A student must determine if the forward reaction of the following equilibrium is endo$-$ or exothermic.

$Co(H_{2}O{)}_6^{2+}(aq)+4C{l}^{-}(aq)CoC{l}_4^{2-}(aq)+6H_{2}O(I),K=1\mathrm{.}7{10}^{-3}$

$Co(H_{2}O{)}_6^{2+}(aq)$ appears pink, while $CoC{l}_4^{2-}(aq)$ appears blue.

The student places a test tube containing the equilibrium mixture in a hot water bath and the solution turns blue. When placed in an ice bath, the mixture becomes pink.

Is the forward reaction endo$-$ or exothermic?

endothermic

exothermic

Why?

If the reaction is $,,$ then heat behaves as a $|.$

Heating the mixture will add heat to the

side of the reaction and drive the reaction

$,$ creating more $---$Select$---,$ compound. The value of $K$

Cooling the mixture will remove heat from the

side of the reaction and drive the reaction

$,$ creating more

compound. The value of