Experiment 4 Chemical EquilibriumSalvador Marrod M. Cruz Institute of Chemistry, College of Science University of the Philippines, Diliman, Quezon City, Philippines Date performed: January 9, 2013 Instructor: Sir Julius Victorius Saluria Introduction In chemical equilibrium, the rate of the forward reaction is the same as the rate of the reverse reaction. When chemical equilibrium is attained at a specific temperature, a constant ratio expression is obtained and is called the equilibrium constant (Keq). The value for Keq is obtained by getting the ratio of the equilibrium concentration of the products raised to their respective stoichiometric coefficients and that of the reactants raised to their respective stoichiometric coefficients (eqn.1).Although for Keq what is being used for the ratios are the activities but since the standard reference state for an aqueous solution and for gases are 1M and 1atm or 1bar respectively and since it is assumed that the solution is ideal; therefore, the concentration or partial pressure (if gaseous) of products and reactants can be used for solving the value for Keq. For reactions involving solids and liquids, the activity used is 1. The value for Keq determines where the equilibrium point lies. If Keq is between 10-2 and 102; therefore, the equilibrium lies between the reactants and products indicating that both products and reactants are present at equilibrium. If Keq is greater than 102; therefore, the forward reaction is favoured and more products are formed at equilibrium. And if Keq is less than 10-2; therefore, the reverse reaction is favoured and more reactants are present at equilibrium. Eqn.1 : for the hypothetical reaction Keq is also related to the spontaneity of a reaction. When a reaction is at equilibrium, its G = 0 and since G = Go + RTlnKeq; therefore, Go = -RTlnKeq. According to the equation, assuming the reaction is at standard condition, if Keq is greater than one, then the forward reaction is spontaneous and if Keq is less than one but strictly greater than 0, then the reverse reaction is spontaneous. And if Keq is equal to one, then the reaction is at equilibrium. Aside from Keq , another ratio similar to that of solving of Keq is used but this time using the initial concentrations and not the equilibrium concentrations. This ratio is known as the reaction quotient (Q). Q determines the direction of the reaction. If Q>Keq then more products and less reactants are present initially and products got consumed and reactant were produced; therefore, if Q>Keq then the reverse reaction is favoured. If Q<Keq then the forward reaction is favoured. And if Q=Keq, then the reaction is already at equilibrium. In chemical equilibrium a system may be disturbed so that the reaction may favour either the forward or the reverse reaction. The Le Chatelier’s principle, which states that “when a stress is applied unto a reaction at equilibrium, then the reaction will adjust unto it to restore equilibrium”, is a principle that can be used to qualitatively predict the reaction direction upon applying stress unto a system. And these stresses that can be applied to a system are changes in heat, pressure (volume) and concentration. a drop of 0. After the formation of the dark blue copper ammonia complex. 1M HCl was added dropwise until the solution turned turbid.0M NaOH were added. 1mL of the stock solution was added. In test tube 3. In test tube two.10 M FeSO4 and 1 mL of 0. The colors were compared to determine the more stable compound when an acid or a base is added. Each in wells 1 and 3.3.10 M AgNO3 were mixed and shook thoroughly in a 4-in test tube. Iron (III) chloride-thiocyanate system: A stock solution was first prepared by adding 2 mL of 0. 2 drops of 2. 5 drops of 0. 2mL of 0. 5 drops of 0.and 4. 1M HCl was continually added dropwise until the solution turned pale blue. Copper-ammonia system: 1 mL of 0. The number of drops from both the precipitation and change in color were recorded. And on test tube three. a drop of the supernate was placed. Color changes were recorded.1M KSCN and 100mL distilled water in a 400mL beaker. 