Analytical solution for kinetics of bimolecular reaction
Assuming that the bimolecular chemical reaction $\ce{A + B -&gt;[\kappa] C}$ has mass action kinetics, we have the following pair of coupled ODEs

$$\begin{array}{rl} \dot a &amp;= - \kappa \, a \, b\\ \dot b &amp;= - \kappa \, a \, b\end{array}$$

where $\kappa &gt; 0$ is the rate constant, $a := [\ce{A}]$ and $b := [\ce{B}]$. Since $\dot a = \dot b$, we have $\frac{\mathrm d}{\mathrm d t} \left( a - b \right) = 0$ and, thus, integrating, we obtain

$$a (t) - b (t) = a_0 - b_0$$

where $a_0 &gt; 0$ and $b_0 &gt; 0$ are the initial concentrations. Since $b (t) = a (t) - (a_0  - b_0)$, the 1st ODE can be decoupled from the 2nd, as follows

$$\dot a = - \kappa \, a \, \left( a - (a_0  - b_0) \right)$$

which can be rewritten in the form

$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$

Assuming that $a_0 \neq b_0$, we have the following partial fraction expansion

$$\left( \frac{1}{a - (a_0  - b_0)} - \frac{1}{a} \right) \mathrm d a = - \kappa \, (a_0  - b_0) \, \mathrm d t$$

Integrating, we obtain

$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a_0 - (a_0  - b_0)} \right) - \ln \left( \frac{a (t)}{a_0} \right) = - \kappa \, (a_0  - b_0) \, t$$

which can be rewritten as follows

$$\ln \left( \frac{a (t) - (a_0  - b_0)}{a (t)} \right) = \ln \left( \frac{b_0}{a_0} \right) - \kappa \, (a_0  - b_0) \, t$$

Exponentiating both sides, we obtain

$$\frac{a (t) - (a_0  - b_0)}{a (t)} = \frac{b (t)}{a (t)} = \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)$$

and, eventually, we obtain

$$\boxed{\begin{array}{rl} &amp;\\ a (t) &amp;= \dfrac{a_0  - b_0}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\\\ b (t) &amp;= \dfrac{(a_0  - b_0) \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}{1 - \left( \frac{b_0}{a_0} \right) \, \exp (- \kappa \, (a_0  - b_0) \, t)}\\ &amp; \end{array}}$$

Taking the limit,

$$\lim_{t \to \infty} a (t) = \begin{cases} a_0 - b_0 &amp; \text{if } a_0 &gt; b_0\\\\ 0 &amp; \text{if } a_0 &lt; b_0\end{cases}$$

$$\\$$

$$\lim_{t \to \infty} b (t) = \begin{cases} 0 &amp; \text{if } a_0 &gt; b_0\\\\  b_0 - a_0 &amp; \text{if } a_0 &lt; b_0\end{cases}$$



What if $a_0 = b_0$?

Previously, we assumed that $a_0 \neq b_0$. If $a_0 = b_0$, then

$$\frac{\mathrm d a}{a \, \left( a - (a_0  - b_0) \right)} = - \kappa \, \mathrm d t$$

becomes

$$-\frac{\mathrm d a}{a^2} = \kappa \, \mathrm d t$$

Integrating, we obtain

$$\frac{1}{a (t)} - \frac{1}{a_0} = \kappa \, t$$

and, eventually, we obtain

$$\boxed{ a (t) = \frac{a_0}{1 + a_0 \, \kappa \, t} = b (t)} $$

In this case, both reactants are eventually exhausted

$$\lim_{t \to \infty} a (t) = \lim_{t \to \infty} b (t) = 0$$