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?

  • $\begingroup$ No. There must be an instance of $\delta^2$. Also, where did the $\frac{d}{dt}(\beta + \gamma)$ thing come from? $\endgroup$ – Ivan Neretin Dec 17 '19 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 Theis Dec 17 '19 at 18:04
  • $\begingroup$ @IvanNeretin: It came from $A \overset{k_2}{\longrightarrow}B + C$? $\endgroup$ – TheLast Cipher Dec 17 '19 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! $\endgroup$ – TheLast Cipher Dec 17 '19 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 $d\gamma\over dt$. $\endgroup$ – Ivan Neretin Dec 17 '19 at 18:11

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

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

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

| improve this answer | |
  • $\begingroup$ we are left with $2\operatorname{\frac{d}{dt}}\delta(t) = k_4\gamma(t) - 2k_6 - k_5\delta(t)^2 \overset{equivalently}{\Longleftrightarrow} \operatorname{\frac{d}{dt}}\delta(t) = \frac{k_4}{2}\gamma(t) - k_6 - \frac{k_5}{2}\delta(t)^2$? $\endgroup$ – TheLast Cipher Dec 18 '19 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 Dec 18 '19 at 11:00
  • $\begingroup$ @Maurice: Why the $2k_4\alpha(t)$ term on the rate of change of $\delta(t)$? $\endgroup$ – TheLast Cipher Dec 20 '19 at 5:23

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.