3  Measures of Reaction Progress

As noted in Chapter 1, the composition of a reacting system changes over time. However, the change in the amount of one reagent relative to the change in the amount of another reagent must be consistent with the stoichiometry of the reactions taking place. Since the changes in the amounts of the reagents are not independent, it is useful to define reaction progress variables that measure how many times each of the reactions have occurred. Given the initial amounts of all of the reagents, the amounts at any later time can then be calculated using these reaction progress variables.

3.1 Describing Single-Reaction Systems

Consider a closed system that initially contains a mixture of reagents. Let $N_{i,0}$ represent the number of molecules of reagent $i$ in that mixture. Suppose that there is only one reaction, $j$, that can occur, and further suppose that in the reaction expression, the stoichiometric coefficients, ${\nu }_{i,j}$, equal the number of $i$ molecules that participate each time molecular reaction event $j$ occurs. If $N_{j,net}$ is the net number of times molecular reaction event $j$ has occurred, then the change in the number of molecules of $i$ is equal to ${\nu }_{i,j}{N}_{j,net}$. (Remember, if $i$ is a reactant, its stoichiometric coefficient will be negative, so the change in the number of molecules will be negative, and the number of $i$ molecules will decrease, as it should. Similarly, if $i$ is a product, the stoichiometric coefficient will be positive and the number of $i$ molecules will increase.)

$$N_i-N_{i,0}={\nu }_{i,j}{N}_{j,net}$$

In real-world systems, the number of molecules is huge, and it is more convenient to work with moles. Dividing both sides of the equation by Avogadro’s number, $N_{Av}$, and rearranging, gives an expression for the moles of $i$ at any later time, given the number of moles of $i$ at the start of the process.

$$n_i=n_{i,0}+{\nu }_{i,j}\frac{N_{j,net}}{N_{Av}}$$

The net number of reaction events divided by Avogadro’s number is called the true extent of reaction j, ${\mathrm{\Xi }}_j$, Equation 3.1. Substitution in the preceding equation yields an expression for the moles of $i$, Equation 3.2.

$$\begin{array}{}\text{(3.1)}& {\mathrm{\Xi }}_j=\frac{N_{j,net}}{N_{Av}}\end{array}$$

$$\begin{array}{}\text{(3.2)}& n_i=n_{i,0}+{\nu }_{i,j}{\mathrm{\Xi }}_j\end{array}$$

That was for a closed system. For an open, steady-state system, the change occurs between the point where the moles enter the process and the point where they leave. When talking about material entering and leaving, it makes more sense to use molar flow rates instead of moles. Doing so leads to Equation 3.3 for an open, steady-state system. It turns out that these equations are accurate no matter how the reaction expression is written, as long as it is balanced. That is, the stoichiometric coefficients in the reaction expression do not have to equal the number of molecules that participate in the molecular reaction event, but the reaction expression does have to be properly balanced. In that case, it is preferable to define the apparent extent of reaction, ${\xi }_j$, as the net number of times the reaction, as written, occurs and to use the apparent extent of reaction in Equation 3.1 and Equation 3.2 instead of the true extent of reaction. Uses of the apparent extent of reaction are illustrated in Examples 3.4.1, 3.4.2, 3.4.3, and 3.4.4.

$$\begin{array}{}\text{(3.3)}& {\dot{n}}_i={\dot{n}}_{i,0}+{\nu }_i{\dot{\mathrm{\Xi }}}_j\end{array}$$

It’s also important to know that the extent of reaction is an extensive quantity. For a closed system it has units of moles, and for an open, steady-state system it has units of moles per time. In the latter case, the extent is related to the net number of times that the molecular reaction event occurs per unit time. Notice that all of the variable symbols in Equation 3.3 except the stoichiometric coefficient have dots over them, indicating they are flow quantities.

There’s a second quantity that is very useful for describing how far a reaction has progressed. It’s called the conversion. It can be either a fraction or a percentage, and it only applies to reactants. Quite simply, the fractional conversion of reactant $i$, denoted as $f_i$, is just the fraction of the starting amount of $i$ that has reacted. It is an intensive quantity, and it does not have any units. The defining equations for closed and open, steady state systems are shown in Equation 3.4 and Equation 3.5. It doesn’t make sense to talk about how much product was converted, so again, conversion only applies to reactants.

$$\begin{array}{}\text{(3.4)}& f_i=\frac{n_{i,0}-n_i}{n_{i,0}}\end{array}$$

$$\begin{array}{}\text{(3.5)}& f_i=\frac{{\dot{n}}_{i,in}-{\dot{n}}_i}{{\dot{n}}_{i,in}}\end{array}$$

$$ where $i$ is a reactant.

If the starting composition is stoichiometric, then the fractional conversion of every reactant will be the same. However, if one reactant is present in excess, its fractional conversion will be smaller than other reactants and will never reach 100%. For this reason, when the starting composition is not stoichiometric, it is preferred (but not required) that the conversion be specified for the limiting reactant because then the conversion will range from zero to 100%, assuming the reaction is irreversible. Of course, if the reaction is reversible, then even the limiting reactant will have a maximum conversion that is less than 100% because the system will reach equilibrium with measurable amounts of the reactants still remaining. The use of fractional conversion is illustrated in Example 3.4.1 and Example 3.4.4.

If it is desired to have a conversion that always ranges from 0 to 100%, the fraction of equilibrium conversion, Equation 3.6 for a closed system or Equation 3.7 for an open, steady-state system, can be used. The fraction of equilibrium conversion, $g_i$ is just the actual conversion relative to the conversion at equilibrium, so it, too, is an intensive variable. The vertical bar in these equations means “evaluated at”, and “eq” is shorthand for equilibrium, so together the vertical bar followed by a subscripted “eq” means the quantity it follows is evaluated at equilibrium. (See Appendix B for a complete list of the sign conventions and notation used in Reaction Engineering Basics.)

$$\begin{array}{}\text{(3.6)}& g_i=\frac{f_i}{f_{i,eq}}=\frac{\frac{n_{i,0}-n_i}{n_{i,0}}}{\frac{n_{i,0}-n_{i,eq}}{n_{i,0}}}=\frac{n_{i,0}-n_i}{n_{i,0}-n_{i,eq}}\end{array}$$

$$\begin{array}{}\text{(3.7)}& g_i=\frac{{\dot{n}}_{i,in}-{\dot{n}}_i}{{\dot{n}}_{i,in}-{\dot{n}}_{i,eq}}\end{array}$$

$$ where $i$ is a reactant.

