4. Given the following equation Cu(H2O)62+ (aq, blue) + 4Br(aq) CuBr42-(aq, gree

Question

4. Given the following equation Cu(H2O)62+ (aq, blue) + 4Br(aq) CuBr42-(aq, green) + 6H20(1) If the mixture begins as a light blue solution, what changes (if any) should be observed after doing each of the following? In each case, explain your answer. a) NaBr is added b) An insoluble copper (I) salt is added c) Pb2* ion was added, which can react with bromide to form PbBr2(s) d) If heating the solution caused the color to become dark blue, is the system most likely to be exothermic or endothermic? Explain

Explanation / Answer

a) NaBr is added : Color will start turning green

As NaBr is added, the concentration of Br- increases, and the reaction will proceed so as to reduce this effect ( Le Chatelier's principle) ; hence the reaction will go on the right side to reduce Br- concentration and the color will soon start turning towards green,

b) An insoluble cupric salt is added :

Due to common ion effect of Cupric ions ,the equilibirium will shift to the right and green color will appear.

c)Pb+2 is added : color will start turning blue

As PbBr2 will be formed which will start reducing concentration of Bromide ions and hence reaction will proceed to increase Bromide ion concentration (by Le Chatelier's principle); hence reaction will proceed backwards and color will start turning blue

d) System is exothermic.

Heat can either be a reactant(endothermic reaction) or a product ( exothermic reaction). If adding heat increases the backward reaction ; it implies the heat is a product in this reaction and hence it is an exothermic system.

This can be understood this way also.

If on heating, the color turned blue, it implies that reaction proceeded backwards on heating. On heating, the reaction will try to reduce heat by going towards endothermic side ( Le Chatelier's principle) ; hence backwards is endothermic ; hence forward reaction is Exothermic.

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