7.2: Multiple Reactions - Conversion, Selectivity and Yield

Most chemical reaction systems of industrial interest produce one or more primary or desirable products and one or more secondary or undesirable products. For example, benzene ($$\ce{C6H6}$$) and propylene ($$\ce{C3H6}$$) undergo reaction in the presence of a catalyst to form both the desired product, isopropyl benzene or cumene ($$\ce{C9H12}$$) and the undesired product, p-diisopropyl benzene ($$\ce{C12H18}$$). This situation is illustrated in Figure $$\PageIndex{1}$$ in which the reactants, benzene and propylene, appear in the exit stream because the reaction does not go to completion. In addition, the undesirable product, p-diisopropyl benzene appears in the exit stream1.

The analysis of this reactor is based on Axioms I and II. For a single entrance (Stream #1) and a single exit (Stream #2), Axiom I takes the form

Axiom I: $- (\dot{M}_{A} )_{1} + (\dot{M}_{A} )_{2} =\mathscr{R}_{ A} , \quad A = 1,2,3,4 \label{6}$

For a system containing only two atomic species, the global form of Axiom II is given by

Axiom II: $\sum_{A = 1}^{A = 4}N_{JA} \mathscr{R}_{ A} =0 , \quad J = 1, 2 \label{7}$

In both the experimental study of the reactor shown in Figure $$\PageIndex{1}$$ and in the operation of that reactor, it is useful to have a number of defined quantities that characterize the performance. The first of these defined quantities is called the conversion which is given by

$\text{ Conversion of reactant }A=\frac{ \text{Total molar rate of consumption of species }A}{\text{ Molar flow rate of species }A\text{ in the feed}} \label{8}$

Since $$\mathscr{R}_{ A}$$ represents the net molar rate of production of species $$A$$, we can use Equation \ref{6} to express the conversion as

$\text{ Conversion of reactant }A=\frac{- \mathscr{R}_{ A} }{ (\dot{M}_{A} )_{1} } \label{9}$

An experimental determination of the conversion of reactant $$A$$ requires the measurement of the molar flow of species $$A$$ into and out of the reactor illustrated in Figure $$\PageIndex{1}$$. If the reaction of benzene and propylene does not go to completion, one might obtain a result given by

$\text{ Conversion of } \ce{C6H6} =\frac{- \mathscr{R}_{ \ce{C6H6} } }{ (\dot{M}_{ \ce{C6H6}} )_{1} } =\frac{(\dot{M}_{ \ce{C6H6}} )_{1} - (\dot{M}_{ \ce{C6H6}} )_{2} }{(\dot{M}_{ \ce{C6H6} })_{1} } =0.68 \label{10}$

This indicates that 68% of the incoming benzene is consumed in the reaction, but it does not indicate how much of this benzene reacts to form the desired product, cumene, and how much reacts to form the undesired product, p-diisoproply benzene. In an efficient reactor, the conversion would be close to one and the amount of undesired product would be small.

A second defined quantity, the selectivity, indicates how one product (cumene for example) is favored over another product (p-diisopropyl benzene for example), and this quantity is defined by

$\text{ Selectivity of D/U}=\frac{ \text{Total molar rate of production of }\mathbf{Desired}\text{ product}}{\text{ Total molar rate of production of } \mathbf{Undesired} \text{ product}} =\frac{\mathscr{R}_{ D} }{\mathscr{R}_{ U} } \label{11}$

For the system illustrated in Figure $$\PageIndex{1}$$, the desired product is cumene ($$\ce{C9H12}$$), and the undesired product is p-diisoproply benzene ($$\ce{C12H18}$$). If the selectivity for this pair of molecules is 0.85 we have

$\text{ Selectivity of }\left(\ce{C9H12} {\bf /} \ce{C12H18} \right)=\frac{\mathscr{R}_{ \ce{C9H12} } }{\mathscr{R}_{ \ce{C12H18} } } =0.85 \label{12}$

and this suggests rather poor performance of the reactor since the rate of production of the undesirable product is greater than the rate of production of the desired product. In an efficient reactor, the selectivity would be large compared to one.

