# Thread: Help with this separable differential equation problem

1. ## Help with this separable differential equation problem

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

d[C]/dt = k[A][B]

Thus, if the initial concentrations are [A] = a moles/L and [B] = b moles/L and we write x = [C]. then we have:

dx/dt = k(a - x)(b - x)

a) Assuming that a != b, find x as a function of t. Use the fact that the initial concentration of C is 0.

b) Find x(t) assuming that a = b. How does this expression for x(t) simplify if it is known that [C] = 1/2a after 20 seconds?

I honestly have no idea where to start.

2. Originally Posted by pantsaregood
In the elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: A + B -> C. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B:

d[C]/dt = k[A][b]

Thus, if the initial concentrations are [A] = a moles/L and [b] = b moles/L and we write x = [C]. then we have:

dx/dt = k(a - x)(b - x)

a) Assuming that a != b, find x as a function of t. Use the fact that the initial concentration of C is 0.

b) Find x(t) assuming that a = b. How does this expression for x(t) simplify if it is known that [C] = 1/2a after 20 seconds?

I honestly have no idea where to start.
I will give you a start then:

$\frac{dx}{dt} = k (a - x) (b - x) \Rightarrow \frac{dt}{dx} = \frac{1}{k} \cdot \frac{1}{(a - x) (b - x)}$.

Now get the partial fraction decomposition of $\frac{1}{(a - x) (b - x)}$.

Now integrate with respect to x.

etc.