1st order opposing reaction having initial concentration of products
 |
| 1st Order Opposing Reaction having initial Concentration of Products |
1st order opposed by 1st order case II
These are the reactions in which both the rate of forward and backward reaction is of first order. Case II involves reactions having initial products.
R ⇄ P
R P
At time t=0 a b
At time t a-x b+x
At equilibrium a-xe b+xe
Rate of forward reaction Rf ∝ [R]
Rf = kf (a-x)
At equilibrium Rf = kf (a-xe)
Rate of backward reaction Rb ∝ [P]
Rb = kb (b+x)
At equilibrium Rb = kb (b+xe)
Net rate will be dx/dt = Rf - Rb
dx/dt = kf (a-x) - kb(b+x) eq.15.1
At equilibrium net rate is zero so Rf = Rb and eq.15.1 becomes
kf (a-xe) = kb (b + xe) eq.15.2
from eq.15.2 value of kb is
Putting the above value of kb in eq.15.1 we got
eq.15.3here, kf, a,b,and xe are constants while xe-x is variable so kf, a, b and, xe will be represented with a new constant K and equation 15.3 will be
Rearranging 
integrating 
-ln (xe-x) = kt +z eq.15.4
to find the value of z from eq.15.4, consider x=0 and t=0 then z will be equal to
z = -lnxe
puting the above value of z in eq.15.4, we got
-ln( xe-x) = kt -lnxe eq.15.5
Eq.15.5 is a straight line equation for 1st order opposing reactions having initial concentration. Rearranging the equation we got,
lnxe -ln (xe-x) = kt
eq.15.6The graph for above equation will be
or 
or 
So, we find value of kf and in our previous topic we find value of kb by the following formula:
kb = slope - kf
Putting the kf value in above equation we got
simplifying:
We know that Kcomposite = kf+kb so putting values of both kf and kb we got
So, Kc=Slope
