Club Chemistry: Chemical Kinetics of Valentine’s Day

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!