Second order reaction with two reactants I'm interested in the second order reaction $\ce{A + B -&gt; P}$. 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 $\ce{P}$?
Assuming mass action kinetics, we have the following differential equation

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

where $\kappa &gt; 0$ is the rate constant. If we denote

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

then

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