A work in progress that will be abandoned as soon as this course is over. I'm not very good at P-Chem, but if you have any questions, feel free to ask. [00 CHEM 308 (1) - THIS ONE]

1. PHYSICAL CHEMISTRY II
CHAPTER 9: Chemical Kinetics I
Rate of Consumption,Formation,and Reaction
Consider the reaction 𝐴 + 3𝐵 → 2𝑌. The numbers of moles of the species at the
beginning of the reaction (at time 𝑡 = 0) are 𝑛 𝐴0
, 𝑛 𝐵0
, 𝑎𝑛𝑑 𝑛 𝑌0
, respectively.
Assuming the reaction thereafter proceeds from left to right, the moles at some time
𝑡 (𝑠) become 𝑛 𝐴0
− 𝜉, 𝑛 𝐵0
− 3𝜉, 𝑎𝑛𝑑 𝑛 𝑌0
+ 2𝜉, respectively, where 𝜉 is the extent of
the reaction in terms of moles.
Now, we can define the moles of reactants and product after time 𝑡 (𝑠) as:
𝑛 𝐴 = 𝑛 𝐴0
− 𝜉, 𝑛 𝐵 = 𝑛 𝐵0
− 3𝜉, 𝑎𝑛𝑑 𝑛 𝑌 = 𝑛 𝑌0
+ 2𝜉, such that 𝑛 𝐴0
, 𝑛 𝐵0
, 𝑎𝑛𝑑 𝑛 𝑌0
are
constants and 𝑛 𝐴 , 𝑛 𝐵, 𝑎𝑛𝑑 𝑛 𝑌 are functions of the reaction extent, i.e. 𝑓( 𝜉).
Furthermore, we may derive these expressions with respect to time and obtain:
𝑑𝑛 𝐴
𝑑𝑡
= 0 −
𝑑𝜉
𝑑𝑡
,
𝑑𝑛 𝐵
𝑑𝑡
= 0 −
3𝑑𝜉
𝑑𝑡
, 𝑎𝑛𝑑
𝑑𝑛 𝑌
𝑑𝑡
= 0 +
2𝑑𝜉
𝑑𝑡
Solving for
𝑑𝜉
𝑑𝑡
gives the rate of reaction in terms of moles:a
𝑑𝜉
𝑑𝑡
= −
𝑑𝑛 𝐴
𝑑𝑡
= −
𝑑𝑛 𝐵
3𝑑𝑡
= +
𝑑𝑛 𝑌
2𝑑𝑡
However, it is more typical to track concentrations than moles. So we may modify
the above expression as follows.
Let 𝑉 be the volume of the reaction chamber (assuming it remains constant
throughout the reaction). Then:
𝑑𝜉
𝑉𝑑𝑡
= −
𝑑𝑛 𝐴
𝑉𝑑𝑡
= −
𝑑 𝑛 𝐵
3𝑉𝑑𝑡
= +
𝑑𝑛 𝑌
2𝑉𝑑𝑡
If we define the extent of the reaction in terms of concentration as 𝑥 =
𝜉
𝑉
and the
concentrations of the reactants and product as [ 𝐴] =
𝑛 𝐴
𝑉
, [ 𝐵] =
𝑛 𝐵
𝑉
, 𝑎𝑛𝑑 [ 𝑌] =
𝑛 𝑌
𝑉
respectively, then the rate of reaction in terms of concentration is:
𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑑𝑡
= −
𝑑[ 𝐵]
3𝑑𝑡
=
𝑑[ 𝑌]
2𝑑𝑡
Specifically, we can define the rate of reaction in terms of concentration with
respect to each species:
𝜐𝐴 = −
𝑑[ 𝐴]
𝑑𝑡
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑠𝑎𝑝𝑝𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑜𝑓 𝐴

2. 𝜐 𝐵 = −
𝑑[ 𝐵]
𝑑𝑡
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑠𝑎𝑝𝑝𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑜𝑓 𝐵
𝜐𝑌 =
𝑑[ 𝑌]
𝑑𝑡
= 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑌
Thus: 𝜐 =
𝑑𝑥
𝑑𝑡
= 𝜐𝐴 =
1
3
𝜐 𝐵 =
1
2
𝜐𝑌 .
Hence, we can reconsider the reaction 𝐴 + 3𝐵 → 2𝑌 in terms of concentration such
that:
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒: [ 𝐴]0, [ 𝐵]0, 𝑎𝑛𝑑 [ 𝑌]0
𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡 𝑎𝑟𝑒: [ 𝐴] = [ 𝐴]0 − 𝑥,[ 𝐵] = [ 𝐵]0 − 3𝑥, 𝑎𝑛𝑑 [ 𝑌] = [ 𝑌]0 + 2𝑥
𝐼𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑟𝑎𝑡𝑒𝑠 𝑎𝑟𝑒:
𝑑[ 𝐴]
𝑑𝑡
= 0 −
𝑑𝑥
𝑑𝑡
,
𝑑[ 𝐵]
𝑑𝑡
= 0 −
3𝑑𝑥
𝑑𝑡
, 𝑎𝑛𝑑
𝑑[ 𝑌]
𝑑𝑡
= 0 +
2𝑑𝑥
𝑑𝑡
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑖𝑠: 𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑑𝑡
= −
𝑑[ 𝐵]
3𝑑𝑡
=
𝑑[ 𝑌]
2𝑑𝑡
The reaction can also be represented graphically:

3. Table 1: Rates w.r.t species at t = 0 and t = any.
Initial rates (at time 𝒕 = 𝟎 𝒔) Rates at any time (𝒕 = 𝒕 𝒔)
𝜐𝐴 = −
𝑑[ 𝐴]0
𝑑𝑡
|
𝑡=0
𝜐𝐴 = −
𝑑[ 𝐴]
𝑑𝑡
|
𝑡
𝜐 𝐵 = −
𝑑[ 𝐵]0
𝑑𝑡
|
𝑡=0
𝜐 𝐵 = −
𝑑[ 𝐵]
𝑑𝑡
|
𝑡
𝜐𝑌 = +
𝑑[ 𝑌]0
𝑑𝑡
|
𝑡=0
𝜐𝑌 = +
𝑑[ 𝑌]
𝑑𝑡
|
𝑡
Table 2: Summary of rate expressions.
Moles at 𝒕 = 𝟎 𝒔 𝑛 𝐴0
𝑛 𝐵0
𝑛 𝑌0
Moles at 𝒕 = 𝒂𝒏𝒚 𝑛 𝐴 = 𝑛 𝐴0
− 𝜉 𝑛 𝐵 = 𝑛 𝐵0
− 3𝜉 𝑛 𝑌 = 𝑛 𝑌0
+ 2𝜉
Moles derived w.r.t time
𝑑𝑛 𝐴
𝑑𝑡
= 0 −
𝑑𝜉
𝑑𝑡
𝑑𝑛 𝐵
𝑑𝑡
= 0 −
3𝑑𝜉
𝑑𝑡
𝑑𝑛 𝑌
𝑑𝑡
= 0 +
2𝑑𝜉
𝑑𝑡
Reaction rate in moles
𝑑𝜉
𝑑𝑡
= −
𝑑𝑛 𝐴
𝑑𝑡
= −
𝑑𝑛 𝐵
3𝑑𝑡
= +
𝑑𝑛 𝑌
2𝑑𝑡
Concentration at 𝒕 = 𝟎 𝒔 [ 𝐴]0 =
𝑛 𝐴0
𝑉
[ 𝐵]0 =
𝑛 𝐵0
𝑉
[ 𝑌]0 =
𝑛 𝑌0
𝑉
Concentration at 𝒕 = 𝒂𝒏𝒚 [ 𝐴] =
𝑛 𝐴
𝑉
[ 𝐵] =
𝑛 𝐵
𝑉
[ 𝑌] =
𝑛 𝑌
𝑉
Concentration derived w.r.t time 𝜐𝐴 = −
𝑑[ 𝐴]
𝑑𝑡
𝜐 𝐵 = −
𝑑[ 𝐵]
𝑑𝑡
𝜐𝑌 =
𝑑[ 𝑌]
𝑑𝑡
Reaction rate in concentration
𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑑𝑡
= −
𝑑[ 𝐵]
3𝑑𝑡
=
𝑑[ 𝑌]
2𝑑𝑡
𝜐 =
𝑑𝑥
𝑑𝑡
= 𝜐𝐴 =
1
3
𝜐 𝐵 =
1
2
𝜐𝑌
Empirical Rate Equations
For the reaction 𝑎𝐴 + 𝑏𝐵 → 𝑦𝑌 + 𝑧𝑍, the rate of consumption of A can be expressed
empirically by an expression of the form 𝜐𝐴 ∝ [ 𝐴] 𝛼[ 𝐵] 𝛽
, called an empirical
expression. We may change this expression to an equation by adding a
proportionality constant k: 𝜐𝐴 = 𝑘 𝐴[ 𝐴] 𝛼 [ 𝐵] 𝛽
, where kA, , and  are independent of
concentration and of time.
Similarly, for product Z, where kZ is not necessarily the same as kA, the rate of
formation of product is 𝜐𝑍 = 𝑘 𝑍[ 𝐴] 𝛼[ 𝐵] 𝛽
.

4. When these equations apply, the rate of reaction must also be given by an equation
of the same form: 𝜐 = 𝑘[ 𝐴] 𝛼[ 𝐵] 𝛽
=
1
𝑎
𝜐𝐴 =
1
𝑏
𝜐 𝐵 =
1
𝑦
𝜐𝑌 =
1
𝑧
𝜐𝑍 .
Thus,
1
𝑎
𝜐𝐴 = 𝑘[ 𝐴] 𝛼[ 𝐵] 𝛽
→ 𝝊 𝑨 = 𝒂𝒌[ 𝑨] 𝜶[ 𝑩] 𝜷
(the rate law in differential form).
If we let 𝑘 𝐴 = 𝑎𝑘, then 𝜐𝐴 = 𝑘 𝐴 [ 𝐴] 𝛼[ 𝐵] 𝛽
. Likewise,
1
𝑏
𝜐 𝐵 = 𝑘[ 𝐴] 𝛼[ 𝐵] 𝛽
→ 𝜐 𝐵 =
𝑏𝑘[ 𝐴] 𝛼[ 𝐵] 𝛽
. If we let 𝑘 𝐵 = 𝑏𝑘, then 𝜐 𝐵 = 𝑘 𝐵[ 𝐴] 𝛼[ 𝐵] 𝛽
.
In these equations, kA, kZ, and k are not necessarily the same, being related by
stoichiometric coefficients; thus if the stoichiometric equation is 𝐴 + 3𝐵 → 2𝑌, then
𝑘 = 𝑘 𝐴 =
1
2
𝑘 𝐵 =
1
3
𝑘 𝑍.
Order of Reaction
The exponent  in the previous equations is known as the order of reaction with
respect to A and can be referred to as a partial order. Similarly, the partial order  is
the order with respect to B. These orders are purely experimental quantities and are
not necessarily integral; that is, they are independent of stoichiometry. The sum of
all the partial orders, 𝛼 + 𝛽 + ⋯, is referred to as the overall order and is usually
given the symbol n.
A first order reaction is one in which the rate is proportional to the first power of
the concentration of a single reactant, such that 𝜐 = 𝑘[ 𝐴]. For example, the
conversion of cyclopropane to propylene is 1st order.
In a second order reaction, the rate must be proportional to the product of two
concentrations [A] and [B] if the reaction simply involves collisions between A and B
molecules. For instance, in the reaction 𝐻2 + 𝐼2 ⇌ 2𝐻𝐼, the rate from left to right is
proportional to the product of the concentrations of the two reactants. That is, 𝜐1 =
𝑘1[ 𝐻2][ 𝐼2], where k1 is a constant at a given temperature. The reaction from left to
right is said to be 1st order in H2, 1st order in I2, and 2nd order overall. The reverse
reaction is also 2nd order; the rate from right to left is proportional to the square of
the concentration of HI: 𝜐−1 = 𝑘−1[ 𝐻𝐼]2
.
Similarly, the kinetics must be third order if a reaction proceeds in one stage and
involves collisions between three molecules, A, B, and C.

5. Examples
For the reaction 𝐴 + 𝐵 𝑘1
⃗⃗⃗⃗⃗ 2𝐶, 𝜐1 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑑𝑡
= −
𝑑[ 𝐵]
𝑑𝑡
=
𝑑[ 𝐶]
2𝑑𝑡
= 𝑘1[ 𝐴] 𝛼[ 𝐵] 𝛽
.
Assuming that  = 1 and  = 1, the forward reaction is 2nd order and 𝜐1 =
𝑘1[ 𝐴]1[ 𝐵]1
.
For the reaction 2𝐶 𝑘−1
⃗⃗⃗⃗⃗⃗ 𝐴 + 𝐵, 𝜐−1 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐶]
2𝑑𝑡
=
𝑑[ 𝐴]
𝑑𝑡
=
𝑑[ 𝐵]
𝑑𝑡
= 𝑘−1[ 𝐶] 𝛾
.
Assuming that  = 2, the forward reaction is 2nd order and 𝜐−1 = 𝑘−1[ 𝐶]2
.
For the reaction 𝐶𝐻3 𝐶𝐻𝑂 → 𝐶𝐻4 + 𝐶𝑂, 𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐶𝐻3 𝐶𝐻𝑂]
𝑑𝑡
=
𝑑[ 𝐶 𝐻4 ]
𝑑𝑡
=
𝑑[ 𝐶𝑂]
𝑑𝑡
=
𝑘[ 𝐶𝐻3 𝐶𝐻𝑂] 𝛼
.
Assuming 𝛼 =
3
2
, the forward reaction is of order
3
2
and 𝜐 = 𝑘[ 𝐶𝐻3 𝐶𝐻𝑂]
3
2.
For the reaction 𝐴 𝑘⃗⃗⃗⃗ 𝐵, 𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑑𝑡
=
𝑑[ 𝐵]
𝑑𝑡
= 𝑘[ 𝐴] 𝛼
.
Assuming that  = 1, the forward reaction is 1st order and 𝜐 = 𝑘[ 𝐴]1
.
Also, −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴]and so
𝑑[ 𝐴]
𝑑𝑡
= −𝑘 𝑑𝑡.
Integrating gives 𝐥𝐧[ 𝑨] = −𝒌𝒕 + 𝑪, where C is a constant. This is the rate law in
integral form.
At 𝑡 = 0 and [ 𝐴] = [ 𝐴]0, ln[ 𝐴]0 = 𝐶.
Therefore, ln[ 𝐴] = −𝑘𝑡 + ln[ 𝐴]0, and ln[ 𝐴] − ln[ 𝐴]0 = −𝑘𝑡, and ln
[ 𝐴]
[ 𝐴]0
= −𝑘𝑡.
Get rid of natural log:
[ 𝐴]
[ 𝐴]0
= 𝑒−𝑘𝑡
. Hence, [ 𝑨] = [ 𝑨] 𝟎 𝒆−𝒌𝒕
is the rate law for the
disappearance of A.
At 𝑡 = 0, [ 𝐴] = [ 𝐴]0 𝑒0
= [ 𝐴]0.
At 𝑡 = ∞, [ 𝐴] = [ 𝐴]0 𝑒−∞
= 0.
In other words, for the reaction 𝐴 𝑘⃗⃗⃗⃗ 𝐵:
The initial concentrations of A and B are [A]0 and 0, respectively.
The concentrations after time 𝑡 = 𝑎𝑛𝑦 are [ 𝐴] = [ 𝐴]0 − 𝑥 and [ 𝐵] = 𝑥.
Hence [ 𝐴] + [ 𝐵] = [ 𝐴]0 and [ 𝐵] = [ 𝐴]0 − [ 𝐴] = [ 𝐴]0 − [ 𝐴]0 𝑒−𝑘𝑡
= [ 𝐴]0(1 − 𝑒−𝑘𝑡).
[ 𝑩] = [ 𝑨] 𝟎(𝟏 − 𝒆−𝒌𝒕
) is the rate law of appearance of B.

