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Kinetics of hydrochloric acid leaching of smithsonite
Authors:Nikhil Dhawan  M Sadegh Safarzadeh  Mustafa Birinci
Affiliation:1.Department of Metallurgical Engineering,PEC University of Technology,Chandigarh,India;2.Laboratory for Leaching and Purification Processes, R&D center,The Iranian Zinc Mines Development Company (IZMDC),Zanjan,Iran;3.Research and Engineering Company for Non-ferrous Metals (RECO) under IZMDC,Zanjan,Iran;4.Mining Engineering Department, Faculty of Engineering,? n?nü University,Malatya,Turkey
Abstract:The dissolution kinetics of smithsonite ore in hydrochloric acid solution has been investigated. As such, the effects of particle size (−180 + 150, −250 + 180, −320 + 250, −450 + 320 μm), reaction temperature (25, 30, 35, 40, and 45°C), solid to liquid ratio (25, 50, 100, and 150 g/L) and hydrochloric acid concentration (0.25, 0.5, 1, and 1.5 M) on the dissolution rate of zinc were determined. The experimental data conformed well to the shrinking core model, and the dissolution rate was found to be controlled by surface chemical reaction. From the leaching kinetics analysis it can be demonstrated that hydrochloric acid can easily and readily dissolve zinc present in the smithsonite ore, without any filtration problems. The activation energy of the process was calculated as 59.58 kJ/mol. The order of the reaction with respect to HCl concentration, solid to liquid ratio, and particle size were found to be 0.70, −0.76 and −0.95, respectively. The optimum leaching conditions determined for the smithsonite concentrate in this work were found to be 1.5 M HCl, 45°C, −180 + 150 μm, and 25 g/L solid to liquid (S/L) ratio at 500 rpm, which correspond to more than 95% zinc extraction. The rate of the reaction based on shrinking core model can be expressed by a semi-empirical equation as:
$1 - \left( {1 - X} \right)^{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-\nulldelimiterspace} 3}} = k_0 \left {HCl} \right]^{0.70} \left( {\frac{S} {L}} \right)^{ - 0.76} r_0^{ - 0.95} \exp \left( {\frac{{ - 59.58}} {{RT}}} \right)t.$1 - \left( {1 - X} \right)^{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-\nulldelimiterspace} 3}} = k_0 \left {HCl} \right]^{0.70} \left( {\frac{S} {L}} \right)^{ - 0.76} r_0^{ - 0.95} \exp \left( {\frac{{ - 59.58}} {{RT}}} \right)t.
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