The fraction of equilibrium conversion may not be convenient to use. For one thing, in order to use it, you have to do additional calculations to determine what the conversion is at equilibrium. In addition, the conversion at equilibrium changes with temperature which means a change in temperature would change the fraction of equilibrium conversion even if no reaction took place.

3.2 Describing Multiple-Reaction Systems

The preceding section considered systems where only one reaction was taking place. When multiple reactions are taking place in the system, additional or different variables are needed to fully describe the reacting system. Before considering such variables it is necessary to review what is meant by “mathematically independent” equations, or in this case, mathematically independent reactions.

To understand why, consider Reactions 3.8, 3.9 and 3.10. Only two of those three reactions are mathematically independent. The third is a linear combination of the others. This can be seen by noting, for example, that the third reaction is the sum of the first two. Put differently, the third reaction is equal to the net effect of the first two reactions occurring sequentially.

$$\begin{array}{}\text{(3.8)}& CO+3H_2\to C{H}_4+H_{2}O\end{array}$$

$$\begin{array}{}\text{(3.9)}& CO+H_{2}O\to C{O}_2+H_2\end{array}$$

$$\begin{array}{}\text{(3.10)}& 2CO+2H_2\to C{H}_4+C{O}_2\end{array}$$

If a system initially contained a mixture of CO and H₂, and then at some later time it was found to contain CO, H₂, CH₄, H₂O and CO₂, it would not be possible to know how much of the CH₄ was produced by Reaction 3.8 and how much was produced by Reaction 3.10. Similarly, it would not be possible to know how much of the CO₂ had been produced by Reaction 3.9 and how much was produced by Reaction 3.10. It is impossible to calculate the true extents of all three reactions if only the initial and final compositions of the system are known.

In general, when some of the reactions occurring in a system are linear combinations of other reactions, then a complete, mathematically independent subset of the reactions captures all of the stoichiometry. A complete mathematically independent subset is a group of reactions taken from all possible reactions where (1) none of the reactions in the subset are a linear combination of other reactions in the subset, and (2) each reaction that is not in the subset is a linear combination of the reactions that are in the subset. Consequently, for a closed system, the change in the number of moles of any reagent, $i$, is equal to the sum of the apparent extents, ${\xi }_{j^{\prime }}$ of the $j^{\prime }$ mathematically independent reactions multiplied by the stoichiometric coefficient, ${\nu }_{i,j^{\prime }}$, of reagent $i$ in independent reaction $j^{\prime }$. This leads to Equation 3.11 for a closed system or to Equation 3.12 for an open, steady-state system. In those equations $j^{\prime }$ indexes the reactions in a complete, mathematically independent subset of the reactions taking place in the system. Note that Equation 3.11 and Equation 3.12 also apply to a system where only one reaction is taking place. In that case, the summation reduces to a single term. Example 3.4.3 illustrates the identification of a complete mathematically independent subset of the reactions taking place in a system, and Example 3.4.4 illustrates the use of the apparent extents of those reactions in the calculation of system composition.

$$\mathrm{\Delta }{n}_i=n_i-n_{i,0}=\sum _{j^{\prime }}{\nu }_{i,j^{\prime }}{\xi }_{j^{\prime }}$$

$$\begin{array}{}\text{(3.11)}& n_i=n_{i,0}+\sum _{j^{\prime }}{\nu }_{i,j^{\prime }}{\xi }_{j^{\prime }}\end{array}$$

$$\begin{array}{}\text{(3.12)}& {\dot{n}}_i={\dot{n}}_{i,in}+\sum _{j^{\prime }}{\nu }_{i,j^{\prime }}{\dot{\xi }}_{j^{\prime }}\end{array}$$

$$ where $j^{\prime }$ indexes the reactions in a complete, mathematically
$$ independent subset of the reactions occurring in the system.

It is important to recognize that the extent of reaction, ${\mathrm{\Xi }}_j$, has physical significance. It is the number times that molecular reaction event $j$ has occurred, expressed in moles. In contrast, the apparent extent of reaction, ${\xi }_{j^{\prime }}$, does not have any physical significance. It is a mathematical construct that is useful for characterizing the composition of a reacting system. Indeed, for a given multiple reaction system, there are typically more than one complete, mathematically independent subsets, and any one of those subsets can be used in Equation 3.11 or Equation 3.12.

In order to use Equation 3.11 or Equation 3.12 it is necessary to determine the number of mathematically independent reactions and then to find at least one such subset. Appendix A presents a very brief and succinct description of how to do this, and Example 3.4.3 illustrates doing so.

If the number of mathematically independent reactions is $N$, then $N$ reaction progress variables are needed to fully characterize the system, given its initial composition. Thus, the apparent extents of a complete, mathematically independent subset of all possible reactions are sufficient to calculate the composition of a system, given its initial composition.

The conversion can be used when describing reaction progress of multiple reaction systems, but it must be used in conjunction with additional measures of reaction progress to fully specify the system composition. When using the conversion as a reaction progress variable in a multiple reaction system, it is necessary to specify the reactant it applies to. Furthermore, that reactant must be initially present in the system. As an example, again consider Reactions 3.8, 3.9 and 3.10. H₂O is a reactant in Reaction 3.9, but it might not initially be present in the system. Initially the system might only contain CO and H₂. H₂O would be generated by Reaction 3.8 and then could participate as a reactant in Reaction 3.9. If one tried to calculate the conversion of H₂O using Equation 3.4, the denominator would equal zero, leading to an infinite conversion. Therefore when multiple reactions are taking place, conversion is only defined for reactants that are initially present.

