I'm interested in the second order reaction $\ce{A + B -> P}$. Mass action kinetics is assumed.

Let $[A]_0 = a$ and $[B]_0 = b$, then $[A] = a - x$ and $[B] = b - x$. The rate law becomes

$$-\frac{\mathrm dx}{\mathrm dt} = -k([A]_0 - x)([B]_0 - x)$$

Why is the right-hand side of the differential equation negative? Shouldn't it just be the left side when we look at the decrease of the reactant $\ce{A}$? Or is it because $x$ is equal to the concentration of $\ce{P}$?

  • 4
    $\begingroup$ The right side can be negative because someone has placed a minus sign before it. Ditto for the left side. To me, both sides would look better without it. $\endgroup$ Dec 10 '17 at 16:38
  • $\begingroup$ But why does it make sense? Usually there is a minus sign only on the left part when you look at the decrease of a reactant.. $\endgroup$
    – AGR
    Dec 10 '17 at 17:28
  • 2
    $\begingroup$ But you don't look at the decrease of a reactant. x is the concentration of P, and it is increasing. $\endgroup$ Dec 10 '17 at 17:31
  • $\begingroup$ I recommend this page: chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/… $\endgroup$
    – DSVA
    Dec 10 '17 at 22:07
  • $\begingroup$ For the problem $\ce{a + B -> P}$ the mathematical equation could use a bit better notation to see what is going on. $$-\dfrac{d\ce{[P]}}{dt} = \ce{-k([A]_o - [P]_t)([B]_o - [P]_t)}$$ $\endgroup$
    – MaxW
    Dec 11 '17 at 8:19

Assuming mass action kinetics, we have the following differential equation

$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = -\kappa \, [\ce{A}] [\ce{B}]$$

where $\kappa > 0$ is the rate constant. If we denote

$$x (t) := [\ce{A}]_0 - [\ce{A}] = [\ce{B}]_0 - [\ce{B}]$$


$$\frac{\mathrm d}{\mathrm d t} [\ce{A}] = \frac{\mathrm d}{\mathrm d t} \left( [\ce{A}]_0 - x \right) = 0 - \dot x = - \dot x = -\kappa \left( [\ce{A}]_0 - x \right) \left( [\ce{B}]_0 - x \right)$$

  • 2
    $\begingroup$ While I personally understand your answer (and up'd it), I would assume, that the last step is a bit too contracted for everyone to follow. (But in its brevity it's beautiful.) $\endgroup$ Dec 11 '17 at 13:01

The main issue I see here is that there's too much fixation on how the equation normally looks. Instead, you would find much more insight into looking at the physical meaning conveyed by the equation.

For example, you can plug in initial and final values to see if the function describes what we expect.

At the beginning of the reaction, $x=0$. Your equation gives $\frac{\mathrm{d}x}{\mathrm{d}t} = k\ce{[A]}_{0}\ce{[B]}_{0}$. This is exactly what's expected, as this is the maximum value of the rate of product formation.

Let's arbitrarily assume that $\ce{[A]}\geq \ce{[B]}$. Then the reaction is done when $x = \ce{[A]}_{0}$. If you plug this value into your equation, you find that $\frac{\mathrm{d}x}{\mathrm{d}t} = 0$ which is also correct.

And as Rodrigo pointed out, the main reason you have sign confusion is that you've chosen to represent the concentration of product as opposed to the concentration of reactant, and since one is produced while the other is consumed, you get an extra minus sign.


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.