The reversible gas-phase elementary reaction is carried out in an isothermal CST

Question

The reversible gas-phase elementary reaction is carried out in an isothermal CSTR with no pressure drop. The feed enters at a volumetric flow rate of upsilon_0 = 5000 dm^3/hour. The feed consists of half ethyl benzene (i.e., A) and half inserts on a molar basis and is well mixed before it enters the reactor (I). The pressure in the reactor is 6 atm (so P_A0 = 3 atm and P_0 = 3 atm, making the entering concentration of ethyl benzene, A, C_A0 = 0.04 mol/dm^3). The molar flow rate of A is F_A0 = 200 mol/hr. At the reaction temperature of 640 degree C, the rate constant, k_A, is 5.92 mol/dm^3 middot hr middot atm. The equilibrium constant, K_P, is 9 atm and the corresponding equilibrium conversion is X_c = 0.84. Reference: Won Jae Lee and Gilbert F. Foment. Ind. Eng Chem. Res. 2008, 47, pp. 9183-9194. (a) Write out each step of the algorithm. (b) Write the rate of reaction, -r_A solely as a function of P_AD X, K_P and k. (c) Calculate the reactor volume necessary to achieve 90% of the equilibrium conversion, X_e. (d) How would the conversion from part (a) be affected if the reactor diameter increased and height decreased but total volume remained the same? Explain.

Explanation / Answer

(a)

FA0=Initial molar flow rate of A=200 mol/hr

CA0=Initial concentration of A=0.04 mol/dm3

v0=volumetric flow rate=5000 dm3/hr

rate constant kA=5.92 mol/dm3.hr.atm

kp=9 atm

T=640 0C

xe=0.84

Now taking balance

FA=FA0(1-Xe)=200*(1-0.84)=32 mol/hr

CA=CA0(1-Xe)=0.04*(1-0.84)=0.0064 mol/dm3

(b)

-rA=kA CA-k2CBCc

A=EB

B=styrene

C=H2

In terms of pressure ,According to ideal gas C= n/V=P/RT

rA=(kA PA-k2PBPc)/RT

We have equilibrium const kp

kp=kA/k2

So

-rA=kA (PA-PBPc/kp)

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