2.1 – Reactive Balances¶

2.1.0 – Learning Objectives¶

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

Understand two of the three reactive balance methods.
Solve two of the three reactive balance methods.

2.1.1 – Introduction¶

Often when we evaluate processes, chemical reactions take place. These chemical reactions add complexity to our system by introducing new unknowns. To reduce these complexities, we can use molecular species balances, atomic balances or the extent of reactions to help solve problems.

2.1.2 – The Production of Ethylene¶

Ethylene is a major chemical product primarily used to form the plastic polyethylene. It is also a natural chemical released by fruits to ripen. You can separate your bananas to help them ripen faster.

Let’s consider the production of ethylene in steady-state. In production, the dehydrogenation of Ethane forms Ethylene:

\[C_2H_{6 \space (g)} \longrightarrow C_2H_{4 \space (g)} + H_{2 \space (g)}\]

Attribution: Said Zaid-Alkailani, Ngai To Lo & UBC [CC BY 4.0 de (https://creativecommons.org/licenses/by/4.0/)]

2.1.3 – Molecular Species Balance¶

The molecular species balances are the balances of all chemical compounds in this reaction. For example:

\[C_2 H_6 \space \text{balance}: \space Accumulation = Input -Output +Generation - Consumption\]

Since the system is in steady state and the reaction does not produce Ethane:

\[Accumulation = 0 , \space Generation = 0\]

Rearranging and applying similar logic to the all of the species gives us 3 equations:

\(C_2 H_6\) balance: \(Input = Output + Consumption\)¶

\[10 \space \frac{kmol}{h} = \dot{n}_{(1, \space C_2 H_6)} + Consumption_{C_2 H_6}\]

\(C_2 H_4\) balance: \(Generation = Output\)¶

\[Generation_{C_2 H_4} = \dot{n}_{(2, \space C_2 H_4)}\]

\(H_2\) balance: \(Generation = Output\)¶

\[Generation_{H_2} = 4 \space \frac{kmol}{h}\]

Since the reaction above has stoichiometric ratio of 1:1:1, we can determine \(Generation_{H_2} = Generation_{C_2 H_4} = Consumption_{C_2 H_6}\). Therefore,

\[\dot{n}_{(2, \space C_2 H_4)} = 4 \space \frac{kmol}{h}\]

\[\dot{n}_{(1, \space C_2 H_6)} = 6 \space \frac{kmol}{h}\]

2.1.4 – Atomic Species Balance¶

The atomic species balances are the balances of all the elemental species in this reaction. The Two species in the above reaction are Hydrogen and Carbon. Atomic species must follow the law of conservation of mass, thus both species take on the equation:

Carbon (C) balance: \(Input = Output\)¶

\[10 \space \frac{kmol \space C_2 H_6}{h} \times \frac{2 \space kmol \space C}{1 \space kmol \space C_2 H_6} = \dot{n}_{(1, \space C_2 H_6)} \times \frac{2 \space kmol \space C}{1 \space kmol \space C_2 H_6} + \dot{n}_{(2, \space C_2 H_4)} \times \frac{2 \space kmol \space C}{1 \space kmol \space C_2 H_4}\]

\[10 \space \frac{kmol \space C}{h} = \dot{n}_{(1, \space C_2 H_6)} + \dot{n}_{(2, \space C_2 H_4)}\]

Hydrogen (H) balance: \(Input = Output\)¶

\[10 \space \frac{kmol \space C_2 H_6}{h} \times \frac{6 \space kmol \space H}{1 \space kmol \space C_2 H_6} = 4 \space \frac{kmol \space H_2}{h} \times \frac{2 \space kmol \space H}{1 \space kmol \space H_2} + \dot{n}_{(1, \space C_2 H_6)} \times \frac{6 \space kmol \space H}{1 \space kmol \space C_2 H_6} + \dot{n}_{(2, \space C_2 H_4)} \times \frac{6 \space kmol \space H}{1 \space kmol \space C_2 H_4}\]

\[52 \space \frac{kmol \space C}{h} = 6 \space \dot{n}_{(1, \space C_2 H_6)} + 4 \space \dot{n}_{(2, \space C_2 H_4)}\]

\[\therefore \space \dot{n}_{(1, \space C_2 H_6)} = 6 \space \frac{kmol}{h}\]

\[\therefore \space \dot{n}_{(2, \space C_2 H_4)} = 4 \space \frac{kmol}{h}\]

In [ ]: