Exp8 Le Chatelier 5d6y5n

Title : Application of Le Chatelier’s Principle to some solution reactions. Aim : To investigates the application of Le Chatelier’s Principle to some solution reactions. Apparatus : 5 mL measuring cylinder, test tubes, 50 mL beaker, Bunsen burner, dropper. Materials : 0.1 mol L⁻¹ potassium chromate(VI) solution [K₂CrO₄], 0.1 mol L⁻¹ potassium dichromate(VI) solution [K₂Cr₂O7 ], (0.2 mol L⁻¹ and 1.0 mol L⁻¹) hydrochloric acid solution [HCL], 0.2 mol L⁻¹ sodium hydroxide solution (NaOH), cobalt (II) chloride hexahydrate peasized [CoCl₂.6H₂O], distilled water. Introduction : Part A : Chromates and dichromates are salts of chromic and dichromic acid. Salts have an intense yellow or orange color, respectively. When solid potassium dichromate (K2Cr2O7) is dissolved in water the resulting solution is orange. The dichromate ion in aqueous solution is in equilibrium with the chromate ion, and this can be shown with the following equation:

This is a dynamic equilibrium and as such is sensitive to the acidity or basicity of the solution. Shifting the equilibrium with pH changes is a classic example of Le Chatelier’s principle at work. Le Chatelier's principle states that if a chemical dynamic equilibrium is disturbed by changing the conditions (concentration, temperature, volume or pressure), the position of equilibrium moves to counteract the imposed change. So if more reactant is added, the equilibrium shifts to the right in order to consume that extra reactant, which results in more product; also if the product is removed from the system, the equilibrium shifts to the right completely increasing the yield. Yellow chromate and orange dichromate are in equilibrium with each other in aqueous solution. The more acidic the solution, the more the equilibrium is shifted to the left towards the dichromate ion. As hydrochloric acid is added to the chromate solution, the yellow color turns to orange. Increasing the hydrogen ion concentration is shifting the equilibrium to the left in accordance with Le Chatelier's principle, where we expect the reaction to try remove some of the H+ we have added by reacting with the CrO42-, and yielding more Cr2O72- which we observe as color change. When sodium hydroxide is added to the dichromate solution, the orange color turns back to yellow, hydroxide ions react with hydrogen ions forming water, driving the equilibrium to the right (OH- removes H+ ions by neutralizing them and the system acts to counteract the change) and further shifting the color. We can observe that the addition of hydroxide ions promotes the conversion of dichromate to chromate.

Acids and bases are added to a system so as to shift the position of a chemical equilibrium. The ions have different colors, so that changes are detected visually. Yellow chromate ion turns orange by addition of acid, while the orange dichromate in reaction with bases turns yellow. The equilibrium depends on the acidity of the solution, so the color in this case is pH dependent. Successive addition of sodium hydroxide and hydrochloric acid causes alternative changes in solution color, during which the color intensity fades due to dilution. The addition of concentrated acids, such as sulfuric acid into chromate/dichromate solution causes further shifting of the equilibrium, and more intense colors, turning the solution to carmine-red. If we add some barium nitrate (Ba(NO3)2) to chromate/dichromate solution, NO3- acts as a spectator ion, not effecting the reaction. On the other hand Ba2+ reacts with the chromate CrO42- forming an insoluble salt BaCrO4, while BaCr2O7 does not precipitate. So the more CrO42- there is, or the higher the pH is, the greater the amount of precipitate would form. Part B : A reversible reaction is a chemical reaction where the reactants form products that, in turn, react together to give the reactants back. Reversible reactions will reach an equilibrium point where the concentrations of the reactants and products will no longer change. A reversible reaction is denoted by a double arrow pointing both directions in a chemical equation. The equation below is the equation for the reversible reaction : Co(H2O)62+(aq) + 4 Cl-(aq) ⇔ CoCl42-(aq) + 6 H2O(l) ∆H +ve pink

blue

Chemical reactions only go so far. The initial rate decreases and a point is reached when no further changes take place. A state of equilibrium exists. Temperature, pressure, concentration, amount of reactants and amount of products all remain constant. The point at which equilibrium is reached is different for each reaction. For some the products are favored; more than half of the reactants have been converted into products. For other reactions the reactants are favored. The concentration of the reactants remains high while that of the products remains low. To push the reaction to the right and create more products, we must understand equilibrium conditions, what is happening at the molecular level and what factors determine the point of equilibrium. As reactions progress, the forward rate decreases while the reverse rate increases. Eventually a point is reached when the forward reaction rate is equal to the reverse reaction rate. Hence, a state of equilibrium is reached.

