Question

Derive the expression for second order reaction with unequal concentration for a reaction `A + B ->` product when (1) if a > b & (2) if b > a, where "a mol/`dm^3`" and "b mol/`dm^3`" are the initial concentrations of A & B respectively?

Answer 1

WARNING! This answer involves calculus!

Explanation:

Assume we have a second order reaction of the type

`"A + B" stackrel(k)(→) "Products"`

Let `x " mol/dm"^3` be the amount reacted in time `t`. Then

`dx/dt = k["A"]["B"]`.

Let `a` and `b` be the initial concentrations of `"A"` and `"B"`, and `a≠b`. Then

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

`int_0^x dx/((a-x)(b-x)) = int_0^tkdt = kt`

We use the method of partial fractions to evaluate the first integral. This gives

`int_0^x dx/((a-x)(b-x)) =int_0^x dx/((b-a)(a-x)) + int_0^x dx/((a-b)(b-x))`

`= 1/(b-a)int_0^xdx/(a-x) + 1/(a-b)int_0^xdx/(b-x)`

We use the method of u-substitution to evaluate the new integrals and get:

`1/(b-a)int_0^xdx/(a-x) + 1/(a-b)int_0^xdx/(b-x)`

`=1/(b-a)ln(a/(a-x))+1/(a-b)ln(b/(b-x))`

`=1/(b-a)(ln(a/(a-x))-ln(b/(b-x)))`

∴ `1/(b-a)ln ((a(b-x))/(b(a-x))) = kt`

We can arrange this to get the integrated rate law:

`ln ((a(b-x))/(b(a-x))) = (b-a)kt`

In terms of the original symbols, the rate law becomes

`ln((["A"]_0["B"])/(["B"]_0["A"]))= (["B"]_0-["A"]_0)kt`

This rate law works for all values of `a≠b`.

However, it avoids negative numbers if `b>a`.

"We leave it as an exercise for the student" to derive a similar expression for `a>b`.