Chapter 12 Self Test

Nonisothermal reactions


NOTE: It will be very beneficial to answering the following questions if you first go through the .

 

1.  The elementary isomerization of A to B was carried out in a packed bed reactor. The following profiles were obtained

Click on the correct true (T) or false (F) answer for this system.

(a) The above profiles could represent an adiabatic system where the addition of inerts to the feed stream will increase conversion.      T     or    F

If inerts were to be added to the system, sketch the exit conversion as a function of the ratio of entering flow rate of inerts to the entering flow rate of A, i.e.

Answer

(b) If the reaction is irreversible, a small decrease in the flow rate will produce a small increase in the conversion.     T     or    F

(c) If the reaction is reversible, a small decrease in the flow rate will produce a small increase in the conversion.     T     or    F

(d) An increase in the feed temperature will increase the conversion.     T     or    F

(e) A decrease in feed temperature will increase conversion.     T     or    F

 

2.  The elementary reaction

takes place in a packed bed reactor. The following profiles were obtained.

Click on the correct true (T) or false (F) answer for this system.
(a) The above profiles could represent an adiabatic system where the addition of inerts will increase the conversion.      T     or    F

If inerts were to be added to the system, sketch the exit conversion as a function of the ratio

Answer

(b) The above profiles could represent a system where decreasing the flow rate will increase the conversion.      T     or    F

(c) The above profiles could represent a system where if the feed temperature is increased, one cannot tell from the above profiles whether or not the conversion will increase or decrease.      T     or    F

  

3.  The following conversion profiles were measured in two PFR's (A and B) for the same reaction

(a) Is the reaction exothermic or endothermic?

(b) Is the reaction reversible or irreversible?

(c) What is the difference between the two reactors?

(d) For reactor A, sketch the conversion profiles for an entering temperature of 300 K and 1000 K and explain.Answer

(e) For reactor A, sketch the exit conversion as a function of inlet feed temperature To. What do you conclude from your plot?

Answer

(f) Discuss each of the regions 1, 2, and 3 in the figure for reactor B.Answer

  

4.  The following temperature profile was observed in a PBR.

(a) Sketch the corresponding equilibrium conversion, Xe and conversion profile assuming the reaction is reversible. Answer

(b) On the above temperature versus catalyst weight plot, sketch the temperature profiles if the feed rate were to be decreased by a factor of 3. Answer

(c) PBR with Heat Exchange

Below are the temperature, equilibrium conversion, Xe, and conversion profiles for an exothermic, reversible reaction in a PBR.

Which of the the above profiles is/are not correct? 

The above profiles are for the PBR with Heat Exchange which comes directly above from this self-test or lecture or .  Answer

  

5.   The elementary isomerization of A to B was carried out adiabatically in a packed bed reactor. The following profiles were obtained when pure A was fed to the reactor

1)   The above profiles could represent an adiabatic system where the addition of inerts to the feed stream will increase the conversion.

2)   If the reaction is irreversible a small decrease in the flow rate will produce a moderate increase in the conversion.

3)   If the reaction is reversible a small decrease in the flow rate will produce a moderate increase in the conversion.

A.  All the above statements are true.
B.   All the above statements are false.
C.  Statements 1 and 2 are true
D.  Statements 1 and 3 are true.
E.   Statements 2 and 3 are false.

Answer

  

6) The elementary isomerization of A to B was carried out in a packed bed reactor. The following profiles were obtained when pure A was fed to the reactor

 

  1. An increase in the feed temperature will increase the conversion.

  2. A decrease in feed temperature will increase the conversion.

  3. There could be a very very very large heat exchanger attached to the reactor with the heat flow given by

A.  All the above statements are true.
B.   All the above statements are false.
C.  Statements 1 and 2 are true.
D.  Statements 1 and 3 are false.
E.   Statements 2 and 3 are false.

 

Answer

   

7.   The elementary, gas phase, isomerization of A to B was carried out in a packed bed reactor. The following profiles were obtained when pure A was fed to the reactor.

Inerts were added to a reactor system while T0, P0, and n0 were kept constant. Which of the following figures the exit conversion is as a function of the qI is correct? i.e. qI (qI = FI0/FA0)

 Note:  ? ? ? represents isothermal conversion for a reactor with 10 kg catalyst.

Answer

 

 8.   The elementary, gas phase reaction

        takes place in a packed bed reactor. The following profiles were obtained

  1. The above profiles could represent an adiabatic system where the addition of a small to moderate amount of inerts will increase the conversion.

