can anyone solve this...need help with this please...?

Question

In a chemical reaction,one molecule of P combines with one molecule of Q to form one molecule R. At the start of the reaction there are a molecules of each of P and Q and no molecules of R. After t hours x molecules of R have been formed. The rate of formation of R at any instant is k times the product of the number of molecules of P and the number of molecules of Q present. Form a differential equation for x and hence find x in terms of a, k and t.

Answer 1

The reaction equation is
P + Q → R
let p,q and r be the number of molecules of P,Q and R respectively. From the reaction equation you can see that the changes of number of molecules are related by.
- Δp = - Δq = x
Thus the amounts at any time t are:
p = a - x
q = a - x
r = 0 + x = x

The rate of formation of r is proportional to p and q:
thus
dr/dt = dx/dt = k · p · q
(k is the rate constant)
←→
dx/dt = k ·(a-x)²

Solve this DE by separation of variables:
1/(a-x)² dx = k dt
←→
∫1/(a-x)² dx = ∫ k dt + c
←→
1/(a-x) = k·t + c

with the initial condition :
for t = 0 → x= 0
1/(a-0) = k·0 + c
←→
c = 1/a

→
1/(a-x) = k·t + 1/a
←→
x = a - 1/(k·t + 1/a)
←→
x = a - 1/(k·t + 1/a)
←→
x = a (1 - 1/(a·k·t + 1)
←→
x = a / (1 + 1/(a·k·t) )

Answer 2

We know that at the start, there are a molecules of P and a molecules of Q. Since for every molecule of P used one molecule of Q is also used, the number of molecules of each type remaining is the same. Further, since exactly one molecule of each type is used in each molecule of R, it follows that if there are x molecules of R, there are a-x molecules of both P and Q. Since the rate of formation is k* (# of molecules of P) * (# of molecules of Q), it follows that:

dx/dt = k(a-x)²

Now all we have to do is solve this equation. Dividing by (a-x)² we have:

1/(a-x)² dx/dt = k

Integrating both sides with respect to t:

1/(a-x) = kt + C

To find the value of C, we note that at t=0, x=0, so we have:

1/a = C

Substituting this value for C into the equation and solving for x:

1/(a-x) = kt + 1/a
1 = (a-x)kt + (a-x)/a
1 = akt - xkt + 1 - x/a
1 = akt + 1 - x(kt+1/a)
x(kt+1/a) = akt
x = akt/(kt+1/a)

Or more aesthetically:

x = a²kt/(akt+1)

And we are done.