0.10 M K3Fe(CN)6 was added to test the presence of Fe2+ if a Prussian blue precipitate is formed.10M K2Cr2O7 were added into each of wells 3 and 4. The supernate of the resulting suspension was then decanted for the ion presence test.10M K2CrO4 were added into each of wells 1 and 2 and 5 drops of 0. The mixture was then centrifuged for the solid Ag to settle in the bottom of the test tube. if a white precipitate settled at the bottom of the test tube.Methodology Silver-Iron(II) system: 1mL of 0.1M CoCl2 were added and then concentrated HCl was added dropwise until the color of the solution changes. 1mL of FeCl3 was added. On test tube one. Fe3+ is present. Cobalt (II) ions system: Two test tubes were labelled 1 and 2. Ag+ is present. if a blood red solution is obtained. In test tube 2. therefore. therefore.0M H2SO4 were added and each in wells 2 and 4. therefore. On test tube two. another 1mL stock solution was added since this served as the standard solution. 1mL of 0. Fe2+ is present. And in test tube 4. and 4.2. In test tube 1.10 M KSCN was added to test for the presence of Fe3+ .2. 1M NH3 was then added dropwise until the solution became turbid. Four test tubes were labelled 1. . In 3 different test tubes.1M FeCl3. a drop of 1. 1mL of KSCN was added. In test tube 1. Chromate-dichromate system: In a spot plate. four wells were labelled 1. 2 drops of 2. In all test tubes. Then 1M NH3 was added dropwise again until a dark blue solution Cu(NH3)42+ was formed.00 M HCl was added to test for the presence of Ag+. 1mL Saturated NaCl solution was added.3.1M CoCl2 was added.1 M CuSO4 was added to a test tube. eqn.eq.eqn. Results and Discussion The experiment performed by the researchers was further categorized into sub-experiments that are intended to define different concepts in equilibrium. To be able to test the presence of the ions Fe2+. An ice bath was prepared and then both the test tubes were immersed again until an observation was recorded.A hot water bath was prepared by boiling about 200mL of water. then the value of Keq lies between 102 and 10-2. if the test yielded products only. a mixture must be obtained from the reaction itself. To separate the precipitate that caused the turbidity and the supernate.1. then the value of Keq must be greater than 102 and if the test yielded both products and reactants.eqn. If the ion test yielded the presence of reactants only. Sub-experiment 1 results Ion Visible Conclusion Net Ionic Equation Result Fe2+ A prussian Fe2+ is (c. Both test tubes were immersed until an observation was recorded.). The first sub-experiment performed by the researchers was to determine qualitatively the relative value of the equilibrium constant (Keq) by testing the ions present in the reaction between silver nitrate (AgNO3) and Iron (II) sulphate (FeSO4) (c. It is so because according to the equation for Keq (eqn. and Ag+.3) solution was present Fe3+(aq) + SCN-(aq) FeSCN2+(aq) obtained + Ag HCl A white Ag+ is (c. a turbid mixture is obtained.4) precipitate present Ag+(aq) + Cl-(aq) AgCl(s) was formed According to the chemical equation. To determine the relative position of Keq whether it is less than 10-2 or greater than 102 or between the two. the reaction between iron (II) sulphate and silver nitrate is a precipitation reaction and at the same time a redox reaction. The two test tubes were then placed in a water bath at room temperature and observations were then recorded.2) blue present 3Fe2+(aq) + 2[Fe(CN)6]3.1 : 3FeSO4(aq) + 3AgNO3(aq) 3Ag(s) + Fe(NO3)3 (aq) + Fe2(SO4)3(aq) Ag(s) + Fe3+(aq) Net Ionic Equation: Fe2+(aq) + Ag+(aq) Table.1. ion tests were performed to determine the presence of reactants or products. Fe3+ .1) the value for Keq will be large if the reaction yielded more products with small amounts of reactants at equilibrium and small if the reaction yielded lesser products with large amounts of reactants at equilibrium. then the value of Keq must be less than 10-2. centrifugation must be performed in order for the precipitate to settle at the bottom of the test tube and for the mixture to be a suspension. Upon combining iron (II) sulphate and silver nitrate.