Consecutive Second and First-order Kinetics
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Scheme:
`A+Bstackrel(k_1) rarrCstackrel(k_2) rarrD`
Differential Equation:
`A'(t)=-k_1AB`
`B'(t)=-k_1AB`
`C'(t)=k_1AB-k_2C`
`D'(t)=k_2C`
Case I: A0 = B0
`A'(t)=B'(t)`
`A_0=A+C+D` and `A=B`
Solutions
`A=A_0/(1+k_1A_0t)`
`B=B_0/(1+k_1B_0t)`
`C(t)=-k_2/k_1 exp(k_2/(A_0k_1)) ["E"_1((1-A_0k_1t)k_2/(A_0k_1))-(A_0k_1)exp(-(1-A_0k_1t)k_2/(A_0k_1))/((1-A_0k_1t)k_2)]exp(-k_2t) +...`
` ... + [k_2/k_1 exp(k_2/(A_0k_1)) ["E"_1(k_2/(A_0k_1))-(A_0k_1)exp(-k_2/(A_0k_1))/(k_2)]]exp(-k_2t)`
Derivation
The differential equation `A'(t)` simplifies using the balance equation `A=B`,
`A'(t)=-k_1AB=-k_1A^2`
This has the separable equation,
`(dA)/A^2=-k_1t`
The solution is obtained directly,
`-[1/A]_0^t=-k_1t rarr -1/A+1/A_0=-k_1t`
This can be solved for A,
`A=A_0/(1+k_1A_0t)`
From the intial conditions, this is also equal to B(t).
We can now solve the differential equation C'(t) using the above solution,
`C'(t)=k_1AB-k_2C=k_1(A_0/(1+k_1A_0t))^2-k_2C`
This can be solved using the integrating factor method. We rearrange it to be of the general form,
`C'+F(t)C=G(t)`
Accordingly, we can define the functions,
`F(t) = k_2`
`G(t)=k_1(A_0/(1+A_0k_1t))^2`
We need to calculate the integrating factor,
`lnI=int F(t)dt` and so `I=exp(k_2t)`
and also the rhs integral,
`int IG(t)dt = k_1A_0^2 int exp(k_2t)/(1+A_0k_1t)^2 dt`
Which we can solve as the exponential integral with a square denominator via u-substitution. Ie.,
`u=1+A_0k_1t` and so, `(du)/(dt)=A_0k_1`, then `(dt)=(du)/(A_0k_1)` and `t=(u-1)/(A_0k_1)`
So the integral becomes,
`int IG(t)dt = A_0 int " " exp(k_2(u-1)/(A_0k_1))/u^2 du`
Next we expand the exponential term,
`exp(k_2(u-1)/(A_0k_1))=exp((k_2u)/(A_0k_1)-k_2/(A_0k_1))=exp((k_2u)/(A_0k_1))exp(-k_2/(A_0k_1))`
Now we perform a second u-substitution,
`w=-uk_2/(A_0k_1)`, so `(dw)/(du)=-k_2/(A_0k_1)`, `du=-dw(A_0k_1)/k_2` and `u=-w(A_0k_1)/k_2`
Hence,
`int IG(t)dt = exp(-k_2/(A_0k_1)) k_2^2/(A_0k_1^2) int " " exp(-w)/w^2 du`
...which can be expressed using the exponential integral identity (square denominator),
` int " " (exp(-w)dw)/(w^2) = "E"_1(w)-exp(-w)/w`
Ie.,
`int IG(t)dt = exp(-k_2/(A_0k_1)) k_2^2/(A_0k_1^2) ["E"_1(w)-exp(-w)/w] + c_i`
Reversing the substitution, `w` and then `u`,
`int IG(t)dt = exp(-k_2/(A_0k_1)) k_2^2/(A_0k_1^2) ["E"_1(-(1+A_0k_1t)k_2/(A_0k_1))+exp((1+A_0k_1t)k_2/(A_0k_1))/((1+A_0k_1t)k_2/(A_0k_1))] + c_i`
Then back to integrating factor method, to solve the initial differential equation we make use of the identity, (note that the variable y, is concentration `C(t)`, and `c_i` is the integration coefficient),
`y=1/(I(x)) ( int I(x)G(x)dx + c_i)`
Thus,
`C(t)=-k_2/k_1 exp(k_2/(A_0k_1)) ["E"_1((1-A_0k_1t)k_2/(A_0k_1))-(A_0k_1)exp(-(1-A_0k_1t)k_2/(A_0k_1))/((1-A_0k_1t)k_2)]exp(-k_2t) + c_iexp(-k_2t)`
Lastly we need to find the value of the integration coefficient, `c_1`. Using the particular value, `C(0)=0` and `t=0` then,
`0=-k_2/k_1 exp(k_2/(A_0k_1)) ["E"_1(k_2/(A_0k_1))-(A_0k_1)exp(-k_2/(A_0k_1))/(k_2)] + c_i`
Hence,
`c_i=k_2/k_1 exp(k_2/(A_0k_1)) ["E"_1(k_2/(A_0k_1))-(A_0k_1)exp(-k_2/(A_0k_1))/(k_2)]`
We then arrive at,
`C(t)=-k_2/k_1 exp(k_2/(A_0k_1)) ["E"_1((1-A_0k_1t)k_2/(A_0k_1))-(A_0k_1)exp(-(1-A_0k_1t)k_2/(A_0k_1))/((1-A_0k_1t)k_2)]exp(-k_2t) +...`
` ... + [k_2/k_1 exp(k_2/(A_0k_1)) ["E"_1(k_2/(A_0k_1))-(A_0k_1)exp(-k_2/(A_0k_1))/(k_2)]]exp(-k_2t)`
Using the mass-balance equations, we can find `D(t)`,
`D=A_0-A-C`