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!

Mitch

1. Noel says:

Isn’t this a bit of a simplification? Getting together is at least a second order reaction. Possibly with several pathways and transition states.

Happy v day everyone!

2. Chemjobber says:

You’re not an organic chemist, are you?

• mitch says:

Just trying to be a well rounded chemist.

3. excimer says:

For me, that reaction always tends to be in equilibrium.

…if you catch my drift…

• Primary Al says:

I looked at a Higgs Boson last summer in Detroit at a Chevy dealership. My only criticism was that it looked like had very little legroom, and the super collider that came with it, they wanted to install on my farm.

Nevertheless, you would be surprised at how many parts it had. The owner’s Parts, Maintenance and Repair manual was two and a half trillion pages long. I didn’t recognize any of it. Neither did they but it was still for sale.

4. Jason says:

Surely the reaction A+B -> AB is more than an elementary, single-step reaction… in my experience, it involves a rather complicated, multi-step mechanism with several potential non-productive pathways!

5. I love it. 🙂