Compound A simultaneously undergoes a pseudo first-order reaction and a second o

Question

Compound A simultaneously undergoes a pseudo first-order reaction and a second order reaction via: K1 A rightarrow B and K2 A+A rightarrow C Obtain an integrated rate expression for this mechanism. You will need to use the method of partial fractions in order to carry out your integration. Show that the result fits the form: ln[[A]_0(K_1 +2k_2[A])/[A](k_1+2k_2[A]_0)] =K_1t Given the partial fractions approach that: 1/K_1A+2K_2A^2=1/A(K_1 +2K_2 A) =a/A+b/K_1+2K_2A Rearrange the expression in a.) and show that it obeys the form: 1/[A]=a.exp[K_1t]-b It has been found that the data below is linear to an equation of the form: ln(1/A+b) = K_1t+ln(c) when b=1 b=2k_2/k_1 and c=K_1 +2K_1 A_0/K_1A_0

Explanation / Answer

-dA/dt= K1[A]+ K2[A]2 = [A] { K1+K2[A]}

-dA/A*(K1+K2A)= dt

Let 1/ A*(K1+K2A)= a/A + b/(K1+K2A) ( a and b are constants to be determined)

                                   = ( a(K1+K2A) +bA)/ (K1+K2A)

                                    = (aK1+aK2A +bA)/ (K1+K2A)

                                     =aK1+A(aK2+b)/ (K1+K2A)

Comparing the coefficients K1a= 1 , a= 1/K1

aK2+b= 0 and b= -K2/K1

So –dA {1/ A*(K1+K2A)}= [1/K1A - K2/K1*(K1+K2A)] –dA

-dA/A*(K1+K2A)= dt becomes

1/K1A – (K2/K1) *(1/K1+K2A)= - dt

When integrated this gives

(1/K1) lnA- (K2/K1)* ln (K1+K2A)/K2= -t+C

lnA- ln (K1+K2A)= -K1t +C’      (1)

where C and C’ are integration constants

at t=0 and A=Ao. This when substituted in A

lnA0- ln(K1+K2A0)= C’

Eq.1 becomes

lnA- ln (K1+K2A)= -K1t+ ln A0- ln (K1+K2A0)

ln[(A0*(K1+K2A)/(A*(K1+K2AO)= k1t

{A0*(K1+K2A)}(A*(K1+K2A0)= exp(k1t)

1/A= exp(K1t)* (K1+K2A0)/ A0*(K1+K2A)

       =exp(K1t)*{ K1+K2A- K2A+K2A0)/ A0*(K1+K2A)

=exp (K1t)* (1/A0- K2(A-A0)/A0*(K1+K2A)

=exp(K1t)a- b

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