- #1

gfd43tg

Gold Member

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Hello,

I went through a problem in my reaction engineering textbook, and solved them correctly by basically mindlessly plugging into the equation, but now I wanted to get some help interpreting the results.

So we had a reversible, elementary gas phase reaction that is isothermal and isobaric in a flow system

##A \leftrightarrows 3C##

The equilibrium constant, ##K_{C}##, as well as initial pressure and initial temperature (same as final T and P) are given in the problem. I went ahead and calculated ##X_{eq,flow} = 0.58## in this situation.

Next, I calculated for a batch reactor that is carried out at constant volume and pressure with the same information given and found ##X_{eq,batch}^{V} = 0.39##. So of course, the conclusion is that ##X_{eq,flow} > X_{eq,batch}^{V}##. Now I want to think about why this is the case. Well, a flow reactor can change it's volume. Since for every mole of A, we are producing 3 moles of C, I figure the volume of the gas is expanding, which by Le Châtelier's principle will favor the side with more moles, hence C. Therefore, the conversion at equilibrium for a flow system will be larger than the batch system at constant volume, without any math to justify it.

Next, they ask about a constant pressure batch reactor, and it turns out ##X_{eq,batch}^{P} = 0.58##, hence ##X_{eq,batch}^{P} = X_{eq,flow}##. I was surprised by this result for a moment. Then, I thought since the pressure is constant, but the volume is not, so using the same argument then equilibrium will lie to the right and favor production of C.

I think one of the issues for the constant volume batch reactor is that one must assume that it is also constant pressure (final pressure is not given in the problem statement). Suppose I had a constant volume batch reactor that was not constant pressure, how might this change the solution? I suppose since for every mole of A, 3 molecules of C are being produced, then the pressure should increase since there will be more molecules slamming into each other, so equilibrium will favor the side with less moles, which is A. Hence, the equilibrium conversion for a isochoric, but not isobaric batch reactor will be less than that of a flow system. It would be even less than an isochoric and isobaric batch reactor.

So now I want to compare ##X_{eq,flow}##, ##X_{eq,batch}^{V}##,##X_{eq,batch}^{P}##, and ##X_{eq,batch}^{V,P}##

I think it would go ##X_{eq,flow} = X_{eq,batch}^{P} > X_{eq,batch}^{V,P} > X_{eq,batch}^{V}##

Now I want to flip this problem the other way, since I noticed none of the textbook problems have the stoichiometric coefficients of the reactants larger than the products, so I figure that would be a good test problem. If we have ##3A \leftrightarrows C##, then how will the ordering change? I use the same arguments (don't know how to compare isobaric and isochoric)

##X_{eq,batch}^{P} = X_{eq,flow} > X_{eq,batch}^{V}##

I want to do the calculation for this problem by making it up myself and convince myself, but I am hesistant because in the original problem ##K_{C} = 0.25 L/mol##, if I were to make it ##3A \leftrightarrows C##, would ##K_{C}## just be the inverse, 4? I figure that I might not get a good result if I just make up an equilibrium constant, and I am unsure how the constant would be affected by the stoichiometry being reversed, so I might be lead astray from the correct conclusion because I don't know how to handle the equilibrium constant..

Sorry for such a long post, but I tend to pick apart every possible scenario and figure out what is going on!

I went through a problem in my reaction engineering textbook, and solved them correctly by basically mindlessly plugging into the equation, but now I wanted to get some help interpreting the results.

So we had a reversible, elementary gas phase reaction that is isothermal and isobaric in a flow system

##A \leftrightarrows 3C##

The equilibrium constant, ##K_{C}##, as well as initial pressure and initial temperature (same as final T and P) are given in the problem. I went ahead and calculated ##X_{eq,flow} = 0.58## in this situation.

Next, I calculated for a batch reactor that is carried out at constant volume and pressure with the same information given and found ##X_{eq,batch}^{V} = 0.39##. So of course, the conclusion is that ##X_{eq,flow} > X_{eq,batch}^{V}##. Now I want to think about why this is the case. Well, a flow reactor can change it's volume. Since for every mole of A, we are producing 3 moles of C, I figure the volume of the gas is expanding, which by Le Châtelier's principle will favor the side with more moles, hence C. Therefore, the conversion at equilibrium for a flow system will be larger than the batch system at constant volume, without any math to justify it.

Next, they ask about a constant pressure batch reactor, and it turns out ##X_{eq,batch}^{P} = 0.58##, hence ##X_{eq,batch}^{P} = X_{eq,flow}##. I was surprised by this result for a moment. Then, I thought since the pressure is constant, but the volume is not, so using the same argument then equilibrium will lie to the right and favor production of C.

I think one of the issues for the constant volume batch reactor is that one must assume that it is also constant pressure (final pressure is not given in the problem statement). Suppose I had a constant volume batch reactor that was not constant pressure, how might this change the solution? I suppose since for every mole of A, 3 molecules of C are being produced, then the pressure should increase since there will be more molecules slamming into each other, so equilibrium will favor the side with less moles, which is A. Hence, the equilibrium conversion for a isochoric, but not isobaric batch reactor will be less than that of a flow system. It would be even less than an isochoric and isobaric batch reactor.

So now I want to compare ##X_{eq,flow}##, ##X_{eq,batch}^{V}##,##X_{eq,batch}^{P}##, and ##X_{eq,batch}^{V,P}##

I think it would go ##X_{eq,flow} = X_{eq,batch}^{P} > X_{eq,batch}^{V,P} > X_{eq,batch}^{V}##

Now I want to flip this problem the other way, since I noticed none of the textbook problems have the stoichiometric coefficients of the reactants larger than the products, so I figure that would be a good test problem. If we have ##3A \leftrightarrows C##, then how will the ordering change? I use the same arguments (don't know how to compare isobaric and isochoric)

##X_{eq,batch}^{P} = X_{eq,flow} > X_{eq,batch}^{V}##

I want to do the calculation for this problem by making it up myself and convince myself, but I am hesistant because in the original problem ##K_{C} = 0.25 L/mol##, if I were to make it ##3A \leftrightarrows C##, would ##K_{C}## just be the inverse, 4? I figure that I might not get a good result if I just make up an equilibrium constant, and I am unsure how the constant would be affected by the stoichiometry being reversed, so I might be lead astray from the correct conclusion because I don't know how to handle the equilibrium constant..

Sorry for such a long post, but I tend to pick apart every possible scenario and figure out what is going on!

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