If the members of group A and group B want to form a union AB it can be described by the following chemical equation.

$$! \text{A} + \text{B} \rightarrow \text{AB} $$

which will have a rate constant of

$$! R = k[\text{A}][\text{B}] $$

Assuming this is an elementary process we can solve for the rate of this reaction by the introduction of a progress variable $$ x $$.

$$! x = ([\text{A}]_0 – [\text{A}]_t) = ([\text{B}]_0 – [\text{B}]_t) $$

Substituting $$ \frac{dx}{dt} $$ for $$ R $$ yields…

$$! \frac{dx}{dt} = k([\text{A}]_0 – x)([\text{B}]_0 – x) $$

And to determine the time behavior we simply integrate.

$$! \int_{x(0)}^{x(t)} \frac{dx}{([\text{A}]_0 – x)([\text{B}]_0 – x)} = k\int_0^t dt $$

Using the method of partial fractions

$$! \int_0^x \frac{dx}{([\text{A}]_0 – [\text{B}]_0)([\text{B}]_0 – x)} – \int_0^x \frac{dx}{([\text{A}]_0 – [\text{B}]_0)([\text{A}]_0 – x)} = k\int_0^t dt $$

Integrating…

$$! -\frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln\left([\text{B}]_0 – x\right)_0^x + \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln\left([\text{A}]_0 – x\right)_0^x = kt $$

Grouping…

$$! \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln\left(\frac{([\text{A}]_0 – x)}{([\text{B}]_0 – x)}\right)_0^x = kt $$

Evaluating this from 0 to $$ x $$

$$! \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln\left(\frac{([\text{A}]_0 – x)}{([\text{B}]_0 – x)}\right) – \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln\left(\frac{([\text{A}]_0 – 0)}{([\text{B}]_0 – 0)}\right) = kt $$

$$! \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln \left(\frac{([\text{A}]_0 – x)}{([\text{B}]_0 – x)}\right) – \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln \left(\frac{[\text{A}]_0}{[\text{B}]_0}\right) = kt $$

Remembering that $$ [\text{A}]_0 – x = [\text{A}]_t $$

$$! \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln \left(\frac{[\text{A}]_t}{[\text{B}]_t} \right) – \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln \left(\frac{[\text{A}]_0}{[\text{B}]_0}\right) = kt $$

Simplifying, we finally have an expression for the union of two reactive groups of people on Valentine’s day.

$$! \frac{1}{([\text{A}]_0 – [\text{B}]_0)}\ln \left(\frac{[\text{A}]_t[\text{B}]_0}{[\text{B}]_t[\text{A}]_0} \right) = kt $$

May the rate constant ($$k$$) be large today!

Mitch