A third defined quantity is the yield of a reactor which is the ratio of the rate of production of a product to the rate of consumption of a reactant. We express the yield for a general case as

$\text{ Yield of }\left(A {\bf /} B\right)=\frac{ \text{Total rate of production of species }A}{\text{ Total rate of consumption of species }B} =\frac{ \mathscr{R}_{ A} }{- \mathscr{R}_{ B} } \label{13}$

If we choose the product to be p-diisopropyl benzene and the reactant to be propylene, the yield takes the form

$\text{ Yield of }\left(\ce{C12H18} {\bf /} \ce{C3H6} \right)=\frac{ \text{Total rate of production of p-diisopropyl benzene }}{\text{ Total rate of consumption of propylene}} =\frac{\mathscr{R}_{ \ce{C12H18} } }{- \mathscr{R}_{ \ce{C3H6} } } \label{14}$

If the yield of p-diisopropyl benzene is 0.15 we have the result given by

$\text{ Yield of }\left(\ce{C12H18} / \ce{C3H6} \right)=\frac{\mathscr{R}_{ \ce{C12H18} } }{- \mathscr{R}_{ \ce{C3H6} } } =\frac{\mathscr{R}_{ \ce{C12H18} } }{(\dot{M}_{ \ce{C3H6} } )_{1} - (\dot{M}_{ \ce{C3H6} } )_{2} } =0.15 \label{15}$

In an efficient reactor, the yield of an undesirable product would be small compared to one, while the yield of a desirable product would be close to one.

The conversion, selectivity and yield of a reactor must be determined experimentally and these quantities will depend on the operating conditions of the reactor (i.e., temperature, pressure, type of catalyst, etc.). In addition, the value of these quantities will depend on their definitions, and within the chemical engineering literature one encounters a variety of definitions. To avoid errors, the definitions of conversion, selectivity and yield must be given in precise form and we have done that in the preceding paragraphs.

Example $$\PageIndex{1}$$: Production of ethylene

Ethylene ($$\ce{C2H4}$$) is one of the most useful molecules in the petrochemical industry2 since it is the building block for poly-ethylene, ethylene glycol, and many other chemical compounds used in the production of polymers. Ethylene can be produced by catalytic dehydrogenation of ethane ($$\ce{C2H6}$$) as shown in Figure $$\PageIndex{2}$$. There we have indicated that the stream leaving the reactor contains the desired product, ethylene ($$\ce{C2H4}$$) in addition to hydrogen ($$\ce{H2}$$), methane ($$\ce{CH4}$$), propylene ($$\ce{C3H6}$$) and some un-reacted ethane ($$\ce{C2H6}$$). As in every example of this type, the reader needs to consider the principle of stoichiometric skepticism that is discussed in Sec. 6.1.1 since small amounts of unidentified molecular species are always present in the output of a reactor.

The experiment illustrated in Figure $$\PageIndex{2}$$ has been performed in which the molar flow rate entering the reactor (Stream #1) is 100 mol/s of ethane. From measurements of the effluent stream (Stream #2), the following information regarding the conversion of ethane, the selectivity of ethylene relative to propylene, and the yield of ethylene are available:

${ C}=\text{ Conversion of }\left(\ce{C2H6} \right)=\frac{-\mathscr{R}_{ \ce{C2H6} } }{(\dot{M}_{ \ce{C2H6} })_1 } =0.2 \label{1a}\tag{1}$

${ S}=\text{Selectivity of }\left(\ce{C2H4} / \ce{C3H6} \right)=\frac{\mathscr{R}_{ \ce{C2H4} } }{\mathscr{R}_{ \ce{C3H6} } } =5 \label{2a}\tag{2}$

${ Y}=\text{ Yield of }\left(\ce{C2H4} / \ce{C2H6} \right)=\frac{\mathscr{R}_{ \ce{C2H4} } }{- \mathscr{R}_{ \ce{C2H6} } } =0.75 \label{3a}\tag{3}$

The global rates of production of individual species are constrained by the stoichiometry of the system that can be expressed in terms of Axiom II. The chemical composition matrix for this reaction is given by

$\text{ Molecular Species}\to \ce{H2} \quad \ce{CH4} \quad \ce{C2H6} \quad \ce{C2H4} \quad \ce{C3H6} \\ \begin{matrix} {carbon} \\ {hydrogen} \end{matrix}\begin{bmatrix} { 0} & { 1} & { 2} & {2} & { 3 } \\ { 2} & { 4} & { 6} & {4} & { 6 } \end{bmatrix} \label{4a}\tag{4}$

and the elements of this matrix are the entrees in $$\left[N_{JA} \right]$$ that can be expressed as

$\left[N_{JA} \right]=\begin{bmatrix} {0} & {1} & {2} & {2} & {3} \\ {2} & {4} & {6} & {4} & {6} \end{bmatrix} \label{5a}\tag{5}$