(aq) precipitate Fe3[Fe(CN)6]2 (s) was formed Fe3+ KSCN A blood red Fe3+ is (c.eqn. After Test Reagent K3Fe(CN)6 . c. Upon adding HCl.eqn. Fe2+ is present in the supernate.eqn.6. concentrated hydrochloric acid (HCl) was added dropwise (c. At the first drop.eqn. Cu2+(aq) + 6NH3(aq) + 2H2O(l) or simply Cu2+(aq) + 4NH3(aq) 2NH4+(aq) + [Cu(NH3)42+](aq) + 2OH-(aq) Cu(NH3)42+(aq) .eqn. At the first part.6).2 Copper-Ammonia system Number of Drops Initial + NH3 Ppt: 1. As the ammonia builds up. the value Keq is between 10-2 and 102. the supernate was used to test for the ions present in it. c. Fe2+ is present in the mixture if a Prussian blue precipitate is obtained upon the addition of potassium ferricyanide (K3Fe(CN)6) into a drop of the supernate due to the formation of Turnbull’s blue (Fe3[Fe(CN)6]2) (c.6).eqn. Cu (aq) + 2NH3(aq) + 2H2O(l) 2NH4+(aq) + Cu(OH)2(s) c. The reverse reaction is favoured here because as the ligands continually turns into NH4+ that will lower the concentration of a reactant and will push the reaction backwards (c.7.6). Fe3+ is present in the supernate. reacts with H+ forming NH4+ thus causing the disappearance of the dark blue copper ammonia complex and forming Cu(OH)2 which is responsible for the turbidity of the solution (c.centrifuging the mixture.3). Fe3+ is present if a blood red solution is obtained when a drop of KSCN is added into a drop of the supernate due to the formation of Iron thiocyanate complex (FeSCN2+ )(c. The ligands.2).eqn.5). And according to the result of the test. therefore. the reverse reaction of c.eqn. Ag+ is present if a white precipitate is formed upon adding hydrochloric acid (HCl) into a drop of the supernate due to the formation of silver chloride (AgCl)(c. According to the result of the test. A precipitate was formed because ammonia is a weak base and it partially dissociates in water and becomes ammonium ion (NH4+) and hydroxide ions (OH-) (c. therefore. an ionic bond will form forming copper hydroxide which is light blue in color and is a solid precipitate.eqn.eqn. the reaction yielded a combination of reactants and products and it neither went to completion nor did not react. And since the equilibrium is not favoured in either the reactants or products.eqn. the reaction between copper and ammonia was observed (c. And as more NH4+ forms.4). the researchers added NH3 to the copper (II) sulphate dropwise. Since copper ions are cations and hydroxide ions are anions. the solution became turbid due to the formation of the light blue copper hydroxide precipitate (c. NH3.5 will be favoured forming more ammonia and Cu2+ that is responsible for the decolorization of the mixture. Ag+ is present in the supernate.eqn. therefore. Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)42+](aq) + 2OH-(aq) Color Light blue Dark blue Pale blue For sub-experiment 2. After the addition of ammonia.eqn.6).7). the copper ions get consumed to Cu(OH)2 and Cu(OH)2 then reacts with excess ammonia resulting into the formation of the dark blue copper-ammonia complex or tetraamminecuprate(II) [Cu(NH3)42+] (c. According to the result of the test. Pale blue: 8 2+ c.eqn.eqn. Table. Since all of the ions are present in the supernate of the mixture.5).5. Dark blue: 18 + HCl Ppt: 1. a strong acid with its anion in the highest oxidation state was added and according to the result of the experiment. Since it dissociates completely. a small amount of HCl can easily neutralize NH3 due to the production of more neutralizing agents (H+) than the basifying agent in NH3 and drive the reaction to the reverse direction because removing a reactant will force the reaction to produce more reactants to regain equilibrium according to Le Chatelier’s Principle. a strong base was added and the species that changed color this time was the potassium dichromate. was placed. it does not completely dissociate in an aqueous solution and does not produce enough OH. a yellow solution.6 represents the chemical reaction responsible for the shift