Question

The reaction between ammonia and formaldehyde to produce hexamine is

4NH3 + 6 HCHO ->(CH2)6N4 + 6 H2O

A 0.5 liter CSTR is used for the reaction. Each reactant is fed to the reactor in a separate stream, at the rate of 1.5x10-3 liter/s each. The ammonia concentration is 4.0 mol/liter and the formaldehyde concentration is 6.4 mol/liter. The reactor temperature is 36C. Calculate the concentration of ammonia and formaldehyde in the effluent stream. In kinetics below A is ammonia and B is formaldehyde.

(-rA) = kCACB^2 mole A/liter-s where k = 1,420 exp (-3090/T)

Answer 1

Find out the limiting reactant of the reaction

From the stoichiometry of the reaction

4 mol/L NH3 require = 6 mol/L HCHO

But we have 6.4 mol/L of HCHO

NH3 is the limiting reactant

A = NH3

B = HCHO

Rate constant k = 1,420 exp (-3090/T)

k = 1,420 exp (-3090/(36+273))

= 0.06446 mol2 L-2 s-1

(-rA) = kCACB^2

(-rA) = kCA0(1-X)[CB0 - 1.5 CA0 X]²

(-rA) = 0.06446 * 4 *(1-X)*[6.4 - 1.5 * 4 * X]²

(-rA) = 0.25784 *(1-X)*[6.4 - 6 * X]²

Molar flow rate of A

FA0 = CA0 * vA = 4 mol/L x 1.5x 10^-3 L/s

= 0.006 mol/s

Design equation for CSTR

V/FA0 = X/-rA

0.5/0.006 = X/(0.25784 *(1-X)*[6.4 - 6 * X]²)

21.486 = X/[(1-X)*(40.96 + 36X² - 76.8X)]

21.486*[(1-X)*(40.96 + 36X² - 76.8X)] = X

21.486*[40.96 + 36X² - 76.8X - 40.96X - 36X³ + 76.8X²] = X

880.06 + 2423.6208X² - 1649.1248 X - 773.496X³ = 0

X = 0.34

Concentration of NH3 in effluent stream

CA = CA0(1-X)

= 4(1-0.34) = 2.64 mol/L

Concentration of HCHO in effluent stream

CB = CBO - 1.5CA0(X)

= 6.4 - 1.5*4*0.34

= 4.36 mol/L