# Introduction to Chemical Engineering Processes/Atom balances

## The concept of atom balances

Let's begin this section by looking at the reaction of hydrogen with oxygen to form water:

$H_{2}+O_{2}\rightarrow H_{2}O$

We may attempt to do our calculations with this reaction, but there is something seriously wrong with this equation! It is not balanced; as written, it implies that an atom of oxygen is somehow "lost" in the reaction, but this is in general impossible. Therefore, we must compensate by writing:

$H_{2}+{\frac {1}{2}}O_{2}\rightarrow H_{2}O$

or some multiple thereof.

Notice that in doing this we have made use of the following conservation law, which is actually the basis of the conservation of mass:

The number of atoms of any given element does not change in any reaction (assuming that it is not a nuclear reaction).

Since by definition the number of moles of an element is proportional to the number of atoms, this implies that ${\dot {n}}_{A,gen}=0$  where A represents any element in atomic form.

## Mathematical formulation of the atom balance

Now recall the general balance equation:

$In-Out+Generation-Consumption=Accumulation$

In this course we're assuming $Accumulation=0$ . Since the moles of atoms of any element are conserved, $generation=0$  and $consumption=0$ . So we have the following balance on a given element A:

For a given element A,

$\Sigma {\dot {n}}_{A,in}-\Sigma {\dot {n}}_{A,out}=0$

 Note: When analyzing a reacting system you must choose either an atom balance or a molecular species balance but not both. Each has advantages; an atom balance often yields simpler algebra (especially for multiple reactions; the actual reaction that takes place is irrelevant!) but also will not directly tell you the extent(s) of reaction, and will not tell you if the system specifications are actually impossible to achieve for a given set of equilibrium reactions.

## Degree of Freedom Analysis for the atom balance

As before, to do a degree of freedom analysis, it is necessary to count the number of unknowns and the number of equations one can write, and then subtract them. However, there are a couple of important things to be aware of with these balances.

• When doing atom balances, the extent of reaction does not count as an unknown, while with a molecular species balance it does. This is the primary advantage of this method: the extent of reaction does not matter since atoms of elements are conserved regardless of how far the reaction has proceeded.
• You need to make sure each atom balance will be independent. This is difficult to tell unless you write out the equations and look to see if any two are identical.
• In reactions with inert species, each molecular balance on the inert species counts as an additional equation. This is because of the following important note:
 Note: When you're doing an atom balance you should only include reactive species, not inerts.

Example:

Suppose a mixture of nitrous oxide ($N_{2}O$ ) and oxygen is used in a natural gas burner. The reaction $CH_{4}+2O_{2}\rightarrow 2H_{2}O+CO_{2}$  occurs in it.

There would be four equations that you could write: 3 atom balances (C, H, and O) and a molecular balance on nitrous oxide. You would not include the moles of nitrous oxide in the atom balance on oxygen.

## Example of the use of the atom balance

Let's re-examine a problem from the previous section. In that section it was solved using a molecular species balance, while here it will be solved using atom balances.

Example:

Consider the reaction of Phosphene with oxygen: $4PH_{3}+8O_{2}\rightarrow P_{4}O_{10}+6H_{2}O$

Suppose a 100-kg mixture of 50% $PH_{3}$  and 50% $O_{2}$  by mass enters a reactor in a single stream, and the single exit stream contains 25% $O_{2}$  by mass. Assume that all the reduction in oxygen occurs due to the reaction. How many degrees of freedom does this problem have? If possible, determine mass composition of all the products.

For purposes of examination, the flowchart is re-displayed here:

### Degree of Freedom Analysis

There are three elements involved in the system (P, H, and O) so we can write three atom balances on the system.

There are likewise three unknowns (since the extent of reaction is NOT an unknown when using the atom balance): the outlet concentrations of $PH_{3},P_{4}O_{10},H_{2}O$

Therefore, there are 3 - 3 = 0 unknowns.

### Problem Solution

Let's start the same as we did in the previous section: by finding converting the given information into moles. The calculations of the previous section are repeated here:

• ${\dot {m}}_{out}={\dot {m}}_{in}=100{\mbox{ kg}}$
• ${\dot {n}}_{PH_{3},in}=0.5*(100{\mbox{ kg}})*{\frac {1{\mbox{ mol}}}{0.034{\mbox{ kg}}}}=1470.6{\mbox{ moles PH}}_{3}{\mbox{ in}}$
• ${\dot {n}}_{O_{2},in}=0.5*(100{\mbox{ kg}})*{\frac {1{\mbox{ mol}}}{0.032{\mbox{ kg}}}}=1562.5{\mbox{ moles O}}_{2}{\mbox{ in}}$
• ${\dot {n}}_{O_{2},out}=0.25*(100{\mbox{ kg}})*{\frac {1{\mbox{ mol}}}{0.032{\mbox{ kg}}}}=781.25{\mbox{ moles O}}_{2}{\mbox{ out}}$

Now we start to diverge from the path of molecular balances and instead write atom balances on each of the elements in the reaction. Let's start with Phosphorus. How many moles of Phosphorus atoms are entering?