6. IntegrationofRate Laws
First Order Reaction:
Consider 𝑎𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠.
At 𝑡 = 0, [ 𝐴] = [ 𝐴]0 and [ 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠] = 0.
At 𝑡 = 𝑎𝑛𝑦, [ 𝐴] = [ 𝐴]0 − 𝑎𝑥, where a, ax, and [A]0 are constants.
The derivative of [A] with respect to time gives us
𝑑[ 𝐴]
𝑑𝑡
= 0 − 𝑎
𝑑𝑥
𝑑𝑡
.
Therefore, 𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑎 𝑑𝑡
= 𝑘[ 𝐴] 𝛼
, where  is the partial order of reaction with
respect to A, determined experimentally. Assuming the reaction is 1st order, 𝛼 = 1.
Thus, 𝜐 = 𝑘[ 𝐴]1
.
To simplify, we can let −
𝑑[ 𝐴]
𝑎 𝑑𝑡
= 𝑘[ 𝐴]and let 𝑘 𝐴 = 𝑎𝑘.
Rearranging therefore gives −
𝒅[ 𝑨]
[ 𝑨]
= 𝒌 𝑨 𝒅𝒕, the rate law in differential form.

7. If we integrate the above equation, we get the following:
−
𝑑[ 𝐴]
[ 𝑨]
= 𝑘 𝐴 𝑑𝑡 → − 𝐥𝐧[ 𝑨] = 𝒌 𝑨 𝒕 + 𝑪, the rate law in integral form.
When 𝑡 = 0, − ln[ 𝐴]0 = 0 + 𝐶. Therefore, 𝑪 = − 𝐥𝐧[ 𝑨] 𝟎.
Substituting gives: −ln[ 𝐴] = 𝑘 𝐴 𝑡 − ln[ 𝐴]0. Rearranging, ln[ 𝐴] − ln[ 𝐴]0 = −𝑘 𝐴 𝑡 and
ln
[ 𝐴]
[ 𝐴]0
= −𝑘 𝐴 𝑡.
We now define the half-life of the reaction to be the time required for [A] to drop to
half its value; that is, 𝒕 = 𝒕 𝟏
𝟐
𝒘𝒉𝒆𝒏 [ 𝑨] =
[ 𝑨] 𝟎
𝟐
.
Thus, for a 1st order reaction, at t½:
ln
1
2
[ 𝐴]0
[ 𝐴]0
= −𝑘 𝐴 𝑡1
2
→ ln
1
2
= −𝑘 𝐴 𝑡1
2
→ − ln 2 →= −𝑘 𝐴 𝑡1
2
→ 𝒕 𝟏
𝟐
=
𝐥𝐧 𝟐
𝒌 𝑨
𝑡1
2
is independent of concentration in 1st order reactions.
Important Integrals
∫
1
𝑥
𝑑𝑥 = ln| 𝑥| (1)
∫ 𝑑𝑥 = 𝑥 (2)
∫ 𝑥 𝑛
𝑑𝑥 =
𝑥 𝑛+1
𝑛+1
(3)
∫( 𝑎 + 𝑏𝑥) 𝑛
𝑑𝑥 =
( 𝑎+𝑏𝑥) 𝑛+1
𝑏( 𝑛+1)
(4)
∫
1
𝑎+𝑏𝑥
𝑑𝑥 =
1
𝑏
ln(| 𝑎 + 𝑏𝑥|) (5)
∫
1
( 𝑎+𝑏𝑥) 𝑛 𝑑𝑥 = −
1
( 𝑛−1) 𝑏( 𝑎+𝑏𝑥) 𝑛−1 (6)

8. Second Order Reaction:
Consider 𝑎𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠.
𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑎 𝑑𝑡
= 𝑘[ 𝐴] 𝛼
, where  is the partial order of reaction with respect to A,
determined experimentally. Assuming the reaction is 2nd order, 𝛼 = 2. Thus, 𝜐 =
𝑘[ 𝐴]2
.
To simplify, we can let −
𝑑[ 𝐴]
𝑎 𝑑𝑡
= 𝑘[ 𝐴]and let 𝑘 𝐴 = 𝑎𝑘.
Rearranging therefore gives
𝒅[ 𝑨]
[ 𝑨] 𝟐 = −𝒌 𝑨 𝒅𝒕, the rate law in differential form.
If we integrate the above equation, we get the following:
−
𝑑[ 𝐴]
[ 𝑨] 𝟐 = 𝑘 𝐴 𝑑𝑡 → −
𝟏
[ 𝑨]
= −𝒌 𝑨 𝒕 + 𝑪, the rate law in integral form.
When 𝑡 = 0, −
1
[ 𝐴]
= 0 + 𝐶. Therefore, 𝑪 = −
𝟏
[ 𝑨] 𝟎
.
Substituting gives: −
1
[ 𝐴]
= −𝑘 𝐴 𝑡 −
𝟏
[ 𝑨] 𝟎
. Rearranging,
1
[ 𝐴]
−
𝟏
[ 𝑨] 𝟎
= 𝑘 𝐴 𝑡.
Thus, for a 2nd order reaction, at t½:
1
1
2
[ 𝐴]0
−
1
[ 𝐴]0
= 𝑘 𝐴 𝑡1
2
→
2
[ 𝐴]0
−
1
[ 𝐴]0
= 𝑘 𝐴 𝑡1
2
→
1
[ 𝐴]0
= 𝑘 𝐴 𝑡1
2
→ 𝒕 𝟏
𝟐
=
𝟏
𝒌 𝑨[ 𝑨] 𝟎
𝑡1
2
depends on [ 𝐴]0 in 2nd order reactions.
Third Order Reaction:
Consider 𝑎𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠.
𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑎 𝑑𝑡
= 𝑘[ 𝐴] 𝛼
, where  is the partial order of reaction with respect to A,
determined experimentally. Assuming the reaction is 3rd order, 𝛼 = 3. Thus, 𝜐 =
𝑘[ 𝐴]3
.
To simplify, we can let −
𝑑[ 𝐴]
𝑎 𝑑𝑡
= 𝑘[ 𝐴]and let 𝑘 𝐴 = 𝑎𝑘.
Rearranging therefore gives
𝒅[ 𝑨]
[ 𝑨] 𝟑 = −𝒌 𝑨 𝒅𝒕, the rate law in differential form.
If we integrate the above equation, we get the following:

9. −
𝑑[ 𝐴]
[ 𝑨] 𝟑 = 𝑘 𝐴 𝑑𝑡 → −
𝟏
𝟐[ 𝑨] 𝟐 = −𝒌 𝑨 𝒕 + 𝑪, the rate law in integral form.
When 𝑡 = 0, −
1
2⌈ 𝐴⌉0
2 = 0 + 𝐶. Therefore, 𝑪 = −
𝟏
𝟐⌈ 𝐴⌉0
2.
Substituting gives: −
𝟏
𝟐[ 𝐴]2 = −𝑘 𝐴 𝑡 −
𝟏
𝟐⌈ 𝐴⌉0
2. Rearranging,
𝟏
𝟐[ 𝐴]2 −
𝟏
𝟐⌈ 𝐴⌉0
2 = 𝑘 𝐴 𝑡.
Thus, for a 3rd order reaction, at t½:
1
2 (
1
2
[ 𝐴]0)
2
−
1
2[ 𝐴]0
2
= 𝑘 𝐴 𝑡1
2
→
1
2 (
1
4
[ 𝐴]2)
−
1
2[ 𝐴]0
2
= 𝑘 𝐴 𝑡1
2
→
1
1
2
[ 𝐴]0
2
−
1
2[ 𝐴]0
2
= 𝑘 𝐴 𝑡1
2
→
2
[ 𝐴]0
2
−
1
2[ 𝐴]0
2
= 𝑘 𝐴 𝑡1
2
→
3
2[ 𝐴]0
2
= 𝑘 𝐴 𝑡1
2
→ 𝒕 𝟏
𝟐
=
𝟑
𝟐[ 𝑨] 𝟎
𝟐
𝑘 𝐴
𝑡1
2
depends on [ 𝐴]0 in 3rd order reactions.
nth Order Reaction:
Consider 𝑎𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠.
𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑎 𝑑𝑡
= 𝑘[ 𝐴] 𝛼
, where  is the partial order of reaction with respect to A,
determined experimentally. Assuming the reaction is nth order, 𝛼 = 𝑛. Thus, 𝜐 =
𝑘[ 𝐴] 𝑛
.
To simplify, we can let −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴] 𝑛
and let 𝑘 𝐴 = 𝑎𝑘.
Rearranging therefore gives
𝒅[ 𝑨]
[ 𝑨] 𝒏 = −𝒌 𝑨 𝒅𝒕, the rate law in differential form.
If we integrate the above equation, we get the following:
𝑑[ 𝐴]
[ 𝑨] 𝒏 = −𝑘 𝐴 𝑑𝑡 →
[ 𝑨]−𝒏+𝟏
−𝒏+𝟏
= −𝒌 𝑨 𝒕 + 𝑪, the rate law in integral form.
When 𝑡 = 0,
[ 𝑨] 𝟎
−𝒏+𝟏
−𝒏+𝟏
= 0 + 𝐶. Therefore, 𝑪 =
[ 𝑨] 𝟎
−𝒏+𝟏
−𝒏+𝟏
.
Substituting gives:
[ 𝑨]−𝒏+𝟏
−𝒏+𝟏
= −𝑘 𝐴 𝑡 +
[ 𝑨] 𝟎
−𝒏+𝟏
−𝒏+𝟏
. Rearranging,
[ 𝑨]−𝒏+𝟏
−𝒏+𝟏
−
[ 𝑨] 𝟎
−𝒏+𝟏
−𝒏+𝟏
= −𝑘 𝐴 𝑡.
Thus, for a nth order reaction, at t½:

10. (
1
2
[ 𝐴]0
−𝑛+1
)
−𝑛 + 1
−
[ 𝐴]0
−𝑛+1
−𝑛 + 1
= −𝑘 𝐴 𝑡1
2
→
((
1
2
)
−𝑛+1
[ 𝐴]0
−𝑛+1
)
−𝑛 + 1
−
[ 𝐴]0
−𝑛+1
−𝑛 + 1
= −𝑘 𝐴 𝑡1
2
→ 𝒕 𝟏
𝟐
=
([ 𝐴]0
−𝑛+1){(
1
2
)
−𝑛+1
− 1}
−𝑛 + 1
= −𝑘 𝐴 𝑡1
2
We can invert the initial concentration term to send it to the number with the index
𝑛 − 1, flip the fraction ½ to also give it the index 𝑛 − 1, and divide both sides of the
equation by −𝑘 𝐴 (to isolate t½ and distribute the negative sign to the denominator,
−𝑛 + 1). Thus:
𝒕 𝟏
𝟐
=
𝟐 𝒏−𝟏
− 𝟏
[ 𝑨] 𝟎
𝒏−𝟏( 𝒏 − 𝟏) 𝒌 𝑨
𝑡1
2
depends on [ 𝐴]0 in nth order reactions.
Example:
Consider the 1st order reaction 𝐴 → 𝐵.
At time 𝑡 = 𝑎𝑛𝑦, [ 𝐴] = [ 𝐴]0 − 𝑥 and [ 𝐵] = 𝑥.
Hence, [ 𝐴] + [ 𝐵] = [ 𝐴]0 and 𝜐 =
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑡
=
𝑑[ 𝐵]
𝑡
= 𝑘[ 𝐴] = 𝑘([ 𝐴]0 − 𝑥).
If −
𝑑[ 𝐴]
𝑡
𝑘[ 𝐴], then [ 𝐴] = [ 𝐴]0 𝑒−𝑘𝑡
or ln
[ 𝐴]0−𝑥
[ 𝐴]0
= −𝑘𝑡.
Therefore, ln
𝑎0−𝑥
𝑎0
= −𝑘𝑡.
Also,
𝑑𝑥
𝑑𝑡
= 𝑘( 𝑎0 − 𝑥)is a separable equation such that
𝑑𝑥
𝑎0 −𝑥
= 𝑘 𝑑𝑡.
Integrating gives −ln| 𝑎0 − 𝑥| = 𝑘𝑡 + 𝐶, where C is an integration constant.
At time 𝑡 = 0: 𝑥 = 0 and − ln| 𝑎0| = 𝐶.
Then −ln| 𝑎0 − 𝑥| = 𝑘𝑡 − ln| 𝑎0| → ln| 𝑎0 − 𝑥| − ln| 𝑎0| = −𝑘𝑡 → ln
𝑎0−𝑥
𝑎0
= −𝑘𝑡.
And if [ 𝐴] + [ 𝐵] = 𝑎0, then [ 𝐵] = 𝑎0 − [ 𝐴] = 𝑎0 − 𝑎0 𝑒−𝑘𝑡
= 𝑎0(1 − 𝑒−𝑘𝑡).
In terms of [B]:
𝑑[ 𝐵]
𝑑𝑡
= 𝑘( 𝑎0 − 𝑥) = 𝑘( 𝑎0 − [ 𝐵]).
This is a separable equation such that:
𝑑[ 𝐵]
𝑎0−[ 𝐵]
= 𝑘 𝑑𝑡.
Integrating gives −ln 𝑎0 − [ 𝐵] = 𝑘𝑡 + 𝐶.
At time 𝑡 = 0: [ 𝐵] = 0 and − ln| 𝑎0| = 𝐶.
Then −ln| 𝑎0 − [ 𝐵]| = 𝑘𝑡 − ln| 𝑎0| → ln| 𝑎0 − [ 𝐵]| − ln| 𝑎0 | = −𝑘𝑡 → ln
𝑎0−[ 𝐵]
𝑎0
=
−𝑘𝑡, as above.