Yield is a reaction progress variable that specifies the amount of one particular product relative to the starting amount of one particular reactant. There are a few different ways to define yield, making it important to know which definition is being used whenever a yield is specified. When multiple reactions are occurring, it is very often true that some of the products are more desirable or valuable than other products. In some such systems, the desired product, $D$, is produced through a combination of reaction steps where it isn’t possible to identify a stoichiometric relationship between $D$ and any one reactant, $i$. Under these circumstances it is common to define the yield as shown in Equation 3.13 for a closed system and in Equation 3.14 for an open, steady-state process. Example 3.4.4 illustrates the use of yield as a reaction progress variable.

$$\begin{array}{}\text{(3.13)}& Y_{D/i}=\frac{n_D}{n_{i,0}}\end{array}$$

$$\begin{array}{}\text{(3.14)}& Y_{D/i}=\frac{{\dot{n}}_D}{{\dot{n}}_{i,in}}\end{array}$$

$$ where $D$ is a product and $i$ is a reactant.

In other reaction systems, the desired product, $D$, is produced from reactant, $i$, in only one reaction, $j$. When this is true, it may be preferable to account for the stoichiometry as shown in Equation 3.15 and Equation 3.16. By doing so, the resulting yield can be expressed as a percentage that ranges from 0 to 100% (assuming reaction $j$ is irreversible).

$$\begin{array}{}\text{(3.15)}& Y_{D/i}^{\prime }=\frac{{\nu }_{i,j}}{{\nu }_{D,j}}\frac{n_D}{n_{i,in}}\end{array}$$

$$\begin{array}{}\text{(3.16)}& Y_{D/i}^{\prime }=\frac{{\nu }_{i,j}}{{\nu }_{D,j}}\frac{{\dot{n}}_D}{{\dot{n}}_{i,in}}\end{array}$$

$$ where $D$ is a product and $i$ is a reactant.

While yield relates the amount of a product to the amount of a reactant, selectivity is a reaction progress variable that relates the amount of one product or group of products to a different product or group of products. If there is one desirable product, $D$, and one undesirable product, $U$, the selectivity for $D$ relative to $U$ can be defined as shown in Equation 3.17 and Equation 3.18. In other instances the selectivity of interest might involve groups of products. Examples include parafins vs. olefins, napthenes vs. aromatics or hydrocarbons with 6 or more carbon atoms (sometimes denotes as C⁶⁺) vs. hydrocarbons with 5 or fewer carbon atoms (sometimes denoted as C^5-). Example 3.4.1 and Example 3.4.4 illustrate the use of selectivity.

$$\begin{array}{}\text{(3.17)}& S_{D/U}=\frac{n_D}{n_U}\end{array}$$

$$\begin{array}{}\text{(3.18)}& S_{D/U}=\frac{{\dot{n}}_D}{{\dot{n}}_U}\end{array}$$

Both yield and selectivity can be used as overall measures or as instantaneous measures. The overall yield or selectivity is the value at the end of a closed process or at the outlet from an open, steady-state process, as defined in Equations 3.17 and 3.18. In a closed process, the yield or selectivity will typically change over time, and in an open, steady-state process it will vary with position within the reactor. An instantaneous yield or selectivity can be defined for that time or position within the reactor. The instantaneous yield or selectivity is related to the rates at which the amounts of the reagents are changing. Reaction rates are defined in Chapter 4, so defining equations for the instantaneous yield or selectivity will be presented in that chapter.

3.3 Mole Tables

A mole table is a convenient tool for summarizing the stoichiometric relationships among the chemical species present in a system. Mole tables can be particularly helpful for those just learning reaction engineering. They are useful when calculating the composition of a system, given its initial composition, and also when converting from one composition variable (concentration, mole fraction, partial pressure, moles, etc.) to another.

A mole table has three columns, as can be seen in Table 3.1. The first column lists each of the chemical species that are present in the system, including any inert reagents that do not participate in any chemical reactions. The second column lists the initial moles (for a closed system) or inlet molar flow rate (for an open, steady-state system) of each species. The third column lists expressions for the moles or molar flow rates at any later time or at any point beyond the inlet to an open system in terms of the apparent extents of reaction. In other words, each row in the third column contains either Equation 3.11 or Equation 3.12 for that row’s species, with the summation expanded and the values of the stoichiometric coefficients substituted. A final row in the table is for total moles or molar flow rate. It lists the sum of the initial values in the second column and the sum of the expressions for moles at any later time in the third column. The construction and use of a mole table is illustrated in Examples 3.4.1, 3.4.3, and 3.4.4.

3.4 Examples

3.4.1 Calculation of Final Composition in a Closed System using a Mole Table

A new catalyst is being developed with the hope of generating ethylene by the oxidative dehydrogenation of ethane, reaction (1). Reaction (2) also takes place at the operating conditions of 1 atm and 150 °C. The process starts with 90% ethane and 10% air (you may take air to contain 79% N₂ and 21% O₂) and ends with a 50% conversion of O₂ and a selectivity of 3 C₂H₄/CO₂. Construct a mole table for this system and use it to calculate the final mole fraction of CO₂.

$$\begin{array}{}\text{(1)}& 2C_2{H}_6+O_2\leftrightarrows 2C_2{H}_4+2H_{2}O\end{array}$$

$$\begin{array}{}\text{(2)}& 2C_2{H}_6+7O_2\leftrightarrows 4C{O}_2+6H_{2}O\end{array}$$

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What this problem asks me to do is something that is often done as one step in the completion of a reaction engineering assigment. The completion of reaction engineering assignments often requires the calculation of the composition of a system given information about its initial composition together with some measure of how much reaction has taken place. I often refer to this kind of task as a reaction progress task.

Almost inevitably, the completion of a reaction progress task entails first using the given information about the initial composition to calculate the initial molar amounts (for closed systems) or inlet molar flow rates (for open, steady-state systems) of every reagent present in the system. The next step involves using these initial amounts and the given measure of reaction progress to calculate the apparent extent of reaction. The amounts of all of the reagents corresponding to the given measure of reaction progress then can be calculated using the known initial composition and the apparent extent of reaction. Finally the desired quantities, e. g. concentrations, mole fractions, partial pressures, etc., corresponding to the given measure of reaction progress can be calculated using their defining equations.

