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Reversible Reactions And Equilibrium - Chemical Kinetics Chemistry Notes | EduRev

Chemistry : Reversible Reactions And Equilibrium - Chemical Kinetics Chemistry Notes | EduRev

The document Reversible Reactions And Equilibrium - Chemical Kinetics Chemistry Notes | EduRev is a part of the Chemistry Course Physical Chemistry.
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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 kA & kB. 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

= –kA[A] + kB[B]
= –kA[A] + kB([A]– [A])
= –[A] (kA + kB) + kB[A]0

Using this relationship we get

[A] (kA + kB) – kB[A]0 = k[A]0 e-( kA+ kB)t

Then

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

It follows that the equilibrium constant of the reaction is

i.e

kC is equilibrium constant in terms of concentration.
At equilibrium, the forward and reverse rates must be same so,

KA[A]eq = kB[B]eq

Problem.  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.

&

Problem.  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]

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Physical Chemistry

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