Derive the expression for second order reaction with unequal concentration for a reaction A + B ->A+B→ product when (1) if a > b & (2) if b > a, where "a mol/dm^3dm3" and "b mol/dm^3dm3" are the initial concentrations of A & B respectively?

1 Answer
Ernest Z.
Aug 14, 2015

WARNING! This answer involves calculus!

Explanation:

Assume we have a second order reaction of the type

"A + B" stackrel(k)(→) "Products"A + Bk−−→Products

Let x " mol/dm"^3x mol/dm3 be the amount reacted in time tt. Then

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

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

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

int_0^x dx/((a-x)(b-x)) = int_0^tkdt = kt∫x0dx(a−x)(b−x)=∫t0kdt=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))∫x0dx(a−x)(b−x)=∫x0dx(b−a)(a−x)+∫x0dx(a−b)(b−x)

= 1/(b-a)int_0^xdx/(a-x) + 1/(a-b)int_0^xdx/(b-x)=1b−a∫x0dxa−x+1a−b∫x0dxb−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)1b−a∫x0dxa−x+1a−b∫x0dxb−x

=1/(b-a)ln(a/(a-x))+1/(a-b)ln(b/(b-x))=1b−aln(aa−x)+1a−bln(bb−x)

=1/(b-a)(ln(a/(a-x))-ln(b/(b-x)))=1b−a(ln(aa−x)−ln(bb−x))

∴ 1/(b-a)ln ((a(b-x))/(b(a-x))) = kt1b−aln(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)ktln(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)ktln([A]0[B][B]0[A])=([B]0−[A]0)kt

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

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

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

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