Chapter 4: Stoichiometry

Matching CA vs. X Curves

Match the following concentration- conversion curves (a-g) with the corresponding reaction and reaction conditions for a flow reactor at constant temperature and pressure.

(a) Graph showing the relationship between CB (concentration of B) and X (conversion). The plot is a straight line with a positive slope from (0, 0) to (1.0, 1.0). (b) Graph showing the relationship between CB (concentration of B) and X (conversion). The plot is a straight line with a negative slope, starting at (0, 1.0) and ending at (1.0, 0). (c) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve starts at (0, 1.0) and decreases non-linearly, approaching zero as X approaches 1.0, with a sudden drop at X = 1.0.
(d) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve starts near zero at X = 0, increases non-linearly, and approaches 0.5 as X approaches 1.0, with a sharp rise at X = 1.0. (e) Graph depicting the relationship between CB (concentration of B) and X (conversion). The curve starts at CB = 1.0 when X = 0, decreases gradually in a nonlinear manner, and approaches zero as X reaches 1.0. (f) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve begins at CB = 1.0 when X = 0 and decreases sharply at first, then gradually flattens out, approaching zero as X approaches 1.0.
(g) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve starts at a low CB value and increases gradually until it plateaus at CB = 0.5 as X approaches 1.0.

 

1) Gas phase, Pure A, CA0=1                         

\(A \longrightarrow B + C\)

A   B   C   D   E   F   G

2) Gas phase, CA0=CB0=1                             

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

A   B   C   D   E   F   G

3) Liquid phase, Pure A, CA0=1                      

\(A \longrightarrow B + C\)

A   B   C   D   E   F   G

4) Gas phase, Stoichiometric Feed, CA0=1/2       

\(A + 2B \longrightarrow C\)

A   B   C   D   E   F   G

5) Gas phase, CA0=CB0=0.5                           

\(A + \frac{1}{2}B \longrightarrow 4C\)

A   B   C   D   E   F   G

Full Solution

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 1

 

\(C_B = \frac{F_B}{\nu} = \frac{F_{A0} X}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} X}{1 + \varepsilon X}, \, \varepsilon = y_{A0} \delta = (1)(1 + 1 - 1) = 1\)

\(C_B = \frac{X}{1 + X} \)


Ans(G)

Back to Problem 1

Problem 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 2

 

\(C_B = \frac{F_B}{\nu} = \frac{F_{B0} \left( \theta_B - X \right)}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} \left( \theta_B - X \right)}{1 + \varepsilon X}, \, \varepsilon = 1(1 + 1 - 1 - 1) = 0, \, \theta_B = \frac{1}{1} = 1\)

\(C_B = (1 - X) \)

Ans(B)

Back to Problem 2

Problem 3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 3

 

\(C_B = \frac{F_B}{\nu} = \frac{F_B}{\nu_0} = \frac{F_{A0} X}{\nu_0} = C_{A0} X\)

Ans(A)

Back to Problem 3

Problem 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 4

 

\(C_B = \frac{F_B}{\nu} = \frac{F_{A0} \left(\theta_B - 2X\right)}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} \left(\theta_B - 2X\right)}{(1 + \varepsilon X)}, \, \varepsilon = \frac{1}{3}(1 - 2 - 1) = -\frac{2}{3}, \, \theta_B = \frac{2}{1} = 2\)

\(C_B = \frac{(1 - X)}{\left(1 - \frac{2}{3}X\right)} \)

Ans(E)

Back to Problem 4

Problem 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 5

\(C_B = \frac{F_B}{\nu} = \frac{F_{A0} \left(\theta_B - 2X\right)}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} \left(\theta_B - \frac{1}{2}X\right)}{1 + \varepsilon X}, \, \varepsilon = \frac{1}{2} \left(4 - 1 - \frac{1}{2}\right) = \frac{5}{4}\)

\(C_B = \frac{\left(1 - \frac{1}{2}X\right)}{\left(1 + \frac{5}{4}X\right)} \)

Ans(C)

Back to Problem 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Match the following concentration- conversion curves (a-g) with the corresponding reaction and reaction conditions for a flow reactor at constant temperature and pressure.

1) Gas phase, Pure A, CA0=1

\(A \longrightarrow B + C\)

2) Gas phase, CA0=CB0=1

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

3) Liquid phase, Pure A, CA0=1

\(A \longrightarrow B + C\)

4) Gas phase, Stoichiometry Feed, CA0=1/2

\(A + 2B \longrightarrow C\)

5) Gas phase, CA0=CB0=0.5

\(A + \frac{1}{2}B \longrightarrow 4C\)

(a) Graph showing the relationship between CB (concentration of B) and X (conversion). The plot is a straight line with a positive slope from (0, 0) to (1.0, 1.0). (b) Graph showing the relationship between CB (concentration of B) and X (conversion). The plot is a straight line with a negative slope, starting at (0, 1.0) and ending at (1.0, 0). (c) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve starts at (0, 1.0) and decreases non-linearly, approaching zero as X approaches 1.0, with a sudden drop at X = 1.0.
(d) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve starts near zero at X = 0, increases non-linearly, and approaches 0.5 as X approaches 1.0, with a sharp rise at X = 1.0. (e) Graph depicting the relationship between CB (concentration of B) and X (conversion). The curve starts at CB = 1.0 when X = 0, decreases gradually in a nonlinear manner, and approaches zero as X reaches 1.0. (f) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve begins at CB = 1.0 when X = 0 and decreases sharply at first, then gradually flattens out, approaching zero as X approaches 1.0.
(g) Graph showing the relationship between CB (concentration of B) and X (conversion). The curve starts at a low CB value and increases gradually until it plateaus at CB = 0.5 as X approaches 1.0.

Solution



1)

\(C_B = \frac{F_B}{\nu} = \frac{F_{A0} X}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} X}{1 + \varepsilon X}, \, \varepsilon = y_{A0} \delta = (1)(1 + 1 - 1) = 1\)

\(C_B = \frac{X}{1 + X} \)

Ans(G)
2)

\(C_B = \frac{F_B}{\nu} = \frac{F_{B0} \left( \theta_B - X \right)}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} \left( \theta_B - X \right)}{1 + \varepsilon X}, \, \varepsilon = 1(1 + 1 - 1 - 1) = 0, \, \theta_B = \frac{1}{1} = 1\)

\(C_B = (1 - X) \)

Ans(B)
3)

\(C_B = \frac{F_B}{\nu} = \frac{F_B}{\nu_0} = \frac{F_{A0} X}{\nu_0} = C_{A0} X\)

Ans(A)
4)

\(C_B = \frac{F_B}{\nu} = \frac{F_{A0} \left(\theta_B - 2X\right)}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} \left(\theta_B - 2X\right)}{(1 + \varepsilon X)}, \, \varepsilon = \frac{1}{3}(1 - 2 - 1) = -\frac{2}{3}, \, \theta_B = \frac{2}{1} = 2\)

\(C_B = \frac{(1 - X)}{\left(1 - \frac{2}{3}X\right)} \)

Ans(E)
5)

\(C_B = \frac{F_B}{\nu} = \frac{F_{A0} \left(\theta_B - 2X\right)}{\nu_0 (1 + \varepsilon X)} = \frac{C_{A0} \left(\theta_B - \frac{1}{2}X\right)}{1 + \varepsilon X}, \, \varepsilon = \frac{1}{2} \left(4 - 1 - \frac{1}{2}\right) = \frac{5}{4}\)

\(C_B = \frac{\left(1 - \frac{1}{2}X\right)}{\left(1 + \frac{5}{4}X\right)} \)

Ans(C)

Back to Chapter 4