Chapter 10: Catalysis and Catalytic Reactors

Example Exam Questions

 

Question 1

\( A + B \rightleftharpoons C \)

A graph showing -r'_A0 on the y-axis and P_C on the x-axis. The curve starts at a high reaction rate and gradually decreases, forming a downward-sloping curve that flattens out, indicating an inverse relationship between the reaction rate and the partial pressure of component C (P_C).

Figure 1: Data from a differential reactor.

Which of the following best describe the data in the above figure?

  1. Species C could be adsorbed on the surface.

  2. The reaction could be reversible.

  3. Both the above statements could be true.

  4. Neither A or B are on the surface.

  5. None of the above could be true.

Go To Solution #1         









Question 2

\( A + B \rightarrow C \)

Assume the reaction is irreversible

Three plots showing -r'_A versus P_C, P_A, and P_B. Figure 1 shows a decreasing curve, Figure 2 shows a linear increase, and Figure 3 shows a saturation curve.

What can you tell from the above figures?

  1. Species A could be adsorbed on the surface but only very weakly adsorbed.

  2. Species A is not adsorbed on the surface.

  3. Species B is on the surface.

  4. Species A is adsorbed on the surface.

  5. Species C is adsorbed on the surface.

  1. 1 and 2 are true.

  2. 1 and 3 are true.

  3. 1 and 4 are false.

  4. 4 and 5 are true.

  5. 2 and 5 are true.

  Go To Solution #2 









  

 Question 3 

 

\( A + B \rightleftharpoons C + D \)

The following data were reported for the reaction

A+B goes to C+D

A  B
Graph A shows -r'_A increasing with P_A and then leveling off. Graph B shows -r'_B increasing with P_Total and leveling off; y_A0 equals y_B0.
C D
Graph C shows -r'_D constant as P_D varies, with P_C fixed at 10. Graph D shows -r'_C decreasing with P_C, with P_D fixed at 0.

Which of the following mechanism is consistent with the above data

(a)  (b) (c)

\( A + S \rightleftharpoons A \cdot S \)

\( B + S \rightleftharpoons B \cdot S \)

\( A \cdot S + B \cdot S \rightleftharpoons C \cdot S + D \cdot S \)

\( C \cdot S \rightleftharpoons C + S \)

\( A + S \rightleftharpoons A \cdot S \)

\( B + S \rightleftharpoons B \cdot S \)

\( A \cdot S + B \cdot S \rightleftharpoons C \cdot S + D \)

\( C \cdot S \rightleftharpoons C + S \)

\( A + S \rightleftharpoons A \cdot S \)

\( A \cdot S + B \rightleftharpoons C \cdot S + D \)

\( C \cdot S \rightleftharpoons C + S \)
(d) (e)

\( B + S \rightleftharpoons B \cdot S \)

\( A + B \cdot S \rightarrow C \cdot S + D \)

\( C \cdot S \rightarrow C + S \)

\( A + S \rightarrow A \cdot S \)

\( B \cdot S \rightleftharpoons B \cdot S \)

\( A \cdot S + B \cdot S \rightarrow C + D \cdot S + S \)

\( C \cdot S \rightarrow C + S \)

\( D \cdot S \rightleftharpoons D + S \)

Go To Solution #3









                                                                                                  

Question 4 

\( A \rightleftharpoons B + C \)

The initial rate of reaction is shown below

(a) (b) (c)
Graph shows -r_A,0 increasing with P_Total and leveling off; P_C and P_B are zero. Graph shows -r_A,0 constant with increasing P_C; P_A and P_B are both 1 atm. Graph shows -r_A,0 decreasing with P_B; P_A is 1 atm and P_C is 0.

         

  1. The reaction is irreversible.

  2. Species B is on the surface.

  3. Species C is on the surface.

  1. 1 and 2 are true.

  2. 1 and 2 are false.

  3. 1 and 3 are false.

  4. 2 and 3 are false.

  5. 2 and 3 are true.

Go To Solution #4                                                                                                 









 

Question 5 

\( A \rightleftharpoons B + C \)

The initial rate of reaction is shown below

Three graphs: (A) -r_A,0 increases with P_Total then levels off (P_C = P_B = 0); (B) -r_A,0 is constant with P_C (P_A = P_B = 1 atm); (C) -r_A,0 decreases with P_B (P_A = 1 atm).

