# Introduction to Chemical Engineering Processes/Why use mole balances?

## Review of Reaction Stoichiometry

Up until now, all of the balances we have done on systems have been in terms of mass. However, mass is inconvenient for a reacting system because it does not allow us to take advantage of the stoichiometry of the reaction in relating the relative amounts of reactants and of products.

Stoichiometry is the relationship between reactants and products in a balanced reaction as given by the ratio of their coefficients. For example, in the reaction:

$C_2H_2 + 2H_2 \rightarrow C_2H_6$

the reaction stoichiometry would dictate that for every one molecule of $C_2H_2$ (acetylene) that reacts, two molecules of $H_2$ (hydrogen) are consumed and one molecule of $C_2H_6$ are formed. However, this does not hold for grams of products and reactants.

Even though the number of molecules in single substance is proportional to the mass of that substance, the constant of proportionality (the molecular mass) is not the same for every molecule. Hence, it is necessary to use the molecular weight of each molecule to convert from grams to moles in order to use the reaction's coefficients.

## Molecular Mole Balances

We can write balances on moles like we can on anything else. We'll start with our ubiquitous general balance equation:

$Input - Output = Accumulation - Generation$

As usual we assume that accumulation = 0 in this book so that:

$Input - Output + Generation = 0$

Let us denote molar flow rates by $\dot{n}$ to distinguish them from mass flow rates. We then have a similar equation to the mass balance equation:

 $\Sigma \dot{n}_{in} - \Sigma \dot{n}_{out} + n_{gen} = 0$

The same equation can be written in terms of each individual species.

There are a couple of important things to note about this type of balance as opposed to a mass balance:

1. Just like with the mass balance, in a mole balance, a non-reactive system has $n_{gen} = 0$ for all species.
2. Unlike the mass balance, the TOTAL generation of moles isn't necessarily 0 even for the overall mole balance! To see this, consider how the total number of moles changes in the above reaction; the final number of moles will not equal the initial number because 3 total moles of molecules are reacting to form 1 mole of products.

Why would we use it if the generation isn't necessarily 0? We use the molecular mole balance because if we know how much of any one substance is consumed or created in the reaction, we can find all of the others from the reaction stoichiometry. This is a very powerful tool because each reaction only creates one new unknown if you use this method! The following section is merely a formalization of this concept, which can be used to solve problems involving reactors.

## Extent of Reaction

In order to formalize the previous analysis of reactions in terms of a single variable, let us consider the generic reaction:

$aA + bB \rightarrow cC + dD$

The Molar Extent of Reaction X is defined as:

 $X = -\frac{\Delta n_A}{a} = -\frac{\Delta n_B}{b} = \frac{\Delta n_C}{c} = \frac{\Delta n_D}{d}$

Since all of these are equivalent, it is possible to find the change in moles of any species in a given reaction if the extent of reaction X is known.

 Note: Though they won't be discussed here, there are other ways in which the extent of reaction can be defined. Some other definitions are dependent on the percent change of a particular substrate, and the stoichiometry is used in a different way to determine the change in the others. This definition makes X independent of the substrate you choose.

The following example illustrates the use of the extent of reaction.

Example:

Consider the reaction $H_2O_2 + O_3 \rightarrow H_2O + 2O_2$. If you start with 50 g of $H_2O_2$ and 25 grams of $O_3$, and 25% of the moles of $O_3$ are consumed, find the molar extent of reaction and the changes in the other components.

Solution: First we need to convert to moles, since stoichiometry is not valid when units are in terms of mass.

$50 \mbox{ g H}_2O_2 * \frac{1 \mbox{ mol}}{34 \mbox{ g}} = 1.471 \mbox{ moles H}_2O_2$

$25 \mbox{ g O}_3 * \frac{1 \mbox{ mol}}{48 \mbox{ g}} = 0.5208 \mbox{ moles O}_3$

Clearly ozone is the limiting reactant here. Since 25% is consumed, we have that:

$\Delta n(O_3) = -0.25*0.5208 = -0.1302 \mbox{ moles O}_3$

Hence, by definition, $X = \frac{-0.1302}{1} = 0.1302$

And then we have $\Delta n(H_2O_2) = -0.1302, \Delta n(H_2O) = 0.1302, \Delta n(O_2) = 2*0.1302 = 0.2604$, all in moles of the appropriate substrate.

## Mole Balances and Extents of Reaction

The mole balance written above can be written in terms of extent of reaction if we notice that the $\Delta n(A)$ term defined above is exactly the number of moles of a generated or consumed by the reaction.

 Note: This is only useful for individual species balances, not the overall mole balance. When doing balances on reactive systems, unlike with non-reactive systems, it is generally easier to use all individual species balances possible, rather than the total mole balance and then all but one of the individual species. This is because the total generation of moles in a reaction is generally not 0, so no algebraic advantage is gained by using the total material balance on the system.

Therefore we can write that:

$n_{A,gen} = \Delta n(A) = -X*a$

where X is the molar extent of reaction and a is the stoichiometric coefficient of A. Plugging this into the mole balance derived earlier, we arrive at the molecular mole balance equation:

 $\Sigma \dot{n}_{A,out} - \Sigma{n}_{A,in} - X*a = 0$ if A is consumed, or +Xa if it is generated in the reaction