Procedure : Part A : The chromate –dichromate equilibrium 1. 1 mL of 0.1 mol L⁻¹ K₂CrO₄ was placed into each of two test tubes and the colour of the solutions was noted. One of the test tubes was used for comparison of colours. 2. 1 mL of 0.1 mol L⁻¹ K₂Cr₂O7 was placed into each of two test tubes and the colour of the solutions was noted. One of the test tubes was used for comparison of colours. 3. 0.2 mol L⁻¹ HCl was added dropwise to one of the test tubes of K₂CrO₄ until a colour change was noted. 4. 0.2 mol L⁻¹ NaOH was added dropwise to the same solution until another colour change was observed. The observations was recorded. 5. 0.2 mol L⁻¹ NaOH was added dropwise to one of the test tubes of K₂Cr₂O7 until a colour change was observed. 6. 0.2 mol L⁻¹ HCl was added dropwise to the same solution until another colour change was occurred. The observations was recorded. Part B : The equilibrium between [Co(H₂O)₆] 2⁺ and [CoCl₄]2⁻ 1. Pea-sized amount of CoCl₂.6H₂O was placed in a test tube and was dissolved in 1-2 mL of distilled water. The colour of solution is due to the [Co(H₂O)₆] 2⁺ ion. 2. A pea-sized amount of CoCl₂.6H₂O was dissolved in another test tube in 1-2 mL of concentrated HCl (1.0 mol L⁻¹). The colour of the solution is due to the [CoCl₄]2⁻ ion. 3. A pea-sized amount of CoCl₂.6H₂O was placed in a small beaker (50 mL), then it was dissolved in 5 mL of water, and the half of the solution was put into each two test tubes. 4. Concentrated HCl was added dropwise to one of the test tubes until a definite colour change is noted. Water was added dropwise to the same test tube until another colour change was occurred. The observations was recorded. 5. The second test tube was heated with a Bunsen burner until almost to boiling and any colour change was noted . The solution was allowed to cool . The observations was recorded.

Results : Part A : Table 1 : Table shows the colour changes in K2CrO4 solution and K2Cr2O7 solution in its original colours and when HCl and NaOH was added. Step Original / initial colour When add HCl When add NaOH

K2CrO4 Yellow Orange Yellow

K2Cr2O7 Orange Yellow Orange

Part B : Table 2 : Table shows the colour changes in CoCl₂.6H₂O when it is dissolve in distilled water ,hydrochloric acid and when it is boil. Step Initial colour When dissolve in distilled water When dissolve in HCl When dissolve in water + HCl + water again When add distilled water + boil

CoCl₂.6H₂O Purple Purple Blue Purple Dark purple and almost blue

Disscussion : Part A : This is the reaction beween chromate ions, CrO₄2⁻ (aq) which are yellow, and dichromate ions Cr₂O72⁻ (aq) which are orange. The reaction that has been investigate is: 2 CrO42- (aq) + 2 H+(aq)

Cr2O72- (aq) + H2O (l)

The procedure involves varying the concentration of the H+ ion in order to see how the concentrations of the yellow and orange species change. First of all, my friends and i are looking only for a change in color. The more yellow the colour, the more CrO42- (aq) is present while the more orange, the more Cr2O72- (aq) is present. In addition, HCl role as a direct source of H+ ions. Therefore, adding HCl is equivalent to increasing the [H+ (aq)] in the reaction and we also added NaOH solution. NaOH removes H+ ions, because of acid-base neutralization. Adding NaOH is equivalent to reducing the [H+ (aq)] in the reaction. When solid potassium chromate, K2CrO4 is dissolved in hydrochloric acid it forms a yellow solution. When solid potassium dichromate, K2Cr2O7 is dissolved in hydrochloric acid the resulting solution is orange. The colors come from the negative ions: CrO42-(aq) and Cr2O72(aq). However, in solution these ions are actually in equilibrium as indicated by the equation:

2 CrO42-(aq) + 2 H+ (aq)

Cr2O72-(aq) + H2O (l)

Yellow

Orange

In a chromate solution, there is also a little bit of dichromate, but the predominant colour will be yellow while in a dichromate solution, there is also a little bit of chromate, but the predominant color will be orange. From Le Chatellier's principle, we know the reaction will try to remove some of the H+ we have added. Some of the added H+ reacts with the CrO42-, removing some of each, and making more Cr2O72-. For the chromate solution, when we add some H+ (aq) and it becomes a much more orange solution and new equilibrium is established. It was noticed, that the H+(aq) is less than it would be if there was no reaction, but greater than it was at the beginning. For the dichromate solution, when we add some H+ (aq) and it becomes a slightly more orange solution and new equilibrium is established. It also was noticed that the H+ is less than it would be if there was no reaction, but greater than it was at the beginning, and there has been a slight increase in the dichromate ion concentration. Moreover, the reason why we add sodium hydroxide solution because it removes H+ ions based of this reaction: H+ (aq) + OH-(aq) → H2O (l) If there was no chemical reaction then adding NaOH should decrease the amount of H+(aq). The original H+in the water would decrease to become H+. Since the H+ is colourless, we wouldn't see any color change. But there is going to be a chemical reaction. From Le Chatelier's principle, we know the reaction will try to restore some of the H+ we have removed. The required H+ comes from turning some of the Cr2O72- back into the CrO42-. When we add some OH- (aq) to chromate solution and it becomes a slightly more yellow solution and new equilibrium is established. The H+ is greater than it would be if there was no reaction, but less than it was at the beginning. When we add some OH- (aq) to the dichromate solution and it becomes a much more yellow solution and new equilibrium is established. The equation is : CrO42-(aq) + 2 H+ (aq)

Cr2O72-(aq) + H2O(l)

The H⁺ is greater than it would be if there was no reaction, but less than it was at the beginning. Le Chatelier's Principle states that if a system is at equilibrium and something is changed so that it is no longer at equilibrium, the system will respond in an effort to counteract that change. If more reactant is added, the equilibrium will shift forward in order to consume some of the extra reactant (since there are more ions available for reaction), resulting in more product. If some of the product is removed from the system, the equilibrium will shift forward to produce more of that product. The equilibrium can be shifted reverse by either adding product to or removing reactant from the system.