11. Example
Consider the 2nd order reaction 2𝐴 𝑘⃗⃗⃗⃗ 𝑃 or 𝐴 + 𝐴 𝑘⃗⃗⃗⃗ 𝑃.
Let −
𝑑[ 𝐴]
2𝑑𝑡
= 𝑘[ 𝐴] 𝛼
= 𝑘[ 𝐴]2
. And let 𝑘 𝐴 = 2𝑘.
Then rearranging gives
𝑑[ 𝐴]
𝑑[ 𝐴]2 = −𝑘 𝐴 𝑑𝑡.
Integrating gives −
1
[ 𝐴]
= −𝑘 𝐴 𝑡 + 𝐶.
At time 𝑡 = 0: −
1
[ 𝐴]0
= 𝐶.
Then −
1
[ 𝐴]
= 𝑘 𝐴 𝑡 −
1
[ 𝐴]0
→
1
[ 𝐴]
−
1
[ 𝐴]0
= 𝑘 𝐴 𝑡, where k has units
1
𝑚𝑜𝑙 𝐿−1 𝑠−1.
At time 𝑡 = 𝑡1
2
:
1
1
2
[ 𝐴]0
−
1
[ 𝐴]0
= 𝑘 𝐴 𝑡1
2
→
2
[ 𝐴]0
−
1
[ 𝐴]0
= 𝑘 𝐴 𝑡1
2
→
1
[ 𝐴]0
= 𝑘 𝐴 𝑡1
2
→ 𝒕 𝟏
𝟐
=
𝟏
[ 𝑨] 𝟎 𝒌 𝑨
.
Consider a 2nd order reaction that has two reactants, such as 𝐴 + 𝐵 𝑘⃗⃗⃗⃗ 𝑃.
We can express the rate as −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴][ 𝐵], but it cannot be integrated as it
contains two different variables.
We still know that at time 𝑡 = 0: [ 𝐴] = [ 𝐴]0, [ 𝐵] = [ 𝐵]0, and [ 𝑃] = 0.
Similarly, at time 𝑡 = 𝑎𝑛𝑦: [ 𝐴] = [ 𝐴]0 − 𝑥, [ 𝐵] = [ 𝐵]0 − 𝑥, and [ 𝑃] = 𝑥.
If we assume that [ 𝐴]0 = [ 𝐵]0, then [ 𝐴] = [ 𝐵] as well.
The rate then becomes −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴]2
, a separable equation.
Thus,
𝑑[ 𝐴]
[ 𝐴]2 = −𝑘 𝑑𝑡, which can be integrated as in the example previous example.
Consider a 3rd order reaction 3𝐴 𝑘⃗⃗⃗⃗ 𝑃.
The rate is −
𝑑[ 𝐴]
3𝑑𝑡
= 𝑘[ 𝐴]3
. Let 𝑘 𝐴 = 3𝑘.
Then rearranging gives
𝑑[ 𝐴]
𝑑[ 𝐴]3 = −𝑘 𝐴 𝑑𝑡.
Integrating gives −
1
2[ 𝐴]2 = −𝑘 𝐴 𝑡 + 𝐶.
At time 𝑡 = 0: −
1
2[ 𝐴]0
2 = 𝐶.
Then −
1
2[ 𝐴]2 = −𝑘 𝐴 𝑡 −
1
2[ 𝐴]0
2 →
1
2[ 𝐴]2 −
1
2[ 𝐴]0
2 = 𝑘 𝐴 𝑡 →
1
[ 𝐴]2 −
1
[ 𝐴]0
2 = 2𝑘 𝐴 𝑡.
At 𝑡 = 𝑡1
2
: GO OVER TO CLARIFY!
Example
Consider the reaction 2𝐴 𝑘⃗⃗⃗⃗ 𝑃 or 𝐴 + 𝐴 𝑘⃗⃗⃗⃗ 𝑃.
Then −
𝑑[ 𝐴]
2𝑑𝑡
= 𝑘[ 𝐴]2[ 𝐵], where [ 𝐴] = [ 𝐴]0 − 𝑥 and [ 𝐵] = [ 𝐵]0 − 𝑥 at any time t.
If we let 2[ 𝐴]0 ≫ [ 𝐴], then 2[ 𝐴]0 − 2𝑥.
Then [ 𝐴] = 2[ 𝐴]0 − 𝑥 → [ 𝐴] = 2([ 𝐴]0 − 𝑥).
And since [ 𝐵] = [ 𝐴]0 − 𝑥, then [ 𝐴] = 2[ 𝐵] (and thus [ 𝐴]0 = 2[ 𝐵]0 and [ 𝐵] =
[ 𝐴]
2
).

12. Therefore, −
𝑑[ 𝐴]
𝑑𝑡
= 2𝑘[ 𝐴]2
= 2𝑘 𝐴
[ 𝐴]2
2
→ −
𝑑[ 𝐴]
𝑑𝑡
= 2𝑘 𝐴[ 𝐴]2
(
1
2
[ 𝐴]) = 𝑘[ 𝐴]3
→
−
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴]3
.
Example
Consider a reaction 𝐴 + 𝐵 + 𝐶 𝑘⃗⃗⃗⃗ 𝑃.
At time 𝑡 = 0, we assume that [ 𝐴] = [ 𝐴]0, [ 𝐵] = [ 𝐴]0, and [ 𝐶] = [ 𝐴]0.
Likewise, at time 𝑡 = 𝑎𝑛𝑦, [ 𝐴] = [ 𝐴]0 − 𝑥, [ 𝐵] = [ 𝐴]0 − 𝑥, and [ 𝐶] = [ 𝐴]0 − 𝑥.
The rate is given by: −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴][ 𝐵][ 𝐶] → −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴]3
→
𝑑[ 𝐴]
[ 𝐴]3 = −𝑘 𝑑𝑡.
Integrating gives −
1
[ 𝐴]2 = −𝑘𝑡 + 𝐶 →
1
[ 𝐴]2 = 𝑘𝑡 + 𝐶.
At 𝑡 = 0,
1
[ 𝐴]0
2 = 𝐶. Therefore,
1
[ 𝐴]2 = 𝑘𝑡 +
1
[ 𝐴]0
2 →
1
[ 𝐴]2 −
1
[ 𝐴]0
2 = 𝑘𝑡.
And at 𝑡 = 𝑡1
2
,
1
[ 𝐴]2 −
1
(
1
2
[ 𝐴]0)
2 = 𝑘𝑡1
2
→
1
[ 𝐴]2 −
1
1
4
[ 𝐴]0
2 = 𝑘𝑡1
2
→ 𝒕 𝟏
𝟐
=
𝟏
𝒌[ 𝑨] 𝟐 −
𝟒
𝒌[ 𝑨] 𝟎
𝟐.
HAVE HOKMABADI CHECK THIS!
Example
Consider the 2nd order reaction 𝐴 + 𝐵 𝑘⃗⃗⃗⃗ 𝑃.
At time 𝑡 = 0, we assume that [ 𝐴] = [ 𝐴]0 and [ 𝐵] = [ 𝐵]0.
Similarly, at time 𝑡 = 𝑎𝑛𝑦, [ 𝐴] = [ 𝐴]0 − 𝑥 and [ 𝐵] = [ 𝐵]0 − 𝑥.
The rate is then given by:
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴][ 𝐵] = 𝑘([ 𝐴]0 − 𝑥 )([ 𝐵]0 − 𝑥).
We will denote [ 𝐴] = 𝑎 and [ 𝐴]0 = 𝑎0.
Then
𝑑𝑥
𝑑𝑡
= 𝑘( 𝑎0 − 𝑥 )( 𝑏0 − 𝑥) →
1
( 𝑎0−𝑥 )( 𝑏0−𝑥)
𝑑𝑥 = 𝑘 𝑑𝑡.
We use partial fractions to solve for a and b.
1 = 𝐴( 𝑏0 − 𝑥) + 𝐵( 𝑎0 − 𝑥)
𝐴𝑏0 + 𝐵𝑎0 − ( 𝐴 + 𝐵) 𝑥 − 1 = 0
Let {
𝐴 + 𝐵 = 0
𝐴𝑏0 + 𝐵𝑎0 − 1 = 0
We can show that 𝐵 =
1
𝑎0−𝑏0
and that 𝐴 =
1
𝑎0−𝑏0
.

13. 1
( 𝑎0 − 𝑥 )( 𝑏0 − 𝑥)
=
−
1
( 𝑎0 − 𝑏0)
( 𝑎0 − 𝑥 )
+
−
1
( 𝑎0 − 𝑏0 )
( 𝑏0 − 𝑥 )
=
1
( 𝑎0 − 𝑏0 )( 𝑎0 − 𝑥)
+
1
( 𝑎0 − 𝑏0 )( 𝑏0 − 𝑥)
= −
1
𝑎0 − 𝑏0
[
1
𝑎0 − 𝑥
−
1
𝑏0 − 𝑥
]
𝑑𝑥
𝑎0 − 𝑥
−
𝑑𝑥
𝑏0 − 𝑥
= −𝑘( 𝑎0 − 𝑏0) 𝑑𝑡
Integrating (integral #6) gives: 𝐥𝐧(| 𝒂 𝟎 − 𝒙|) − 𝐥𝐧(| 𝒃 𝟎 − 𝒙|) = −𝒌( 𝒂 𝟎 − 𝒃 𝟎) 𝒕.
Review
𝐴 + 𝐵 → 𝑃
At time 𝑡 = 𝑎𝑛𝑦: 𝑎 = 𝑎0 − 𝑥 and 𝑏 = 𝑏0 − 𝑥, so that
𝑑𝑥
𝑑𝑡
= −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴][ 𝐵] =
𝑘( 𝑎0 − 𝑥)( 𝑏0 − 𝑥).
Thus,
𝑑𝑥
( 𝑎0−𝑥 )( 𝑏0−𝑥)
= −𝑘 𝑑𝑡.
Using the integral ∫
𝑑𝑥
( 𝑎+𝑏𝑥 )( 𝑐+𝑔𝑥)
=
1
𝑎𝑔−𝑏𝑐
ln |
𝑐+𝑔𝑥
𝑎+𝑏𝑥
|, we get:
𝑑𝑥
( 𝑎0 − 𝑥 )( 𝑏0 − 𝑥)
= −
1
−𝑎0 + 𝑏0
ln |
𝑏0 − 𝑥
𝑎0 − 𝑥
| = −𝑘𝑡 + 𝐶
Or
1
−( 𝑎0−𝑏0)
ln |
𝑏0−𝑥
𝑎0−𝑥
| = −𝑘𝑡 + 𝐶.
At 𝑡 = 0, −
1
( 𝑎0−𝑏0)
ln
𝑏0
𝑎0
= 𝐶 → 𝐶 =
1
−( 𝑎0−𝑏0)
ln
𝑏0
𝑎0
.
Therefore,
1
−( 𝑎0−𝑏0)
ln |
𝑏0−𝑥
𝑎0−𝑥
| = −𝑘𝑡 −
1
( 𝑎0−𝑏0)
ln
𝑏0
𝑎0
.
Multiplying both sides by ( 𝑎0 − 𝑏0) gives
Rearranging gives −ln |
𝑏0−𝑥
𝑎0−𝑥
| = −( 𝑎0 − 𝑏0) 𝑘𝑡 − ln
𝑏0
𝑎0
Thus ln|
𝑏0−𝑥
𝑎0−𝑥
| = ( 𝑎0 − 𝑏0) 𝑘𝑡 + ln
𝑏0
𝑎0
And − ln |
𝑎0−𝑥
𝑏0−𝑥
| = ( 𝑎0 − 𝑏0) 𝑘𝑡 + ln
𝑏0
𝑎0
Hence ln |
𝑎0−𝑥
𝑏0−𝑥
| + ln
𝑏0
𝑎0
= −( 𝑎0 − 𝑏0) 𝑘𝑡
And −
1
( 𝑎0−𝑏0)
ln
𝑏0( 𝑎0−𝑥)
𝑎0( 𝑏0−𝑥)
+ ln
𝑏0
𝑎0
= 𝑘𝑡
WAIT FOR CORRECTION

14. Reversible 1st Order Reactions
Consider the reaction: 𝐴 𝑘⃡⃗⃗ 𝐵.
This overall reaction is composed of two elementary reactions:
 𝐴 𝑘1
⃗⃗⃗⃗ 𝐵 for which the rate is: −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘1[ 𝐴]1
→
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴]; and
 𝐵 𝑘−1
⃗⃗⃗⃗⃗⃗ 𝐴 for which the rate is:
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐵]
The overall rate is
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴]1
+ 𝑘1[ 𝐵]1
.
Before we can integrate, we must express B in terms of A.
[ 𝐴] = [ 𝐴]0 − 𝑥
[ 𝐵] = 𝑥
} [ 𝐴] + [ 𝐵] = [ 𝐴]0 → [ 𝐵] = [ 𝐴]0 − [ 𝐴]
Hence
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴]1
+ 𝑘1([ 𝐴]0 − [ 𝐴])
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴]1
+ 𝑘1[ 𝐴]0 − 𝑘1[ 𝐴]
𝑑[ 𝐴]
𝑑𝑡
= −( 𝑘1 + 𝑘−1)[ 𝐴]+ 𝑘−1[ 𝐴]0
Rearranging gives:
𝑑[ 𝐴]
( 𝑘1+𝑘−1)[ 𝐴]−𝑘−1[ 𝐴]0
= −𝑑𝑡
Integrating (integral #6) gives:
1
𝑘1+𝑘−1
ln{( 𝑘1 + 𝑘−1)[ 𝐴] − 𝑘−1[ 𝐴]0} = −𝑡 + 𝐶.
At 𝑡 = 0:
1
𝑘1+𝑘−1
ln{( 𝑘1 + 𝑘−1)[ 𝐴]0 − 𝑘−1[ 𝐴]0} = 𝐶
1
𝑘1 + 𝑘−1
ln{ 𝑘1[ 𝐴]0 + 𝑘−1[ 𝐴]0 − 𝑘−1[ 𝐴]0} = 𝐶
1
𝑘1 + 𝑘−1
ln{ 𝑘1[ 𝐴]0} = 𝐶
Hence
1
𝑘1+𝑘−1
ln{( 𝑘1 + 𝑘−1)[ 𝐴] − 𝑘−1[ 𝐴]0} = −𝑡 +
1
𝑘1+𝑘−1
ln{ 𝑘1[ 𝐴]0}
1
𝑘1 + 𝑘−1
ln{( 𝑘1 + 𝑘−1)[ 𝐴] − 𝑘−1[ 𝐴]0} −
1
𝑘1 + 𝑘−1
ln{ 𝑘1[ 𝐴]0} = −𝑡
1
𝑘1 + 𝑘−1
〈ln{( 𝑘1 + 𝑘−1)[ 𝐴] − 𝑘−1[ 𝐴]0} − ln{ 𝑘1[ 𝐴]0}〉 = −𝑡
ln{( 𝑘1 + 𝑘−1)[ 𝐴] − 𝑘−1[ 𝐴]0} − ln{ 𝑘1[ 𝐴]0} = −( 𝑘1 + 𝑘−1) 𝑡
𝐥𝐧 {
( 𝒌 𝟏 + 𝒌−𝟏)[ 𝑨]− 𝒌−𝟏[ 𝑨] 𝟎
𝒌 𝟏[ 𝑨] 𝟎
} = −( 𝒌 𝟏 + 𝒌−𝟏) 𝒕

