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Kinetic Study and Mathematical Model of Hemimorphite Dissolution in Low Sulfuric Acid Solution at High Temperature
Authors:Hongsheng Xu  Chang Wei  Cunxiong Li  Zhigan Deng  Minting Li  Xingbin Li
Affiliation:1. State Key Laboratory of Complex Nonferrous Metal Resources Cleaner Utilization, Faculty of Metallurgical and Energy Engineering, Kunming University of Science and Technology, Kunming, 650093, P.R. China
Abstract:The dissolution kinetics of hemimorphite with low sulfuric acid solution was investigated at high temperature. The dissolution rate of zinc was obtained as a function of dissolution time under the experimental conditions where the effects of sulfuric acid concentration, temperature, and particle size were studied. The results showed that zinc extraction increased with an increase in temperature and sulfuric acid concentration and with a decrease in particle size. A mathematical model able to describe the process kinetics was developed from the shrinking core model, considering the change of the sulfuric acid concentration during dissolution. It was found that the dissolution process followed a shrinking core model with “ash” layer diffusion as the main rate-controlling step. This finding was supported with a linear relationship between the apparent rate constant and the reciprocal of squared particle radius. The reaction order with respect to sulfuric acid concentration was determined to be 0.7993. The apparent activation energy for the dissolution process was determined to be 44.9 kJ/mol in the temperature range of 373 K to 413 K (100 °C to 140 °C). Based on the shrinking core model, the following equation was established: $$ 1.21\ln \left( {1 - 0.83x} \right) - \left {1.02\ln \frac{{0.35 + \left( {1 - x} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern-0pt} 3}}} - 0.59\left( {1 - x} \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-0pt} 3}}} }}{{0.35 + \left( {1 - x} \right)^{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern-0pt} 3}}} + 1.18\left( {1 - x} \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-0pt} 3}}} }} + 3.52\arctan \left( {1.96\left( {1 - x} \right)^{{{1 \mathord{\left/ {\vphantom {1 3}} \right. \kern-0pt} 3}}} - 0.58} \right)} \right] + 2.06 = 42,192.59{\text{e}}^{{ - \frac{44,900}{{{\text{R}}T}}}} t. $$
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