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I'm considering the following reaction:

$\ce{ZnSO4} + 2 \ce{KCl} \ce{->} \ce{ZnCl2} + \ce{K2SO4}$

The high solubility of $\ce{ZnCl2}$ would favour the $\ce{ZnSO4}$ & $\ce{KCl}$ direction of the reaction, but I'm wondering if I could turn this around by making the reaction occur in a flow of boiling hot water.

The idea would be to subject (initially dry) $\ce{ZnSO4}$ & $\ce{KCl}$ stationary phase to a flow of water in the 80-100°C range. The comparative solubility of $\ce{ZnCl2}$ at this temperature being high, I'm wondering if it being comparatively flushed away more than the other species could help drive the reaction forwards in the opposite direction?

I wouldn't get pure $\ce{ZnCl2}$ of course but I'd be happy if I could get a 90% $\ce{ZnCl2}$ solution from dry $\ce{ZnSO4}$ & $\ce{KCl}$.

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    $\begingroup$ There is no reaction, nor is there a need for one. A solution of ZnSO4 and KCl is the same as solution of ZnCl2 and K2SO4. $\endgroup$ – Ivan Neretin Aug 12 at 18:07
  • $\begingroup$ Is it not called a double-replacement reaction? $\endgroup$ – Hans Aug 12 at 18:23
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    $\begingroup$ It would be, if some of the participating salts were insoluble. That's not the case, though. $\endgroup$ – Ivan Neretin Aug 12 at 18:25
  • $\begingroup$ If one of the salts can be otherwise partly isolated from the reaction medium, is it not somewhat equivalent to one being insoluble? Could the "reaction" proceed by one of the products being comparatively carried away many times more than the reagents combined? $\endgroup$ – Hans Aug 12 at 18:29
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    $\begingroup$ It is lower, not higher solubility, that might separate one of the products from the reaction mixture and thus drive the reaction that way. With all four salts rather well-soluble, we're out of luck. $\endgroup$ – Ivan Neretin Aug 12 at 18:47

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