  2. The above profiles could represent an adiabatic system where decreasing the flow rate will increase the conversion.

  3. The above profiles could represent an adiabatic system where if the feed temperature is increased, one cannot tell from the above profiles whether or not the conversion will increase or decrease.

  4. There could be a heat exchanger on the reactor for which the heat flow is

A.  All the above statements are true.
B.   All the above statements are false.
C.  Statements 1 and 2 are true
D.  Statements 2 and 4 are true.
E.   Statements 2 and 3 are false.

Answer 

9.   The following conversion profiles were measured in two PFRs, (A and B) for the same reaction

Which of the following statements is true?

  1. The reaction is exothermic and reversible.

  2. The reaction is exothermic and irreversible.

  3. The reaction is endothermic and irreversible.

  4. The reaction is endothermic and reversible.

  5. Cannot tell for sure if any, some, or all the above are true.

  Answer

10.   The exit conversion from an adiabatic PFR is shown below as a function of θI for constant ν0, T0 and P0

  1. The reaction could be a second order endothermic reaction.

  2. The reaction could be a second order exothermic reaction.

  3. The reaction could be a first order exothermic reaction.

  4. The reaction could be a first order endothermic reaction.

  5. The reaction could be an irreversible zero order exothermic reaction.

  1. Statements 1 and 2 are false.

  2. Statements 3 and 4 are false.

  3. Statements 3 and 5 are false.

  4. Statements 3 and 4 are true.

  Answer

11.   The exit conversion from an adiabatic PFR is shown below as a function of θI for constant ν0, T0 and P0

  1. The reaction could be a second order endothermic reaction.

  2. The reaction could be a second order exothermic reaction.

  3. The reaction could be a first order exothermic reaction.

  4. The reaction could be a first order endothermic reaction.

  5. The reaction could be an irreversible zero order exothermic reaction.

  1. Both statements 1 and 2 are true.

  2. Both statements 1 and 5 are true.

  3. Both statements 2 and 3 are false.

  4. Both statements 1 and 4 are false.

  5. Both statements 4 and 5 are true.

  Answer

12. The exit conversion from an adiabatic PFR is shown below as a function of θI for constant ν0, T0 and P0

     Which of the following statements is false?

  1. The reaction could be a second order endothermic reaction.

  2. The reaction could be a second order exothermic reaction.

  3. The reaction could be a first order exothermic reaction.

  4. The reaction could be an irreversible zero order exothermic reaction.

  5. All of the above are false.

  Answer

13. The equilibrium conversion is shown below as a function of catalyst weight

  1. The reaction could be first order endothermic and carried out adiabatically.

  2. The reaction is first order endothermic and reactor is heated along the length with Ta being constant.

  3. The reaction is second order exothermic and cooled along the length of the reactor with Ta being constant. 

  4. The reaction is second order exothermic and carried out adiabatically.

  1. Both 1 and 2 are false.

  2. Both 2 and 3 are false.

  3. Both 3 and 4 are false.

  4. Both 1 and 4 are false.

  5. Both 2 and 4 are false.

  Answer

14. The equilibrium conversion is shown below as a function of catalyst weight

 

  1. The reaction is first order endothermic and carried out adiabatically.

  2. The reaction is first order endothermic and reactor is heated along the length with Ta being constant.

  3. The reaction is second order exothermic and cooled along the length of the reactor with Ta being constant.

  4. The reaction is second order exothermic and carried out adiabatically.

  1. Both 1 and 2 are true.

  2. Both 2 and 3 are true.

  3. All are true.

  4. Both 3 and 4 are false.

  5. Both 2 and 4 are false.

  Answer

15. The equilibrium conversion is shown below as a function of catalyst weight

      Which of the following statements is false?

  1. The reaction could be first order endothermic and carried out adiabatically.

  2. The reaction could be first order endothermic and reactor is heated along the length with Ta being constant.

  3. The reaction could be second order exothermic and cooled along the length of the reactor with Ta being constant.

  4. The reaction could be second order exothermic and carried out adiabatically.

  5. The reaction could be first order endothermic with very high heating rate.

  Answer

16. The equilibrium conversion is shown below as a function of catalyst weight

  1. The reaction could be first order endothermic and carried out adiabatically.

  2. The reaction could be could be first order endothermic and reactor is heated along the length with Ta being constant.

  3. The reaction could be second order exothermic and cooled along the length of the reactor with Ta being constant.

  4. The reaction could be second order exothermic and carried out adiabatically.

  1. Both 1 and 4 are false.

  2. Both 1 and 3 are true.

  3. Both 2 and 4 are true.

  4. Both 3 and 4 are false.

  5. All are true.

  Answer

17. The equilibrium conversion and temperature are shown below as a function of catalyst weight for three sets of conditions