15. At equilibrium:
[ 𝐴] 𝑒𝑞 + [ 𝐵] 𝑒𝑞 = [ 𝐴]0 and
𝑑[ 𝐴]
𝑑𝑡
= 0 since [A] is a constant.
Hence
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴] 𝑒𝑞 + 𝑘−1 [ 𝐵] 𝑒𝑞 = 0.
Therefore, 𝑘1[ 𝐴] 𝑒𝑞 = 𝑘−1[ 𝐵] 𝑒𝑞, which is in accordance with the equilibrium
condition that the forward rate should equal the reverse rate.
Rearranging this gives
[ 𝑩] 𝒆𝒒
[ 𝑨] 𝒆𝒒
=
𝒌 𝟏
𝒌−𝟏
= 𝒌 𝑪, which we will name the equilibrium rate
constant, a chemical thermodynamics term.
{
[ 𝐴] 𝑒𝑞 + [ 𝐵] 𝑒𝑞 = [ 𝐴]0 … … …… … …(1)
[ 𝐵] 𝑒𝑞
[ 𝐴] 𝑒𝑞
=
𝑘1
𝑘−1
… … … …… … …… … …(2)
We solve for [ 𝐴] 𝑒𝑞 and [ 𝐵] 𝑒𝑞 by substitution.
Rearrange (2): [ 𝐵] 𝑒𝑞 =
𝑘1
𝑘−1
[ 𝐴] 𝑒𝑞……(3)
Substitute the above into (1): [ 𝐴] 𝑒𝑞 +
𝑘1
𝑘−1
[ 𝐴] 𝑒𝑞 = [ 𝐴]0
Factor out [ 𝐴] 𝑒𝑞: [ 𝐴] 𝑒𝑞 (1 +
𝑘1
𝑘−1
) = [ 𝐴]0
Make k-1 the common denominator: [ 𝐴] 𝑒𝑞 (
𝑘−1+𝑘1
𝑘−1
) = [ 𝐴]0
Make [ 𝐴] 𝑒𝑞 the subject: [ 𝑨] 𝒆𝒒 =
𝒌−𝟏[ 𝑨] 𝟎
𝒌−𝟏+𝒌 𝟏
Note that [ 𝐴]0 =
[ 𝐴] 𝑒𝑞( 𝑘−1+𝑘1)
𝑘−1
Substitute into (3): [ 𝐵] 𝑒𝑞 =
𝑘1
𝑘−1
{
𝑘−1[ 𝐴]0
𝑘−1+𝑘1
}
The k-1 term cancels to give: [ 𝑩] 𝒆𝒒 =
𝒌 𝟏[ 𝑨] 𝟎
𝒌−𝟏+𝒌 𝟏
.
To prove these solutions for [ 𝐴] 𝑒𝑞 and [ 𝐵] 𝑒𝑞 hold, we substitute them into (2) and
cancel common terms:
[ 𝐵] 𝑒𝑞
[ 𝐴] 𝑒𝑞
=
𝑘1[ 𝐴]0
𝑘−1 + 𝑘1
∙
𝑘−1 + 𝑘1
𝑘−1[ 𝐴]0
=
𝑘1
𝑘−1
= 𝑘 𝐶
Furthermore, we can substitute our new expression for [ 𝐴]0 into the original
integral equation, ln {
( 𝑘1+𝑘−1)[ 𝐴]−𝑘−1[ 𝐴]0
𝑘1[ 𝐴]0
} = −( 𝑘1 + 𝑘−1) 𝑡:

16. ln {
( 𝑘1 + 𝑘−1)[ 𝐴] − 𝑘−1 (
[ 𝐴] 𝑒𝑞( 𝑘−1 + 𝑘1)
𝑘−1
)
𝑘1 (
[ 𝐴] 𝑒𝑞 ( 𝑘−1 + 𝑘1)
𝑘−1
)
} = −( 𝑘1 + 𝑘−1) 𝑡
The term ( 𝑘−1 + 𝑘1) is common to both numerator terms as well as the
denominator and can be canceled out; k-1 in the numerator can also be canceled
out. This gives:
ln {
[ 𝐴] − [ 𝐴] 𝑒𝑞
𝑘1
𝑘−1
[ 𝐴] 𝑒𝑞
} = −( 𝑘1 + 𝑘−1) 𝑡
Review
Consider the reaction: 𝐴 𝑘⃡⃗⃗ 𝐵.
This overall reaction is composed of two elementary reactions: 𝐴 𝑘1
⃗⃗⃗⃗ 𝐵 and 𝐵 𝑘−1
⃗⃗⃗⃗⃗⃗ 𝐴.
The overall rate in differential form is:
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴] + 𝑘1[ 𝐵].
The integral form is: ln {
( 𝑘1+𝑘−1)[ 𝐴]−𝑘−1[ 𝐴]0
𝑘1[ 𝐴]0
} = −( 𝑘1 + 𝑘−1) 𝑡. (Eq. 1)
This simplifies to give: ln {
[ 𝐴]−[ 𝐴] 𝑒𝑞
[ 𝐴]0−[ 𝐴] 𝑒𝑞
} = −( 𝑘1 + 𝑘−1) 𝑡. (Eq. 2)
We can show that:
1. [ 𝐴] =
𝑘−1[ 𝐴]0
𝑘1+𝑘−1
{1 +
𝑘1
𝑘−1
𝑒−( 𝑘1+𝑘−1) 𝑡
}:
Show that:
𝑘1
𝑘−1
[ 𝐴] 𝑒𝑞 = [ 𝐴] − [ 𝐴] 𝑒𝑞.
[ 𝐵] 𝑒𝑞
[ 𝐴] 𝑒𝑞
=
𝑘1
𝑘−1
→ [ 𝑩] 𝒆𝒒 =
𝒌 𝟏
𝒌−𝟏
[ 𝑨] 𝒆𝒒
[ 𝐵] 𝑒𝑞 + [ 𝐴] 𝑒𝑞 = [ 𝐴]0 → [ 𝑩] 𝒆𝒒 = [ 𝑨] 𝟎 − [ 𝑨] 𝒆𝒒
These two expressions for [ 𝐵] 𝑒𝑞 must be equivalent:
𝑘1
𝑘−1
[ 𝐴] 𝑒𝑞 = [ 𝐴]0 − [ 𝐴] 𝑒𝑞

17. At time 𝑡 = 0: [ 𝐴] =
𝑘−1[ 𝐴]0
𝑘1+𝑘−1
{1 +
𝑘1
𝑘−1
}
Make k-1 the common denominator: [ 𝐴] =
𝑘−1[ 𝐴]0
𝑘1+𝑘−1
{
𝑘−1+𝑘1
𝑘−1
}
The terms 𝑘−1 and 𝑘1 + 𝑘−1 both cancel out in the numerator and denominator
to give: [ 𝐴] = [ 𝐴]0 at time 𝑡 = 0, which is true.
2. [ 𝐵] = [ 𝐴]0 − [ 𝐴] =
𝑘1[ 𝐴]0
𝑘1+𝑘−1
{1 − 𝑒−( 𝑘1+𝑘−1) 𝑡
}:
At time 𝑡 = 0: [ 𝐵] =
𝑘1[ 𝐴]0
𝑘1+𝑘−1
{1 − 1}
Therefore: [ 𝐵] =
𝑘1[ 𝐴]0
𝑘1+𝑘−1
{0}
This boils down to [ 𝐵] = 0 at time 𝑡 = 0, which is true.
Hence both equations are valid.
Relative rate constants in a reversible 1st order reaction
Consider the reaction 𝐴 𝑘⃡⃗⃗ 𝐵, composed of two elementary reactions: 𝐴 𝑘1
⃗⃗⃗⃗ 𝐵 and
𝐵 𝑘−1
⃗⃗⃗⃗⃗⃗ 𝐴.
If we suppose that 𝒌−𝟏 ≃ 𝟎, then the forward reaction predominates and the
integrated rate law for the overall reaction (Eq. 1) changes as follows:
ln {
( 𝑘1 + 𝑘−1 )[ 𝐴]− 𝑘−1[ 𝐴]0
𝑘1[ 𝐴]0
} = −( 𝑘1 + 𝑘−1) 𝑡
ln {
𝑘1[ 𝐴]
𝑘1[ 𝐴]0
} = −𝑘1 𝑡
𝐥𝐧
[ 𝑨]
[ 𝑨] 𝟎
= −𝒌 𝟏 𝒕 = [ 𝑨]
The equation for the concentration of product B changes as follows:
[ 𝐵] = [ 𝐴]0 − [ 𝐴] =
𝑘1[ 𝐴]0
𝑘1 + 𝑘−1
{1 − 𝑒−( 𝑘1+𝑘−1) 𝑡
}
[ 𝐵] =
𝑘1[ 𝐴]0
𝑘1 + 𝑘−1
{1 − 𝑒−( 𝑘1+𝑘−1) 𝑡
}
[ 𝐵] =
𝑘1[ 𝐴]0
𝑘1
{1 − 𝑒−( 𝑘1) 𝑡
}
[ 𝑩] = [ 𝑨] 𝟎{𝟏 − 𝒆−𝒌 𝟏 𝒕
}

18. Hence we can write a new expression for [ 𝐴]0:
[ 𝑨] 𝟎 = [ 𝑨]+ [ 𝑩] = 𝐥𝐧
[ 𝑨]
[ 𝑨] 𝟎
+ [ 𝑨] 𝟎{𝟏 − 𝒆−𝒌 𝟏 𝒕
}
If, similarly, we suppose that 𝒌 𝟏 ≫ 𝒌−𝟏; that is, that 𝒌 ≃ 𝟎, then [ 𝐴] 𝑒𝑞 = 0 and Eq.
2 becomes:
ln {
[ 𝐴] − [ 𝐴] 𝑒𝑞
[ 𝐴]0 − [ 𝐴] 𝑒𝑞
} = −( 𝑘1 + 𝑘−1) 𝑡
𝐥𝐧
[ 𝑨]
[ 𝑨] 𝟎
= −𝒌 𝟏 𝒕
Dynamic Equilibrium in a Reversible 1st Order Reaction
When a reversible 1st order reaction reaches dynamic equilibrium, the following
conditions are true:
 [ 𝐴] 𝑒𝑞 + [ 𝐵] 𝑒𝑞 = [ 𝐴]0

[ 𝐵] 𝑒𝑞
[ 𝐴] 𝑒𝑞
=
𝑘1
𝑘−1
= 𝑘 𝐶

19. Consecutive 1st Order Reactions
Consider the reaction 𝐴 𝑘1
⃡⃗⃗⃗⃗ 𝐵 𝑘2
⃡⃗⃗⃗⃗⃗ 𝐶.
At time 𝑡 = 0, [ 𝐴] = [ 𝐴]0, [ 𝐵] = 0, and [ 𝐶] = 0.
At time 𝑡 = 𝑎𝑛𝑦, [ 𝐴] = [ 𝐴]0 − 𝑥, [ 𝐵] = 𝑦, and [ 𝐶] = 𝑧.
As we define 𝑥 = 𝑦 + 𝑧, we can say that [ 𝐴] + [ 𝐵] + [ 𝐶] = [ 𝐴]0 − 𝑥 + 𝑦 + 𝑧 = [ 𝐴]0
The rates with respect to each species are as follows:
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴],
𝑑[ 𝐵]
𝑑𝑡
= −𝑘1[ 𝐴] − 𝑘2[ 𝐵], and
𝑑[ 𝐶]
𝑑𝑡
= 𝑘2[ 𝐵].
With some manipulation, we can formulate equations for the concentrations of A, B,
and C.
[ 𝑨] = [ 𝑨] 𝟎 𝒆−𝒌 𝟏 𝒕
(1)
𝑑[ 𝐵]
𝑑𝑡
= 𝑘1[ 𝐴] − 𝑘2[ 𝐵] (2)
𝑑[ 𝐶]
𝑑𝑡
= 𝑘2[ 𝐵] (3)
We can substitute [ 𝐴] = [ 𝐴]0 𝑒−𝑘1 𝑡
into the expression for
𝑑[ 𝐵]
𝑑𝑡
as follows:
𝑑[ 𝐵]
𝑑𝑡
= 𝑘1[ 𝐴]0 𝑒−𝑘1 𝑡
− 𝑘2[ 𝐵]
𝑑[ 𝐵]
𝑑𝑡
+ 𝑘2[ 𝐵] = 𝑘1[ 𝐴]0 𝑒−𝑘1 𝑡
(4)
We can also differentiate the expression [ 𝐵] 𝑒 𝑘2 𝑡
with respect to t by using the
product rule [
𝑑
𝑑𝑥
(𝑔( 𝑥) 𝑓( 𝑥)) = 𝑔( 𝑥) 𝑓′( 𝑥)+ 𝑓( 𝑥) 𝑔′( 𝑥)] and then factor out the
common term, 𝑒 𝑘2 𝑡
:
𝑑
𝑑𝑡
{[ 𝐵] 𝑒 𝑘2 𝑡} = 𝑒 𝑘2 𝑡
𝑑[ 𝐵]
𝑑𝑡
+ 𝑘2 𝑒 𝑘2 𝑡[ 𝐵]
𝑑
𝑑𝑡
{[ 𝐵] 𝑒 𝑘2 𝑡} = 𝑒 𝑘2 𝑡
(
𝑑[ 𝐵]
𝑑𝑡
+ 𝑘2[ 𝐵])
The term in parentheses above is equivalent to (4) and so we can substitute and use
the laws of exponents to simplify:
𝑑
𝑑𝑡
{[ 𝐵] 𝑒 𝑘2 𝑡} = 𝑒 𝑘2 𝑡( 𝑘1[ 𝐴]0 𝑒−𝑘1 𝑡)
𝑑
𝑑𝑡
{[ 𝐵] 𝑒 𝑘2 𝑡} = 𝑘1[ 𝐴]0 𝑒( 𝑘2−𝑘1) 𝑡

