Question

Cobalt(II) cation can combine with other small ions besides the chloride anion studied in Lab 2. Consider the following equilibrium reaction and the experimental data collected for it. Co(H2O)6]2 (aq) 4 SCN-(aq) ⇌ [Co(SCN)4]2-(aq) 6 H2O(l) Experimental data: Placed in a hot water bath, the solution turns blue, which is the color of the [Co(SCN)4]2-(aq) complex. Placed in an ice bath, the solution turns light pink, which is the color of the Co(H2O)6]2 (aq) complex. Should the "heat" term be placed on the reactant side with the [Co(H2O)6]2 or on the product side with the [Co(SCN)4]2-Write a balanced chemical equation that describes the equilibrium reaction of the hexaaquocobalt(II) and tetrachlorocobalt(II) complex ions, in which the tetrachlorocobalt(II) species is a product. To do this, think about the following questions: what are you adding to the hexaaquocobalt(II) complex to convert it to the tetrachlorocobalt(II) complex? What molecule is displaced when the tetrachlorocobalt(II) complex is made?

Answer 1

Answer:

"Heat" should appear on the left-hand side of the equation (i.e. the side with $\rm [Co(H_2O)_6]^{2+}$ and $\rm SCN^{-}$.

$\rm [Co(H_2O)_6]^{2+}\; (aq) + 4\; SCN^{-1}\;(aq) \rightleftharpoons [Co(SCN)_4]^{2-}\; (aq) + 6\; H_2O\; (aq)$.

Explanation:

The hot-water bath increases the temperature of the solution. That's an external change to this equilibrium. By Le Chatelier's Principle, when there's an external change on an equilibrium, the equilibrium will shift in the direction that minimizes the change.

In this case, to minimizes the increase in temperature, the equilibrium needs to shift in the direction that absorbs heat. The question states that the direction that changes $\rm [Co(H_2O)_6]^{2+}$ to $[Co(SCN)_4]^{2-}$ is the one that absorbs heat. As a result, "heat" should be placed on that side of the equation that comes with $\rm [Co(H_2O)_6]^{2+}$.

Cobalt is a transition metal. The charge on its ions can be very dense. (High charge density.) That allows it to act as a Lewis acid and accept lone pairs. When that happens, the ion would be connected to the lone pair donors (called ligands) to produce a coordination complex. Many of these complexes are colored.

In this question, both $\rm [Co(H_2O)_6]^{2+}$ and $[Co(SCN)_4]^{2-}$ are coordination complexes. One lone pair donor can displace another. That's how the blue [Co(SCN)_4]^{2-}[/tex] ions and the pink $\rm [Co(H_2O)_6]^{2+}$ ions convert to each other.

• $\rm SCN^{-}$ ions displace water molecules in $\rm [Co(H_2O)_6]^{2+}$ ions to produce $\rm [Co(SCN)_4]^{2-}$ ions.
• Water molecules displace $\rm SCN^{-}$ ions in $\rm [Co(SCN)_4]^{2-}$ ions to produce $\rm [Co(H_2O)_6]^{2+}$ ions.'

The internal structure of the ligands ($\rm SCN^{-}$ ions and water molecules) did not change in this reaction. Hence, treat each of them as a single atom. Apply the conservation rules to balance the equation.