1. The rate law is

(A)

\(-r_A = \frac{k P_A}{1 + K_A P_A + K_B P_B}\)

(B)

\(-r_A = \frac{k P_A^2}{\left(1 + K_A P_A + K_C P_C\right)^2}\)

(C)

\(-r_A = \frac{k P_A^2}{\left(1 + K_A P_A + K_B P_B\right)^2}\)

(D)

\(-r_A = \frac{k P_A}{1 + K_A P_A + K_C P_C}\)

  Go To Solution #5  










Question 6

For the reaction \( A \rightleftharpoons B + C \) from Self Tests 12 and 13 we found

\( -r_A = \frac{k P_A^2}{(1 + K_A P_A + K_C P_C)^2} \)

which mechanism is consistent with the rate law

(A)

\( A + S \rightleftharpoons A \cdot S \)

\( A \cdot S \rightleftharpoons B \cdot S + C \)

\( B \cdot S \rightleftharpoons B + S \)

(B)

\( A + S \rightleftharpoons A \cdot S \)

\( A \cdot S \rightarrow B + C \cdot S \)

\( C \cdot S \rightleftharpoons C + S \)

(C)

\( A + S \rightleftharpoons A \cdot S \)

\( A \cdot S + A(g) \rightarrow B + C \cdot S \)

\( C \cdot S \rightleftharpoons C + S \)

(D)

\( A + S \rightleftharpoons A \cdot S \)

\( A \cdot S + A \cdot S \rightarrow B \cdot S + C \cdot S \)

\( C \cdot S \rightleftharpoons C + S \)

(E)

\( A + S \rightleftharpoons A \cdot S \)

\( A \cdot S + A \cdot S \rightarrow B \cdot S + C + S \)

\( B \cdot S \rightleftharpoons B + S \)

   Go To Solution #6  


 







































Solution #1

If the reaction is reversible the increase in PC will decrease the rate. If C is on the surface it will be in the denomination of the rate law and thus increasing PC will decrease the rate.

Ans: C

Back to Problem #1

Problem #2

Back to Chapter 13  

 




















Solution #2

 A is not on the surface or if it is KAPA << 1.  B is on the surface and we see as we increase PB, the rate reaches a plateau. The curves from A, B, and C partial pressure suggest a rate law of the for

Figure 1

\(-r'_A \sim \frac{1}{P_C}\)
Figure 2

\(-r_A \sim P_A\)
Figure 3

\(-r_A \sim \frac{P_B}{1 + K_B P_B}\)
Combining

\(-r'_A = \frac{k P_A P_B}{1 + K_B P_B + K_C P_C}\)
Possible Mechanism

\( B + S \rightleftharpoons B \cdot S \)

\( r_{AD} = k_A \left[ P_B C_V - \frac{C_{BS}}{K_B} \right] \)

\( A + B \cdot S \rightleftharpoons C \cdot S \)

\( r_S = k_S \left[ P_A C_A C_{BS} - \frac{C_{CS}}{K_S} \right] \)

\( C \cdot S \rightleftharpoons C + S \)

\( r_D = k_D \left[ C_{CS} - K_C P_C C_V \right] \)

\( C_{BS} = K_B P_B C_V \)

\( C_{CS} = K_C P_C C_V \)

\( r_S = k_S K_B \left[ P_A P_B - P_C \right] C_V \)

\( C_t = C_V + C_{BS} + C_{CS} = C_V (1 + K_B P_B + P_C K_C) \)

\( r_S = \frac{k (P_A P_B - P_C)}{K_P} \cdot \frac{1}{1 + K_B P_B + K_C P_C} \)

From the rate law we know species B and C are adsorbed on the surface. From the data you can’t tell whether or not C is adsorbed on the surface.

