[jl]

I am given the following equation:

Co(Cl)₄^2-(aq) + 6H₂O <---> Co(OH₂)₆²⁺(aq) + 4Cl^-(aq)

I know that the solution is blue at elevated temperatures and pink at or below room temperature.  I also know that Co(OH₂)₆²⁺ is pink and that
Co(Cl)₄^2- is blue. 

I am asked to predict the color of the reaction mixture at 85 degrees, if the forward reaction is endothermic and to explain why.  I think that if it is endothermic then it will be blue, and pink if it is exothermic due to the color of the reactants and products.  Also, if it is endothermic, then it is absorbing heat, so the energy will be on the reactants side and vice versa.

Unfortunately this is for a lab class that is weeks ahead of my lecture so I am forced to teach myself these concepts.  I get the idea of how concentration, temperature, and catalysts can change the equilibrium in a hypothetical situation.  However, I have not seen it applied without dealing with strict values.  Anyone care to explain to get me off the ground and running?

[Yggdrasil]

Also, if it is endothermic, then it is absorbing heat, so the energy will be on the reactants side and vice versa.

This is a good starting point.  For an endothermic reaction, you can think of heat as a reactant.  Therefore, at high temperature, you have excess heat and you can treat this excess heat as an excess reactant.  So, for an endothermic reaction, do you think the addition of heat will favor the forward or the backward reaction?

[jl]

Let me see if I understand this. 

 My thoughts are that it is blue because the temperature is above room temp.  Also, it appears blue because there is more Co(Cl)₄^2- than Co(OH₂)₆²⁺ present.  Since there is more Co(Cl)₄^2-, then the equilibrium will be shifting towards the right.  The additional heat is absorbed by the products, so the reaction is exothermic.

[charco]

If the reaction as stated:

Co(Cl)42-(aq) + 6H2O <---> Co(OH2)62+(aq) + 4Cl-(aq)

becomes blue as more energy is available...  it moves to the LHS as heat is added (get more [CoCl₄]^2-)

So heat must be released as it goes to the RHS - this is the direction of EXOthermic change

[justin122289]

If you have an endothermic reaction, as mentioned before, adding heat is like adding a reactant to the system. So, adding heat will cause equilibrium to shift to the right (products) to achieve equilibrium again. (Remember that the reverse is true for exothermic reactions. I'm not quite sure what colors each compound produces, but whichever one is on the right is the color that will become more abundant.

[pizza1512]

Can someone explain why adding heat to a reversible reaction that has a exothermic forward reaction would move the equilibrium so to say 'to the left'?

[justin122289]

Can someone explain why adding heat to a reversible reaction that has a exothermic forward reaction would move the equilibrium so to say 'to the left'?

When you have an exothermic reaction, that means that heat is being released as the reactants turn into products. So, you could say that the reactants produce the procucts PLUS heat --heat can be treated as a product. Therefore, adding heat will cause the equilibrium to shift to the left to consume excess heat. The opposite is true for endothermic equilibriums. Heat can be treated as a reactant, and the addition of it shifts equilibrium to the right.