in equilibrium because when NH3 is added into the reaction. Then the reason why HCl can decolorize the solution with fewer drops than colorizing it with NH3 is because HCl is a strong acid which means that it dissociates completely in an aqueous solution.that must be fully consumed by the Cu2+ ions to produce Cu(OH)2 immediately after a few drops. an orange solution. potassium dichromate (K2Cr2O7).7). Table. while adding H+ into the reaction will reduce the amount of NH3and produce more NH4+ which is responsible for forming more of the reactants (c. more drops are needed to drive the reaction forward to completion. potassium chromate (K2CrO4). Since NH3 is a weak base.3 Chromate-Dichromate system Well number 1 2 3 4 Reagents H2SO4 and K2CrO4 NaOH and K2CrO4 H2SO4 and K2Cr2O7 NaOH and K2Cr2O7 Visible result Yellow to orange No change No change Orange to yellow Net ionic equation (c.The sum of c.2) to the Cu2+ solution to obtain the dark blue copper ammonia complex (Tetraamminecuprate(II)) and only 9 drops for HCl (table.eqn. . therefore.eqn.eqn9): Cr2O72-(aq) +2OH-(aq) 2CrO42-(aq) + H2O(l) On wells 1 and 2. the species that changed color was the potassium chromate. The strong base and the strong acid used in the experiment were sodium hydroxide (NaOH) and sulphuric acid (H2SO4) respectively.5 and c. And on even numbered wells.8): 2CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) NA NA (c.2) for the dark blue complex to decolorize because NH3 is a weak base. It is important that an acid with its anion in its highest oxidation state must be used because using an acid with an anion that is not in its highest oxidation state will have its anion oxidized by the chromate or dichromate and may cause several interferences in the reaction that might result to a failure since chromate and dichromate ions are oxidizing agents. After consuming all of Cu2+ ions then will the complex appear if there is an excess in NH3. And the amount removed from the reactant because of HCl is greater than that added a while ago to the reactants due to NH3. On odd numbered wells.eqn. H2SO4 was used to acidify the medium and NaOH was used to basify the medium in order to determine which one is the more stable species against the other and to be able to disturb the equilibrium reaction because the reaction is pH dependent and H2SO4 was also used because H2SO4 is already in its maximum oxidation state which means that it can no longer be oxidized by the reaction. the researchers added 19 drops of NH3(table. is placed and on wells 3 and 4. more products are formed.eqn. In the experiment. eqn. then the concentration of the reactant OH.11). Also. The forward reaction is favoured here due to an increase in the concentration of a reactant and a decrease in concentration of a product which is under the Le Chatelier’s principle.to form FeCl4-. yields more products in order to restore equilibrium and the products formed by this reaction are Cr2O72.eqn. therefore.is the more stable species because in the equilibrium state. adding H+. which is a reactant.eqn.10 and 11) The reaction of iron (III) chloride (FeCl3) and potassium thiocyanate (KSCN) yielded an iron thiocyanate complex or pentaaquathiocyanatoferrate(III) which is a blood red complex (c. Cr2O72. Upon the addition of Fe3+.eqn. the solution’s color shifted from orange to dark orange because of the production of more iron thiocyanate complex that is responsible for the reddish color.9). And since the remaining Fe3+ ions in the reaction reacted with Cl.8) Cr2O72.eqn.10) will be disturbed because reactant concentration will be reduced therefore the reverse reaction will be favoured in order for that reaction to restore equilibrium.and H2O. In the experiment. And as the .is the more stable species in an acidic medium. the solution’s color shifted from orange to dark orange. the solution formed a colorless complex ([FeCl4]-) upon reacting with FeCl3 thereby competing with the production of FeSCN2+ (c. Since Cr2O72. Table.10: Fe3+(aq) + SCN-(aq) Color Dark orange Dark orange Light orange or pale yellow [FeSCN]2+(aq) [Fe(H2O)5SCN]2+ (aq) Shift in Equilibrium Right Right Left or Fe3+(aq) + 5H2O(l) + SCN-(aq) c.According to the balanced equation (c.12: 2Fe3+(aq) + SCN-(aq) + 4Cl-(aq) [FeSCN]2+(aq) + [FeCl4]-(aq) (sum of c.eqn. It means that the forward reaction was favoured upon the addition of Fe3+ thereby producing more iron thiocyanate complex that made the color darker.will increase thereby favouring the forward reaction.10). Since according to the net ionic equation. the initial color of the solution involving the reaction is orange because it was diluted with excess water. the reaction producing the FeSCN2+ (c.eqn. the solution became light orange or pale yellow because the iron(III)(Fe3+) ions interacted with chloride ions (Cl-) to form another complex which is colorless named as tetrachloroferrate(III) which made it even more colorless than the original solution because instead of forming [FeSCN]2+ .eqn. Reagents added to KSCN + FeCl3 system Fe3+ SCNSaturated NaCl c. Upon the addition of SCN-.9) in the acidified medium (c. when the medium is basified (c.eqn.eqn.4 FeCl3-KSCN system.eqn. the experiment results were correct because the net ionic equation of the reaction itself predicted the same change that occurred in the experiment.11: Fe3+(aq) + 4Cl-(aq) [FeCl4]-(aq) c. And when saturated NaCl was added.is the one formed when the medium is acidic.8 and c. therefore. the more stable species in an acidic medium is K2Cr2O7 and the more stable species in a basic medium is K2CrO4 and the experiment yielded an orange compound in an acidic medium and a yellow compound in a basic medium which also corresponds to K2Cr2O7 and K2CrO4 respectively. (aq) 6H2O(l) + [CoCl4]2-(aq) Color Pink Purple Blue Pink Purple or [Co(H2O)6]2+(aq) + 4Cl-(aq) For the cobalt ions system. if a reaction is endothermic.5 Cobalt (II) ions system Condition Before adding HCl After adding HCl At boiling water temperature At ice water temperature At room temperature c.and will make the solution colorless. therefore. In the experiment.eqn. . a blue colored compound (c. then adding heat to the reaction will favour the forward reaction and when heat is removed. the initial color of the compound CoCl2 was pink and when HCl was added. the solution turned pink meaning the reaction shifted towards the production of CoCl2 and since the solution is neither purple nor blue. the researchers reacted CoCl2 or Co2+.Fe3+ reacts with Cl-.was formed at an amount much greater than CoCl2 because the solution is neither purple nor pink. the concentration of FeSCN2+ will decrease to produce more Fe3+ that will react with Cl. equilibrium is established between the reactants and products because Ag+. And in room temperature. the solution turned purple due to the presence of CoCl42. the color of the solution became blue. The results were satisfying because the result predicted theoretically was the same as the result yielded experimentally.eqn. which means that CoCl42. For the first reaction. with HCl to produce a cobalt chloride complex or tetrachlorocobaltate(II) ([CoCl4]2-). a pink colored compound.is present in the heated solution and a very large amount of CoCl2 is present in a cold solution.13: Co2+(aq) + 4Cl-(aq) [CoCl4]2. therefore. Fe2+. When the solution was placed in an ice water bath. some are very difficult to predict experimentally due to human errors and systematic errors.is present in a very small amount. the reverse reaction is favoured. Table. and Fe3+ are all present in the solution meaning. It would be better if another test would be added to determine the probable value of the equilibrium constant expression and to be able to determine the spontaneity of the reaction. the solution is purple. and according to Le Chatelier’s principle. Conclusion / Recommendations The experiment performed starting from the reaction between silver nitrate and iron (II) sulphate up to the system involving cobalt ions yielded satisfying results.13). equilibrium is very difficult to pinpoint whether it is slightly inclined towards the reactants or towards the products. meaning Co2+ and CoCl42. CoCl42. It will be colorless because more of the colorless compound will be produced and the compound with the red color will be reduced thereby changing making the color of the solution clearer and yellowish.are both present together in the solution. although.along with Co2+ thereby combining pink with blue When the solution was heated. the reaction is endothermic. Since a very large amount of CoCl42. It is so because heat is treated as a part of the reaction. 