  1. The figures could correspond to an exothermic reversible reaction with too high of a cooling rate.

  2. The figures could correspond to an exothermic irreversible reaction with too high of a cooling rate.

  3. The figures could correspond to an endothermic reversible reaction with too high of a heating rate.

  4. The figures could correspond to an endothermic irreversible reaction with too high of a heating rate.

  1. Both 2 and 4 are false.

  2. Both 1 and 3 are false.

  3. Both 3 and 4 are true.

  4. Both 1 and 4 are true.

  5. Both 1 and 2 are true.

  Answer

18. The conversion from an adiabatic PFR is shown below as a function of temperature

      Which of the following is false

  1. 90% conversion could be achieved for a first order reaction by increasing the inerts by a factor of 100 or greater for entering temperature of 700 K.

  2. 90% conversion could be achieved for a second order reaction by adding a very large heat exchanger with a very large area for an entering temperature of 700 K.

  3. For a small but fixed reactor length, one is more likely to get close to the adiabatic equilibrium conversion for an entering temperature of 500 K than an entering temperature of 900 K.

  4. The reaction could be a second order endothermic reaction carried out with a very large heat exchanger attached to the reactor.

  5. If the reaction is carried out adiabatically, the temperature in a CSTR will drop more for an entering temperature of 900 K then that for an entering temperature of 500 K.

  Answer

19.

  1. The above reaction could be exothermic.

  2. The above reaction could be endothermic.

  3. The above reaction could be first order.

  4. The above reaction could be second order.

  5. The above reaction is not carried out adiabtically.

  1. Both 1 and 5 are false.

  2. Both 2 and 5 are false.

  3. 1, 2 and 3 are true.

  4. 3, 4 and 5 are true.

  5. 1, 2, 3, and 4 are true.

  Answer

20.

  1. The above reaction could be carried out adiabatically.

  2. The above reaction could be exothermic.

  3. The above reaction could be endothermic.

  4. The temperature profile could go through a maximum.

  5. The temperature profile could go through a minimum.

  1. 1, 2, 3, and 4 are true.

  2. 1, 2, 5, and 5 are false.

  3. 1, 4 and 5 are true.

  4. 2, 3, 4, and 5 are true.

Answer

21. 

  1. The above reaction could be adiabatic.

  2. The above reaction could be exothermic with cooling.

  3. The above reaction could be endothermic with heating.

  4. The above reaction could be second order

  1. Both 1 and 2 are true.

  2. Both 2 and 3 are true.

  3. Both 2 and 4 are true.

  4. Both 3 and 4 are true.

  5. Both 1 and 4 are true.

Answer

22. 

  1. The above reaction could be reversible.

  2. The above reaction could be irreversible.

  3. The above reaction could be exothermic.

  4. The above reaction could be endothermic.

  5. The above reaction could be adiabatic.

  1. 1, 2, 3, and 4 are true.

  2. 1, 3, and 4 are true.

  3. 1, 3, 4, and 5 are true.

  4. Only 1 and 3 are true.

  5. Both 2 and 5 are false.

Answer


 

 

 

 

 

 

Solutions

Problem 1

1 (a) TRUE. If it is an adiabatic system, then it has to be endothermic, because the temperature decreases and the heat of reaction is positive. Increasing inerts increases the temperature, increasing k, increasing the rate and hence increasing coversion.

Adiabatic Energy Balance                                     (1-1)

The reaction could have essentially stopped at 4 kg for one of two reasons.

A. First, the reaction could have stopped because the temperature drops sufficiently to cause k to reach a very very small value so that the rate is so small and reaction does not proceed farther.  That is the reaction becomes frozen.

B. The second reason the reaction could have stopped is because it reaches equilibrium (as the temperature drops, so does Xe).

                                                                        (1-5)

Case I: An Irreversible First-Order Endothermic Reaction

Combined Mole Balance, Rate Law, and Stoichiometry

                                             (1-2)          

Note fixed Tτ0, if the total molar flow rate and temperature and pressure are constant (hence Cτ0) the ν0 is constant, consequently the presence of inerts does not effect this equation except by increasing the temperature which increases k. Adding inerts to the system will cause the temperature not to drop as much so that the rate will be faster and the conversion greater.

      

In which case question (a) is true.