20. Now we can rearrange this equation and integrate using [∫( 𝑒 𝑎𝑥) 𝑑𝑥 =
1
𝑎
𝑒 𝑎𝑥
+ 𝐶]:
∫ 𝑑{[ 𝐵] 𝑒 𝑘2 𝑡} = ∫ 𝑘1[ 𝐴]0 𝑒( 𝑘2−𝑘1) 𝑡
𝑑𝑡
[ 𝐵] 𝑒 𝑘2 𝑡
=
𝑘1[ 𝐴]0
𝑘2 − 𝑘1
𝑒( 𝑘2−𝑘1) 𝑡
+ 𝐶
At time 𝑡 = 0, the equation becomes: 0 =
𝑘1[ 𝐴]0
𝑘2−𝑘1
+ 𝐶 → 𝐶 = −
𝑘1[ 𝐴]0
𝑘2−𝑘1
.
Hence, at time 𝑡 = 𝑎𝑛𝑦: [ 𝐵] 𝑒 𝑘2 𝑡
=
𝑘1[ 𝐴]0
𝑘2−𝑘1
𝑒( 𝑘2−𝑘1) 𝑡
−
𝑘1[ 𝐴]0
𝑘2−𝑘1
.
Factoring the common term
𝑘1[ 𝐴]0
𝑘2−𝑘1
, rearranging, and simplifying gives an expression
for [B]:
[ 𝐵] 𝑒 𝑘2 𝑡
=
𝑘1[ 𝐴]0
𝑘2 − 𝑘1
{𝑒( 𝑘2−𝑘1) 𝑡
− 1}
[ 𝐵] =
𝑘1[ 𝐴]0
𝑘2 − 𝑘1
{
𝑒( 𝑘2−𝑘1) 𝑡
− 1
𝑒 𝑘2 𝑡
}
[ 𝐵] =
𝑘1[ 𝐴]0
𝑘2 − 𝑘1
{
𝑒 𝑘2 𝑡
− 𝑒 𝑘1 𝑡
− 𝑒 𝑘2 𝑡
𝑒 𝑘2 𝑡
}
[ 𝐵] =
𝑘1[ 𝐴]0
𝑘2 − 𝑘1
{
−𝑒 𝑘1 𝑡
𝑒 𝑘2 𝑡
}
[ 𝑩] =
𝒌 𝟏[ 𝑨] 𝟎
𝒌 𝟐 − 𝒌 𝟏
{𝒆−𝒌 𝟏 𝒕
− 𝒆 𝒌 𝟐 𝒕
}
Finally, as we have stated that [ 𝐴] + [ 𝐵]+ [ 𝐶] = [ 𝐴]0, we can solve for [C] as
follows:
Rearrange to make [C] the subject: [ 𝐶] = [ 𝐴]0 − [ 𝐴] − [ 𝐵]
Substitute equations for [A] and [B]:[ 𝐶] = [ 𝐴]0 − [ 𝐴]0 𝑒−𝑘1 𝑡
− (
𝑘1[ 𝐴]0
𝑘2−𝑘1
{ 𝑒−𝑘1 𝑡
− 𝑒 𝑘2 𝑡})
Factor [ 𝐴]0: [ 𝐶] = [ 𝐴]0 {1 − 𝑒−𝑘1 𝑡
−
𝑘1
𝑘2−𝑘1
{ 𝑒−𝑘1 𝑡
+ 𝑒 𝑘2 𝑡}}
FINAL ANSWER SHOULD BE [ 𝐶] = [ 𝐴]0 {1 −
𝑘2
𝑘2−𝑘1
𝑒−𝑘1 𝑡
+
𝑘1
𝑘2−𝑘1
𝑒−𝑘2 𝑡
}

21. The Rate-Determining Step(RDS)
Suppose now that 𝑘2 ≫ 𝑘1. This would mean that B forms C much more quickly than
A forms B. Therefore, [ 𝐵] remains small and almost constant throughout the
reaction. That is, whenever a molecule of B is formed, it decays rapidly into C.
Hence the equation for [ 𝐵] ([ 𝐵] =
𝑘1[ 𝐴]0
𝑘2−𝑘1
{ 𝑒−𝑘1 𝑡
− 𝑒−𝑘2 𝑡}) becomes [ 𝑩] =
𝒌 𝟏[ 𝑨] 𝟎
𝒌 𝟐
𝒆−𝒌 𝟏 𝒕
.
We say that [ 𝐵] is very small and remains constant over the course of the reaction.
Similarly, [ 𝐶] = [ 𝐴]0 {1 −
𝑘2
𝑘2−𝑘1
𝑒−𝑘1 𝑡
+
𝑘1
𝑘2−𝑘1
𝑒−𝑘2 𝑡
} becomes [ 𝑪] = [ 𝑨] 𝟎{𝟏 − 𝒆−𝒌 𝟏 𝒕
},
and [ 𝑨] = [ 𝑨] 𝟎 𝒆−𝒌 𝟏 𝒕
.
All three of these equations depend only on k1, thus proving that the 1st step
determines the rate (or is the rate-determining step) of the reaction.
We say that a steady state approximation holds for species B, since [B] is constant
and small over the course of the reaction.
Therefore, the rate with respect to B is given by:
𝑑[ 𝐵]
𝑑𝑡
= 0 = 𝑘1[ 𝐴] − 𝑘2[ 𝐵].
Hence 𝑘1[ 𝐴] = 𝑘2[ 𝐵] and [ 𝐵] =
𝑘1
𝑘2
[ 𝐴].
Similarly, the rate with respect to C is:
𝑑[ 𝐶]
𝑑𝑡
= 𝑘2[ 𝐵].
Substituting the final expression above for B gives:
𝑑[ 𝐶]
𝑑𝑡
= 𝑘2
𝑘1
𝑘2
[ 𝐴] = 𝑘1[ 𝐴]
Finally, we substitute for [A] to obtain:
𝑑[ 𝐶]
𝑑𝑡
= 𝑘1[ 𝐴]0 𝑒−𝑘1 𝑡
.
This is an equation in [A] only and can be integrated to provide a new equation for
[C] as follows.
Rearrange: 𝑑[ 𝐶] = 𝑘1[ 𝐴]0 𝑒−𝑘1 𝑡
𝑑𝑡
Integrate using [∫( 𝑒 𝑎𝑥) 𝑑𝑥 =
1
𝑎
𝑒 𝑎𝑥
+ 𝐶]: [ 𝐶] =
𝑘1[ 𝐴]0 𝑒−𝑘1 𝑡
−𝑘1
+ 𝐶
Simplify: [ 𝐶] =
𝑘1[ 𝐴]0 𝑒−𝑘1 𝑡
−𝑘1
+ 𝐶 → [ 𝐶] = −[ 𝐴]0 𝑒−𝑘1 𝑡
+ 𝐶
At time 𝑡 = 0, therefore: 0 = −[ 𝐴]0 + 𝐶 → 𝐶 = [ 𝐴]0.
Thus: [ 𝐶] = −[ 𝐴]0 𝑒−𝑘1 𝑡
+ [ 𝐴]0 = [ 𝐴]0(−𝑒−𝑘1 𝑡
+ 1)
Or: [ 𝐶] = [ 𝐴]0{1 − 𝑒−𝑘1 𝑡}, the same expression attained via the steady state
approximation.

22. ReactionMechanisms
Consider the reaction 2𝑆𝑂2 + 𝑂2 → 2𝑆𝑂3.
The nitrogen monoxide-catalyzed oxidation of 𝑆𝑂2 gas gives the overall
stoichiometry of the reaction but does not tell us the process, or mechanism, by
which the reaction actually occurs. This reaction has been postulated to occur by the
following two-step process:

23. 𝑆𝑡𝑒𝑝 1: [ 𝑂2 + 2𝑁𝑂 → 2𝑁𝑂2] 𝑆1 = 1
𝑆𝑡𝑒𝑝 2: [ 𝑁𝑂2 + 𝑆𝑂2 → 𝑁𝑂 + 𝑆𝑂3] 𝑆2 = 1
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛: 𝑂2 + 2𝑆𝑂2 → 2𝑆𝑂3
The number of times a given step in the mechanism occurs for each occurrence of
the overall reaction is called the stoichiometric number S of the step.
Do not confuse the stoichiometric number S of a mechanistic step with the
stoichiometric coefficient of a chemical species.
Example
The gas-phase decomposition of 𝑁2 𝑂5 has overall reaction:
𝑆𝑡𝑒𝑝 1: [ 𝑁2 𝑂5 ⇌ 𝑁𝑂2 + 𝑁𝑂3] 𝑆1 = 2
𝑆𝑡𝑒𝑝 2: [ 𝑁𝑂2 + 𝑁𝑂3 → 𝑁𝑂 + 𝑂2 + 𝑁𝑂2] 𝑆2 = 1
𝑆𝑡𝑒𝑝 3: [ 𝑁𝑂+ 𝑁𝑂3 → 2𝑁𝑂2] 𝑆3 = 1
𝑂𝑣𝑒𝑟𝑎𝑙𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛: 2𝑁2 𝑂5 → 4𝑁𝑂2 + 𝑂2
Species like 𝑁𝑂3 and 𝑁𝑂, which do not appear in the overall reaction, are called
reaction intermediates. Each step in the mechanism is called an elementary reaction.
A simple reaction consists of a single elementary step. A complex/composite
reaction consists of two or more elementary steps.
The Diels-Alder addition of ethylene to butan-1,3-diene to give cyclohexene is
believed to be simple, occurring as a simple step:
𝐶𝐻2 = 𝐶𝐻2 + 𝐶𝐻2 = 𝐶𝐻𝐶𝐻 = 𝐶𝐻2 → 𝐶6 𝐻10
… 𝑤ℎ𝑒𝑟𝑒 𝜐 = 𝑘[ 𝐶𝐻2 = 𝐶𝐻2][ 𝐶𝐻2 = 𝐶𝐻𝐶𝐻 = 𝐶𝐻2]
Rate-Determining Step Determination
Consider the composite reaction 𝐴 𝑘1
⃗⃗⃗⃗⃗⃗ 𝐵 𝑘2
⃗⃗⃗⃗⃗⃗ 𝐶.
If we assume that 𝑘2 ≫ 1, then the overall reaction becomes 𝐴 𝑘1
⃗⃗⃗⃗⃗⃗ 𝐶.
The concentration of B, [ 𝐵], remains very small and constant during the course of
the reaction, such that
𝑑[ 𝐵]
𝑑𝑡
= 0 = 𝑘1[ 𝐴] − 𝑘2[ 𝐵]. Hence, 𝑘1[ 𝐴] = 𝑘2[ 𝐵] and [ 𝐵] =
𝑘1
𝑘2
[ 𝐴].
We also know that
𝑑[ 𝐶]
𝑑𝑡
= 𝑘2[ 𝐵]. Thus,
𝑑[ 𝐶]
𝑑𝑡
= 𝑘2 (
𝑘1
𝑘2
[ 𝐴]) = 𝑘1[ 𝐴].
Finally,
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴].

24. This shows that the rate of appearance of A depends on the 1st elementary step,
which – by virtue of the large size of 𝑘2 – is assumed to be the slower, rate-
determining step.
Example
Consider the reaction 𝐴 𝑘1, 𝑘−1
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝐵 𝑘2
⃗⃗⃗⃗⃗⃗ 𝐶.
We define: 𝑘 =
𝑘1 𝑘2
𝑘−1+𝑘2
.
We know that:
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴] + 𝑘−1 [ 𝐵]
𝑑[ 𝐵]
𝑑𝑡
= 𝑘1[ 𝐴]− 𝑘−1[ 𝐵] − 𝑘2[ 𝐵] = 0 ( 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛)
𝑑[ 𝐶]
𝑑𝑡
= −𝑘2[ 𝐵]
Therefore:
𝑘1[ 𝐴] = 𝑘−1[ 𝐵] + 𝑘2[ 𝐵] = [ 𝐵]( 𝑘−1 + 𝑘2) → [ 𝑩] =
𝒌 𝟏[ 𝑨]
𝒌−𝟏 + 𝒌 𝟐
𝑑[ 𝐶]
𝑑𝑡
= −𝑘2[ 𝐵] = −𝑘2 (
𝑘1
𝑘−1 + 𝑘2
) [ 𝐴] =
𝑘1 𝑘2
𝑘−1 + 𝑘2
[ 𝐴] →
𝒅[ 𝑪]
𝒅𝒕
= 𝒌[ 𝑨]
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴] + 𝑘−1 [ 𝐵]
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴] + 𝑘−1 (
𝑘1[ 𝐴]
𝑘−1 + 𝑘2
)
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴] +
𝑘1 𝑘−1
𝑘−1 + 𝑘2
[ 𝐴]
𝑑[ 𝐴]
𝑑𝑡
=
−𝑘1( 𝑘−1 + 𝑘2)[ 𝐴]+ 𝑘1 𝑘−1[ 𝐴]
𝑘−1 + 𝑘2
𝑑[ 𝐴]
𝑑𝑡
=
−𝑘1 𝑘−1[ 𝐴] − 𝑘1 𝑘2[ 𝐴] + 𝑘1 𝑘−1[ 𝐴]
𝑘−1 + 𝑘2
𝑑[ 𝐴]
𝑑𝑡
=
−𝑘1 𝑘2[ 𝐴]
𝑘−1 + 𝑘2
𝒅[ 𝑨]
𝒅𝒕
= −𝒌[ 𝑨]
The equation
𝑑[ 𝐴]
𝑑𝑡
= −𝑘[ 𝐴] becomes 𝑑[ 𝐴] = −𝑘[ 𝐴] 𝑑𝑡 and can be integrated to give:
[ 𝑨] = [ 𝑨] 𝟎 𝐞−𝒌𝒕
.
Thus,
𝑑[ 𝐶]
𝑑𝑡
= 𝑘[ 𝐴] = 𝑘[ 𝐴]0 𝑒−𝑘𝑡
.