2011. the whole experiment yielded the same results that can be predicted theoretically by applying different principles and calculations involved in chemical equilibrium. For the cobalt ions system. and the basic medium yielded a yellow compound. R.will yield to a colorless solution.For the second reaction. For some other experiments. For testing all the stresses under Le chatelier’s principle. Toronto.). a standard must be set up to be able to predict the results carefully and more accurately. But results could have been better if more tests were performed and if the value of Keq for some reactions will be obtained exactly than predicting it because a Keq that lies between 10-2 and 102 is difficult to describe. the experiments performed were limited because the changes in volume and pressure for a reaction was not performed and it is highly recommended that an experiment portraying the changes in equilibrium when the volume and pressure are both changed in a gaseous reaction so that the principle will be more believable. Adding Fe3+ and adding SCN. It is difficult to compare due to the closeness of the number of drops before the solution decolorized and turned dark blue because the term dark blue used is quite vague because a dark blue solution was obtained when about 12 drops was added but it was not yet considered to be a dark blue complex due to the turbidity of the precipitate.. General Chemistry: Principles and Modern Applications (10th ed. The Central Science (11th ed. the experiment yielded satisfying results that are very close to the theoretically predicted result. Ont: Pearson Canada. Lemay. the result of the experiment is the same as the predicted results due to the presence of the complex ion theoretically and experimentally as well as the decolorization of the solution. especially its spontaneity. And the results were the same as the theoretically predicted result by means of the chemical equation. But overall. Jurong. which is K2Cr2O7. et al. et al. Chemistry. pp 656-685 [2] Brown. What is difficult to compare is the number of drops because a dark blue solution was obtained after about 12 drops only for the addition of NH3 yet the reaction is not yet complete due to the turbidity caused by the copper hydroxide precipitate. But overall.will create a dark orange solution while adding Cl.). T. The only problem is that colors are sometimes confusing because certain compounds can have the same color as another and a stage in equilibrium with a color may have the same color as the compound in the equilibrium but this time diluted. References [1] Petrucci. Singapore: Pearson Education South Asia PTE.2009. the experimentally observed data was the same as the theoretically predicted result by means of a chemical equation and the Le Chatelier’s principle. For the Iron (III) chloride – potassium thiocyanate system. pp 626-656 . For the experiment as a whole.. the experiment yielded the same results that were predicted theoretically. which is K2CrO4. For the chromate-dichromate system the experiment suggested that K2CrO4 is more stable in a basic medium and K2Cr2O7 is more stable in an acidic medium because acidic mediums yielded an orange compound. the experiment result of this part was satisfying because it follows the Le Chatelier’s principle. Retrieved from http://www. G < 0 because the reaction is spontaneous.eqn. For part B. Based on the principles of equilibrium. Retrieved from http://pages.Department of Chemistry.which made the reverse process spontaneous. therefore.which has a value of K>102 because FeCl4. the value of G for the forward reaction or the production of Cu(NH3)42+ is negative K > 1 and if K > 1 then the reaction is spontaneous.