Case II: Reversible 1st order Endothermic Reaction

Fixed Tτ0

At equilibrium

                                                                                    (1-6)

Endothermic

      

For no inerts (Line A)

                                                                        (1-7)

Along the energy balance line, the temperature will continue to decrease until the reaction has reached essentially equilibrium, and the adiabatic equilibrium conversion.

For inerts (Line B)

                                                            (1-8)

Again, increasing the amount of inerts increases the slope of the line, the energy balance line (B) causing the temperature not to drop as much as the case with no inerts and as shown. In addition, the equilibrium conversion is increased.

1 (a) TRUE If the reaction is an endothermic, reversible, and first-order then,

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1 (b) TRUE. For an irreversible reaction, decreasing the flow rate will always increase conversion. Even if it's on the plateau, the conversion still increases ever so slightly. The reactants spend a longer time in the reactor.

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1 (c) FALSE. If the exit conversion is essentially the equilibrium conversion (e.g., X = 0.999Xe), then the reactants spending more time in the reactor as a result of a decrease in the flow rate will not affect conversion, because the reaction has reached equilibrium.

If the reaction had not reached equilibrium, then decreasing the flow rate could increase the conversion.

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1 (d) TRUE. Higher entering temperature, higher rate, higher equilibrium conversion, therefore greater extent of reaction (greater conversion).

      

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1 (e) FALSE. If it is an endothermic, reversible reaction, then it will have a lower equilibrium conversion at a lower temperature (in addition, the lower temperature will cause the specific rate, k, to be lower), resulting in a slower rate and thus smaller conversion. 

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Problem 2

2 (a) TRUE. If it is an adiabatic system, then it has to be exothermic. Also, because both the temperature and conversion reach a plateau, equilibrium has been reached. Adding inerts increases the slope of the energy balance line.

                                                 

Irreversible
Reversible

          (2-1)

Case 1 - Rapid Reaction: Equilibrium reached even at isothermal temperature, To.

The addition of inerts will lower the exit temperature and hence will increase both the equilibrium conversion and the exit conversion, provided the rate is sufficiently large to always closely approach equilibrium, even at To, as shown in the figure below. As more and more inerts continue to be added, the reactor approaches isothermal condition, To.

Irreversible rapid reaction, equilibrium achieved.

Case 2 - Slow Reaction: Equilibrium not achieved at isothermal temperature, To.

We now consider the case when the reaction is slow at the start (i.e. T = T0) but as it proceeds down the reactor the temperature increases as does the rate and the conversion until it finally takes off (see = 0 below).

If we were to add inerts to the point that the reactor approached isothermal conditions and the reaction rate was very small, then under these conditions very little conversion would be achieved (as in theta=4 above).   

For an intermediate rate of reaction we could have

Suppose the reaction was an irreversible second-order reaction

Adding inerts will increase the temperature for endothermic reaction of any order. Consequently, k will increase and the rate will increase with increasing inerts, but only to a point.

Combined Mole Balance, Rate Law, and Stoichiometry       wav mp3

                                                (1-3)

              

 

                                           (1-4) 

We see that at very large values of the rate becomes very small and consequently very little conversion is achieved. Consequently there is an optimum in the amount of inerts for a 2nd order reaction.

 

There would be an optimum value of thetaI for which the exit conversion is a maximum. As we initially increase thetaI we decrease the maximum temperature reached, therefore increasing Xe. However, as we increase thetaI further, the rate decreases and we don't approach Xe.

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2 (b) FALSE. Decreasing the flow rate will not change the exit condition because it is an equilibrium condition.

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2 (c) FALSE. We see that for the plot of entering temperature shown, equilibrium has already been reached in the reactor.  If feed temperature is increased, equilibrium conversion is decreased. Therefore, one can tell whether the exit conversion will increase or decrease.

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Problem 3

3 (a) EXOTHERMIC. Temperature increases reaching a plateau in reactor A at the adiabatic equilibrium conversion.

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3 (b) REVERSIBLE. In reactor A, the conversion reaches a plateau (i.e.X=Xe) well below complete conversion, X=1.0, indicating a reversible reaction.

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3 (c) A is adiabatic. B has a heat exchanger. As heat is removed in Reactor B, temperature drops, shifting equilibrium to the right to increase conversion.

By equilibrium being reached we are saying X = 0.99Xe.

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3 (d)

In curve A (1000 K), the equilibrium conversion is virtually reached near the reactor entrance. However due to the high temperature, the equilibrium conversion is very low. For curve B (300 K), this very low temperature results in an extremely low specific reaction rate (k). Consequently, the reaction is so slow the reaction never takes off, resulting in a very small conversion.  Therefore, there is an optimum inlet temperature.  You cannot tell whether or not conversion will increase because you do not know which side of the optimum you are on.