25. Although the reaction is 𝐴 𝑘1, 𝑘−1
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝐵 𝑘2
⃗⃗⃗⃗⃗⃗ 𝐶, the overall reaction is 𝐴 𝑘⃗⃗⃗⃗ 𝐶, due to the
steady state approximation. Hence:
[ 𝐴] + [ 𝐶] = [ 𝐴]0 → [ 𝐶] = [ 𝐴]0 − [ 𝐴] = [ 𝐴]0 − [ 𝐴]0 𝑒−𝑘𝑡
→ [ 𝑪] = [ 𝑨] 𝟎(𝟏 − 𝒆−𝒌𝒕
)
Example
Consider the reaction 𝐴 𝑘1, 𝑘−1
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝐼 𝑘2, 𝑘−2
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝐵.
We know that:
𝑑[ 𝐴]
𝑑𝑡
= −𝑘1[ 𝐴] + 𝑘−1 [ 𝐼]
𝑑[ 𝐵]
𝑑𝑡
= 𝑘2[ 𝐼] − 𝑘−2[ 𝐵]
𝑑[ 𝐼]
𝑑𝑡
= 𝑘1[ 𝐴] − 𝑘−1[ 𝐼]− 𝑘2[ 𝐼]+ 𝑘−2[ 𝐵] = 0 ( 𝑠𝑡𝑒𝑎𝑑𝑦 𝑠𝑡𝑎𝑡𝑒 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛)
Therefore:
𝑘1[ 𝐴] + 𝑘−2[ 𝐵] = 𝑘−1[ 𝐼] + 𝑘2[ 𝐼]
𝑘1[ 𝐴] + 𝑘−2[ 𝐵] = [ 𝐼]( 𝑘−1 + 𝑘2)
[ 𝑰] =
𝒌 𝟏[ 𝑨] + 𝒌−𝟐[ 𝑩]
𝒌−𝟏 + 𝒌 𝟐
Thus:
𝑑[ 𝐵]
𝑑𝑡
= 𝑘2 (
𝑘1[ 𝐴]+ 𝑘−2[ 𝐵]
𝑘−1 + 𝑘2
) − 𝑘−2[ 𝐵]
𝑑[ 𝐵]
𝑑𝑡
=
𝑘1 𝑘2[ 𝐴] + 𝑘−2 𝑘2[ 𝐵]
𝑘−1 + 𝑘2
− 𝑘−2[ 𝐵]
𝑑[ 𝐵]
𝑑𝑡
=
𝑘1 𝑘2[ 𝐴] + 𝑘−2 𝑘2[ 𝐵] − 𝑘−2( 𝑘−1 + 𝑘2)[ 𝐵]
𝑘−1 + 𝑘2
𝑑[ 𝐵]
𝑑𝑡
=
𝑘1 𝑘2[ 𝐴] + 𝑘−2 𝑘2[ 𝐵] − 𝑘−1 𝑘−2[ 𝐵] − 𝑘−2 𝑘2[ 𝐵]
𝑘−1 + 𝑘2
𝑑[ 𝐵]
𝑑𝑡
=
𝑘1 𝑘2[ 𝐴] + 𝑘−2 𝑘2[ 𝐵] − 𝑘−1 𝑘−2[ 𝐵] − 𝑘−2 𝑘2[ 𝐵]
𝑘−1 + 𝑘2
𝑑[ 𝐵]
𝑑𝑡
=
𝑘1 𝑘2[ 𝐴] − 𝑘−1 𝑘−2 [ 𝐵]
𝑘−1 + 𝑘2
If we let 𝑘 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 = 𝑘 𝑓 =
𝑘1 𝑘2
𝑘−1+𝑘2
and 𝑘 𝑏𝑎𝑘𝑤𝑎𝑟𝑑 = 𝑘 𝑏 =
𝑘−1 𝑘−2
𝑘−1+𝑘2
, then:
𝑑[ 𝐵]
𝑑𝑡
=
𝑘1 𝑘2
𝑘−1 + 𝑘2
[ 𝐴] −
𝑘−1 𝑘−2
𝑘−1 + 𝑘2
[ 𝐵]
𝒅[ 𝑩]
𝒅𝒕
= 𝒌 𝒇[ 𝑨]− 𝒌 𝒃[ 𝑩]
The reaction can therefore be simplified to 𝐴 𝑘 𝑓, 𝑘 𝑏
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝐵.

26. And if
𝑑[ 𝐵]
𝑑𝑡
= 𝑘 𝑓[ 𝐴] − 𝑘 𝑏[ 𝐵], then
𝒅[ 𝑨]
𝒅𝒕
= −𝒌 𝒇[ 𝑨]+ 𝒌 𝒃[ 𝑩].
Furthermore, if 𝑘 𝑓 ≫ 𝑘 𝑏, then 𝐴 𝑘𝑓
⃗⃗⃗⃗⃗⃗ 𝐵. Likewise, 𝑘 𝑓 ≪ 𝑘 𝑏, then 𝐴 𝑘 𝑏
⃡⃗⃗⃗⃗⃗⃗ 𝐵.
Example

27. Kinetics of Relaxation
Consider the reaction 𝐴 𝑘1, 𝑘−1
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑍.
We know that:
 [ 𝐴]0 = [ 𝐴] + [ 𝑍]
 𝑥 = [ 𝑍]
 [ 𝐴] = [ 𝐴]0 − [ 𝑍] = 𝑎0 − 𝑥
Hence, the rate is given by
𝑑𝑥
𝑑𝑡
= 𝑘1[ 𝐴] − 𝑘−1[ 𝑍].
At equilibrium:
𝑑𝑥
𝑑𝑡
= 𝑘1( 𝑎0 − 𝑥) − 𝑘−1 𝑥 = 0.
If 𝑥 𝑒 is the concentration of Z at equilibrium (i.e. if 𝑥 𝑒 = [ 𝑍] 𝑒𝑞), then:
𝑑𝑥
𝑑𝑡
= 𝑘1( 𝑎0 − 𝑥 𝑒) − 𝑘−1 𝑥 𝑒 = 0
We define the change in concentration of x as: ∆𝑥 = 𝑥 = 𝑥 𝑒 → 𝑥 = ∆𝑥 + 𝑥 𝑒. Thus:
𝑑∆𝑥
𝑑𝑡
= 𝑘1[ 𝑎0 − (∆𝑥 + 𝑥 𝑒)] − 𝑘−1(∆𝑥 + 𝑥 𝑒)
𝑑∆𝑥
𝑑𝑡
= 𝑘1 𝑎0 − 𝑘1(∆𝑥 + 𝑥 𝑒) − 𝑘−1(∆𝑥 + 𝑥 𝑒)
𝑑∆𝑥
𝑑𝑡
= 𝑘1 𝑎0 − 𝑘1∆𝑥 − 𝑘1 𝑥 𝑒 − 𝑘−1∆𝑥 − 𝑘−1 𝑥 𝑒
𝑑∆𝑥
𝑑𝑡
= 𝑘1( 𝑎0 − 𝑥 𝑒)− ( 𝑘1 + 𝑘−1)∆𝑥 − 𝑘−1 𝑥 𝑒
𝑑∆𝑥
𝑑𝑡
= −( 𝑘1 + 𝑘−1)∆𝑥 + 𝑘1( 𝑎0 − 𝑥 𝑒)− 𝑘−1 𝑥 𝑒
𝒅∆𝒙
𝒅𝒕
= −( 𝒌 𝟏 + 𝒌−𝟏)∆𝒙
This is the rate law in differential form.
Rearranging gives
𝑑∆𝑥
∆𝑥
= −( 𝑘1 + 𝑘−1) 𝑑𝑡.
Integrating gives: ln ∆𝑥 = −( 𝑘1 + 𝑘−1 ) 𝑡 + 𝐶.
At time 𝑡 = 0: ln∆𝑥0 = 0 + 𝐶 → 𝐶 = ln ∆𝑥0. Therefore:
ln ∆𝑥 = −( 𝑘1 + 𝑘−1) 𝑡 + ln ∆𝑥0
ln ∆𝑥 − ln∆𝑥0 = −( 𝑘1 + 𝑘−1) 𝑡
𝐥𝐧
∆𝒙
∆𝒙 𝟎
= −( 𝒌 𝟏 + 𝒌−𝟏) 𝒕
This is the rate in the form 𝑦 = 𝑎𝑥 + 𝑏, where 𝑦 = ln
∆𝑥
∆𝑥0
, 𝑎 = 𝑠𝑙𝑜𝑝𝑒 = −( 𝑘1 + 𝑘−1),
and 𝑥 = 𝑡.

28. A graph of ln
∆𝑥
∆𝑥0
𝑣𝑠. 𝑡 would therefore look like:
A plot of 𝑥 𝑣𝑠. 𝑡, however, would give a completely different graph:
The Principle of the Temperature Jump Technique
The relaxation time  is defined as the time corresponding to
(∆𝑥)0
∆𝑥
= 𝑒 and 𝑡 = 𝜏.
Therefore, ln
(∆𝑥)0
∆𝑥
= ( 𝑘1 + 𝑘−1) 𝑡 becomes ln 𝑒 = ( 𝑘1 + 𝑘−1) 𝜏.
And so: 1 = ( 𝑘1 + 𝑘−1) 𝜏 and 𝝉 =
𝟏
( 𝒌 𝟏+𝒌−𝟏)
.
Thus, the negative reciprocal of the slope (−
1
𝑠𝑙𝑜𝑝𝑒
) of a graph of ln
∆𝑥
∆𝑥0
𝑣𝑠. 𝑡 is equal
to the relaxation time .
N.B The equilibrium constant is given by 𝐾𝑒𝑞 =
𝑘1
𝑘−1
.

29. Example
Consider the reaction 𝐴 + 𝐵 𝑘1, 𝑘−1
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑍.
Let [ 𝑍] = 𝑥, [ 𝐴] = [ 𝐴]0 − 𝑥, and [ 𝐵] = [ 𝐵]0 − 𝑥.
The rate is
𝑑[ 𝑍]
𝑑𝑡
=
𝑑𝑥
𝑑𝑡
= 𝑘1[ 𝐴][ 𝐵]− 𝑘−1[ 𝑍] = 𝑘1( 𝑎0 − 𝑥)( 𝑏0 − 𝑥) − 𝑘−1 𝑥.
If we let ∆𝑥 = 𝑥 − 𝑥 𝑒, then:
𝑑∆𝑥
𝑑𝑡
=
𝑑𝑥
𝑑𝑡
− 0
𝑑∆𝑥
𝑑𝑡
=
𝑑𝑥
𝑑𝑡
= 𝑘1( 𝑎0 − 𝑥)( 𝑏0 − 𝑥) − 𝑘−1 𝑥
𝑑∆𝑥
𝑑𝑡
= 𝑘1 𝑎0 𝑏0 − 𝑘1 𝑎0 𝑥 − 𝑘1 𝑏0 𝑥 + 𝑘1 𝑥2
− 𝑘−1 𝑥
𝑑∆𝑥
𝑑𝑡
= 𝑘1 𝑎0 𝑏0 − ( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1) 𝑥 + 𝑘1 𝑥2
At equilibrium,
𝑑∆𝑥
𝑑𝑡
=
𝑑𝑥
𝑑𝑡
= 0.
Therefore, 𝑘1 𝑎0 𝑏0 = ( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1) 𝑥 + 𝑘1 𝑥2
.
If 𝑥 = ∆𝑥 + 𝑥 𝑒, then:
𝑑∆𝑥
𝑑𝑡
= 𝑘1 𝑎0 𝑏0 − ( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1)(∆𝑥 + 𝑥 𝑒) + 𝑘1(∆𝑥 + 𝑥 𝑒)2
𝑑∆𝑥
𝑑𝑡
= 𝑘1 𝑎0 𝑏0 − 𝑘1 𝑎0∆𝑥 − 𝑘1 𝑎0 𝑥 𝑒 − 𝑘1 𝑏0∆𝑥 − 𝑘1 𝑏0 𝑥 𝑒 − 𝑘−1∆𝑥 − 𝑘−1 𝑥 𝑒
+ 𝑘1(∆𝑥2
+ 2∆𝑥 ⋅ 𝑥 𝑒 + 𝑥 𝑒
2)
Terms only in 𝑥 𝑒 cancel out.
𝑑∆𝑥
𝑑𝑡
= 𝑘1 𝑎0 𝑏0 − 𝑘1 𝑎0∆𝑥 − 𝑘1 𝑎0 𝑥 𝑒 − 𝑘1 𝑏0∆𝑥 − 𝑘1 𝑏0 𝑥 𝑒 − 𝑘−1∆𝑥 − 𝑘−1 𝑥 𝑒
+ 𝑘1(∆𝑥2
+ 2∆𝑥 ⋅ 𝑥 𝑒 + 𝑥 𝑒
2)
𝑑∆𝑥
𝑑𝑡
= −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1)∆𝑥 + 𝑘1(∆𝑥 + 2𝑥 𝑒)∆𝑥
If the displacement from equilibrium is only slight, then 𝑥 ≈ 𝑥 𝑒, so that 𝑥 + 𝑥 𝑒 may
be written as 2𝑥 𝑒 𝑥 + 𝑥 𝑒 = 2𝑥 𝑒.
𝑑∆𝑥
𝑑𝑡
= −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1)∆𝑥 + 2𝑘1 𝑥 𝑒∆𝑥
𝑑∆𝑥
𝑑𝑡
= −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒)∆𝑥
Rearranging gives
𝑑∆𝑥
∆𝑥
= −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒) 𝑑𝑡.

30. Integrating gives ln∆𝑥 = −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒) 𝑡 + 𝐶.
At time 𝑡 = 0, ∆𝑥 = ∆𝑥0 and therefore ln∆𝑥0 = 0 + 𝐶 → 𝐶 = ln ∆𝑥0.
Thus:
ln ∆𝑥 = −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒) 𝑡 + ln ∆𝑥0
ln ∆𝑥 − ln∆𝑥0 = −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒) 𝑡
ln
∆𝑥
∆𝑥0
= −( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒) 𝑡
ln
∆𝑥0
∆𝑥
= ( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒) 𝑡
The relaxation time  is defined as that corresponding to
∆𝑥0
∆𝑥
= 𝑒.
ln 𝑒 = ( 𝑘1 𝑎0 + 𝑘1 𝑏0 + 𝑘−1 − 2𝑘1 𝑥 𝑒) 𝜏 → 𝝉 =
𝟏
𝒌 𝟏 𝒂 𝟎 + 𝒌 𝟏 𝒃 𝟎 + 𝒌−𝟏 − 𝟐𝒌 𝟏 𝒙 𝒆
The dissociation of a weak acid, 𝐻𝐴 + 𝐻2 𝑂 → 𝐻3 𝑂+
+ 𝐴−
, can be represented as
𝐴 𝑘1, 𝑘−1
⃡⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 𝑌+ 𝑍.
The rate constants k1 and k-1 cannot be determined by conventional methods but
can be determined by the T-jump technique.
We can prove that the relaxation time is given by 𝜏 =
1
𝑘1+2𝑘−1 𝑥 𝑒
, where xe is the
concentration of the ions (Y and Z) at equilibrium.
ASK HOKMABADI FOR THE HW BACK IN ORDER TO SEE THE ANSWER TO THIS.
The Bohr Hydrogen or Hydrogen-like (isoelectronic to hydrogen) Atom
Consider a hydrogen bulb shining light upon a concave lens in front of a triangular
pris. If a screen is placed behind the prism, discrete bands of colored and invisible
light (see below).

31. As the capacitor plates in a hydrogen bulb accumulate charge, the negatively
charged plate emits a beam of electrons that bombards hydrogen atoms, which in
turn release photons of light as they are excited.
Compare hydrogen to some hydrogen-like atoms:
H 1 proton 1 electron
He+ 2 protons 1 electron
Li2+ 3 protons 1 electron
… … 1 electron
Let 𝑍 = 𝑎𝑡𝑜𝑚𝑖𝑐 𝑛𝑢𝑚𝑏𝑒𝑟, 𝑚 𝑝 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎 𝑝𝑟𝑜𝑡𝑜𝑛, 𝑚 𝑒 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑛 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛, and
+𝑧𝑒 = 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑛𝑢𝑐𝑙𝑒𝑢𝑠.
𝑚 𝑝 = 1836𝑚 𝑒
1. The attractive force between negative and positive charges (protons and
electrons) is the normal type of Coulombic attraction, such that: 𝐹 ∝
(+𝑧𝑒)(−1𝑒)
𝑟2 ,
where ∝=
1
4𝜋𝜀0
, 𝜀0 is the permittivity of the vacuum. Therefore:
𝐹 =
−𝑧𝑒2
(4𝜋𝜀0) 𝑟2
2. The electron can remain in any particular state (fixed radius) indefinitely
without radiating its energy, provided that the electron’s angular momentum is
𝐿 = 𝑚𝑣𝑟 (Equation 20.10), where 𝐿 = 𝑚𝑣𝑟 =
𝑛ℎ
2𝜋
(quantum number, 𝑛 =
1, 2, 3,…, Planck’s constant, ℎ = 6.626 × 10−34
𝑚2
𝑘𝑔/𝑠 ( 𝐽/𝑠), 𝑚 = 𝑚 𝑒, 𝑣 =
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛, 𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠).
Hence 𝐿 ∝ 𝑛, and possible values of L are:
ℎ
2𝜋
,
2ℎ
2𝜋
=
ℎ
𝜋
,
3ℎ
2𝜋
,
4ℎ
2𝜋
=
2ℎ
𝜋
, …
Also, L has the same units as h: joule per second.
Note: The total energy of the atom is equal to the sum of the kinetic energy of
orbiting electrons and the potential energy of electrons attracted to protons.

