Reversible Reactions & Equilibrium | Physical Chemistry PDF Download

Introduction

A reversible chemical reaction is one in which the products, as soon as they are formed, react to produce the original reactants. At equilibrium, the two opposing reactions go on at equal rates, or velocities, and hence there is no net change in the amounts of substances involved.

Reversible reactions and Equilibrium

Consider the following reaction in which the forward reaction is first order in A, and the back reaction is first order in B:
 
The forward and back rate constant are k_A & k_B. Then rate law are


Only reactant is present at t = 0 and the concentration of reactant and product for t > 0 must be equal to the initial concentration of reactant. [A]₀ = [A] + [B]
then
 
= –k_A[A] + k_B[B]
= –k_A[A] + k_B([A]₀ – [A])
= –[A] (k_A + k_B) + k_B[A]₀

∵
Using this relationship we get
 


[A] (k_A + k_B) – k_B[A]₀ = k_A [A]₀ e^{-( k_A+ k_B)t}
 


Then
As t → ∝, the concentration reach their equilibrium values then

&  
It follows that the equilibrium constant of the reaction is
 
i.e.  
k_C is equilibrium constant in terms of concentration.
At equilibrium, the forward and reverse rates must be same so,
K_A[A]_eq = k_B[B]_eq
 

Examples

Q.1. Using the following equation
A → C
mechanism,
  
(a)  Find rate of reaction?
(b)  Find rate of reaction when (i) is fast.
Sol. (a) From the rate law



[B] is intermediate then we apply SSA then we get
0 = k₁[A] – k₂[B] – k₃[B]

then

(b) When (i) is fast then
k₁[A] = k₂[B]
i.e.
&

Q.2. Using the following equation 
2NO₂ + F₂ → 2NO₂F
Mechanism
 

Find the rate of reaction.
Sol. From the rate law


= 0 (F is intermediate)
∴

= 2k₁[F₂][NO₂]