Consider the following system of gas-phase reactions: 3 1/2 1/2 k -0.004(mol/dm
ID: 1074073 • Letter: C
Question
Consider the following system of gas-phase reactions: 3 1/2 1/2 k -0.004(mol/dm min k, C A k -0.3 min 2 A ry k CA 0.25 dm3/mol min A Y B is the desired product, and X and Y are foul pollutants that are expensive to get rid of. The specifie reaction rates are at 27oC. The reaction system is to be operated at 27°C and 4 atm. Pure A enters the system at a volumetric flow rate of 10 dm3/min. (a) Sketch the instantaneous selectivities (S x,Say, and S rBorx+rr)) as a function of the concentration of CA. (b) Consider a series of reactors. What should be the volume of the first reactor? (c) What are the effluent concentrations of A, B, X, and Y from the first reactor?Explanation / Answer
a) Selectivity tells us how one product is favored over another when multiple reactions take place.
SB/X = rB/rX = k2CA/ k1CA1/2 = k2. CA1/2 / k1
SB/Y = rB/rY = k2CA / k3 CA2 = k2 CA-1/ k3
SB/XY = rB/rX + rY =
k2CA / k1CA1/2 +k3 CA2
b) PV = nRT
T = 300.5 K, P = 4 atm, V = ? n = 10 /22.4 = 0.446, R = 0.0821
4*V = 0.446 * 0.0821* 300.5
V = 2.75 L
C) Initial concentration CA = moles/volume = 0.446/2.75 = 0.1621 moles/L
rx = 0.004 * (0.1621)^1/2= 0.0016
rB = 0.3 * 0.1621 = 0.048
ry = (0.1621)^2 * 0.25 = 0.0065
net rate of disappearence of A = rx +rB +rY = 0.0016+0.048 +0.0065 = 0.056 mol/min
residence time = volume /velocity = 2.75 /10 = 0.275 min.
concentration of effluent A = 0.275 *0.056 = 0.0154, 0.1621 -0.0154 = 0.1467 mol
total B, X and Y = 0.0154 moles