• Inlet: Only $PH_{3}$  provides P, so the inlet moles of P are just $1*1470.6=1470.6{\mbox{ moles P in}}$
• Outlet: There are two ways phosphorus leaves: as unused $PH_{3}$  or as the product $P_{4}O_{10}$ . Therefore, the moles of $PH_{3}$  out are $1*n_{PH_{3},out}+4*n_{P_{4}O_{10},out}$ . Note that the 4 in this equation comes from the fact that there are 4 Phosphorus atoms in every mole of $P_{4}O_{10}$ .

Therefore the atom balance on Phosphorus becomes:

Phosphorus

$1*n_{PH_{3},out}+4*n_{P_{4}O_{10},out}=1470.6$

Similarly, on Oxygen we have:

• Inlet: $2*n_{O_{2},in}=2*1562.5=3125{\mbox{ moles O}}_{2}$
• Outlet: $2*n_{O_{2},out}+10*n_{P_{4}O_{10},out}+1*n_{H_{2}O,out}=1562.5+10*n_{P_{4}O_{10},out}+1*n_{H_{2}O,out}$
Oxygen

$1562.5+10*n_{P_{4}O_{10},out}+1*n_{H_{2}O,out}=3125$

Finally, check to see if you can get the following Hydrogen balance as a practice problem:

Hydrogen

$2*n_{H_{2}O,out}+3*n_{PH_{3},out}=4411.8$

Solving these three linear equations, the solutions are:

$n_{PH_{3},out}=1080,n_{H_{2}O,out}=586,n_{P_{4}O_{10},out}=97.66$  moles

All of these answers are identical to those obtained using extents of reaction. Since the remainder of the solution to that problem is identical to that in the previous section, the reader is referred there for its completion.

## Example of balances with inert species

Sometimes it's more difficult to choose which type of balance you want, because both are possible but one is significantly easier than the other. As an example, lets consider a basic pollution control system.

Example:

Suppose that you are running a power plant and your burner releases a lot of pollutants into the air. The flue gas has been analyzed to contain 5% $SO_{2}$ , 3% $NO_{2}$ , 7% $O_{2}$  and 15% $CO_{2}$  by moles. The remainder was determined to be inert.

Local regulations require that the emissions of sulfur dioxide be less than 200 ppm (by moles) from your plant. They also require you to reduce nitrogen dioxide emissions to less than 50 ppm. You decide that the most economical method for control of these for your plant is to utilize ammonia-based processes. The proposed system is as follows:

1. Put the flue gas through a denitrification system, into which (pure) ammonia is pumped. The amount of ammonia pumped in is three times as much as would theoretically be needed to use all of the nitrogen dioxide in the flue gas.
2. Allow it to react a specified amount of time.
3. Pump it into a desulfurization system. Nothing new is injected here, it just has a different catalyst than the denitrification, and the substrates are at a different temperature and pressure.

The reactions that occur are:

1. $2NO_{2}+4NH_{3}+O_{2}\rightarrow 3N_{2}+6H_{2}O$
2. $H_{2}O+2NH_{3}+SO_{2}\rightarrow (NH_{4})_{2}SO_{3}$

If your plant makes $130{\frac {ft^{3}}{s}}$  of flue gas at $T=900K$  and $P=2{\mbox{ atm}}$ , how much ammonia do you need to purchase for each 8-hour shift? How much of it remains unused? Why do we want to have a significant amount of excess ammonia?

Assume that the flue gas is an ideal gas. Recall the ideal gas law, $PV=nRT$ , where $R=0.0821{\frac {L*atm}{mol*K}}$ .

### Step 1: Flowchart

Flowcharts are becoming especially important now as means of organizing all of that information!

### Step 2: Degrees of Freedom

Let's consider an atomic balance on each reactor.

• Denitrification system: 9 unknowns (all concentrations in stream 3, and ${\dot {n}}_{2}$ .) - 3 atom balances (N, H, and O) - 3 inert species ($CO_{2},SO_{2},inerts$ ) - 1 additional info (3X stoichiometric feed) = 2 DOF
• Desulfurization system: 15 unknowns - 4 atom balances (N, H, O, and S) - 5 inerts ($CO_{2},O_{2},NO_{2},N_{2},inerts$ ) = 6 DOF
• Total = 2 + 6 - 8 shared = 0 DOF, hence the problem has a unique solution.

We can also perform the same type of analysis on molecular balances.

• Denitrification system: 10 unknowns (now the conversion $X_{1}$  is also unknown) - 8 molecular species balances - 1 additional info = 1 DOF
• Desulfurization system: 16 unknowns (now the conversion $X_{2}$  is unknown) - 9 balances = 7 DOF.
• Total = 1 + 7 - 8 shared = 0 DOF.

Therefore the problem is theoretically solvable by both methods.

