Two-step Consecutive First Order

`Astackrel(k_1) rarrB stackrel(k_2) rarrC`

Differential Equations

`A'(t)=-k_1A`

`B'(t)=k_1A-k_2B`

`C'(t)=k_2B`

Mass balance equation

`A_0=A+B+C`

Initial conditions:

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

Case I: k₁≠k₂

Time-dependant solutions:

`A=A_0exp(-k_1t)`

`B=(k_1A_0)/(k_2-k_1) (exp(-k_1t)-exp(-k_2t))`

`C=(k_2k_1A_0)/(k_2-k_1) (exp(-k_2t)/k_2 - exp(-k_1t)/k_1 ) + A_0`

Graph:

Derivations

The immediately evaluable differential equation is A’(t),

`(dA)/A=-k_1dt`

Which has the solution

`A=A_0exp(-k_1t)`

Using this solution, we can make the differential equation, B’(t), separable,

`B'(t)=k_1A_0exp(-k_1t)-k_2B`

We rearrange the equation, such that it is of this particular standard form,

`(dy)/(dx)+F(x)y=G(x)`

We obtain,

`B'(t)+k_2B=k_1A_0exp(-k_1t)`

…which can be solved by means of the integrating factor method. For this part of the derivation, we temporarily rename B=y and t=x, as to coincide with the above standard form. By inspection we obtain definitions for F and G, and then evaluate I,

`F=k_2` and `G=k_1A_0exp(-k_1x)`

`I=exp(int Fdx) = exp(k_2x)`

As per the integrating factor method method, to solve the differential equation we set the equation up as below,

`Iy=int IGdx +c`

So we first need to find the integral. Using,

`IG=k_1A_0exp(-k_1x)exp(k_2x)=k_1A_0exp((k_2-k_1)x)`

Then,

`int IGdx = k_1A_0 int exp((k_2-k_1)x)dx`

Which we can evaluate using u-substitution, where,

`u=(k_2-k_1)x` and `(du)/(dx)=k_2-k_1`

and so we have an expression for the integral,

`int IGdx = (k_1A_0)/(k_2-k_1) int exp(u)du = (k_1A_0)/(k_2-k_1) exp((k_2-k_1)x)`

…which is equal to Iy

`Iy=exp(k_2x)y=(k_1A_0)/(k_2-k_1) exp((k_2-k_1)x) + c`

We solve this for y,

`y=(k_1A_0)/(k_2-k_1) exp(-k_1x) + cexp(-k_2x)`

Then to find the coefficient of integration, we use the particular value y(0)=0.

`0=(k_1A_0)/(k_2-k_1) exp(-k_1 0) + cexp(-k_2 0)=(k_1A_0)/(k_2-k_1) + c`

Which gives us the value of c,

`c=-(k_1A_0)/(k_2-k_1)`

Substituting this in we obtain the exact solution,

`y=(k_1A_0)/(k_2-k_1) (exp(-k_1x)-exp(-k_2x))`

We rename the variables y=B and x=t, and obtain the final expression for B(t),

`B=(k_1A_0)/(k_2-k_1) (exp(-k_1t)-exp(-k_2t)`

Finally, we can address the time-differential equation, C’(t). To begin we substitute in the expression B derived above,

`C'(t)=k_2B=(k_2k_1A_0)/(k_2-k_1) (exp(-k_1t)-exp(-k_2t))`

The expression is separable and so can be integrated immediately,

`C=(k_2k_1A_0)/(k_2-k_1) (int exp(-k_1t)dt - int exp(-k_2t)dt + c)`

To obtain the solution,

`C=(k_2k_1A_0)/(k_2-k_1) ( exp(-k_2t)/k_2 - exp(-k_1t)/k_1 + c )`

Using the particular value, C(0)=0,

`0=(k_2k_1A_0)/(k_2-k_1) ( exp(-k_2 0)/k_2 - exp(-k_1 0)/k_1 + c) = (k_2k_1A_0)/(k_2-k_1) (1/k_2 - 1/k_1 + c)`

Thus we can express c,

`c=1/k_1-1/k_2=(k_2-k_1)/(k_1k_2)`

With this definition for c, we can simplify the solution and obtain an exact expression for C(t),

`C=(k_2k_1A_0)/(k_2-k_1) ( exp(-k_2t)/k_2 - exp(-k_1t)/k_1 ) +A_0`

`kappa=`
graphing k=2.1