General Kinetics Of Chain Reactions
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| General Kinetics Of Chain Reactions |
Consider a reaction A → P
This reaction is supposed to complete in following steps.
Step 1: Chan initiation
r1 = k1 [A]
Step 2: Chain propagation 
r2 = k2 [R][A]
Step 3: Chain termination
r3 = k3 [R]
In above steps, the rate determining step (slowest step) is the second one so,
Rate of reaction = dx/dt = k2 [R][A] eq. 20.1
Here [R] is highly unstable, highly reactive and its concentration determination is very difficult. To determine the concentration of [R], equation 20.1 will be modified by replacing [R] with a stable specie. According to steady state hypothesis:
"At equilibrium, the rate of formation of an intermediate specie is equal to its rate of concentration. So, net rate of intermediate specie at equilibrium is zero.
Net rate 
rate of formation - rate of decompositionrate of formation = rate of decomposition eq.20.2
Rate of formation of [R]
R is produced in first and second step so, rate of formation of R will be equal to sum of both.
Rate of formation of [R] = r1 + r2
Rate of formation of [R] = k1 [A] + k2 [n-1][R][A]
Rate of decomposition of [R] = k3 [R]
Now by eq.20.2 we got,
k1 [A] + k2 (n-1) [R][A] = k3 [R]
k1 [A] = k3 [R] - k2 (n-1) [R] [A]
k1 [A] = k2 (1-n)[R] [A] + k3 [R]
k1 [A] = [R] [k2 (1-n) [A] + k3]
[R] = 
eq.20.3
eq.20.3Now putting this value of [R] in eq. 20.1 we got,
eq.20.4Now we have successfully replaced the unstable [R] and now the rate of reaction can be found from eq. 20.4.
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