32. Mechanical Stability
Mechanical stability requires that the attractive and centripetal forces cancel each
other; that is:
−
𝑧𝑒2
(4𝜋𝜀0) 𝑟2 +
𝑚𝑣2
𝑟
= 0 →
𝑧 𝑒2
(4𝜋𝜀0) 𝑟2 =
𝑚𝑣2
𝑟
→ (4𝜋𝜀0) 𝑚𝑣2
𝑟 = 𝑧𝑒2
(Eq. 20.13)
From the quantization of Eq. 20.10 ( 𝐿 = 𝑚𝑣𝑟), we can write that:
𝑚2
𝑣2
𝑟2
=
𝑛2
ℎ2
4𝜋2 → 𝑣2
=
𝑛2
ℎ2
𝑚2 𝑟2 (Eq. 20.15)
…where ℎ =
ℎ
2𝜋
and 𝑚𝑣𝑟 = ℎ.
Substitute for v2 in Eq. 20.13:
(4𝜋𝜀0) 𝑚 (
𝑛2
ℎ2
𝑚2 𝑟2
) 𝑟 = 𝑧𝑒2
→ 𝑟 =
𝑛2
ℎ2 (4𝜋𝜀0)
𝑚𝑧𝑒2
=
𝑛2
𝑧
(
ℎ2 (4𝜋𝜀0)
𝑚𝑒2
) =
𝑛2
𝑧
𝑎0
Therefore: {𝒓 =
𝒏 𝟐
𝒛
𝒂 𝟎}, where 𝑛 = 1,2, 3, …, 𝑎0 =
ℎ2(4𝜋𝜀0)
𝑚𝑒2 = 5.29 × 10−11
𝑚.
Figure 1: Proof of a0.
This proves that the distance of the electron from the nucleus is quantized.
Note that for the hydrogen atom, 𝑧 = 1 and thus 𝑟𝑛 = 𝑛2
𝑎0.

33. Proof that kinetic energy (T), potential energy (U), and total energy are also
quantized:
𝐸𝑡𝑜𝑡 = 𝑇 + 𝑈
The force is given by: −𝐹 ≡
𝑑𝑈
𝑑𝑟
→ 𝑑𝑈 = −𝐹 𝑑𝑟 = −(
−𝑧𝑒2
(4𝜋𝜀0) 𝑟2) 𝑑𝑟 = (
𝑧𝑒2
4𝜋𝜀0
)
𝑑𝑟
𝑟2
Integrating gives: 𝑈 = − (
𝑧𝑒2
4𝜋𝜀0
)
1
𝑟
+ 𝐶
When 𝑟 → ∞, 𝑈 = 0and 𝐶 = 0.
Then 𝑼 = −
𝒛𝒆 𝟐
( 𝟒𝝅𝜺 𝟎) 𝒓
.
The kinetic energy is is given by: 𝑇 =
1
2
𝑚𝑣2
.
Substitute Eq. 20.13: 𝑇 =
1
2
𝑚𝑣2
=
1
2
(
𝑧𝑒2
(4𝜋𝜀0) 𝑟
). That is: 𝑻 =
𝟏
𝟐
(
𝒛𝒆 𝟐
( 𝟒𝝅𝜺 𝟎) 𝒓
).
Therefore: 𝐸𝑡𝑜𝑡 = 𝑇 + 𝑈 =
1
2
(
𝑧𝑒2
(4𝜋𝜀0) 𝑟
) − (
𝑧𝑒2
(4𝜋𝜀0) 𝑟
) = −
1
2
(
𝑧𝑒2
(4𝜋𝜀0) 𝑟
).
We see that Etot is negative and inversely proportional to r.
The final expression for the (total) energy is obtained by substituting the value of r
as expressed in Eq. 20.17 below:
𝑟 =
𝑛2
ℎ2(4𝜋𝜀0)
𝑚𝑧𝑒2
and
𝑟𝑛 =
𝑛2
𝑎0
𝑧
…where 𝑎0 =
ℎ(4𝜋𝜀0)
𝑚𝑒2 5.29 × 10−11
𝑚 and ℎ =
ℎ
2𝜋
= 0.529Å
Therefore: 𝐸𝑡𝑜𝑡 = 𝐸 𝑛 = −
1
2
𝑧𝑒2
(4𝜋𝜀0)
∙
1
𝑟
= −
1
2
𝑧 𝑒2
(4𝜋𝜀0)
∙
𝑚𝑧 𝑒2
𝑛2ℎ2 (4𝜋𝜀0)
= −
𝑚𝑧2
𝑒4
2𝑛2 ℎ2(4𝜋𝜀0)2 =
−
𝑧2
𝑛2 (
𝑚𝑒4
2ℎ2 (4𝜋𝜀0)2) = −
𝑧2
𝑛2 𝑅 = −
𝑧2
𝑛2 𝑅 𝐻
That is: 𝑬 𝒏 = −
𝒛 𝟐
𝒏 𝟐 𝑹 𝑯. (Eq. 20.17)
…where 𝑅( 𝐻) =
𝑚𝑒4
2ℎ2 (4𝜋𝜀0)2 = 𝑅𝑦𝑑𝑏𝑢𝑟𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 2.179 × 10−18
𝐽.

34. The Bohr Theory/Model of the Hydrogen Atom
Let 𝑧 = 1.
Let 𝑚𝑣 = ( 𝑙𝑖𝑛𝑒𝑎𝑟) 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚, 𝑚𝑣𝑟 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚, and 𝐿 = 𝑚𝑣𝑟 =
𝑛ℎ
2𝜋
,
where 𝑛 = 1, 2, 3, …, 𝑟𝑛 = 𝑛2
𝑎0, and 𝐸 𝑛 = −
𝑅 𝐻
𝑛2 .
The potential energy is 𝑈 = −
𝑒2
(4𝜋𝜀0) 𝑟
= −
2𝑅 𝐻
𝑛2 .
The kinetic energy is 𝑇 =
1
2
𝑒2
(4𝜋𝜀0) 𝑟
=
𝑅 𝐻
𝑛2 .
The total energy is 𝐸𝑡𝑜𝑡 = 𝐸 𝑛 = −
𝑅 𝐻
𝑛2

35. The energy required to move an electron from 𝑛 = 1 to 𝑛 = ∞ is called its ionization
energy (IE).
𝐼𝐸 = 𝐸𝑓 − 𝐸𝑖 = −
𝑅 𝐻
(∞)2
—
𝑅 𝐻
(1)2
= 0 + 𝑅 𝐻 = 𝑅 𝐻 = 2.179 × 10−18
𝐽
𝐼𝐸/𝑚𝑜𝑙 = 𝑅 𝐻 𝑁𝐴 = 2.179 × 10−18
𝐽(6.022 × 10−23
𝑚𝑜𝑙−1)
= 13.11758 × 102
𝑘𝐽/𝑚𝑜𝑙
∆𝐸 = ℎ𝑣0 → 𝑣0 =
∆𝐸
ℎ
=
2.179 × 10−18
𝐽
6.626 × 10−34 𝐽𝑠
= 3.289 × 1015
𝐻𝑧
The term 𝑣0 is called the threshold frequency.
Hydrogen-like Atoms
Angular momentum is an integral multiple of
ℎ
2𝜋
, such that:
𝐿 = 𝑚𝑣𝑟 =
𝑛ℎ
2𝜋
, where 𝑛 = 1, 2, 3, …
From this, we can get that:
i. 𝑟𝑛 =
𝑛2
𝑎0
𝑧
𝑤ℎ𝑒𝑟𝑒 𝑎0 = 0.529 Å = 𝐵𝑜ℎ𝑟 𝑟𝑎𝑑𝑖𝑢𝑠
ii. 𝐸 𝑛 = − (
𝑧2
𝑛2 ) 𝑅 𝐻, 𝑤ℎ𝑒𝑟𝑒 𝑅 𝐻 = 2.179 × 10−18
𝐽 = 𝑅𝑦𝑑𝑏𝑢𝑟𝑔 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
iii. 𝐿𝑒𝑡 𝐸 𝑛1
= −𝑅 𝐻 (
𝑧2
𝑛1
2 ) 𝑎𝑛𝑑 𝐸 𝑛2
= −𝑅 𝐻 (
𝑧2
𝑛2
2 )
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:∆𝐸 = 𝐸 𝑛2
− 𝐸 𝑛1
= [−𝑅 𝐻 (
𝑧2
𝑛2
2 )]— 𝑅 𝐻 (
𝑧2
𝑛1
2 )
∆𝑬 = 𝑹 𝑯 𝒛 𝟐
(
𝟏
𝒏 𝟏
𝟐 −
𝟏
𝒏 𝟐
𝟐 )
iv. ∆𝐸 = ℎ𝑣, since this change in energy is due to absorption or emission of
photons.
For a hydrogen atom, 𝑧 = 1. Therefore: ℎ𝑣 = 𝑅 𝐻 (
1
𝑛1
2 −
1
𝑛2
2 ).
𝑐 = 𝜆𝑣 → 𝑣 =
𝑐
𝜆
= 𝑐
1
𝜆
, 𝑤ℎ𝑒𝑟𝑒
1
𝜆
= 𝑣̅ = 𝑤𝑎𝑣𝑒𝑛𝑢𝑚𝑏𝑒𝑟 ( 𝑐𝑚−1)
Therefore: 𝑣 = 𝑐𝑣̅.
And ℎ𝑐𝑣̅ = 𝑅 𝐻 (
1
𝑛1
2 −
1
𝑛2
2 ) → 𝒗̅ =
𝑹 𝑯
𝒉𝒄
𝑹 𝑯 (
𝟏
𝒏 𝟏
𝟐 −
𝟏
𝒏 𝟐
𝟐)

36. If we let 𝑛1 = 1 and 𝑛2 ≫ 2, then: 𝑣̅ =
𝑅 𝐻
ℎ𝑐
(1 −
1
𝑛2
2 ). This is the Lyman series (UV).
If we let 𝑛1 = 2 and 𝑛2 ≫ 3, then: 𝑣̅ =
𝑅 𝐻
ℎ𝑐
(
1
4
−
1
𝑛2
2 ). This is the Balmer series (visible).
If we let 𝑛1 = 3 and 𝑛2 ≫ 4, then: 𝑣̅ =
𝑅 𝐻
ℎ𝑐
(
1
9
−
1
𝑛2
2 ). This is the Paschen series (IR).
De Broglie Wavelength
The De Broglie wavelength is given by: 𝝀 =
𝒉
𝒑
=
𝒉
𝒎𝒗
, where 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦.
Since 𝑚𝑣𝑟 =
𝑛ℎ
2𝜋
→ 𝑚𝑣 =
𝑛ℎ
2𝜋𝑟
, then 𝜆 =
ℎ
𝑛ℎ
2𝜋𝑟⁄
=
2𝜋𝑟
𝑛
; that is, 𝝀 =
𝟐𝝅𝒓
𝒏
.
Example
If 𝑛 = 3, then 𝜆 =
2𝜋𝑟
3
.
In the diagram above, 𝑟𝑛 =
𝑛2
𝑎0
𝑧
=
(32) 𝑎0
1
= 9𝑎0.
This comes from the standing wave shown below:
Joining the ends of the string, a and b, creates the pattern and circle above.

37. {
𝜆 =
2𝜋𝑟
𝑛
𝑟𝑛 =
𝑛2
𝑎0
𝑧
→ 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒: 𝜆 𝑛 =
2𝜋
𝑛
(
𝑛2
𝑎0
𝑧
) → 𝝀 𝒏 =
𝒏
𝒛
( 𝟐𝝅𝒂 𝟎)
When 𝑛 = 1, 𝝀 𝟎 = 𝟐𝝅𝒂 𝟎 = 𝑩𝒐𝒉𝒓 𝒘𝒂𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟑. 𝟑𝟐𝟒 × 𝟏𝟎−𝟏𝟎
𝒎.
Therefore, 𝜆 𝑛 =
𝑛
𝑧
( 𝜆0).
When 𝑛 = 1 and 𝑧 = 1, 𝜆1 = 𝜆0.
When 𝑛 = 2 and 𝑧 = 1, 𝜆2 = 2𝜆0.
Now, 𝜆𝑣 = v → 𝜆 𝑛 𝑣 𝑛 = v 𝑛, where 𝑣 = 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 and v = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦.
If 𝑚𝑣𝑟 =
𝑛ℎ
2𝜋
, then v =
𝑛ℎ
2𝜋𝑚𝑟
.
Since 𝑟 =
𝑛2
𝑎0
𝑧
, then:
v =
𝑛ℎ
2𝜋𝑚 (
𝑛2 𝑎0
𝑧
)
=
𝑧𝑛ℎ
(2𝜋𝑚) 𝑛2 𝑎0
=
𝑧ℎ
𝑛(2𝜋𝑚𝑎0)
=
𝑧
𝑛
(
ℎ
2𝜋𝑚𝑎0
)
And so:
𝐯 𝟎 =
𝒉
𝟐𝝅𝒎𝒂 𝟎
= 𝑩𝒐𝒉𝒓 𝒔𝒑𝒆𝒆𝒅 = 𝟎. 𝟎𝟎𝟎𝟐𝟏𝟖𝟖 × 𝟏𝟎 𝟖
𝒎𝒔−𝟏
…when 𝑛 = 1 and 𝑧 = 1.
Generally: 𝐯 𝒏 =
𝒛
𝒏
𝐯 𝟎
When 𝑛 = 1 and 𝑧 = 1, then v1 = v0.
When 𝑛 = 2 and 𝑧 = 1, then v2 =
v0
2
.
When 𝑛 = ∞ and 𝑧 = 1, then v∞ =
v0
∞
= 0.
Therefore, we can say that:
𝝀 𝟎 𝒗 𝟎 = 𝐯 𝟎 → 𝒗 𝟎 =
𝐯 𝟎
𝝀 𝟎
= 𝑩𝒐𝒉𝒓 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 = 𝟔. 𝟓𝟖𝟑 × 𝟏𝟎 𝟓
𝑯𝒛
Then we can find the speed for other atoms:
v 𝑛 =
𝑧
𝑛
v0
v1 = v0 for H ( 𝑠𝑖𝑛𝑐𝑒 𝑛 = 1 𝑎𝑛𝑑 𝑧 = 1)
v1 = 2v0 for He2+ ( 𝑠𝑖𝑛𝑐𝑒 𝑛 = 1 𝑎𝑛𝑑 𝑧 = 2)
v1 = 102v0 for Lr102+ ( 𝑠𝑖𝑛𝑐𝑒 𝑛 = 1 𝑎𝑛𝑑 𝑧 = 103)
For what value of z does v1 match the speed of light?
𝑧 =
v1
v0
=
𝑐
v0
=
2.998 × 108
𝑚𝑠−1
0.0002188 × 108 𝑚𝑠−1
= 13697
No element with atomic number (z) 13697.

