An irreversible first-order, liquid-phase reaction is carried out in a 18.-L adi

Question

An irreversible first-order, liquid-phase reaction is carried out in a 18.-L adiabatic CSTR. The concentration of reactant in the feed is 3.0 M and the volumetric flow rate 60 cm3/sec, with no product in the feed. The density and specific heat of the reaction mixture are constant at 1.0 g/cm3 and 1.0 cal/g-°C, respectively. The molar heat of reaction is = -50 kcal/mol, and the rate constant has a pre-exponential factor of 4.48×106 s-1 and an activation energy of 15 kcal/mol.

Determine the possible conversions and temperatures corresponding to steady states for a feed temperature of 25 °C. Which are most likely to be stable steady states?

Explanation / Answer

V = 18 L = 18000 cm3

CAo = 3 M

F = 60 cm3/s = 60*10-3 L/s = 0.06 L/s

D = 1g/cm3 = 1000 kg/m3

Cp = 1 cal/gC = 4.186 kJ/molK

HR = -50 kcal/mol

k = 4.48*106 s-1

E = 15 kcal/mol

T = 25 C

Equation for CSTR is:

V/F = XA/-rA

V = volume of reactor

F = volumetric flowrate

XA = conversion

-rA = rate expression

its a first order liquid phase reaction

-rA = kCA

CA = CAo*(1-XA)

= 3*(1-XA)

now put values in equation

18000/60 = XA / (4.48*106*3*(1-XA))
solve for XA to get

XA = 0.996

From arrhenius equation

ln(K2/K1) = (HR/R)* ((1/T1) - (1/T2))

K1 = reaction constant at 298 K = 4.48*106

K2 = K1*eE/RT

K2 = 4.48*106 *e(15000/(1.986*T2)

HR = molar heat of reaction

R = gas constant = 1.987 cal/molK

T1 = 25 C = 298K

T2 = to find

substitute values to get:

ln(4.48*106 *e(15000/(1.986*T2) / 4.48*106) = (-50000/1.986)*((1/298)-(1/T2))

solve for T2 to get

T2 = 208.6 K

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