I like to begin by summarizing the information provided in the problem statement. As I started to do so, I noticed that the system initially contains 10% air. Multiplying that 10% by the fraction of O₂ found in air (21%), gives the mole fraction of O₂. The same is true for N₂, except its fraction in air is 79%.

Given: $P=1\text{ atm}$, $T=150\text{ °C}$, $f_{O_2}=0.5$, $S_{C_2{H}_4/C{O}_2}=3.0$

Given Initial Composition: $y_{C_2{H}_6,0}$ = 0.9, $y_{O_2,0}$ = 0.21(0.1), $y_{N_2,0}$ = 0.79(0.1)

Requested: $y_{C{O}_2}$

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A mole table can be a useful tool when performing reaction progress tasks, particularly when one is just beginning to study reaction engineering. This problem tells me to construct a mole table, so I’ll do that first. I know that when I construct a mole table, I’ll need to use Equation 3.11. I also know that to use Equation 3.11, I need to identify a complete, mathematically independent subset of the reactions that are taking place. Since there are only two reactions taking place in this problem, I can do that by inspection. The reactions produce different products, so there is no way that one can be a simple multiple of the other, and therefore reactions (1) and (2) are mathematically independent. Consequently, in Equation 3.11, $j^{\prime }$ will range from 1 to 2.

$$n_i=n_{i,0}+\sum _{j^{\prime }}{\nu }_{i,j^{\prime }}{\xi }_{j^{\prime }}\Rightarrow n_{i,0}+{\nu }_{i,1}{\xi }_1+{\nu }_{i,2}{\xi }_2$$

With that knowledge, I can construct the mole table. There are six reagents in this system: C₂H₆, O₂, C₂H₄, H₂O, CO₂, and N₂. The first column of the mole table lists one of these species in each row. A row is added at the bottom listing “Total.” The second column lists variables representing the initial molar amounts of each reagent. I can use $n_{\text{total},0}$ to represent $n_{C_2{H}_6,0}$ + $n_{O_2,0}$ + $n_{C_2{H}_4,0}$ + $n_{H_{2}O,0}$ + $n_{C{O}_2,0}$ + $n_{N_2,0}$.

The equation above can then be used to construct the entries in each row of the third column of the mole table. For example, the stoichiometric coefficient of C₂H₆ in reaction (1), ${\nu }_{C_2{H}_6,1}$, is -2, and its stoichiometric coefficient in reaction (2), ${\nu }_{C_2{H}_6,2}$, is also equal to -2. Letting $i$ equal C₂H₆, substituting these values, and expanding the summation yields an expression for the moles of C₂H₆ at any later time.

$$\begin{array}{rl}{n}_{C_2{H}_6}& =n_{C_2{H}_6,0}+{\nu }_{C_2{H}_6,1}{\xi }_1+{\nu }_{C_2{H}_6,2}{\xi }_2\\ & =n_{C_2{H}_6,0}-2{\xi }_1-2{\xi }_2\end{array}$$

In a similar manner, the rows in the third column corresponding to each of the species are completed. For the last row, the expressions in all the other rows are simply added together to complete the mole table.

A mole table for this system is presented in Table 3.1.

Table 3.1: Mole Table for Example 3.4.1.

Species	Initial Amount	Later Amount
C2H6	$n_{C_2{H}_6,0}$	$n_{C_2{H}_6,0}-2{\xi }_1-2{\xi }_2$
O2	$n_{O_2,0}$	$n_{O_2,0}-{\xi }_1-7{\xi }_2$
C2H4	$n_{C_2{H}_4,0}$	$n_{C_2{H}_4,0}+2{\xi }_1$
H2O	$n_{H_{2}O,0}$	$n_{H_{2}O,0}+2{\xi }_1+6{\xi }_2$
CO2	$n_{C{O}_2,0}$	$n_{C{O}_2,0}+4{\xi }_2$
N2	$n_{N_2,0}$	$n_{N_2,0}$
Total	$n_{\text{total},0}$	$n_{\text{total},0}+{\xi }_1+{\xi }_2$

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The problem directs me to use Table 3.1 to calculate the final mole fraction of CO₂. Looking at the mole table and at the given information, I can see that there is no way to calculate the initial molar amounts that appear in the mole table. However, quantity that is provided in the problem statement is an intensive quantity. When no extensive quantities are provided, I can assume the value of any one extensive quantity and use it as the basis for my calculations. Since the initial mole fractions are given, it makes sense to choose the total moles initially present in the system as a basis.

Basis: $n_{\text{total},0}=1\text{ mol}$.

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As I noted above, the first step in completing a reacton progress task is to calculate the initial molar amounts. Having chosen a basis, this is straightforward. The initial percentages of ethane (90%) and air (10%) add up to 100%, meaning that there was no C₂H₄, H₂O, or CO₂ initially present. The total initial moles can then be calculated.

Necessary Equations

$$\begin{array}{}\text{(3)}& n_{C_2{H}_6,0}=y_{C_2{H}_6,0}{n}_{\text{total},0}\end{array}$$

$$\begin{array}{}\text{(4)}& n_{O_2,0}=y_{O_2,0}{n}_{\text{total},0}\end{array}$$

$$\begin{array}{}\text{(5)}& n_{N_2,0}=y_{N_2,0}{n}_{\text{total},0}\end{array}$$

$$\begin{array}{}\text{(6)}& n_{C_2{H}_4,0}=n_{H_{2}O,0}=n_{C{O}_2,0}=0\end{array}$$

$$\begin{array}{}\text{(7)}& n_{\text{total},0}=n_{C_2{H}_6,0}+n_{O_2,0}+n_{C_2{H}_4,0}+n_{H_{2}O,0}+n_{C{O}_2,0}+n_{N_2,0}\end{array}$$

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The second step in completing a reacton progress task is using these initial amounts and the given measures of reaction progress to calculate the apparent extents of reaction. In this problem there are two apparent extents of reaction, and the problem provides two measures of reaction progress, namely the O₂ conversion and the selectivity for C₂H₄ over CO₂. Thus, I can write the defining equations for these two measures of reaction progress.