[3] Silberberg. K>1. For part B. therefore. For the addition of Fe3+ and SCN-. therefore.com/acad/webtext/chemeq/ [6]Foundations to Chemistry . And since adding Cl.). (n. then because G = -RTlnKeq because Go = G = -RTlnKeq.edu/ladon/chemeq. For part D. But since there is another reaction reacting simultaneously with c.ox. And when H+ is added. then the reverse reaction is non-spontaneous.htm Appendix Answers to questions: 1..d.and 102 because the reactants and products are present significantly.eqn.10 (c.pushes the reaction backwards (c.10). the magnitude of the equilibrium constant lies between 102.eqn.chem. And the addition of Cl. has K<1 which means that it is nonspontaneous and it has a G whose value is positive (for c. explain the observations in part A.eqn. . Since the reverse reaction has Keq < 1.d. Boston: McGrawHill 2006.11 has a K>102.uk/vrchemistry/chemicalequilibrium/html/page01.eqn.10 and since c.10).chem1.html [5]Chemical Equilibrium. pp 723-756 [4]Chemical Equilibria. and the statement that if the forward reaction is spontaneous. and D.d. the value of Keq at the beginning lies between 10-2 and 102 but as Fe3+ and SCN. What can you infer about the signs of their G? Assuming that G is equal to Go and the reaction is at standard condition.towson. (n. Retrieved from http://www.and Fe3+ ions are being added.). the value of Keq is greater than 102. And equilibrium concentration must be calculated in this situation to determine if the forward reaction is more favoured than the other or the other way around. What can you infer about the magnitude of the equilibrium constants for these reactions? For part A. Matter and the Universe". Chemistry: The Molecular Nature of Matter and Change.eqn.ac. the forward reaction will be favoured thereby producing more products until the reaction lies on the product side making the Keq value greater than 102.11) then there is a production of FeCl4. the reverse reaction is nonspontaneous and the value of G is positive. the addition of 19 drops of NH3 produces products and the reactants are quite not significant anymore.). B. the reaction is spontaneous and the sign of G is negative. the reverse reaction occurs and the value for Keq will be <10-2 because it is the reciprocal of the Keq of the forward reaction. therefore. (n. but since c.11 occurs simultaneously with c. the value for Keq will be reciprocated resulting to a Keq value of less than 10-2.concentration increases as SCN.is being formed as the reaction progresses and equilibrium in this reaction is favoured towards the product side.eqn. the sign of G is difficult to predict because it can be either positive or negative but the sign of G approaches 0 as the value of Keq approaches 1. M.adapted from "Chemistry. For part A. 77x1012)(37) Ktotal = 3.77x10-43 )( ) 2H+(aq) + 2OH-(aq) 2H2O(l) 2HCrO4-(aq) 2CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + 2OH-(aq) 2CrO42-(aq) + H2O(l) 4. adding H+ will shift the equilibrium towards the right of the reaction or towards the side of the reaction with Cr2O72.for the ion with the oxygen extracted from it will attach to another CrO42. .77x1012 K1 = 3. The reaction below competes with the formation of the [Fe(H2O)5SCN]2+ complex: Fe3+(aq) + 4Cl-(aq) Yellow FeCl4-(aq) Colorless Explain what will happen to the color of a dilute solution containing FeCl4.0 x 10 -14 Cr2O72-(aq) + H2O(l) b.because according to the net ionic equation.ion to produce a Cr2O72ion.if: a.to form AgCl).2 x 10 -7 K2 = 37 K3 = 1. Cr2O72-(aq) + 2OH-(aq) Cr2O72-(aq) + H2O(l) 2CrO42-(aq) + H2O(l) 2HCrO4-(aq) = = = Ktotal = ( ) ( Ktotal = ( ) Ktotal = 2.2. (Ag+ reacts with Cl. the reaction tends to produce more Cr2O72.and adding 2H+ will extract an oxygen atom from one of two molecules of CrO42. 2CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l) 2CrO42-(aq) + 2H+(aq) 2HCrO4-(aq) 2CrO42-(aq) + 2H+(aq) 2HCrO4-(aq) Cr2O72-(aq) + H2O(l) ( )2 = ( K2 = 37 ) = (9. What was the effect of the acidity of the medium on the chromate and the dichromate ions in the solution? When the medium was acidified. A solution containing Ag+ was added.61x1014 ( ) = 9. 