operatorname{\frac{d}{dt}}\delta(t) = k_4\gamma(t) - 2k_6 - k_5\delta(t)^2 \overset{equivalently}{\Longleftrightarrow} \operatorname{\frac{d}{dt}}\delta(t) = \frac{k_4}{2}\gamma(t) - k_6 - \frac{k_5}{2}\delta(t)^2$? $\endgroup$ Commented Dec 18, 2019 at 3:03
  • $\begingroup$ I appended the last equation from your another answer and cleaned up syntax a little. Please note there is a gray edit button on the bottom of the posts, which allows to commit changes to the Q&As, which is a preferred way over adding a new post if the addition is marginal. Also, please avoid breaking math expressions in the middle with dollar signs for no reason: $x + y$ is correct, $x$ + $y$ is wrong. Math operators are usually typed in upright typeface to avoid confusion with variables. $\endgroup$
    – andselisk
    Commented Dec 18, 2019 at 11:00
  • $\begingroup$ @Maurice: Why the $2k_4\alpha(t)$ term on the rate of change of $\delta(t)$? $\endgroup$ Commented Dec 20, 2019 at 5:23

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