Abstract

In this study, using coconut fibers as raw material, activated carbon fibers were prepared via carbonization and KOH activation processes. The morphology, composition, specific surface area, pore structure and thermal stability of the resulting activated carbon fibers were systematically characterized. It was found that the activation process increases the specific surface area of carbon fibers to a greater extent via formation of a large number of micropores (0.7–1.8 nm) and a certain amount of slit-shaped mesopores (2–9 nm). The specific surface area and the pore volume of the activated carbon fibers reach 1556 m2 g−1 and 0.72 cm3 g−1, respectively. The activation process can also decompose the tar deposits formed after the carbonization process by pyrolysis, making the surface of the activated carbon fibers smoother. To study the adsorption properties of the as-prepared activated carbon fibers, the adsorption capacities and adsorption kinetics of various organic dyes including methylene blue, Congo red and neutral red were investigated. The adsorption capacities of the dyes increased with the increasing initial dye concentrations, and varied greatly with the pH value of the system. In methylene blue and neutral red systems, the adsorption capacities reach the maximum at pH 9, and in the Congo red system, it reaches the maximum at pH 3. The adsorption capacities of the activated carbon fibers in methylene blue, Congo red and neutral red systems reached equilibrium at 150, 120, and 120 min, and the maximum adsorption capacities were 21.3, 22.1, and 20.7 mg g−1, respectively. The kinetics of the adsorption process was investigated using three models including pseudo-first-order, pseudo-second-order and intraparticle diffusion models. The results indicated that the dynamic adsorption processes of coconut-based activated carbon fibers to methylene blue, Congo red and neutral red were all in accordance with the second-order kinetic model, and the equations are as follows: t/Qt = 0.1028 + t/21.3220, t/Qt = 0.1128 + t/21.5982 and t/Qt = 0.0210 + t/20.6612.

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