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3 (e)

There is an optimum inlet temperature. Therefore you cannot say whether or not conversion will increase or decrease if you increase T0.  It depends on which side of the optimum you are on.

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3 (f)

1. Very rapid rate and equilibrium is approached at 100 dm3. If the reactor were adiabatic, then conversion would remain the same after V=100 dm3.

2. Reactor begins to cool down shifting equilibrium to the right, conversion slowly increases.

3. Reactants almost exhausted and low temperature results in a slow rate and conversion levels off.

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Problem 4

4 (a) As the temperature increases, so does the reaction rate until the equilibrium conversion is approached at which point the reaction slows down. As the temperature drops, Xe increases, the reaction continues to shift to the right, thereby increasing conversion.

 

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4 (b)

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4 (c) Heat Exchange

Solution: Higher flow rates will cause the temperature profile to flatten out more near the entrance to the reactor, so (A) is correct, as is (D). (B) and (C) are not correct because for Fao=1, the temperature begins to decrease, Xe begins to increase, so the conversion should increase but instead X reaches a plateau. 

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Solution 5.   The elementary isomerization of A to B was carried out adiabatically in a packed bed reactor. The following profiles were obtained when pure A was fed to the reactor

1)   The above profiles could represent an adiabatic system where the addition of inerts to the feed stream will increase the conversion.

2)   If the reaction is irreversible a small decrease in the flow rate will produce a small increase in the conversion.

3)   If the reaction is reversible a small decrease in the flow rate will produce a small increase in the conversion.

A.  All the above statements are true.
B.   All the above statements are false.
C.  Statements 1 and 2 are true
D.  Statements 1 and 3 are true.
E.   Statements 2 and 3 are false.

Solution: Ans: E

1)   True. If it is an adiabatic system then it has to be endothermic because the temperature decreases. Increasing inerts increases the exit temperature and hence the conversion is higher.

2)   False. The reaction has reached a sufficiently low temperature the reaction is essentially frozen. Thus, changing the flow rate will not produce a moderate increase the temperature or conversion.

3)   False. If the exit condition is an equilibrium condition, then a small change in the flow rate will not affect the equilibrium condition. Hence, it will not change the conversion. If the reaction is frozen, as in 2, then neither the conversion nor temperature will increase significantly.

 Back to Problem 5

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Solution 6.   The elementary isomerization of A to B was carried out in a packed bed reactor. The following profiles were obtained when pure A was fed to the reactor

 

  1. An increase in the feed temperature will increase the conversion. (True)

  2. A decrease in feed temperature will increase the conversion.(False)

  3. There could be a very very very large heat exchanger attached to the reactor with the heat flow given by   (False)

A.  All the above statements are true.
B.   All the above statements are false.
C.  Statements 1 and 2 are true.
D.  Statements 1 and 3 are false.
E.   Statements 2 and 3 are false.

 

Explanation:  The reaction is endothermic because temperature drops.

1)   The higher temperature, the higher equilibrium conversion, higher rate, therefore greater conversion.

      

2)   If it is an endothermic reaction then it should have smaller equilibrium conversion at low temperature and also a smaller k resulting in a slower rate and smaller conversion.

3)   Because final temperature is 500 K and reaction is either frozen (i.e., the temperature is so low the reaction rate is virtually zero or it is in equilibrium. The ambient temperature of the heat exchanger cannot be 400.

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Solution 7.   The elementary isomerization of A to B was carried out in a packed bed reactor. The following profiles were obtained when pure A was fed to the reactor

  Inerts were added to a reactor system while T0, P0, and n0 were kept constant. Which of the following figures the exit conversion is as a function of the qI is correct? i.e. qI (qI = FI0/FA0)

Note:  ? ? ? represents isothermal conversion for a reactor with 10 kg catalyst.

Solution: Ans: C.  Inerts provide sensible heat to raise temperature. The combined mole balance rate law and stoichiometry,

is independent of amount of inerts.

temperature will be higher because inerts supply sensible heat.  The higher temperature the higher rate and the greater the conversion.

