First & Second Order Reactions | Physical Chemistry PDF Download

(2)  First-order reaction

Consider the following elementary reaction
A  → P
If the reaction is first order with respect to [A], the rate law expression is

k is rate constant


If t = 0, the init ial concentration is [A]0 and the concentration at t = t, is [A], then integrating yields

[A] = [A]₀ e^–kt…(i)
or …(ii)

Using this idea, the concentration of product with time for this first-order reaction is :

[P] + [A] = [A]₀
[P] = [A]₀ – [A]
[P] = [A]₀ – [A]₀ e^–kt 
[P] = [A]₀ (1 – e^–kt) ….(iii)

Graph representation of first order reaction

[A] = [A]₀ e^–kt

Plot of concentration vs time.

ln [A] = –kt + ln [A]₀

Plot of log [A] vs time.

t_1/2 i.e. half life t ime of first order reaction

when t = t_1/2; then [A] =

ln 2 = kt_1/2

Problem. The half life for the first order decomposition of N₂O₅ is 2.05 � 10⁵ s. How long will it take for a sample of N₂O₅ to decay to 60% of its initial value?

Sol. We know that,

The time at which the sample has decayed to 60% of its initial value then

T = 1.51 x 10⁴ s
Problem. Find the t_3/4 i.e. 3/4 life time of first order reaction.

Sol.

Integrated rate law expression is

when t = t_3/4 than              

then         

ln 4 = kt_3/4

(3)  Second-order reaction: (Type I) 

Consider the following elementary reaction,

If the reaction is second order with respect to [A], the rate law expression is rate =

k is rate constant

 

⇒

If t = 0, the init ial concentration is [A]₀ and the concentration at t = t, is [A], then integration yields

 
…(i)

The concentration of product with time for second order reaction

 

or 
then        
…(ii)

t_1/2 i.e. Half-life t ime of second order reaction (type I)

when t = t_1/2 then

⇒  

Second-order reaction (Type II) 

Second order reactions of t ype II invo lves two different reactants A and B, as fo llows

Assuming that the reaction is first order in both A and B, the reaction rate is

If t = 0 then the init ial concentration are [A]₀ & [B]₀ and the concentration at t = t, are [A] & [B].

The loss of reactant i.e. the formation of product is equal to

[A]₀ – [A] = [B]₀ – [B] = [P]
[B]₀ – [A]₀ + [A] = [B]
then

⇒

the integration yield

let Δ = [B]₀ – [A]
The solution to the integral involving [A] is given by

Using this so lut ion to the integral, the integrated rate law expressio n beco mes

⇒ 

 Graph representation of second order reaction of type I

Y = mx + C

Plot of concentration vs time 

(4)  n^th order reaction where n ≥ 2 :

An nth order reaction may be represented as

the rate law is,

where k is rate constant for nth order reaction

⇒

If at t = 0, the init ial concentration is [A]₀ and the concentration at t = t, is [A], then integration yields

Let

…(1)
t_1/2 i.e. Half life time of nth order reaction

Where t = t_1/2 then

 

 …(2)

i.e.  …(3)

Thus we can say that t_1/2 of the reaction is inversely proportional to the init ial concentration of reactant, except first order reaction.

So, for a first order reaction (n = 1), t_1/2 is independent on [A]₀ for a second order reaction (n = 2), t_1/2 is dependent on [A]₀

for a nth order reaction 

Note : For the elementary reaction, the order of reaction is equal to the molecularity of the reaction.

Problem. Find the rate law for the following reaction.

Sol.

Rate law is 

Problem. Find the rate law for the following reaction.

Sol.  

 (1)  

(2) 

  
(3) = k₁[A] + k₂[A] = (k₁ + k₂)[A]

Problem. Find the rate law for the following reaction

Sol. 

(1) 
(2)  = k₁[A] – k₂[B] – k₃[B]
               = k₁[A] – (k₂ + k₃) [B]
(3) 
 (4)