Ans: E

Back to Problem #2

Problem #3

Back to Chapter 13

 

Solution #3

\( A + B \rightleftharpoons C + D \)

Recall:
\( -r'_A = -r'_B = r'_C = r'_D \)
Figure (a) suggests:
\( -r'_A \sim \frac{P_A}{1 + K_A P_A + \dots} \quad \therefore A \text{ is on the surface} \)
Figure (b) suggests:

\( P_A = y_A P_T \)

\( P_B = y_B P_T \)

\( P_C = y_C P_T \)

\( -r'_A \sim \frac{P_T^2}{[1 + (K_A + K_B)P_T + \dots]^2} \sim \frac{P_T^2}{1 + K P_T^2} \)

Low \( P_T \):

\( -r'_A \sim P_T^2 \)

 -r'_A vs PT (low)

High \( P_T \):

\( -r'_A \sim \frac{P_T^2}{P_T^2} \sim k \)

 -r'_A vs PT (high)
Combining: -r'_A vs PT combined
\( -r'_A \sim \frac{P_A P_B}{[1 + K_A P_A + K_B P_B + \dots]^2} \)
Figure (c) suggests:
\( -r'_A \neq f(P_D) \Rightarrow \text{Reaction is irreversible and D is not on the surface} \)
Figure (d) suggests:

\( -r'_A \sim \frac{1}{1 + K_C P_C + \dots} \)

\( \therefore C \text{ is on the surface} \)
Combining all the above:
\( -r'_A = \frac{k P_A P_B}{\left[ 1 + K_A P_A + K_B P_B + K_C P_C \right]^2} \)
Therefore (b) is consistent:

\( A + S \rightleftharpoons A \cdot S \quad \Rightarrow C_A \cdot S = K_A P_A C_V \)

\( B + S \rightleftharpoons B \cdot S \quad \Rightarrow C_B \cdot S = K_B P_B C_V \)

\( A \cdot S + B \cdot S \rightarrow C \cdot S + D + S \)

\( -r'_A = r_S = k_S \left[ C_A \cdot S \cdot C_B \cdot S \right] = k_S K_A K_B P_A P_B C_V^2 \)

\( C_C \cdot S = K_C P_C C_V \)

\( C_T = C_V + C_A \cdot S + C_B \cdot S + C_C \cdot S \)

\( = C_V (1 + K_A P_A + K_B P_B + K_C P_C) \)

\( C_V = \frac{C_T}{1 + K_A P_A + K_B P_B + K_C P_C} \)

\( -r'_A = \frac{k P_A P_B}{K} \cdot \frac{1}{(1 + K_A P_A + K_B P_B + K_C P_C)^2} \)

Back to Problem #3

Problem 4

Back to Chapter 13

 

Solution #4

1.   The reaction is irreversible because when PB = 1 atm, increasing the product C does not change the rate. (TRUE)

2.   Species B is on the surface because in 1 we showed that the reaction is irreversible, therefore the only way the rate could decrease were if were in the denominator of the rate law. If a species is in the denominator of the rate law it’s on the surface. (TRUE)

3.   Species C is not on the surface because increasing PC does not affect the rate.(FALSE)
Ans: A

Back to Problem #4

Problem #5

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Solution #5

A and B are on the surface. C is not on the surface.

\( -r_A = \frac{k P_A^2}{(1 + K_A P_A + K_B P_B)^2} \)

\( P_A = y_A P_T \)

\( -r_A = \frac{k y_A^2 P_T^2}{(1 + K_A y_A P_T)^2} \)

Low PT

\( -r_A \sim P_T^2 \)

 Graph shows -r_A increasing sharply with total pressure P_T.

High PT

\( -r_A \sim \frac{k}{K_A} \)

 Graph shows -r_A constant with increasing total pressure P_T.

Combining

 Graph shows -r_A increasing with P_T and then leveling off.

\( -r_A \sim \frac{1}{\left[ 1 + \dots + K_C P_C \right]^2} \)

 Graph shows -r_A decreasing with increasing P_C.


Ans: (c)

Back to Problem #5

Problem #6

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Solution #6

\( -r_A = r_S = k_S C_A^2 S \)

\( C_A \cdot S = K_A P_A C_V \)

\( C_B \cdot S = K_B P_B C_V \)

\( C_T = C_V + K_A P_A C_V + K_B P_B C_V \)

\( C_V = \frac{C_T}{[1 + K_A P_A + K_B P_B]^2} \)

\( -r_A = \frac{k P_A^2}{[1 + K_A P_A + K_B P_B]^2} \)

Ans: (e)

Back to Problem #6

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