$$\begin{array}{}\text{(8)}& f_{O_2}=\frac{n_{O_2,0}-n_{O_2}}{n_{O_2,0}}=\frac{n_{O_2,0}-(n_{O_2,0}-{\xi }_1-7{\xi }_2)}{n_{O_2,0}}=\frac{{\xi }_1+7{\xi }_2}{n_{O_2,0}}\end{array}$$

$$\begin{array}{}\text{(9)}& S_{C_2{H}_4/C{O}_2}=\frac{n_{C_2{H}_4}}{n_{C{O}_2}}=\frac{n_{C_2{H}_4,0}+2{\xi }_1}{n_{C{O}_2,0}+4{\xi }_2}\end{array}$$

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I can see that if I substitute the initial molar amount of O₂ into equations (8) and (9) along with the expressions for the molar amounts of O₂, C₂H₄, and CO₂ from the mole table, I will have two equations with two unknowns, ${\xi }_1$ and ${\xi }_2$. So at this point I can calculate the apparent extents of reactions (1) and (2).

All that remains to complete this reaction progress task is to calculate the requested final mole fraction of CO₂. To do that I simply need the defining equation for a mole fraction in a closed system, Equation 1.9.

$$\begin{array}{}\text{(10)}& y_{C{O}_2}=\frac{n_{C{O}_2}}{n_{\text{total}}}\end{array}$$

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I can substitute the expressions for the molar amount of CO₂ and the total molar amount from the mole table into equation (10). Then I can substitute the initial molar amounts and the apparent extents of reactions (1) and (2) and calculate the requested mole fraction of CO₂.

Calculations

1. Substitute the given values and the basis into all other equations.
2. Calculate $n_{C_2{H}_6,0}$, $n_{O_2,0}$, $n_{N_2,0}$, $n_{C_2{H}_4,0}$, $n_{H_{2}O,0}$, $n_{C{O}_2,0}$, and $n_{\text{total},0}$ using equations (3) through (7), and substitute the results into all other equations.
3. Solve equations (8) and (9) for ${\xi }_1$ and ${\xi }_2$.
4. Calculate the mole fraction of CO₂ using equation (10).

Result

Performing the calculations as described above shows that the final mole fraction of CO₂ is 0.00321.

3.4.2 Calculation of Outlet Composition from an Open System using the Conversion and Extent of Reaction

Suppose reaction (1) was studied in an open, steady-state, isothermal, isobaric system. The flow into the system contained 0.5 mol min⁻¹ of N₂O₅ and 0.5 mol min⁻¹ of N₂ at 600 K and 5 MPa. Generate an expression for the concentration of NO₂ in terms of the molar flow rate of N₂O₅, the inlet molar flows, the temperature, and the pressure. If the outlet flow contains 30% N₂, calculate the outlet concentration of NO₂.

$$\begin{array}{}\text{(1)}& 2N_2{O}_5\leftrightarrows 4N{O}_2+O_2\end{array}$$

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This is what I call a reaction progress task; it is something that is typically done as one step in the completion of a reaction engineering assighment. As I noted in the preceding example, the completion of reaction engineering assignments often requires the calculation of the composition of a system given information about its initial composition together with some measure of how much reaction has taken place. In this case an expression for the composition is requested instead of a value.

Generally here are the steps for completing a reaction progress task: (1) use the given information about the initial composition to calculate the initial molar amounts (for closed systems) or inlet molar flow rates (for open, steady-state systems) of every reagent present in the system, (2) use the resulting initial amounts and the given measure of reaction progress to calculate the apparent extent of reaction, (3) use the apparent extents of reaction to calculate (or here, write expressions for) the final molar amounts or flow rates, and (4) use those results to calculate the requested quantity (or write an expression for it).

I will start by summarizing the information profided in the problem statement. The problem only mentions N₂O₅ and N₂ as flowing into the system, so I’ll assume the inlet flows of NO₂ and O₂ are zero.

Given: $T=600\text{ K}$, $P=5\text{ MPa}$, $y_{N_2,out}=0.3$.

Given Initial Composition: ${\dot{n}}_{N_2{O}_5,in}=0.5{\text{ mol min}}^{-1}$, ${\dot{n}}_{N_2,in}=0.5{\text{ mol min}}^{-1}$, ${\dot{n}}_{N{O}_2,in}=0$, ${\dot{n}}_{O_2,in}=0$.

Requested: an expression for $C_{N{O}_2}$ and its value.

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I could write a mole table, but with only one reaction taking place, I think I can get by without one. Expressions for the molar flow rates of N₂O₅, NO₂, and O₂ are straightforward, Equation 3.12. Looking at the reaction, I can see that each time the reaction occurs as written the total number of moles increases by 3, from 2 to 5, so ${\dot{n}}_{\text{total}}={\dot{n}}_{\text{total},in}+3\dot{\xi }$.

Necessary Equations

$$\begin{array}{}\text{(2)}& {\dot{n}}_{N_2{O}_5}={\dot{n}}_{N_2{O}_5,in}-2\dot{\xi }\end{array}$$

$$\begin{array}{}\text{(3)}& {\dot{n}}_{N{O}_2}={\dot{n}}_{N{O}_2,in}+4\dot{\xi }\end{array}$$

$$\begin{array}{}\text{(4)}& {\dot{n}}_{O_2}={\dot{n}}_{O_2,in}+\dot{\xi }\end{array}$$

$$\begin{array}{}\text{(5)}& {\dot{n}}_{N_2}={\dot{n}}_{N_2,in}\end{array}$$

$$\begin{array}{}\text{(6)}& {\dot{n}}_{\text{total}}={\dot{n}}_{\text{total},in}+3\dot{\xi }\end{array}$$

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In this problem, two measures of the final composition are mentioned. The first is the concentration of NO₂, and the other is the mole fraction of N₂. I’ll write the defining equations for both, and for the concentration, I’ll also express the volumetric flow rate in terms of the molar amounts using the ideal gas equation. Then, I’ll substitute equations (2) through (6) for all of the molar flow rates.

