Consider the following consecutive, second order, irreversible reactions are carried out in a batch reactor:

A + S Yield(k1) X

X + S yield (k2) Y

One mole of A and two moles of S are initially charged to the batch reactor. Find the mole fraction of X remaining in solution after 75% of A is consumed. Assume k2=2k1.

I know some differential equations are needed but I'm not sure how to set this up.


1 Answer 1



The rate of change of amount (concentration) of any given component is a differential equation.

$$ r_j = \frac{dC_j}{dt} $$

The empirical rate law expression for a reaction defines how reaction rate depends on concentrations in the reaction expression.

$$ r_j = \pm k_{rxn} \Pi\ C_j^{n_j} $$

Reversible reactions have forward and backward terms. Irreversible reactions have only forward terms. The reaction orders $n_j$ are the stoichiometric coefficients of the component.

For a system of $N$ components, we will have $N$ independent differential equations.

Specific Example

For component X in your system, you can obtain this expression.

$$ r_X = \frac{dC_X}{dt} = k_1 C_A C_S - k_2 C_X C_S $$


This gives first insight as to how to set up the problem. You will not need to consider the production rate of component Y. You will have three independent, first order, coupled differential equations. How this system is to be solved is entirely a different question of its own right. I recommend a numerical approach with graphical output. Two packages that take this approach are below.


Kinetic Reaction Demo

Options using MatLab or Mathematica are likely posted on their sites as well.


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