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]0 − x)([B]0 − x) 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.