Yellow chromate ion and orange dichromate ion are in equilibrium with each other in aqueous solution. The more acidic the solution, the more the equilibrium is shifted to favour the dichromate ion. As hydrochloric acid is added to the potassium chromate solution, the yellow colour turns to orange. When sodium hydroxide is added to the potassium chromate solution, the orange colour turns back to yellow. The sodium hydroxide reacts with hydrogen ions, removing them from the solution. When one reactant is removed from an equilibrium system, the equilibrium shifts reverse, in this case forming the yellow chromate ion again. Part B : The following equilibrium is observed: Co(H2O)62+(aq) + 4 Cl-(aq) <=> CoCl42-(aq) + 6 H2O(l) ∆H +ve pink

blue

The Co(H2O)62+ complex is pink, and the CoCl42- complex is blue. This reaction is endothermic as written, so adding heat causes the equilibrium constant to shift to the right. This, correspondingly, makes the solution blue.When the H₂O is added, Cl- is removed from solution. This shifts the equation back to the left, and the solution turns pink again. When HCl is added, there is more Cl- in solution, so the equilibrium is shifted to the right, and the solution turns blue. All of the above effects are variations of LeChatelier's principle. Conclusion : Yellow chromate ion and orange dichromate ion are in equilibrium with each other in aqueous solution. The more acidic the solution, the more the equilibrium is shifted to favour the dichromate ion. As hydrochloric acid is added to the potassium chromate solution, the yellow colour turns to orange. When sodium hydroxide is added to the potassium chromate solution, the orange colour turns back to yellow. The sodium hydroxide reacts with hydrogen ions, removing them from the solution. When one reactant is removed from an equilibrium system, the equilibrium shifts reverse, in this case forming the yellow chromate ion again. Furthermore, this reaction is exothermic reaction. In addition, the reaction in part B is reversible reaction. Then it is endothermic reaction.

Questions : 1. for the observed colour change when HCl solution is added to the K₂CrO₄ solution. The yellow colour of the solution turns to orange colour. 2.When NaOH solution is added to a solution containing HCl, the concentration of H⁺ ions is reduced because of the neutralisation reaction : H⁺ (aq) + OH⁻ (aq) → H₂O (l) for the colour change that occurred when NaOH solution and then HCl solution were added to the K₂Cr₂O7 solution. Firstly, when NaOH was added into the K₂Cr₂O7 solution the colour remains unchanged which is orange colour while when the HCl was added after that, the solution turns from orange to yellow. 3. for the colour changes that occurred when concentrated HCl and then water were added to the solution containing the [ Co(H₂O)₆]2⁺ ion. When concentrated HCl was added to the solution containing the [ Co(H₂O)₆]2⁺ ion, the solution will changes from purple colour to blue colour. Then, when water was added to the same solution it will turns back to purple in colour. 5. (a) From your observations of the colour change that occurred when the solution containing the [ Co(H₂O)₆]2⁺ ion was heated , predict whether the reaction, as written is exothermic or endothermic. The reaction is endothermic. (b) Explain your choice. It is because in accordance with Le Chatelier’s principle, when the temperature is raised, the position of the equilibrium will move to the right, forming more of the blue complex ion at the expense of the pink species.Adding concentrated hydrochloric added raises the chloride ion concentration, causing the equilibrium to move to the right, in accordance with Le Chatelier. Adding water lowers the chloride ion concentration, moving the equilibrium in the opposite direction As an extension it is possible to show that it is the Cl– ions in the hydrochloric acid that shift the equilibrium by adding a spatula of sodium chloride instead to the pink solution. This produces a bluer colour, but this may take some time because the salt is slow to dissolve.

References : 1. Answers Corporation, 2013. Name CoCl₂.6H₂O , [online] Available at : < http://wiki.answers.com/Q/Name_CoCl2_6H2O > [Accessed on 14 April 2013] 2. David Dice, 1998. Le chatelier’s Principle, [online] Available at : < http://www.digipac.ca/chemical/equilibrium/dichromate/l_procedure.htm> [Accessed on 14 April 2013] 3. Chem-toddler, 2001. Chromate/Dichromate, [online] Available at < http://www.chem-toddler.com/chemical-equilibrium/chromatedichromate.html > [Accessed on 14 April 2013] 4. Geoffrey Neuss, 2007. IB Chemistry Course Companion, Oxford Press. 5. A .Caroline, M. Chris, O. Steve, 2011. Cambridge, Chemistry For The IB Diploma. Cambridge University Press, UK.