2nd Order Type Parallel Reactions Case II
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| 2nd Order Type Parallel Reactions Case II |
Second order type parallel reactions case II refers to those reaction in which concentration of both reactants are not same. Consider
A + B ⇄ P1 + P2
At time t=0 a + b 0 0
At time t a-x b-x x x
Rate of formation of P1 is R1 ∝ [A] [B] R1 = k1 (a-x) (b-x) eq.18.1
Rate of formation of P2 is R2 ∝ [A] [B] R2 = k2 (a-x) (a-b) eq.18.2
Net rate of reaction will be: dx/dt =R1 + R2
As k1 + k2 = Kt so, 
Integrating the above equation we got, 
eq.18.3
Now suppose that
eq.18.4By the method of partial fraction multiply (a-x) (b-x) on both sides. We will get
1 = A (b-x) + B (a-x) eq.18.5
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To find value of A, put x = a and to find the value of B, put x = b in eq.18.5 one by one.
For A 1 = A (b-a) So,
For B 1 = B (a-b) So, 
Now put the values of both A and B in eq.18.5.
or to make it more easy
Integrating both sides and multiplying dx on both sides
eq.18.6Now observe that left hand side of equation 18.3 and 18.6 is same so their right hand side is also same.
eq.18.7To find the value of z, consider that x = 0 and t = 0 and put in eq.18.7. Then z will be
putting the value of z in eq.18.7 we got.
eq.18.8Eq.181.8 is the rate equation for 2nd order type parallel reactions having different initial concentration of reactants.
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