$$\begin{array}{}\text{(7)}& C_{N{O}_2}=\frac{{\dot{n}}_{N{O}_2}}{\dot{V}}=\frac{P{\dot{n}}_{N{O}_2}}{{\dot{n}}_{total}RT}=\frac{4P\dot{\xi }}{RT({\dot{n}}_{\text{total},in}+3\dot{\xi })}\end{array}$$

$$\begin{array}{}\text{(8)}& y_{N_2}=\frac{{\dot{n}}_{N_2}}{{\dot{n}}_{total}}=\frac{{\dot{n}}_{N_2,in}}{{\dot{n}}_{total,in}+3\dot{\xi }}\end{array}$$

Expression for $C_{N{O}_2}$

The problem asks for an expression for the concentration of NO₂ in terms of the molar flow rate of N₂O. Equation (2) can be solved to obtain an expression for the apparent extent of reaction in terms of the molar flow rate of N₂O. The result can be substituted into equation (7), yielding the requested expression for the concentration of NO₂, equation (9).

$$\dot{\xi }=\frac{{\dot{n}}_{N_2{O}_5,in}-{\dot{n}}_{N_2{O}_5}}{2}$$

$$\begin{array}{}\text{(9)}& \begin{array}{rl}{C}_{N{O}_2}& =\frac{4P\dot{\xi }}{RT({\dot{n}}_{\text{total},in}+3\dot{\xi })}\\ & =\frac{4P}{RT}\frac{\frac{{\dot{n}}_{N_2{O}_5,in}-{\dot{n}}_{N_2{O}_5}}{2}}{{\dot{n}}_{N_2{O}_5,in}+{\dot{n}}_{N_2,in}+3\frac{{\dot{n}}_{N_2{O}_5,in}-{\dot{n}}_{N_2{O}_5}}{2}}\\ & =\frac{4P}{RT}\frac{({\dot{n}}_{N_2{O}_5,in}-{\dot{n}}_{N_2{O}_5})}{(5{\dot{n}}_{N_2{O}_5,in}+2{\dot{n}}_{N_2,in}-3{\dot{n}}_{N_2{O}_5})}\end{array}\end{array}$$

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To find the value of the apparent extent of reaction, I need to use the given measure of reaction progress, which in this case is the final mole fraction of N₂.

Calculation of $C_{N{O}_2}$

1. Subtitute the given and known quantities into all equations.
2. Solve equation (8) to find the value of $\dot{\xi }$.
3. Use the resulting apparent extent of reaction to calculate $C_{N{O}_2}$ using equation (7).

Results

The requested expression for the concentration of NO₂ is given in equation (9). Performing the calculations as described above yields a value of 0.535 mol/L for the concentration of NO₂ at the time when the outlet contains 30% N₂.

3.4.3 Identification of a Complete, Mathematically Independent Subset of a Group of Reactions

A chemical system initially contains CO and H₂. At system conditions, reactions (1) through (6) occur. Identify a complete, mathematically independent subset of those reactions and construct a mole table for the system.

$$\begin{array}{}\text{(1)}& CO+2H_2\leftrightarrows C{H}_{3}OH\end{array}$$

$$\begin{array}{}\text{(2)}& CO+3H_2\leftrightarrows C{H}_4+H_{2}O\end{array}$$

$$\begin{array}{}\text{(3)}& C{H}_4+H_{2}O\leftrightarrows C{H}_{3}OH+H_2\end{array}$$

$$\begin{array}{}\text{(4)}& CO+H_{2}O\leftrightarrows C{O}_2+H_2\end{array}$$

$$\begin{array}{}\text{(5)}& C{O}_2+4H_2\leftrightarrows C{H}_4+2H_{2}O\end{array}$$

$$\begin{array}{}\text{(6)}& C{O}_2+3H_2\leftrightarrows C{H}_{3}OH+H_{2}O\end{array}$$

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What this problem asks me to do is something that is sometimes done as one step in the completion of a reaction engineering assigment. I often refer to this kind of task as an independent reactions task, since completing it involves identifying a complete, mathematically independent subset of the reactions that are taking place.

I know from math courses I’ve taken that identifying mathematically independent equations involves constructing a matrix, and the same is true for identifying independent chemical reactions. Each column in the matrix corresponds to one of the reagants and each row corresponds to one of the reactions. The matrix element at any one row and column is the stoichiometric coefficent of the reagent corresponding to that column in the reaction corresponding to that row.

Here the reaction matrix will have six rows, one for each reaction, and six columns, one for each reagent. I’ll let the first column corresponds to CO, the second to H₂, the third to CH₃OH, the fourth to CH₄, the fifth to H₂O, and the last to CO₂. Letting the first row correspond to reaction (1), I see that the stoichiometric coefficient of CO in that reaction is -1, and so there is a -1 in the first column of row 1 (see equation (7) below). The stoichiometric coefficeint of H₂ in the first reaction is -2, so there is a -2 in the second column of row 1. The stoichiometric coefficient of CH₃OH in reaction (1) is +1, so there is a 1 in the third column of row 1. CH₄, H₂O, and CO₂ do not appear in reaction (1), so there are zeros in the fourth, fifth and sixth columns of row 1. The remaining rows are filled in analogously.

The rank of this matrix will then equal the number of mathematically independent reactions. I’ll use computer software to find the rank.

Reaction Matrix

Equation (7) shows a reaction matrix for this problem. The first row corresponds to reaction (1), the second row to reaction (2), and so on. The first column corresponds to CO, the second to H₂, the third to CH₃OH, the fourth to CH₄, the fifth to H₂O, and the last to CO₂.

$$\begin{array}{}\text{(7)}& \left[\begin{array}{cccccc}-1& -2& 1& 0& 0& 0\\ -1& -3& 0& 1& 1& 0\\ 0& 1& 1& -1& -1& 0\\ -1& 1& 0& 0& -1& 1\\ 0& -4& 0& 1& 2& -1\\ 0& -3& 1& 0& 1& -1\end{array}\right]\end{array}$$

Mathematics software can be used to find that the rank of the matrix in equation (7) is equal to 3. This means that there will be 3 equations in any complete, mathematically independent subset of reactions (1) through (6).