Back to Problem 7 

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Solution 8.   The elementary, gas phase reaction

        takes place in a packed bed reactor. The following profiles were obtained

     

  1. The above profiles could represent an adiabatic system where the addition of a small to moderate amount of inerts will increase the conversion. (True)

  2. The above profiles could represent an adiabatic system where decreasing the flow rate will increase the conversion. (False)

  3. The above profiles could represent an adiabatic system where if the feed temperature is increased, one cannot tell from the above profiles whether or not the conversion will increase or decrease. (False)

  4. There could be a heat exchanger on the reactor for which the heat flow is (True)

A.  All the above statements are true.
B.   All the above statements are false.
C.  Statements 1 and 2 are true
D.  Statements 2 and 4 are true.
E.   Statements 2 and 3 are false.

Explanation:

1) TRUE  () If it is an adiabatic system, then it has to be exothermic. Addition of a moderate amount of inerts will lower the exit temperature increase the equilibrium conversion and hence will increase the conversion. If a very large amount of inerts are added then the conversion could decrease since the combined M.B., R.L., and Stoich is

and,  goes to zero as θI ? ?

In addition, k will decrease as inerts are added because the temperature is decreased.

2) FALSE Decreasing the flow rate will not change the exit condition because it is an equilibrium condition.

3) FALSE One can tell becuase equilibrium is achieved early in the reactor if feed temperature is increased, equilibrium conversion is decreased.

4) TRUE The ambient temperature is 500 K, the same as the final equilibrium temperature. Consequently if the reaction is endothermic, heating the reactor will increase the rate, Xe and X.

Answer: E

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Solution 9.   The following conversion profiles were measured in two PFRs, (A and B) for the same reaction

Which of the following statements is true?

  1. TRUE The reaction is exothermic and reversible. 

  2. FALSE The reaction is exothermic and irreversible (equilibrium is reached).

  3. FALSE The reaction is endothermic and irreversible (temperature increases, exothermic).

  4. FALSE The reaction is endothermic and reversible (temperature increases, exothermic).

  5. Cannot tell for sure if any, some, or all the above are true.

 Solution:  A is true, at high temperature equilibrium achieved early in reactor.

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Solution 10.   The exit conversion from an adiabatic PFR is shown below as a function of θI for constant ν0, T0 and P0

  1. The reaction could be a second order endothermic reaction. False.

  2. The reaction could be a second order exothermic reaction. False.

  3. The reaction could be a first order exothermic reaction. True.

  4. The reaction could be a first order endothermic reaction. True.

  5. The reaction could be an irreversible zero order exothermic reaction. False.

  6. Statements 1 and 2 are false.

  7. Statements 3 and 4 are false.

  8. Statements 3 and 5 are false.

  9. Statements 3 and 4 are true.

Explanation:

1)  FALSE () , as θI becomes very large, X decreases.

2)  FALSE () , as θI becomes very large, X decreases.

3)  TRUE () , does not appear in this combined mole balance, rate law, and stoichiometry, but as increases, temperature decreases, k decreases, and Kc increases as does Xe.

Case I: k very very large. Equilibrium is always achieved in the column.


 

Even at T = T0 down the reactor, equilibrium is reached. So it could be a first order exothermic reaction. TRUE

Case II: k is moderate. Equilibrium is not always achieved.

 

k is so low at T0, reaction never really takes off. In this case the answer would be FALSE. But because Case I is possible, the overall answer is TRUE.

4) TRUE () , as &thetaI increases, so does T, Xe and X.

5) FALSE () ,  conversion increases as &thetaI increases.

Case I: E is very small so that decreasing the temperature by adding inerts does not change k very much at all. Thus,

Case II: Large E, decreasing the temperature by adding inerts decreases T and k at the exit. Smaller k, smaller conversion.

 

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Solution 11.   The exit conversion from an adiabatic PFR is shown below as a function of θI for constant ν0, T0 and P0

    1. The reaction could be a second order endothermic reaction.  True.
    2. The reaction could be a second order exothermic reaction. True.
    3. The reaction could be a first order exothermic reaction. True.
    4. The reaction could be a first order endothermic reaction. False.
    5. The reaction could be an irreversible zero order exothermic reaction. False.
    6. Both statements 1 and 2 are true.
    7. Both statements 1 and 5 are true.
    8. Both statements 2 and 3 are false.
    9. Both statements 1 and 4 are false.
    10. Both statements 4 and 5 are true.

Explanation:

answer 6

True. In region 1 increasing qI increases temperature and X. In region 2,. Large decrease, therefore X decreases.

2)   True. In region 1, adding inserts could decrease temperatures and increase equilibrium conversion. In region 2. Large decrease, therefore X decreases.

    3) True.  does not appear in this combined mole balance, rate law, and stoichiometry, but as increases, temperature decreases, k decreases, and Kc increases as does Xe.