### Step 3: Units

The only weird units in this problem (everything is given in moles already so no need to convert) are in the volumetric flowrate, which is given in ${\frac {ft^{3}}{s}}$ . Lets convert this to ${\frac {moles}{s}}$  using the ideal gas law. To use the law with the given value of R is is necessary to change the flowrate to units of ${\frac {L}{s}}$ :

$130{\frac {ft^{3}}{s}}*{\frac {28.317{\mbox{ L}}}{ft^{3}}}=3681.2{\frac {L}{s}}$  $P{\dot {V}}={\dot {n}}RT\rightarrow 2*3681.2={\dot {n}}_{1}(0.0821)(900)$

${\dot {n}}_{1}=99.64{\frac {moles}{s}}$

Now that everything is in good units we can move on to the next step.

### Step 4: Devise a plan

We can first determine the value of ${\dot {n}}_{2}$  using the additional information. Then, we should look to an overall system balance.

Since none of the individual reactors is completely solvable by itself, it is necessary to look to combinations of processes to solve the problem. The best way to do an overall system balance with multiple reactions is to treat the entire system as if it was a single reactor in which multiple reactions were occurring. In this case, the flowchart will be revised to look like this:

Before we try solving anything, we should check to make sure that we still have no degrees of freedom.

Atom Balance

There are 8 unknowns (don't count conversions when doing atom balances), 4 types of atoms (H, N, O, and S), 2 species that never react, and 1 additional piece of information (3X stoichiometric), so there is 1 DOF. This is obviously a problem, which occurs because when performing atom balances you cannot distinguish between species that react in only ONE reaction and those that take part in more than one.

In this case, then, it is necessary to look to molecular-species balances.

Molecular-species balance

In this case, there are 10 unknowns, but we can do molecular species balances on 9 species $(SO_{2},NO_{2},NH_{3},N_{2},O_{2},CO_{2},H_{2}O,(NH_{4})_{2}SO_{3},inerts)$  and have the additional information, so there are 0 DOF when using this method.

Once we have all this information, getting the information about stream 3 is trivial from the definition of extent of reaction.

### Step 5: Carry Out the Plan

First off we can determine ${\dot {n}}_{2}$  by using the definition of a stoichiometric feed.

${\dot {n}}_{NO_{2},in}=0.03*99.64=2.9892{\frac {mol}{s}}$

The stoichiometric amount of ammonia needed to react with this is, from the reaction,

${\frac {4{\mbox{ moles NH}}_{3}}{2{\mbox{ moles NO}}_{2}}}*2.9892=5.96{\frac {moles{\mbox{ NH}}_{3}}{s}}$

Since the problem states that three times this amount is injected into the denitrification system, we have:

${\dot {n}}_{2}=17.88{\frac {moles}{s}}$

Now, we are going to have a very complex system of equations with the 9 molecular balances. This may be a good time to invest in some equation-solving software.

See if you can derive the following system of equations from the overall-system flowchart above.

$NH_{3}:{\dot {n}}_{4}*x_{NH_{3},4}=17.88-4*X_{1}-2*X_{2}$

$SO_{2}:{\dot {n}}_{4}*2*10^{-4}=0.05*99.64-X_{2}$
$NO_{2}:{\dot {n}}_{4}*5*10^{-5}=0.03*99.64-2*X_{1}$
$N_{2}:{\dot {n}}_{4}*x_{N_{2},4}=3*X_{1}$
$O_{2}:{\dot {n}}_{4}*x_{O_{2},4}=0.07*99.64-X_{1}$
$H_{2}O:{\dot {n}}_{4}*x_{H_{2}O,4}=6*X_{1}-X_{2}$
$CO_{2}:{\dot {n}}_{4}*x_{CO_{2},4}=0.15*99.64$
$(NH_{4})_{2}(SO_{3}):{\dot {n}}_{4}*x_{(NH_{4})_{2}SO_{3},4}=X_{2}$
$Inerts:{\dot {n}}_{4}*(1-2*10^{-4}-5*10^{-5}-x_{NH_{3},4}-x_{N_{2},4}-x_{O_{2},4}-x_{H_{2}O,4}-x_{CO_{2},4}-x_{(NH_{4})_{2}SO_{3},4})=0.7*99.64$

Using an equation-solving package, the following results were obtained:

$X_{1}=1.492{\mbox{ moles}}$

$X_{2}=4.961{\mbox{ moles}}$
${\dot {n}}_{3}=105.62{\frac {mol}{s}}$
$x_{NH_{3},4}=0.01884$
$x_{N_{2},4}=0.04238$
$x_{O_{2},4}=0.05191$
$x_{H_{2}O,4}=0.03778$
$x_{CO_{2},4}=0.1415$
$x_{(NH_{4})_{2}SO_{3},4}=0.04697$
$x_{I}=1-\Sigma {\mbox{ (other components) }}=0.6606$

#### Stream 3

Now that we have completely specified the composition of stream 4, it is possible to go back and find the compositions of stream 3 using the extents of reaction and feed composition. Although this is not necessary to answer the problem statement, it should be done, so that we can then test to make sure that all of the numbers we have obtained are consistent. To do:finish this