Use of this atomic matrix with Axiom II as given by Eq. $$(6.2.10)$$ leads to

Axiom II: $\begin{bmatrix} {0} & {1} & {2} & {2} & {3} \\ {2} & {4} & {6} & {4} & {6} \end{bmatrix}\begin{bmatrix} {\mathscr{R}_{ \ce{H2} } } \\ {\mathscr{R}_{ \ce{CH4} } } \\ {\mathscr{R}_{ \ce{C2H6} } } \\ {\mathscr{R}_{ \ce{C2H4} } } \\ {\mathscr{R}_{ \ce{C3H6} } } \end{bmatrix}=\begin{bmatrix} {0} \\ {0} \end{bmatrix} \label{6a}\tag{6}$

At this point the atomic matrix can be expressed in row reduced echelon form leading to the global pivot theorem given by Eq. $$(6.4.8)$$. We express this result in the form

Global Pivot Theorem: $\begin{bmatrix} {\mathscr{R}_{ \ce{H2} } } \\ {\mathscr{R}_{ \ce{CH4} } } \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ -2 & -2 & -3 \end{bmatrix} \begin{bmatrix} {\mathscr{R}_{ \ce{C2H6} } } \\ {\mathscr{R}_{ \ce{C2H4} } } \\ {\mathscr{R}_{ \ce{C3H6} } } \end{bmatrix} \label{7a}\tag{7}$

and carry out the matrix multiplication to obtain

$\mathscr{R}_{ \ce{H2} } = \mathscr{R}_{ \ce{C2H6} } + 2\mathscr{R}_{ \ce{C2H4} } + 3\mathscr{R}_{ \ce{C3H6} } \label{8a}\tag{8}$

$\mathscr{R}_{ \ce{CH4} } = -2 \mathscr{R}_{ \ce{C2H6} } - 2\mathscr{R}_{ \ce{C2H4} } - 3 \mathscr{R}_{ \ce{C3H6} } \label{9a}\tag{9}$

The five net rates of production that appear in Eqs. \ref{6a} though \ref{9a} are represented in the degree of freedom analysis given in Table $$\PageIndex{1}$$. There we see that there are zero degrees of freedom and we have a solvable problem.

Table $$\PageIndex{1}$$: Degrees-of-freedom for production of ethylene from ethane

Stream & Net Rate of Production Variables compositions (5 species & 2 streams) $$N \times M = 5 \times 2 =10$$ $$M=2$$ $$N=5$$ 17 $$N=5$$ $$M=2$$ $$T=2$$ Eqs. \ref{7a} and \ref{8a} 9 4 Stream #1 1 Stream #1 0 3 Eqs. \ref{1a}, \ref{2a}, and \ref{3a} 8 0

At this point we direct our attention to Equation \ref{8a} and make use of the information provided in Eqs. \ref{1a}, \ref{2a}, and \ref{3a} in order to express the net molar rate of production of hydrogen as

$\mathscr{R}_{ \ce{H2} } = \left[-1 + 2Y + 3\frac{ Y}{ S} \right] C(\dot{M}_{ \ce{C2H6} } )_{1} \label{10a}\tag{10}$

Moving on to Equation \ref{9a} we again utilize Eqs. \ref{1a}, \ref{2a}, and \ref{3a} with Equation \ref{9a} in order to obtain.

$\mathscr{R}_{ \ce{CH4} } =\left[2{ C} - 2{ Y} - \frac{ 3Y}{ S} \right] C(\dot{M}_{ \ce{C2H6} } )_{1} \label{11a}\tag{11}$

From the experimental values of the conversion, yield and selectivity indicated by Eqs. \ref{1a} through \ref{3a}, along with the molar flow rate of Stream #1 given by,

$(\dot{M}_{ \ce{C2H6} } )_{1} ={ 100 \ mol/s} \label{13a}\tag{12}$

we can solve Eqs. \ref{10a} and \ref{11} to determine the global rates of production for methane and hydrogen.

$\mathscr{R}_{ \ce{H2} } = 19.0 \ { mol/s}, \quad \mathscr{R}_{ \ce{CH4} } = { 1.0 \ mol/s} \label{14a}\tag{13}$

Given the global rates of production for $$\ce{CH4}$$ and $$\ce{H2}$$, the values for the global rates of production of the other species can be easily calculated and this is left as an exercise for the student.