Copper, present as an impurity in liquid Pb, can be removed by adding PbS to the

Question

Copper, present as an impurity in liquid Pb, can be removed by adding PbS to the Cu-Pb alloy and allowing the exchange reaction: 2Cu(s) + PbS(s) = Cu2S(s) + Pb(l) to come to equilibrium. The solid sulfides are mutally immiscible, Pb is insoluble in solid Cu, and the Cu liquids, below 850 C, can be represented by

where XCu is the solubility of Cu in liquid Pb. If Cu obeys Henry's law in liquid Pb, calculate the extent to which Cu can be removed from liquid Pb by this process at 800 C. Would the extent of purification of the lead be increased by increasing or by decreasing the temperature?

Answer: "XCu =0.017, Increasing T decreases the extent to which Cu is removed"

Can I please get the steps to arrive at this solution?

Explanation / Answer

As per the given equation

log XCu = -3500/T + 2.261

when T = 800C

log XCu = -4.375 + 2.261 = -2.114

XCu = 10-2.114 = 0.0077

As per the given relation, increasing the temp T would increase XCu. Since the solubility of Cu would increase this should result in an increase in the purification of lead with increasing temperature.

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