Performing material balances on reactive systems is slightly
more complex than for non-reactive systems

N_{2} + 3H_{2} -> 2NH_{3}

While we can still do a balance on total mass, this is often not
particularly useful (as we would like to know component
compositions). What differs now is that a balance on total moles
will *not* work (in general).

While the book notes three types of reactive balances, there are really only two types that are clearly distinct: a molecular species/extent of reaction balance; and a balance on atomic species.

For simplicity, this course will only focus on reactive processes that are open systems which are run at steady state. This yields a simplified balance equation of:

$0 = \dot M_I - \dot M_O + G - C$

In this balance method we will examine each molecular species individually, and will typically rearrange our balance equation to be in the form:

$\dot M_O = \dot M_I + G - C$

The G and C terms will then come directly from a combination of
stoichiometry and the concept of the *extent of
reaction*.

DEFINITION

The **extent of reaction** ($\xi$) is a
method of quantifying how many "times" a reaction has occurred. It
has units of moles/time and numerically, it is chosen such that the
stoichiometric coefficient times $\xi$ is equal to the quantity of
species reacted.

Using the above definition of the extent of reaction, it is a simple matter to write G and C in terms of this quantity simply using stoichiometry. In anticipation of performing reactive balances, we will switch to using molar quantities ($\dot n$'s rather than $\dot M$'s) so that we get:

$\dot n_O = \dot n_I + g - c$

where we have used lowercase g and c to note molar quantities.

EXAMPLE

Let's write the expressions for our figure above:

$\dot n_{{N_2}_O} = 8 kmol/h = \dot n_{{N_2}_I} + g_{N_2} - c_{N_2} = 10kmol/h + 0 - 1\xi$

$\dot n_{{H_2}_O} = 15 kmol/h = \dot n_{{H_2}_I} + g_{H_2} - c_{H_2} = \dot n_{{H_2}_I} + 0 - 3\xi$

$\dot n_{{NH_3}_O} = 4 kmol/h = \dot n_{{NH_3}_I} + g_{NH_3} - c_{NH_3} = 0 + 2\xi - 0$

NOTE:

Typically "reactants" have g=0, while "products" have a zero initial concentration and have c=0. This obviously becomes more complex when multiple reactions take place.

Using the extent of reaction method for systems with multiple reactions involves including a new value of $\xi$ for each reaction, and then calculating c and g as the sum of applicable $\xi$'s.

EXAMPLE

Consider the reaction network:

A -> B

2B -> C

If we assign $\xi_1$ to the first reaction and $\xi_2$ to the second, we get expressions that look like:

$\dot n_{A_O} = \dot n_{A_I} + g_A - c_A = \dot n_{A_I} + 0 - \xi_1$

$\dot n_{B_O} = \dot n_{B_I} + g_B - c_B = \dot n_{B_I} + \xi_1 - 2\xi_2$

$\dot n_{C_O} = \dot n_{C_I} + g_C - c_C = \dot n_{C_I} + \xi_2 - 0$

OUTCOME

Write extent of reaction expressions from stoichiometry

Due to the introduction of the new variables $\xi$ our degrees of freedom analysis must change:

- Number of unknowns
- + (yes, plus!) the number of independent reactions
- - number of independent species
- - other equations (physical constraints, process specifications, etc.)

NOTE:

Equations are not independent if they can be obtained through algebraic manipulation of the other (already deemed independent) equations! (This includes multiplying an equation by a constant factor.)

TEST YOURSELF:

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carbon dioxide and water:

CH_{4} + 0_{2} -> HCHO +
H_{2}0

CH_{4} + 20_{2} -> CO_{2} +
2H_{2}0

The feed to the reactor contains equimolar amounts of methane and oxygen. Assume a basis of 100 mol feed/s.

Draw and label a flowchart. Derive expressions for the product stream components in terms of the extents of reaction, $\xi_1$ and $\xi_2$.

The fractional conversion of methane is 0.900 and the fractional yield of formaldehyde is 0.855. Calculate the molar composition of the reactor output stream and the selectivity of the formaldehyde production relative to carbon dioxide production.