In the matrix shown in equation (8) the first row corresponds to reaction (1) and the second row to reaction (2). The rank of this matrix is found to equal 2, so reactions (1) and (2) are mathematically independent.

$$\begin{array}{}\text{(8)}& \left[\begin{array}{cccccc}-1& -2& 1& 0& 0& 0\\ -1& -3& 0& 1& 1& 0\end{array}\right]\end{array}$$

In the matrix shown in equation (9) a third row, corresponding to reaction (3), was added to the matrix in equation (8). The rank of the matrix in equation (9) is 2. This means that reaction (3) is not independent; it is a linear combinations of reactions (1) and (2).

$$\begin{array}{}\text{(9)}& \left[\begin{array}{cccccc}-1& -2& 1& 0& 0& 0\\ -1& -3& 0& 1& 1& 0\\ 0& 1& 1& -1& -1& 0\end{array}\right]\end{array}$$

Reaction (3) was removed from the matrix and replaced with reaction (4), resulting in the matrix shown in equation (10). The rank of the matrix shown in equation (10) is 3, so the reactions corresponding to the rows of that matrix, in this case reactions (1), (2), and (4), represent a complete, mathematically independent subset of reactions (1) through (6).

$$\begin{array}{}\text{(10)}& \left[\begin{array}{cccccc}-1& -2& 1& 0& 0& 0\\ -1& -3& 0& 1& 1& 0\\ -1& 1& 0& 0& -1& 1\end{array}\right]\end{array}$$

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Having identified a complete, mathematically independent subset of the reactions, I can now construct a mole table as requested in the problem statement. There are six reagents in this system: CO, H₂, CH₃OH, CH₄, H₂O, and CO₂. As in Example 3.4.1, the first column of the mole table lists these reagents along with a row at the bottom for “Total.” The second column lists variables representing the initial molar amounts of each reagent and the third column lists expressions for the molar amounts of the reagents at any later time in terms of the apparent extents of the independent reactions. I need to use Equation 3.11 for the entries in the third column of of the mole table

$$n_i=n_{i,0}+\sum _{j^{\prime }}{\nu }_{i,j^{\prime }}{\xi }_{j^{\prime }}$$

Final Answer

While there are six reactions taking place in this system, only three reactions are mathematically independent. Therefore the mole table only uses the apparent extents of three reactions. Knowing that reactions (1), (2), and (4), represent a complete, mathematically independent subset of reactions (1) through (6), the apparent extents of those reactions were used to construct Table 3.2.

Table 3.2: Mole Table for Example 3.4.3

Species	Initial Amount	Later Amount
CO	$n_{CO,0}$	$n_{CO,0}-{\xi }_1-{\xi }_2-{\xi }_4$
H2	$n_{H_2,0}$	$n_{H_2,0}-2{\xi }_1-3{\xi }_2+{\xi }_4$
CH3OH	$n_{C{H}_{3}OH,0}$	$n_{C{H}_{3}OH,0}+{\xi }_1$
CH4	$n_{C{H}_4,0}$	$n_{C{H}_4,0}+{\xi }_2$
H2O	$n_{H_{2}O,0}$	$n_{H_{2}O,0}+{\xi }_2-{\xi }_4$
CO2	$n_{C{O}_2,0}$	$n_{C{O}_2,0}+{\xi }_4$
Total	$n_{\text{total},0}$	$n_{\text{total},0}-2{\xi }_1-2{\xi }_2$

Notes

1. Reactions (1), (2), and (4) are not the only complete, mathematically independent subset. For example, reactions (1), (4), and (5) is also a complete, mathematically independent subset and the mole table could have been written using ${\xi }_1$, ${\xi }_4$, and ${\xi }_5$.
2. The solution presented here used a reaction matrix. Appendix A describes identifying independent reactions using the stoichiometric coefficient matrix where the columns correspond to the reactions and the rows to the reagents. The reaction matrix is the transpose of the stoichiometric coefficient matrix, and the rank of a matrix equals the rank of its transpose. For that reason, either the method used here or the method described in Appendix A can be used.

3.4.4 Calculation of the Final Composition of a Closed System using the Conversion, Yield, Selectivity and Apparent Extents of Reaction

A chemical system initially contains 34 mol CO and 66 mol H₂. At system conditions, reactions (1) through (6) occur. At the end of the process, the CO conversion is 40%, the yield of CH₃OH is 10.7% (mol CH₃OH per initial mol CO), and the selectivity for methanol vs. methane is 0.38 mol CH₃OH per mol CH₄. Calculate the number of moles of each species at the end of the process.

$$\begin{array}{}\text{(1)}& CO+2H_2\leftrightarrows C{H}_{3}OH\end{array}$$

$$\begin{array}{}\text{(2)}& CO+3H_2\leftrightarrows C{H}_4+H_{2}O\end{array}$$

$$\begin{array}{}\text{(3)}& C{H}_4+H_{2}O\leftrightarrows C{H}_{3}OH+H_2\end{array}$$

$$\begin{array}{}\text{(4)}& CO+H_{2}O\leftrightarrows C{O}_2+H_2\end{array}$$

$$\begin{array}{}\text{(5)}& C{O}_2+4H_2\leftrightarrows C{H}_4+2H_{2}O\end{array}$$

$$\begin{array}{}\text{(6)}& C{O}_2+3H_2\leftrightarrows C{H}_{3}OH+H_{2}O\end{array}$$

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This is a reaction progress task. Generally here are the steps for completing a reaction progress task: (1) use the given information about the initial composition to calculate the initial molar amounts of every reagent present in the system, (2) use the resulting initial amounts and the given measures of reaction progress to calculate the apparent extents of reaction, (3) use the apparent extents of reaction to calculate the final molar amounts, and (4) use those results to calculate the requested quantities.

I’ll begin by summarizing the information provided in the problem statement. The system described here is the same as in Example 3.4.3, so I can use the mole table from that example, Table 3.2.

Given: $f_{CO}=0.4$, $Y_{C{H}_{3}OH}=0.107$, and $S_{C{H}_{3}OH/C{H}_4}=0.38$.

