# Mixed Second Order Kinetics

### Patrick McLaurin

We’ll first solve for the first-order integrated rate law using variable substitution so the variable substitution method can be understood for the following reaction:

A → products

The differential rate equation is obtained by setting the infinitesimal average rate equal to the rate law.

 d[A] dt
= k[A]

What follows is the variable transformation to x, which is an extent of reaction as measured by how much [A] has been consumed.

 [A] = [A]0 − x
 [dA] dt
=
 d[A]0 dt
−
 dx dt
= 0 −
 dx dt
= −
 dx dt

Substituting into the differential rate equation:

 dx dt
= k ([A]0 − x

Rearranging and integrating:

 dx [A]0 − x
= k dt
 −ln([A]0 − x) |x0xf = kt |t0tf
 −ln([A]0 − xf) + ln([A]0 − x0) = k(tf − t0)

Where x0 = 0 at t0 = 0.

 −ln([A]0 − xf) + ln[A]0 = ktf

Substituting the definition of [A] for xf = x at time tf = t gives:

 −ln[A] + ln[A]0 = kt

And we obtain the normal first-order integrated rate law:

 ln[A] = ln[A]0 − kt

We’ll now solve for the second-order integrated rate law using variable transformation for the reaction:

A + B → products
 d[A] dt
= −
 d[B] dt
= k[A][B]

Variable transformation to x:

 [A] = [A]0 − x
 [B] = [B]0 − x
 [dA] dt
=
 d[A]0 dt
−
 dx dt
= 0 −
 dx dt
= −
 dx dt

Substituting into the differential rate law:

 dx dt
= k ([A]0 − x)([B]0 − x

Rearranging and integrating:

 dx ([A]0 − x)([B]0 − x)
= k dt     (1)

In the special case where [A]0 = [B]0 (and therefore [A] = [B] at all times), the solution becomes effectively the same as for a second order reaction of the type "2 A → products", except without the multiplicative factor of 2 in the kt term.

 dx ([A]0 − x)2
= k dt
 1 [A]0 − x
|x0xf = kt
 1 [A]0 − xf
−
 1 [A]0
= kt
 1 [A]
=
 1 [A]0
+ kt

In the specific case where [A]0 ≠ [B]0, partial-fraction decomposition allows us to separate the left-hand side into a sum of fractions.

 1 ([A]0 − x)([B]0 − x)
=
 Y ([A]0 − x)
+
 Z ([B]0 − x)
(2)

Multiplying by ([A]0x)([B]0x) on both sides:

 1 = Y([B]0 − x) + Z([A]0 − x)

Distributing:

 1 = Y[B]0 − Yx + Z[A]0 − Zx

Collecting terms:

 1 = Y[B]0 + Z[A]0 − (Y+Z)x

The left-hand side must be equal to the right-hand side.

There are no terms in x on the left-hand side so the x terms must be equal to 0x = 0.

 (Y+Z)x = 0
 Y = −Z

Now we just need to figure out what Z is using the terms that are not in x:

 1 = Y[B]0 + Z[A]0
 1 = −Z[B]0 + Z[A]0
 1 = ([A]0 − [B]0)Z
Z =
 1 ([A]0 − [B]0)

Y = −
 1 ([A]0 − [B]0)
=
 1 ([B]0 − [A]0)

This allows us to write the original fraction from equation 1 as the sum of two fractions from equation 2:

 1 ([A]0 − x)([B]0 − x)
=
 1 ([A]0 − x)([B]0 − [A]0)
+
 1 ([B]0 − x)([A]0 − [B]0)

The solution of the differential rate equation is now easier:

 1 ([A]0 − x)([B]0 − [A]0)
+
 1 ([B]0 − x)([A]0 − [B]0)
= k dt
 1 ([B]0 − [A]0)
 1 ([A]0 − x)
+
 1 ([A]0 − [B]0)
 1 ([B]0 − x)
= k dt
 1 ([B]0 − [A]0)
(−ln([A]0 − x) |x0xf) +
 1 ([A]0 − [B]0)
(−ln([B]0 − x) |x0xf) = kt
 −ln([A]0 − xf) + ln([A]0 − x0) ([B]0 − [A]0)
+
 −ln([B]0 − xf) + ln([B]0 − x0) ([A]0 − [B]0)
= kt
 −ln[A] + ln[A]0 ([B]0 − [A]0)
+
 −ln[B] + ln[B]0 ([A]0 − [B]0)
= kt
 −ln[A] + ln[A]0 ([B]0 − [A]0)
−
 −ln[B] + ln[B]0 ([B]0 − [A]0)
= kt
 −ln[A] + ln[A]0 + ln[B] − ln[B]0 ([B]0 − [A]0)
= kt
 −ln[A] + ln[A]0 + ln[B] − ln[B]0 = ([B]0 − [A]0)kt
ln
 [A]0 [B]0
+ ln
 [B] [A]
= ([B]0 − [A]0)kt
ln
 [B] [A]
= −ln
 [A]0 [B]0
+ ([B]0 − [A]0)kt
ln
 [B] [A]
= ln
 [B]0 [A]0
+ ([B]0 − [A]0)kt

Or if you wanted to reverse the sign on everything:

ln
 [A] [B]
= ln
 [A]0 [B]0
+ ([A]0 − [B]0)kt

This document was translated from LATEX by HEVEA.