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?

1 Answer
Aug 14, 2015

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#.