Given Initial Composition: $n_{CO,0}=34\text{ mol}$, $n_{H_2,0}=66\text{ mol}$, $n_{C{H}_{3}OH,0}=n_{C{H}_4,0}=n_{H_{2}O,0}=n_{C{O}_2,0}=0\text{ mol}$.

Requested: $n_{CO}$, $n_{H_2}$, $n_{C{H}_{3}OH}$, $n_{C{H}_4}$, $n_{H_{2}O}$, and $n_{C{O}_2}$.

Mole table (from Section 3.4.3)

Species	Initial Amount	Later Amount
CO	$n_{CO,0}$	$n_{CO,0}-{\xi }_1-{\xi }_2-{\xi }_4$
H2	$n_{H_2,0}$	$n_{H_2,0}-2{\xi }_1-3{\xi }_2+{\xi }_4$
CH3OH	$n_{C{H}_{3}OH,0}$	$n_{C{H}_{3}OH,0}+{\xi }_1$
CH4	$n_{C{H}_4,0}$	$n_{C{H}_4,0}+{\xi }_2$
H2O	$n_{H_{2}O,0}$	$n_{H_{2}O,0}+{\xi }_2-{\xi }_4$
CO2	$n_{C{O}_2,0}$	$n_{C{O}_2,0}+{\xi }_4$
Total	$n_{\text{total},0}$	$n_{\text{total},0}-2{\xi }_1-2{\xi }_2$

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Knowing the initial molar amounts, I can proceed to calculate the extents of reaction using the given measures of reaction progress. Here I’m given the conversion, yield and selectivity. Starting from their definitions (Equations 3.4, 3.13, and 3.17) the given values can be related to the apparent extents of reaction using the mole table, Table 3.2.

Necessary Equations

$$\begin{array}{}\text{(7)}& \begin{array}{rl}{f}_{CO}& =\frac{n_{CO,0}-n_{CO}}{n_{CO,0}}\\ & =\frac{n_{CO,0}-(n_{CO,0}-{\xi }_1-{\xi }_2-{\xi }_4)}{n_{CO,0}}\\ & =\frac{{\xi }_1+{\xi }_2+{\xi }_4}{n_{CO,0}}\end{array}\end{array}$$

$$\begin{array}{}\text{(8)}& Y_{C{H}_{3}OH}=\frac{n_{C{H}_{3}OH}}{n_{CO,0}}=\frac{n_{C{H}_{3}OH,0}+{\xi }_1}{n_{CO,0}}\end{array}$$

$$\begin{array}{}\text{(9)}& S_{C{H}_{3}OH/C{H}_4}=\frac{n_{C{H}_{3}OH}}{n_{C{H}_4}}=\frac{n_{C{H}_{3}OH,0}+{\xi }_1}{n_{C{H}_4,0}+{\xi }_2}\end{array}$$

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There are three unknown quantities in equations (7) through (9), so they can be solved to find the values of ${\xi }_1$, ${\xi }_2$, and ${\xi }_4$. The resulting apparent extents of reaction can then be substituted in the expressions in the third column of Table 3.2 to compute the molar amounts of each of the reagents.

Calculations

1. Subtitute the given and known quantities into all equations.
2. Solve the equations (7) through (9) for ${\xi }_1$, ${\xi }_2$, and ${\xi }_4$.
3. Use the resulting apparent extents of reaction to alculate $n_{CO}$, $n_{H_2}$, $n_{C{H}_{3}OH}$, $n_{C{H}_4}$, $n_{H_{2}O}$, and $n_{C{O}_2}$ using the expressions in the third column of the mole table.

Results

Performing the calculations as described yields the final molar amounts shown in Table 3.3.

Table 3.3: Final Molar Amount of Each Reagent

Reagent	Final Moles
CO	20.400
H2	30.400
CH3OH	3.640
CH4	9.570
H2O	9.190
CO2	0.388

3.5 Symbols Used in Chapter 3

Symbol	Meaning
$eq$	Subscript denoting an equilibrium value.
$f_i$	Fractional conversion of reactant $i$.
$g_i$	Fraction of equilibrium conversion of reactant $i$.
$i$	Subscript denoting a fluid phase reagent.
$j$	Subscript denoting a reaction occurring in the system.
$j^{\prime }$	Subscript denoting a reaction from a complete, mathematically independent subset of the reactions occurring in the system.
$n$	Number of moles; a subscripted $i$ denotes the number of moles of reagent $i$; an additional subscripted $0$ denotes the initial moles.
$\dot{n}$	Molar flow rate; a subscripted $i$ denotes the molar flow rate of reagent $i$; an additional subscripted $in$ denotes the inlet molar flow rate.
$y_i$	Gas phase mole fraction of reagent $i$; an additional subscripted $0$ denotes the initial mole fraction; an additional subscripted $in$ denotes the inlet mole fraction.
$C_i$	Molar concentration of reagent $i$.
$N_{Av}$	Avogadro’s number.
$N_i$	Number of molecules of reagent $i$; an additional subscripted $0$ denotes the initial number of molecules of $i$.
$N_{j,net}$	Net number of times reaction event $j$ has occurred.
$P$	Pressure.
$R$	Ideal gas constant.
$S_{D/U}$	Overall selectivity for product or products, $D$, relative to product or products, $U$.
$T$	Temperature.
$\dot{V}$	Volumetric flow rate.
$Y_{D/i}$	Overall yield of product $D$ from reactant $i$.
$Y_{D/i}^{\prime }$	Overall stoichiometric yield of product $D$ from reactant $i$.
${\nu }_{i,j}$	Stoichiometric coefficient of reagent $i$ in reaction $j$; if only one reaction is taking place the index, $j$ is optional.
${\xi }_{j\text{-}ind}$	Apparent extent of reaction $j\text{-}ind$.
${\dot{\xi }}_{j\text{-}ind}$	Apparent extent of reaction $j\text{-}ind$ in an open, steady-state system.
${\mathrm{\Xi }}_j$	True extent of reaction $j$.
${\dot{\mathrm{\Xi }}}_j$	True extent of reaction $j$ in an open, steady-state system.