    3) True.  does not appear in this combined mole balance, rate law, and stoichiometry, but as increases, temperature decreases, k decreases, and Kc increases as does Xe.

    Case I: k very very large. Equilibrium is always achieved in the column.

     

    Even at T = T0 down the reactor, equilibrium is reached.

    Case II: k is moderate. Equilibrium is not always achieved.

 

    k is so low at T0, reaction never really takes off.

4)   False. For first order  independent of inserts except in k and KC. Increase θI increase T, k, and X up until we reach isothermal conditions then X will be independent of &thetaI.

5)   False. ,  conversion increases as θI increases.

Case I: E is very small so that decreasing the temperature by adding inerts does not change k very much at all. Thus,

Case II: Large E, decreasing the temperature by adding inerts decreases T and k at the exit. Smaller k, smaller conversion.

 

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Solution 12. The exit conversion from an adiabatic PFR is shown below as a function of θI for constant constant ν0, T0 and P0

Which of the following statements is false?

  1. The reaction could be a second order endothermic reaction. True
  2. The reaction could be a second order exothermic reaction. True
  3. The reaction could be a first order exothermic reaction. True
  4. The reaction could be an irreversible zero order exothermic reaction. True
  5. All of the above are false.

A) True if DHRX is very small such that temperature does not change upon adding inerts and, then for large Kc, X could decrease as qI increases.

B) True adding inerts could decrease temperature and therefore k and X.

C) True., increase inerts, could decrease T, k and X

D) False.





 Increases with qI

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Solution 13. The equilibrium conversion is shown below as a function of catalyst weight

                                                 

  1. The reaction could be first order endothermic and carried out adiabatically. True.
  2. The reaction is first order endothermic and reactor is heated along the length with Ta being constant. False.
  3. The reaction is second order exothermic and cooled along the length of the reactor with Ta being constant.  False.
  4. The reaction is second order exothermic and carried out adiabatically. True.

 

  1. Both 1 and 2 are false.
  2. Both 2 and 3 are false.
  3. Both 3 and 4 are false.
  4. Both 1 and 4 are false.
  5. Both 2 and 4 are false.

 

Explanation:

1)   As temperature drops so does KC and Xe. However X continues to increase until Xe is reached.

     

2)   Flat profile rules out heating or cooling.

3)   If cooling, temperature would decrease. Xe would increase at some point if there were heating or cooling.

4)  

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Solution 14. The equilibrium conversion is shown below as a function of catalyst weight

                                               

  1. The reaction could be first order endothermic and carried out adiabatically.

    2. False? No reason for Xe to increase if adiabatic.

  2. The reaction could be first order endothermic and reactor is heated along the length with Ta being constant. True.

  3. The reaction could be second order exothermic and cooled along the length of the reactor with Ta being constant. True.

  4. The reaction could be second order exothermic and carried out adiabatically.

    False ? No reason for Xe to increase if adiabatic.

  1. Both 1 and 2 are true.
  2. Both 2 and 3 are true.
  3. All are true.
  4. Both 3 and 4 are false.
  5. Both 2 and 4 are false.

Answer: B

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Solution 15. The equilibrium conversion is shown below as a function of catalyst weight

                                                

      Which of the following statements is false?

A.  False. The reaction could be first order endothermic and carried out adiabatically. Temperature will drop.

B.  False. The reaction could be first order endothermic and reactor is heated along the length with Ta being constant. If 1st order endothermic, then

  C.  False. The reaction could be second order exothermic and cooled along the length of the reactor with Ta being constant. If 2nd order exothermic, then

D.  False. The reaction could be second order exothermic and carried out adiabatically. If 2nd order exothermic and adiabatic, then

E.  True. The reaction could be first order endothermic with very high heating rate. If heating rate such that temperature always increases, then

    Back to Problem 15

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Solution 16. The equilibrium conversion is shown below as a function of catalyst weight

 

                                                

1)  False. The reaction could be first order endothermic and carried out adiabatically. If adiabatic, then

2)   True. The reaction could be first order endothermic and reactor is heated along the length with Ta being constant.

3)   True. The reaction could be second order exothermic and cooled along the length of the reactor with Ta being constant.

4)   False. The reaction could be second order exothermic and carried out adiabatically.

      1. Both 1 and 4 are false.
      2. Both 1 and 3 are true.
      3. Both 2 and 4 are true.
      4. Both 3 and 4 are false.
      5. All are true.

