Question

EXAMPLE 1000 L back-sti CHCOOCHET The OLE 4 (HOMEWORK): Acetate is saponified in back-stirred flow reactor. DOCH + NaOH → C2H5OH + CH3COONa rate of reaction 0.1(L.mol'.s>')[CH3COOC2H5 ][NaOH). There is no noe in density. The flowrate of initial feed solution is 100 L.s”; initial concentration is CH.COOC2H5]=[NaOH)=1 mol.L-?. Initially, there is no product. Determine the concentrations of reactants and r = 0.1(L.mor1 change in Ip = 100 product in output stream.

Answer 1

CH₂ 100GHs + Naol a Colon & CH₂ Coolna o ra 0.1. [C1 1000Hs] [Naon ra Ool CAG date of decomposition of A = date of decomposition of 3 - 0.1 for back-stired flow reacter- T = SA SA U. 0.162 - volume offection mixture y lumtr flow rate Ao initial con 1oooh 100L/ 0.1 loxo. 19 + GA - 1 = 0 =0 A = -1 zity not possible Cance of reactents & product = 0.618 = -1.618 in outlet stream 19-0.618 mel 17 (CH₂ 100GHs] = 0.618 melal TG = 0.618 melk I [NaOH = 0.618 melko 16 M50H ) = 0.389 melle / E T=07618 = 0.38194 Ir Ha cooNa] = 0.382 molk G = G = 0.382 mol/l