First, free energy data

The components considered for sulfur and the data used are listed in Table 1. Data containing the copper component under consideration are shown in Table 2 below.

Table 1 Free energy of each substance in the S-H 2 O system

substance

S 2 -

HS -

H 2 S(aq)

SO 4 2 -

HSO 4 -

H 2 O(1)

△G Θ ∕kJ

+111.4

+12.08

-27.83

-744.53

-755.91

-237.129

Table 2 Free energy data for the formation of various copper-containing components

substance

Cu +

Cu 2 +

CuO

Cu 2 O

CuO 2 2 -

HcuO -

Cu 2 S

CuS

△G Θ ∕kJ

49.98

65.49

-129.7

-146.0

-183.6

-258.5

-86.2

-53.6

Second, the equilibrium equation

Third, the advantage zone map and its analysis

Figure 1 is a diagram of the Cu-S-H 2 O dominant zone in the solution with a sulfur-containing activity of 0.1 mol ∕L and a copper ion activity of 10 -3 and 10 -6 mol ∕L. . The Cu-H 2 O system dominant zone map of the same activity value has a stable zone free of sulfides, and the boundary between Cu 2 + and Cu, ie, equilibrium 7, extends to the axis of pH=0. That is, CuS and Cu 2 S region are non-occurrence of the region in FIG advantage when the presence area of the metal copper sulfide stationary phase, and extends slightly Cu 2 + region. There is no Cu + stable region in the figure, so |Cu + | is less than 10 -6 mol∕L at 298.15K. The low activity ion balance produced by copper oxide and H 2 CuO 2 at pH=13 or higher.

Figure 1 Cu-S-H 2 O dominant area map

(298K, S and H 2 S activity is 10 -1 , Cu ion activity is 10 -3 or 10 -6 )

If FIG. 2 on the activity of sulfur species to elemental sulfur is 10 -1 to 1 Videos stable region, all it within the stability region of CuS. Therefore, elemental sulfur can coexist thermodynamically with CuS at 298.15 K, but does not coexist with Cu 2 S. The figure clearly indicates that under acidic oxidation conditions, the chalcopyrite Cu 2 S and copper blue CuS can theoretically be leached to produce a solution containing Cu 2 + but not containing Cu + , and the contained sulfur is oxidized to Elemental sulfur, or oxidized to sulfate at a higher potential E. The conditions required for the reaction to proceed are that the potential E value in the solution is maintained at a relatively high value so that no H 2 S or its dissociated ions are present in the solution. The boundary line between HSO 4 - and SO 4 2 - is also plotted on the graph, because the equilibrium line between Cu 2 S and Cu 2 + represented by equilibrium (k) and formula (l) changes the slope at this pH. The equilibrium line between Cu 2 S and CuS represented by the equilibrium formula (q) and the formula (r) also changes the slope, but is hardly seen on the graph.

Figure 2 S-H 2 O dominant area map

(298K, S and H 2 S activities are 10 -1 or 10 -4 respectively )

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