2.3 – Solving Reactive Balances In-depth

2.3.0 – Learning Objectives

By the end of this section you should be able to:

Solve simple reactive balances using the molecular and atomic reactive balances.

2.3.1 – Introduction

We will solve the example in the diagram 4.71 using the molecular species and atomic species balance. Python will be implemented to perform calculations. This section goes in-depth of the logic used to solve the problem that is addressed before.

2.3.2 – Problem statement

Recall the block diagram in figure 4.7-1

$C_{2} H_{6} ⟶ C_{2} H_{4} + H_{2}$

![](../figures/Module-2/ethane-reaction.png)

What are the mol flows of the exit streams of ethane $C_{6} H_{6}$ and ethylene $C_{4} H_{4}$ ?

2.3.3 – Solving using the molecular species balance

First, we identify our molecular species; for us, this is Hydrogen, Ethane, and Ethylene.

Next, we write out their respective balance equations. Note how the diagram 4.7-1 only shows the output variables ( $\dot{n_{1}}, \dot{n_{2}}$ )

$C_{2} H_{6} b a l a n c e : 100 {\frac{k m o l}{m i n}}_{i n p u t} = \dot{n_{1}} {\frac{k m o l}{m i n}}_{O u t p u t} + C_{2} H_{6 C o n s u m e d}$

$C_{2} H_{4} b a l a n c e : C_{2} H_{4 g e n e r a t e d} = \dot{n_{2}} {\frac{k m o l}{m i n}}_{O u t p u t}$

$H_{2} b a l a n c e : 40 {\frac{k m o l}{m i n}}_{g e n e r a t e d} = 40 {\frac{k m o l}{m i n}}_{O u t p u t}$

To further relate these equations, use the stoichiometrey of the dehydrogenation reaction which gives a relationship between the consumed and generated terms:

$C_{2} H_{6} \to C_{2} H_{4} + H_{2}$

Since there is a 1:1 ratio between the Generation of Hydrogen to $C_{2} H_{4}$ :

$40 \frac{k m o l}{m i n} H_{2 g e n e r a t e d} * \frac{1 \frac{k m o l}{m i n} C_{2} H_{4}}{1 \frac{k m o l}{m i n} H_{2}} = 40 \frac{k m o l}{m i n} C_{2} H_{4 g e n e r a t e d}$

Since there is a 1:1 ratio between the generation of hydrogen and the consumption of Ethane, $C_{2} H_{6 c o n s u m e d} = 40 \frac{k m o l}{m i n}$ .

Substituting the $C_{2} H_{6 C o n s u m e d}$ and the $C_{2} H_{4 g e n e r a t e d}$ into their respective equations, we can see that the outputs of ethane and ethylene $(\dot{n_{1}}, \dot{n_{2}})$ are $60 \frac{k m o l}{m i n}$ and $40 \frac{k m o l}{m i n}$ respectively.

$C_{2} H_{6} b a l a n c e : 100 {\frac{k m o l}{m i n}}_{i n p u t} = \dot{n_{1}} {\frac{k m o l}{m i n}}_{O u t p u t} + 40 \frac{k m o l}{m i n} C_{2} H_{6 O u t p u t}$

$\dot{n_{1}} = 60 {\frac{k m o l}{m i n}}_{O u t p u t}$

$C_{2} H_{4} b a l a n c e : C_{2} H_{4 g e n e r a t e d} = \dot{n_{2}} {\frac{k m o l}{m i n}}_{O u t p u t} = 40 \frac{k m o l}{m i n} C_{2} H_{4 g e n e r a t e d}$

2.3.4 – Solving using the Atomic species balance

Let's recall the Atomic balances:

$C a r b o n (C) b a l a n c e : I n p u t = O u t p u t$

$H y d r o g e n (H) b a l a n c e : I n p u t = O u t p u t$

First the Atomic balances are broken down to collect any molecular species containing the atomic species (Carbon and Hydrogen).

The carbon balance becomes: $C_{2} H_{6 i n p u t} = C_{2} H_{6 o u t p u t} + C_{2} H_{4 o u t p u t}$

The hydrogen balance becomes:

$C_{2} H_{6 i n p u t} = C_{2} H_{6 o u t p u t} + C_{2} H_{4 o u t p u t} + H_{2 o u t p u t}$

We then isolate the atomic species from the molecular species. Another way of thinking is: How many of N atoms are there in this molecule? For example, there are 2 carbon atoms for every 1 ethane molecule.

The carbon balance becomes:

$C_{2} H_{6 i n p u t} * \frac{2 C \frac{k m o l}{m i n}}{1 C_{2} H_{6} \frac{k m o l}{m i n}} = C_{2} H_{6 o u t p u t} * \frac{2 C \frac{k m o l}{m i n}}{1 C_{2} H_{6} \frac{k m o l}{m i n}} + C_{2} H_{4 o u t p u t} * \frac{2 C \frac{k m o l}{m i n}}{1 C_{2} H_{6} \frac{k m o l}{m i n}}$

The hydrogen balance becomes:

$C_{2} H_{6 i n p u t} * \frac{6 H \frac{k m o l}{m i n}}{1 C_{2} H_{6} \frac{k m o l}{m i n}} = C_{2} H_{6 o u t p u t} * \frac{6 H \frac{k m o l}{m i n}}{1 C_{2} H_{6} \frac{k m o l}{m i n}} + C_{2} H_{4 o u t p u t} * \frac{4 H \frac{k m o l}{m i n}}{1 C_{2} H_{6} \frac{k m o l}{m i n}} + H_{2 o u t p u t} * \frac{2 H \frac{k m o l}{m i n}}{1 H_{2} \frac{k m o l}{m i n}}$

Note from diagram 4.7-1, the input and output values are substituted for their respective integers and variables. Remember that ethane and ethene outputs are defined as ( $\dot{n_{1}}, \dot{n_{2}}$ )

rewriting the balances

The carbon balance becomes: $200 \frac{k m o l}{m i n} = \dot{2 n_{1}} + \dot{2 n 2}$

The hydrogen balance becomes:

$600 \frac{k m o l}{m i n} = 6 \dot{n_{1}} + 4 \dot{n_{2}} + 80 \frac{k m o l}{m i n}$