What happens when CoCl2 is added to the following system?
Co(H20)6 + Cl + heat <-> CoCl4 + 6H20?
When CoCl2 is added to the system represented by the equation:
Co(H2O)6 + Cl + heat ↔ CoCl4 + 6H2O
It will react with the other components of the system, leading to a shift in the equilibrium position of the reaction. Let's break down the reaction to understand what happens:
On the left side of the reaction, we have Co(H2O)6, which is a hydrated cobalt(II) ion, Cl, a chloride ion, and heat, which is a source of energy.
On the right side of the reaction, we have CoCl4, which is a tetrachlorocobaltate(II) ion, and 6H2O, which represents water molecules.
The addition of CoCl2 to the system provides additional cobalt(II) ions (Co2+), which can react with the chloride ions (Cl-) to form CoCl4, forming tetrachlorocobaltate(II) ions.
The increase in the concentration of cobalt(II) ions and chloride ions will cause a shift in the equilibrium position of the reaction in accordance with Le Chatelier's principle. Specifically, the equilibrium will shift to the right, favoring the formation of tetrachlorocobaltate(II) ions (CoCl4), and water molecules (H2O).
In summary, when CoCl2 is added to the system, it will react with the other components, increasing the concentration of cobalt(II) ions and chloride ions, causing a shift in the equilibrium position to the right, resulting in the formation of CoCl4 and water.