       

    Answer A

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    Solution 17. The equilibrium conversion and temperature are shown below as a function of catalyst weight for three sets of conditions

     

                        

  1. The figures could correspond to an exothermic reversible reaction with too high of a cooling rate.  True. k(T) smaller \ X smaller
  2. The figures could correspond to an exothermic irreversible reaction with too high of a cooling rate. True. k(T) smaller \ X smaller
  3. The figures could correspond to an endothermic reversible reaction with a high heating rate. True. T higher, k larger, Xe larger, heating rate of (3) > (2) > (1). For endothermic reactions
  4. The figures could correspond to an endothermic irreversible reaction with a high heating rate. True. For endothermic reactions X always increases as T increases.

 

  1. Both 2 and 4 are false.
  2. Both 1 and 3 are false.
  3. Both 3 and 4 are true.
  4. Both 1 and 4 are true.
  5. All are true.

Answer: E

 

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Solution 18. The conversion from an adiabatic PFR is shown below as a function of temperature

 

                               

      Which of the following is false

  1. 90% conversion could be achieved for a first order reaction by increasing the inerts by a factor of 100 or greater for entering temperature of 700 K. True.
  2. 90% conversion could be achieved for a second order reaction by adding a very large heat exchanger with a very large area for an entering temperature of 700 K. True.
  3. For a small but fixed reactor length, one is more likely to get close to the adiabatic equilibrium conversion for an entering temperature of 500 K than an entering temperature of 900 K. False.
  4. The reaction could be a second order endothermic reaction carried out with a very large heat exchanger attached to the reactor. False.
  5. If the reaction is carried out adiabatically, the temperature in a CSTR will drop more for an entering temperature of 900 K then that for an entering temperature of 500 K. True.

Explanation:

 1.   

 

2.   Eventually becomes isothermal as the product of UA continually increases. Same as (1) above.

     3.  High T0, higher rate, equilibrium conversion reached near entrance to reactor.

4.  False. If very large heat exchanger, the energy balance term will be essentially vertical.

5.   . Higher T0, higher rate, higher X greater temperature drop.

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Solution 19.

      1. The above reaction could be exothermic.
      2. The above reaction could be endothermic.
      3. The above reaction could be first order.
      4. The above reaction could be second order.
      5. The above reaction is not carried out adiabatically.

       

      1. Both 1 and 5 are false.
      2. Both 2 and 5 are false.
      3. 1, 2 and 3 are true.
      4. 3, 4 and 5 are true.
      5. 1, 2, 3, and 4 are true.

       

    Answer: D

    Explanation:

    1).  True.

2)   True.

3)   True.

4).  True.

5)   False ? if  then

or

     

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Solution 20.

 

 

  1. The above reaction could be carried out adiabatically.
  2. The above reaction could be exothermic.
  3. The above reaction could be endothermic.
  4. The temperature profile could go through a maximum.
  5. The temperature profile could go through a minimum.
  1. 1, 2, 3, and 4 are true.
  2. 1, 2, 5, and 5 are false.
  3. 1, 4 and 5 are true.
  4. 2, 3, 4, and 5 are true.

Explanation:

1).  False. Xe cannot increase if it is adiabatic.

2)

3)  True.

3)   True.

4).  True. If exothermic


5)   True. If endothermic

Here both  and  are negative

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Solution 21.

  1. The above reaction could be adiabatic.
  2. The above reaction could be exothermic with cooling.
  3. The above reaction could be endothermic with heating.
  4. The above reaction could be second order
  1. Both 1 and 2 are true.
  2. Both 2 and 3 are true.
  3. Both 2 and 4 are true.
  4. Both 3 and 4 are true.
  5. Both 1 and 4 are true.

Answer: D

Explanation:

  1. False. Xe could not increase if adiabatic.
  2. False. Xe would follow slope temperature curve.
  3. True.
  4. True.

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Solution 22.

  1. The above reaction could be reversible.
  2. The above reaction could be irreversible.
  3. The above reaction could be exothermic.
  4. The above reaction could be endothermic.
  5. The above reaction could be adiabatic.
  1. 1, 2, 3, and 4 are true.
  2. 1, 3, and 4 are true.
  3. 1, 3, 4, and 5 are true.
  4. Only 1 and 3 are true.
  5. Both 2 and 5 are false.

Explanation:

  1. True: Heat Added to Reactor at high rate
  2. True: Heat Added to Reactor at high rate
  3. True: Heat Added to Reactor at high rate
  4. True: Heat Added to Reactor at high rate
  5. False. Ta ambient temperature changes.

Back to Problem 22

 

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