Question

Chemistry 102 EXPERİMENT3 Equilibrium& Le Chatelier's Principle tive Qualita of this experiment is to study qualitative aspects of chemical systems in dynamic equilibrium. in doing this experiment, you will learm the operation of Le Chateliers principle. n Vslowly add 1 to 2 drops of dilute 6 M sodium hydroxide solution to a solution of iodine in a test tube and observe any color change. In an aqueous solution of iodine, the following equilibrium is established 12(aq)+ H200 2 H (aq)+I (aq) + I0 (aq) Any color in the solution can be regarded as entirely due to the presence of 2 (aq). The OH- (aq) ions from the sodium hydroxide solution will react with the H+ (a) ions present to form largely unionizerd water. Taking this into account, explain the observed color change. What should be added to swing the equilibrium in the opposite direction? Carry out a test to see if this is correct. eds KSCN (2) Mix together equal volumes of 1 x 102 M solutions of iron (III) chloride and potassium thiocyanate. The blood red coloration that results is due to the formation of a new ion as follows: Fe (aq)+SCN (aq)FeSCN(aq) What would be the expected effect on the intensity of the blood-red color of adding a more concentrated solution of: Ja iron (II) chloride potassium thiocyanate to separate portions of the equilibrium mixture? Test your prediction by dividing the equilibrium mixture into halves and observing separately the effects of adding drops of 1 M solutions of iron (III) chloride and of potassium thiocyanate. (3) (A) Cobalt (I) salts form complex compounds readily. The pink aqueous solutions of cobalt (I) salts are characteristic ofthe complex ion [Co(H2Ok广, in which the cobalt (II) ion is bonded to six water molecules in an octahedral arrangement. If Cl ions are present in sufficient concentration, they will compete effectively with the water molecules for bonding to the cobalt ion to form a deep blue solution of CoCl". The resulting equilibrium may be written as follows: pink blue To 5.0 mL of a 0.4 M Co solution, add two successive 2.0-mL increments of concentrated 12 M hydrochloric acid. What would be the expected effect of adding water to the resulting equilibrium mixture? Test your prediction by adding 5.0-mL increments of distilled water to the contents of your test tube.

Need help for answering first part(1) color change is due to presence? And what was the reason for the equilibrium turn to reverse direction? And also last 2 questions.!

Answer 1

1) The equilibrium reaction in an aqueous solution of iodine (I₂) is shown as

I₂ (aq) + H₂O (l) <=====> 2 H⁺ (aq) + I^- (aq) + IO^- (aq) …..(1)

The solution of I₂ is dark brown. When NaOH is added dropwise to the aqueous solution of I₂, the dark color of the solution fades to light yellow.

The color change is due to the addition of NaOH.

NaOH ionizes in aqueous solution to form OH^-. OH^- combines with H⁺ to form water which remains undissociated.

H⁺ (aq) + OH^- (aq) --------> H₂O (l)

The formation of water removes H⁺ from the equilibrium mixture (1). The equilibrium constant for the equilibrium reaction (1) is given as

K = [H⁺]²[I^-][IO^-]/[I₂]

The removal of H⁺ from the equilibrium mixture (1) reduces the numerator in the above expression. However, K is an equilibrium constant and must remain constant at a particular temperature. In order to keep K constant, the denominator must decrease proportionately. The denominator in the above expression can decrease only when I₂ converts to I^- and IO^-. Thus, the equilibrium reaction proceeds to the right, forming more H⁺ and hence, the dark color of the solution fades to yellow.

3a) The balanced chemical equation for the reaction is written as

[Co(H₂O)₆]²⁺ (aq) + 4 Cl^- (aq) <======> CoCl₄^2- (aq) + 6 H₂O (l) …..(2)

The equilibrium constant for the reaction is given as

K = [CoCl₄^2-]/[Co(H₂O)₆²⁺][Cl^-]⁴

where the square braces denote molar concentrations.

When concentrated HCl is added to an aqueous solution of [Co(H2O)₆]²⁺, the color of the solution slowly changes from pink to blue due to the reaction in (2).

When water is added to the equilibrium system in (2), the color of the solution changes from blue to pink. This is due to the fact that H₂O combines with [CoCl₄]^2- to regenerate [Co(H₂O)₆]²⁺.

b) The reaction in (2) above is endothermic, i.e, heat must be supplied to the reactants to form the products. Therefore, heat is can be considered a hypothetical reactant in the reaction.

There is no change in the color of the equilibrium system at room temperature.

When the reaction system is placed in a icebath, the temperature of the system decreases. Therefore, it can be assumed that the hypothetical reactant heat is decreased by placing the reaction system in an icebath. Therefore, the denominator in the expression for K must decrease. However, to keep K constant, the numerator decreases proportionately, i.e, [CoCl₄]^2- converts to [Co(H₂O)₆]²⁺ and thus, the solution turns pink.

When the reaction system is heated to a high temperature by placing in boiling water, the hypothetical reactant heat is increased and thus, the reaction proceeds to the right to counteract the increased temperature. Therefore, the solution turns blue.