Second-order Kinetics (the Bimolecular Reaction)

Scheme:

`A+Bstackrel(k) rarrC`

Differential Equation:

`A'(t)=-kAB`

Mass balance equation:

`A_0=A+C` and `B_0=B+C`

Initial conditions:

`B(0)=B_0`, `A(0)=A_0` and `C(0)=C_0`

Case I: A₀=B₀

Since A₀=B₀, the two mass balance equations are equal. Hence, A=B, and the differential equation simplifies to:

`A'(t)=-kA^2`

The equation is now separable,

`(dA)/A^2=-kdt`

Which has the solution,

`-kt=int(dA)/A^2=-[1/A]_(A_0)^(A_t)=-1/A_t+1/A_0`

This is then solved for `A(t)`,

`A(t)=A_0/(1+A_0kt)`

Graph:

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Case II: A₀≠B₀

 With no simplification of the mass balances equations possible, it is most direct to solve the differential equation for the product, ie.,

`C'(t)=k(A_0-C)(B_0-C)`

The equation is separable,

`(dC)/((A_0-C)(B_0-C))=kdt`

Whilst it is probably possible to evaluate this integral using integration by parts, it is perhaps easier to first apply a partial fraction decomposition,

`1/((A_0-C)(B_0-C))=P/(A_0-C)+Q/(B_0-C)`

So,

`1/((A_0-C)(B_0-C))=P(B_0-C)/((A_0-C)(B_0-C))+Q(A_0-C)/((A_0-C)(B_0-C))`

Hence,

`1=P(B_0-C)+Q(A_0-C)`

The values of P and Q are found by obtaining the zero’s for the expression,
ie., setting C=B₀ the P term becomes zero, and we obtain,

`1=Q(A_0-B_0)`

Then setting C=A₀ the Q term becomes zero, and we obtain,

`1=P(B_0-A_0)`

Therefore,

`1/(A_0-B_0)=Q`

and

`1/(B_0-A_0)=P`

Using these values, the decomposition yields,

`1/((A_0-C)(B_0-C))=1/(B_0-A_0) 1/(A_0-C)+1/(A_0-B_0) 1/(B_0-C)`

`=1/(A_0-B_0) (1/(B_0-C)-1/(A_0-C))`

The integral is now much easier to evaluate,

`kdt=1/(A_0-B_0) ((dC)/(B_0-C)-(dC)/(A_0-C))`

We exploit the distributive property of integration, and evaluate each term individually,

`kt=1/(A_0-B_0) ( int_0^(C_t) (dC)/(B_0-C) - int_0^(C_t) (dC)/(A_0-C) )`

Using u-substitution,

`u=B_0-C` and `(du)/(dC)=-1`

We obtain,

`int_0^(C_t) (dC)/(B_0-C) =int_0^(C_t) (du)/u =-[lnu]_0^(C_t) =-[ln(B_0-C_t)-ln(B_0)]=-ln((B_0-C_t)/B_0)`

Applying the same procedure to the second integral we obtain,

`int_0^(C_t) (dC)/(A_0-C) =int_0^(C_t) (du)/u =-[lnu]_0^(C_t) =-[ln(A_0-C_t)-ln(B_0)]=-ln((A_0-C_t)/A_0)`

Combining these two results into the integral we arrive at the integrated rate equation, 

`kt=1/(A_0-B_0) [ln((A_0-C_t)/A_0) - ln((B_0-C_t)/B_0)]`

We can solve this for C,

`C(t)=(B_0A_0(exp(kt(A_0-B_0))-1)) / (A_0exp(kt(A_0-B_0))-B_0)`