Chapter 9.3, Problem 40E

In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: $\text{A + B}\to \text{C}$. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B:

$\frac{d\left[\text{C}\right]}{dt}=k\left[\text{A}\right]\left[\text{B}\right]$

(See Example 2.7.4.) Thus, if the initial concentrations are $\left[\text{A}\right]=a\text{ moles/L}$ and $\left[\text{B}\right]=b\text{ moles/L}$ and we write $x=\left[\text{C}\right]$, then we have

$\frac{dx}{dt}=k\left(a-x\right)\left(b-x\right)$

(a) Assuming that $a\ne b$, find x as a function of t. Use the fact that the initial concentration of C is 0.

(b) Find $x\left(t\right)$ assuming that $a=b$. How does this expression for $x\left(t\right)$ simplify if it is known that $\left[\text{C}\right]=\frac{1}{2}a$ after 20 seconds?