38. The Photoelectric Effect
Consider the horizontal, parallel, clean metal plates (see below) connected by a
circuit containing an ammeter. Light of fixed intensity and varying frequency is
shone upon the lower plate at an angle. Above a given threshold frequency,
electrons jump from the lower to the upper plate, traveling through the circuit to
generate a measurable current.
In other words, a metal with a clean surface is placed in vacuum and illuminated
with light of a known frequency. If the frequency is greater than a particular
minimum, or threshold frequency, electrons are instantly liberated fron the metal
surface at a rate proportional to the light intensity (increasing the intensity does not
liberate more electrons).
The energy is given by 𝐸 =
1
2
𝑚𝑣2
= 𝕍𝑒.
The total energy is conserved; therefore, ℎ𝑣 = 𝑊 + 𝕍𝑒
…where 𝑚 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑚𝑎𝑠𝑠, 𝑣 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑒 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑐ℎ𝑎𝑟𝑔𝑒, 𝕍 =
𝑠𝑜𝑝𝑝𝑖𝑛𝑔 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 (1 𝑣𝑜𝑙𝑡 = 1 𝑗𝑜𝑢𝑙𝑒 𝑝𝑒𝑟 𝑐𝑜𝑢𝑙𝑜𝑚𝑏; 1𝑉 = 1𝐽/𝐶), and 𝑊 =
𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑝𝑜𝑡𝑒𝑛𝑖𝑡𝑙𝑎 𝑜𝑓 𝑚𝑒𝑡𝑎𝑙 𝑜𝑟 𝑤𝑜𝑟𝑘 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
Note that 𝕍𝑒 = 1.60206 × 10−19
𝐽 per electron.
Conservation of energy requires that:
ℎ𝑣 = 𝑊 +
1
2
𝑚𝑣2
= 𝑊 + 𝕍𝑒
ℎ𝑣0 = 𝑊
ℎ𝑣 = ℎ𝑣0 + +𝕍𝑒 → 𝑣 = 𝑣0 +
1
ℎ
𝕍𝑒
…where 𝑣0 = 𝑡ℎ𝑒 𝑡ℎ𝑟𝑒𝑠ℎ𝑜𝑙𝑑 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦.
𝟏 𝒆𝕍 = 𝟏. 𝟔𝟎𝟐 × 𝟏𝟎−𝟏𝟗
𝑱 = 𝒆𝒏𝒆𝒓𝒈𝒚 𝒑𝒆𝒓 𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏 and 𝑬 𝑲 = 𝕍𝒆

39. A plot of the frequency against the stopping voltage is shown below:
The equation of the graph 𝑣 = 𝑣0 +
1
ℎ
𝕍𝑒 is of the form 𝑦 = 𝑎 + 𝑏𝑥.

40. Refinement of the Bohr Model
Consider the system below consisting of two masses – the nucleus (with charge ze
and mass mn) and the electron (with charge –ze and mass me):
The moment of inertia gives rise to the relationship below:
{
𝑚 𝑒 𝑟𝑒 = 𝑚 𝑛 𝑟𝑛
𝑟 = 𝑟𝑒 + 𝑟𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
The top equation rearranges to 𝑟𝑒 =
𝑚 𝑛 𝑟 𝑛
𝑚 𝑒
and 𝑟𝑛 =
𝑚 𝑒 𝑟 𝑒
𝑚 𝑛
.
The bottom equation rearranges to 𝑟𝑒 = 𝑟 − 𝑟𝑛 and 𝑟𝑛 = 𝑟 − 𝑟𝑒.
Then:
𝑚 𝑒( 𝑟 − 𝑟𝑛) = 𝑚 𝑛( 𝑟 − 𝑟𝑒)
𝑚 𝑒 𝑟 − 𝑚 𝑒 𝑟𝑛 = 𝑚 𝑛 𝑟 − 𝑚 𝑛 𝑟𝑒
𝑚 𝑛 𝑟𝑒 = 𝑚 𝑛 𝑟 − 𝑚 𝑒 𝑟 + 𝑚 𝑒 𝑟𝑛
𝑚 𝑛 𝑟𝑒 = 𝑚 𝑛 𝑟 − 𝑚 𝑒 𝑟 + 𝑚 𝑒( 𝑟 − 𝑟𝑒)
𝑚 𝑛 𝑟𝑒 = 𝑚 𝑛 𝑟 − 𝑚 𝑒 𝑟 + 𝑚 𝑒 𝑟 − 𝑚 𝑒 𝑟𝑒
𝑚 𝑛 𝑟𝑒 + 𝑚 𝑒 𝑟𝑒 = 𝑚 𝑛 𝑟
𝑟𝑒 ( 𝑚 𝑛 + 𝑚 𝑒) = 𝑚 𝑛 𝑟
𝑟𝑒 =
𝑚 𝑛 𝑟
𝑚 𝑛 + 𝑚 𝑒
𝒓 𝒆 = (
𝒎 𝒏
𝒎 𝒏 + 𝒎 𝒆
) 𝒓
𝑚 𝑒( 𝑟 − 𝑟𝑛) = 𝑚 𝑛( 𝑟 − 𝑟𝑒)
𝑚 𝑒 𝑟 − 𝑚 𝑒 𝑟𝑛 = 𝑚 𝑛 𝑟 − 𝑚 𝑛 𝑟𝑒
𝑚 𝑒 𝑟𝑛 = 𝑚 𝑒 𝑟−𝑚 𝑛 𝑟 + 𝑚 𝑛 𝑟𝑒
𝑚 𝑒 𝑟𝑛 = 𝑚 𝑒 𝑟−𝑚 𝑛 𝑟 + 𝑚 𝑛( 𝑟 − 𝑟𝑛 )
𝑚 𝑒 𝑟𝑛 = 𝑚 𝑒 𝑟−𝑚 𝑛 𝑟 + 𝑚 𝑛 𝑟 − 𝑚 𝑛 𝑟𝑛
𝑚 𝑒 𝑟𝑛 + 𝑚 𝑛 𝑟𝑛 = 𝑚 𝑒 𝑟
𝑟𝑛 ( 𝑚 𝑒 + 𝑚 𝑛) = 𝑚 𝑒 𝑟
𝑟𝑛 =
𝑚 𝑒 𝑟
𝑚 𝑒 + 𝑚 𝑛
𝒓 𝒏 = (
𝒎 𝒆
𝒎 𝒆 + 𝒎 𝒏
) 𝒓
The moment of inertia about the center of mass is given by:
𝑰 = ∑ 𝒎𝒊 𝒓 𝒊
𝟐
𝒊
Then:
𝐼 = 𝑚 𝑒 𝑟𝑒
2
+ 𝑚 𝑛 𝑟𝑛
2
= 𝑚 𝑒 {(
𝑚 𝑛
𝑚 𝑒 + 𝑚 𝑛
) 𝑟}
2
+ {(
𝑚 𝑒
𝑚 𝑒 + 𝑚 𝑛
) 𝑟}
2
Expanding gives:
𝐼 =
𝑚 𝑒 𝑚 𝑛
2
𝑟2
( 𝑚 𝑒 + 𝑚 𝑛)2
+
𝑚 𝑛 𝑚 𝑒
2
𝑟2
( 𝑚 𝑒 + 𝑚 𝑛)2
=
𝑚 𝑒 𝑚 𝑛 𝑟2( 𝑚 𝑛 + 𝑚 𝑒)
( 𝑚 𝑒 + 𝑚 𝑛)2
=
𝑚 𝑒 𝑚 𝑛 𝑟2
𝑚 𝑛 + 𝑚 𝑒
=
𝑚 𝑒 𝑚 𝑛
𝑚 𝑛 + 𝑚 𝑒
𝑟2
If we we define the reduced mass as 𝜇 =
𝑚 𝑒 𝑚 𝑛
𝑚 𝑛+𝑚 𝑒
, then:
𝑰 = 𝝁𝒓 𝟐
And since 𝑚 𝑛 =
1.0079 𝑔/𝑚𝑜𝑙
6.022 × 1023 𝑚𝑜𝑙 −1 = 1.674 × 10−24
𝑔, then 𝑚 𝑛 ≫ 𝑚 𝑒, and we can
approximate that:
𝜇 =
𝑚 𝑒 𝑚 𝑛
𝑚 𝑛 + 𝑚 𝑒
→
𝑚 𝑒 𝑚 𝑛
𝑚 𝑛
= 𝑚 𝑒 → 𝝁 = 𝒎 𝒆

41. Foundations of Quantum Mechanics
Two vectors 𝐴 and 𝐵⃗ can be added or subtracted as follows:
A vector whose magnitudeis unity is called a unit vector.
SKIPPED SOME STUFF

42. Operators
An operator is a symbol that indicates that a particular operation is being performed
on what follows the operator. For example, the square root operatoris √ and can
operate as follows: √4 = 2. The differential operator
𝑑
𝑑𝑥
operates as in
𝑑
𝑑𝑥
( 𝑥2) = 2𝑥.
We denote operators in quantum mechanics by using the symbol hat, ̂ , as in the
operator 𝑃̂.
Examples
If 𝑃̂ =
𝑑
𝑑𝑥
and 𝑓( 𝑥) = 𝑥4
, then 𝑃̂ 𝑓( 𝑥) =
𝑑
𝑑𝑥
𝑓( 𝑥) =
𝑑
𝑑𝑥
( 𝑥4) = 4𝑥3
.
Then 𝑃̂ 𝑃̂ 𝑓( 𝑥) = 𝑃̂[𝑃̂ 𝑓( 𝑥)] = 𝑃̂[4𝑥3] = 12𝑥2
And 𝑃̂ 𝑃̂ 𝑃̂ 𝑓( 𝑥) = 𝑃̂[12𝑥2] = 24𝑥
Example
If 𝑅̂ = 𝑥2
, then 𝑃̂ 𝑄̂[ 𝑓( 𝑥)] = 𝑃̂ 𝑄̂[ 𝑓( 𝑥)]
Reactions Having No Order
Not all reactions behave in the manner aforementioned, and the term order should
not be used for those that do not. Instead, such reactions are usually enzyme-
catalyzed and frequently follow a law of the form:
𝜐 =
𝑉[ 𝑆]
𝐾 𝑚 + [ 𝑆]
In the above equation, V and Km are constants, while [S] is a variable known as the
substrate concentration. This equation does not correspond to a simple order, but
under two limiting conditions, an order may be assigned:
i. If the substrate concentration is sufficiently low, so that [ 𝑆] ≪ 𝐾 𝑚, then the law
becomes 𝜐 =
𝑉[ 𝑆]
𝐾 𝑚
, and the reaction is then 1st order with respect to S; or
ii. If [S] is sufficiently large, so that [ 𝑆] ≫ 𝐾 𝑚, then the law becomes 𝜐 = 𝑉, and
the reaction is said to be 0th order (meaning the rate is independent of [S]).
Consider a reaction 𝐴 𝑘⃗⃗⃗⃗ 𝑃.
Then −
𝑑[ 𝐴]
𝑑𝑡
= 𝑘[ 𝐴] 𝛼
= 𝑘[ 𝐴]0
= 𝑘.
This is separable such that 𝒅[ 𝑨] = −𝒌 𝒅𝒕, the rate law in differential form.
Integrating gives [ 𝐴] = −𝑘𝑡 + 𝐶.
At time 𝑡 = 0: [ 𝐴]0 = 𝐶.

43. Then [ 𝐴] = −𝑘𝑡 + [ 𝐴]0 → [ 𝑨]− [ 𝑨] 𝟎 = −𝒌𝒕, the rate law in integral form.
At 𝑡 = 𝑡1
2
:
1
2
[ 𝐴]0 − [ 𝐴]0 = −𝑘𝑡1
2
→ −
1
2
[ 𝐴]0 = −𝑘𝑡1
2
→ 𝒕 𝟏
𝟐
=
[ 𝑨] 𝟎
𝟐𝒌
, where k has units
𝑚𝑜𝑙 𝐿−1
𝑠−1
.
Rate Constants and Rate Coefficients
The constant k used in such rate equations is known as the rate constant or rate
coefficient, depending on whether the reaction is believed to be elementary or to
occur in more than one stage (respectively).
The units of k vary with the order of the reaction, as shown in the table below.
Order
Rate Equation
Units of k Half-life, t½
Differential Form Integrated Form
0
𝑑𝑥
𝑑𝑡
= 𝑘 𝑘 =
𝑥
𝑡
𝑚𝑜𝑙 𝑑𝑚−3 𝑠−1
𝑎0
2𝑘
1
𝑑𝑥
𝑑𝑡
= 𝑘( 𝑎0 − 𝑥) 𝑘 =
1
𝑡
ln (
𝑎0
𝑎0 − 𝑥
) 𝑠−1 ln 2
𝑘
2
𝑑𝑥
𝑑𝑡
= 𝑘( 𝑎0 − 𝑥)2 𝑘 =
1
𝑡
ln(
𝑥
𝑎0( 𝑎0 − 𝑥)
) 𝑑𝑚3 𝑚𝑜𝑙−1 𝑠−1
1
𝑘𝑎0
2
𝑑𝑥
𝑑𝑡
= 𝑘( 𝑎0 − 𝑥)( 𝑏0 − 𝑥)
for reactants with
different concentrations
𝑘 =
1
𝑡( 𝑎0 − 𝑏0)
ln (
𝑏0( 𝑎0 − 𝑥)
𝑎0( 𝑏0 − 𝑥)
) 𝑑𝑚3 𝑚𝑜𝑙−1 𝑠−1
–
n
𝑑𝑥
𝑑𝑡
= 𝑘( 𝑎0 − 𝑥) 𝑛 𝑘 =
1
𝑡( 𝑛 − 1)
[
1
( 𝑎0 − 𝑥) 𝑛−1 −
1
𝑎0
𝑛−1
] 𝑚𝑜𝑙1−𝑛 𝑑𝑚3𝑛−3 𝑠−1 2 𝑛−1 − 1
𝑘( 𝑛 − 1) 𝑎0
𝑛−1