How should I start answering this chemical substance question?

Question

Two substances, A and B, are being converted into a single compound C. In the laboratory it has been shown that, for these substances, the following law of conversion holds: the time rate of change of the amount x of compound C is proportional to the product of the amounts of unconverted substances A & B. If at time t = 0 there are C_AO mols/li of substance A and C_BO mols of substance B, and none of compound C present, find x(t). Assume that 1 mole of C is formed from the combination of 1 mole of A and one mole of B.

Answer 1

Start by setting up equations.

I prefer to write $A_0$ and $B_0$ for $C_{AO}$ and $C_{BO}$, respectively, and $C$ for $x$.

$\dfrac{dC}{dt}=-\dfrac{dA}{dt}=-\dfrac{dB}{dt}=kAB$

$A+C=A_0$

$B+C=B_0$

Therefore $\dfrac{dC}{dt}=k(A_0-C)(B_0-C)=k(A_0B_0-(A_0+B_0)C+C^2)$.

This is a separable differential equation, which can be solved with standard methods.

Answer 2

Starting from @J. W. Tanner's answer, the easy way to solve $$\dfrac{dC}{dt}=k(A_0-C)(B_0-C)$$ is to write it as $$\dfrac{dt}{dC}=\frac 1{k(A_0-C)(B_0-C)}=\frac{1}{k (A_0-B_0)}\left( \frac 1{C-A_0}- \frac 1{C-B_0}\right)$$ which gives (since $C_0=0$ $$t =\frac{1}{k (A_0-B_0)}\,\log \left(\frac{C-A_0}{C-B_0}\right)$$ from which it is easy to express $C(t)$.