3. From the following reactions: HCrO4-(aq) CrO42-(aq) + H+(aq) 2HCrO4-(aq) Cr2O72-(aq) + H2O(l) H+(aq) + OH-(aq) H2O(l) Derive the equilibrium constant expressions for: a.rather than CrO42. eqn. the reaction will be the same as that of adding NaCl because HCl is a strong electrolyte. a precipitate will be formed because FePO4 is not soluble in water.(IF FeSCN2+ is included in the solution) The solution will become dark orange with a white precipitate due to the formation of AgCl and shifting of the reaction towards the left side because the concentration of a reactant (Cl-) decreased.making the solution colorless.11) and the Fe3+ from c. d.11).will decrease. therefore. (IF FeSCN2+ is not included) the color of the solution will be yellow because adding Ag+ will consume Cl. the reaction will produce more reactants by consuming the products to form more Cl. Fe3+ concentration will decrease and the equilibrium will be disturbed and more FeCl4.10 will reduce thereby reducing the concentration of FeSCN2+ because equilibrium will shift to the left thereby producing more FeCl4. But the effect of H3PO4 is not that significant because it is a weak acid and it only dissociates partially in a solution yielding few PO43. FeCl4. therefore.will be consumed in order to produce more Fe3+ in order to regain equilibrium. Concentrated H3PO4 was added When H3PO4 was added.ions will produce more FeCl4. Since FeCl4. therefore.is present in a certain amount and upon addition of Ag+. c.since NaCl is a strong electrolyte.eqn.and Fe3+ to restore equilibrium. Q > Keq because initially. 5.and to push the reaction that produces FeSCN2+ (c. adding Cl.which leads to a decrease in the concentration of a reactant. b. Compare the values of Q to Keq in each of the four conditions described in number 4.to form the red complex FeSCN-and the simultaneous reactions will continue to counteract the stress being applied unto it. And since another reaction takes place with it simultaneously. the reaction will shift towards the reactant side by creating more Cl. the reaction will produce more reactants to restore equilibrium because .eqn. by Le chatelier’s principle. therefore. therefore. therefore. a.(c.ions that will react with Fe3+. Fe3+ will be produced and the Fe3+ produced will react with SCN. NaCl solution was added (For solutions either with or without FeSCN2+ ) The solution will become colorless or pale yellow due to formation of more FeCl4. it dissociates completely in water thereby producing more chloride ions that will push the reaction forward to the produce FeCl4.(c.eqn.and Fe3+ ions that will cause the yellowish color of the solution.10) backwards to produce more Fe3+ that will be converted to FeCl4. Concentrated HCl was added (For both solutions either with or without FeSCN2+) When HCl was added. the color of the solution will be yellow due to more Fe3+ ions. the concentration of the product was greater than that of the reactants and after the reaction.concentration Cl. d. Q > Keq.decreases upon addition of Ag+. Q < Keq because the forward reaction is favoured and the amount of products formed at equilibrium is greater than that of the initial point of the reaction yielding a greater ratio of the equilibrium concentrations of products and reactants than that of the initial ratios of the products and reactants. c. . And since the reverse reaction is favoured. the ratio of product and reactant concentrations at equilibrium will be smaller than that of the product and reactant concentrations initially. And because the reverse reaction was the one favoured in this situation. b. Q < Keq because the forward reaction is favoured because additional reactants are added then the reaction will go forward thereby having a larger amount of products formed and resulting into a greater ratio rather than the initial concentration of the products and reactants. the product’s concentration decreased as that of the reactants increased. Q > Keq because initially. therefore. therefore.