I know that anhydrous $\ce{CaSO4}$ is used as drying agent.

I also know that anhydrous $\ce{CaSO4}$ (dead burnt plaster) is prepared by heating gypsum $\ce{CaSO4.2H2O}$ at a temperature above $\pu{393K}$, which is an irreversible process.

How does $\ce{CaSO4}$ perform the drying if it can't accept any $\ce{H2O}$ molecules? Is there some other mechanism here? Or do I have some wrong information fed in my brain?

  • $\begingroup$ How are you not answering your own question? Don't you think you just described the two opposite processes (water absorption and volatilization)? $\endgroup$ Mar 14 '18 at 5:24

I think your problem is with the idea of "irreversible" process.

The term irreversible here means that, at $\pu{393 K}$ and in the conditions where this is performed, the reaction

$$\ce{CaSO4.2H2O <=> CaSO4 + 2 H2O}$$

is pushed towards products so much, that it can be considered "one-way" only:

$$\ce{CaSO4.2H2O -> CaSO4 + 2 H2O}$$

This is dependent on the temperature, and likely other conditions: for example, water exists as a gas at this temperature at a pressure of $\pu{1 atm}$ or lower. If you remove some of the water while heating (such as vacuum, or by circulating dry air around to get the water out), then you further push the equilibrium to the right (and make the backwards reaction impossible since water is no longer present, thus making the process irreversible).

Once you have your dry $\ce{CaSO4}$, it can be cooled down to room temperature (in a dessicator, to avoid it absorbing water). At temperatures below $\pu{393 K}$, the reaction will work backwards:

$$\ce{CaSO4 + 2 H2O -> CaSO4.2H2O}$$

Thus, your dry $\ce{CaSO4}$ will absorb water from the environment, becoming a drying agent. You may regenerate it by reheating it to $\pu{393 K}$, and so on and so forth.


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