The irreversible endothermic vapor-phase reaction follows an elementary rate law CH3COCH3 CH2CO + CH4 A

Question:

The irreversible endothermic vapor-phase reaction follows an elementary rate law
CH3COCH3 → CH2CO + CH4
A → B + C and is carried out adiabatically in a 500-dm3 PFR. Species A is fed to the reactor at a rate of 10 mol/min and a pressure of 2 atm. An inert stream is also fed to the reactor at 2 atm, as shown in Figure P11-6B. The entrance temperature of both streams is 1100 K.FAD 10 mol/min =  10 CA01

Additional information:k= exp (34.34)-34222/T)1/s Cp/ 200 J/mol. K (T in degrees Kelvin, K) CPA = 700 J/mol K Cpc = 80 J/mol - K CPB

a. First derive an expression for CA01 as a function of CA0 and ΘI.
b. Sketch the conversion and temperature profiles for the case when no inerts are present. Using a dashed line, sketch the profiles when a moderate amount of inerts are added. Using a dotted line, sketch the profiles when a large amount of inerts are added. Qualitative sketches are fine. Describe the similarities and differences between the curves.
c. Sketch or plot and then analyze the exit conversion as a function of ΘI. Is there a ratio of the entering molar flow rates of inerts (I) to A (i.e., ΘI = FI0/FA0) at which the conversion is at a maximum? Explain why there “is” or “is not” a maximum.
d. What would change in parts (b) and (c) if reactions were exothermic and reversible with ΔHRx∘ = –80 kJ/mol and KC = 2 dm3/mol at 1100 K?
e. Sketch or plot FB for parts (c) and (d), and describe what you find.

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: