LAB 11: RATES & EQUILIBRIUM: REACTION RATES AND LECHATELIER S PRINCIPLE

Transcription

1 LAB 11: RATES & EQUILIBRIUM: REACTION RATES AND LECHATELIER S PRINCIPLE PURPOSE: To observe and describe factors that affect the rate of a chemical reaction. To observe the concentration effects on a system at equilibrium. SAFETY CONCERNS: Always wear safety goggles. Handle and dispose of broken glass safely. Acids and bases are corrosive and can severely damage tissue. If acid or base is spilled, flood the area with water and notify the instructor immediately. If acid or base contacts the skin, remove any clothing covering the skin and flush with copious amounts of water. If acid or base contacts the eyes, flush with water for 20 minutes. Wash your hands with soap and water after handling acids and bases. Solutions of hydrogen peroxide, especially greater than 3%, should be handled with great care. If spilled, immediately flood the area with water and wash the liquid down the drain. Be careful not to spill any on your person, books, or clothing. Concentrations greater than 3% may cause white patches or blisters and chemical burns on the skin. REACTION RATES: The rate of a chemical reaction is the speed at which a reaction occurs. Some reactions, like the rusting of an iron nail, take years to complete. Other reactions, like the explosion of dynamite, are over in a fraction of a second. Reaction Energy: In order to react, two molecules must physically collide with each other; the more frequent the collisions, the faster the reaction. The molecules must collide with enough energy (the activation energy) to overcome the repelling forces between the molecules. If molecules do not collide with energy at least as large as the activation energy, they will not react. The greater the number of collisions meeting or exceeding the activation energy, the faster the reaction. You will investigate several factors that affect the rate of a chemical reaction: the concentration of the reactants, the temperature of the reactants, and the effect of adding a catalyst. These factors determine the frequency and energy of reactant collisions. Concentration: In this experiment you will study the reaction rates of acids with metals under different conditions. You will vary the concentration of reactants by using acids of differing molarity (M). If two solutions have equal volumes, the one having a greater molarity, M, will have a greater number of solute molecules. When a metal reacts with an acid, vigorous bubbling, due to the liberation of hydrogen gas, takes place at the surface of the metal. The existence of hydrogen gas bubbles is evidence that a reaction is occurring. The presence of hydrogen can be verified by placing a burning wooden splint above the surface of reactants. A whistling or popping sound indicates the presence of hydrogen. The lighted splint furnishes the activation energy needed to initiate the reaction between hydrogen and oxygen. 2H 2(g) + O 2(g) 2H 2 O (g) hydrogen oxygen water The intensity of the sound is determined by the concentrations of oxygen and hydrogen. Pure hydrogen will produce a muffled "whoof," while the right mixture of hydrogen and air gives a loud "pop." Thus, CH105 Lab 11: Rates & Equilibrium (F15) 1

2 the intensity of the sound of the reaction between hydrogen and oxygen cannot be used to judge the rate of the reaction between the metal and hydrochloric acid. The rate of reaction between metal and hydrochloric acid can be qualitatively judged by observing how rapidly the metal produces bubbles of gas. A more quantitative determination can be made by timing the number of minutes before the metal disappears. CATALYSIS: A catalyst is a substance that increases the rate of a chemical reaction, not by increasing the number or energy of molecular collisions, but by lowering the activation energy required for the reaction. The catalyst interacts with the reactants to create an alternative path or route for the reaction--a route with lower activation energy. Reactant molecules are more likely to collide with sufficient energy and the reaction rate is greater in the presence of a catalyst. The reaction under study in this experiment is the decomposition (breaking apart) of hydrogen peroxide, H 2 O 2. Hydrogen peroxide slowly decomposes as it sits in its bottle, gradually turning to water and oxygen gas. The unbalanced equation: H 2 O 2(aq) H 2 O (l) + O 2(g) This reaction rate is accelerated by a catalyst. The presence of oxygen, which indicates that decomposition has taken place, can be easily verified. A glowing wooden splint is slowly burning in the oxygen of air. If the glowing splint is placed in a higher concentration of oxygen, the rate of combustion will increase and the wood splint will burst into flames. Hydrogen peroxide is an oxidizing agent--it removes electrons from other substances. It destroys organic compounds that come in contact with it. A 3% solution of hydrogen peroxide is used as an antiseptic for small cuts of the skin--it oxidizes germs to death. When placed on an open wound, the hydrogen peroxide bubbles; the bubbles are oxygen gas. The exposed blood contains an enzyme (peroxidase) that catalyzes the decomposition of hydrogen peroxide. The oxygen bubbles come from the hydrogen peroxide, not from the catalyst. A catalyst is neither used up in a reaction nor irreversibly altered; a catalyst is a condition of the reaction and, if included in an equation, is written above the equation arrow. The unbalanced equation: Peroxidase cat H 2 O 2(aq) H 2 O (l) + O 2(g) When a red spot is suspected of being blood, forensic chemists can use this reaction in a presumptive test for blood. The release of oxygen from hydrogen peroxide by the peroxidase activity causes an indicator to change color. This is not a definitive test for blood since other substances can also catalyze the decomposition of hydrogen peroxide. Hydrogen peroxide is used in higher concentrations (6%) to bleach (oxidize) hair pigments in many hair-dyeing systems. Still higher concentrations (90%) are sometimes used as oxidizers in rocket propulsion. 2 CH105 Lab 11: Rates & Equilibrium (F15)

3 CHEMICAL EQUILIBRIUM AND LECHATELIER S PRINCIPLE: Some chemical reactions, like the reaction of metal with acid or the decomposition of hydrogen peroxide, "go to completion"--every molecule of reactant is eventually converted to product. Other chemical reactions do not go to completion--the reactant is never totally changed into product. The later type of reaction is a chemical equilibrium. Chemical equilibriums do not go to completion because of a reverse reaction that competes with the forward reaction. In an equilibrium, the forward reaction occurs at the same rate as the reverse reaction. Let's consider a hypothetical chemical reaction in which the reactants are represented by A and B, and the products by C and D. At the beginning of the reaction, only A and B are present, but as the reaction proceeds, C and D are formed. The concentrations of A and B decrease, while the concentrations of C and D increase. The forward reaction may be represented by the equation: A + B C + D If C or D is not removed from the reaction site, they can react to form the original substances A and B: C + D A + B As the concentrations of A and B decrease, the forward reaction slows down. As the concentrations of C and D increase the reverse reaction speeds up. Eventually the rate of both reactions becomes equal and an equilibrium is established: A + B C + D In other words, after a period of time, a mixture of products and reactants exists. On the microscopic level, A and B are constantly changing to C and D in the forward reaction and C and D are constantly changing to A and B in the reverse reaction. However, the concentrations of A, B, C, and D remain constant because the forward and reverse reactions occur at the same rate. The composition of this equilibrium mixture will remain constant unless the concentrations of reactants or products are changed by the experimenter. Increasing the concentration of A or B causes the forward reaction to proceed faster, forming additional C and D, until the rate of the reverse reaction increases enough to again establish a state of equilibrium. In the new equilibrium the concentrations of C and D, represented by [C] and [D], are greater than in the original equilibrium. The nineteenth century French chemist Henri Louis LeChậtelier described this increased concentration of A or B as a stress. LeChậtelier's Principle states that if stress is applied to an equilibrium system, the system will respond by altering the equilibrium in such a way as to minimize the stress. Increasing the concentration of A or B is a stress that can be relieved by removing some of the added material and converting A and B to C and D. The equilibrium "shifts to the right" and a new equilibrium is established with higher concentrations of C and D. Similarly, addition of C or D to a system at equilibrium will cause the equilibrium to "shift to the left" and the new equilibrium position will have higher concentrations of A and B than the original equilibrium. You can also cause equilibrium shifts by removing reactants or products (decreasing their concentrations). Thus, removal of A or B from the equilibrium system will cause a shift to the left. Removal of C or D will cause a shift to the right. CH105 Lab 11: Rates & Equilibrium (F15) 3

4 In this experiment you will examine the following equilibrium system: FeCl 3(aq) + 3 NH 4 SCN (aq) Fe(SCN) 3(aq) + 3 NH 4 Cl (aq) iron(iii) chloride ammonium thiocyanate iron(iii) thiocyanate ammonium chloride (pale yellow) (colorless) (deep red) (colorless) The shifts in equilibrium can be easily followed in each case because the color of reactants is different than the color of products. PROCEDURES: ACTIONS: I. REACTION RATES: A. Concentration Effects: 1. Obtain 3 pieces of Magnesium ribbon (Mg) of equal length, about 2-3 cm long. 1 Clean and polish the Mg with steel wool to remove any oxide coating Obtain 3 clean test tubes with rubber stoppers and label them #1-3. Pour acids into the tubes as follows: Tube #1: 10 ml of 1.0 M HCl Tube #2: 10 ml of 2.0 M HCl Tube #3: 10 ml of 3.0 M HCl 3. Have a clock with a second hand or a stop watch available to time the reactions of Mg and acids that you are about to perform. 4. As close to the same time as possible, place the cleaned pieces of Mg into each of the tubes #1-3, stopper loosely 3, and note the start time of the reactions. 5. With the clock still running carefully bring a burning wood splint to the mouth of one of the tubes. Quickly remove the stopper and test the flammability and verify the identity of the gas. 6. Record the amount of time for each sample of Mg to completely react and disappear At the end of the reaction determine whether the reaction is exothermic or endothermic by touching the bottom of the test tube. NOTES: 1 The exact length of the Mg ribbon is not important; however you want all 3 pieces to be of equal length for comparison purposes. 2 Metals slowly react with oxygen in the air to form oxides. Iron reacts with oxygen over time to form rust, iron(iii) oxide; 4Fe + 3O 2 2Fe 2 O 3 Likewise, magnesium over time may acquire a coating of magnesium oxide that can interfere with our experiment. 2Mg + O 2 2MgO 3 The reaction of hydrochloric acid with magnesium produces hydrogen gas. If stoppers are on too tight then pressure will build and they will be popped off. 1 M HCl 2 M HCl 3 M HCl 4 Magnesium, Mg, reacts with hydrochloric acid, HCl, to produce Magnesium chloride, MgCl 2, and Hydrogen gas, H Rank the rates of reactions from slowest to fastest. 4 CH105 Lab 11: Rates & Equilibrium (F15)

5 B. Temperature Effects: 1. Prepare a hot water bath by filling a 250 ml beaker about ½ full of water and heating on a hot plate. Do not allow it to boil. 5 You might use a thermometer to verify the temperatures. NaHCO 3 2. Prepare an ice water bath by filling a 250 ml beaker about ½ full of ice and water. 3. Place 10 ml of vinegar in each of two large test tubes one labeled H (for hot) and the other C (for cold). 4. Place tube H into the hot water bath and tube C into the ice bath and allow them to stay long enough to insure that each vinegar solution has reached the bath temperature Obtain 2 clean dry test tubes and into each place equal amounts (1 scoop or 2 spatula tips) of sodium bicarbonate, NaHCO Remove test tubes H and C from their temperature baths and place them in a test tube rack. Measure the temperature in each test tube. 7. At the same time, add a sample of NaHCO 3 to each tube H and C, and observe and compare the fizzing (bubbles) in each Determine which reaction (H or C) clears first. C. Catalyst Effects: 1. Divide 10 mls of 3% hydrogen peroxide, H 2 O 2 into 3 test tubes labeled 1-3 and loosely stopper them. Keep tube #1 as a room temperature control for comparison. Place tube #2 in a hot water bath. H Hot HC 2H 3O 2 C Cold HC 2H 3O 2 6 Caution: Move your papers out of the way as this reaction may get messy. Acetic Acid, HC 2 H 3 O 2, reacts with sodium bicarbonate, NaHCO 3, to give sodium acetate, NaC 2 H 3 O 2, and carbonic acid, H 2 CO 3. The carbonic acid immediately falls apart to give carbon dioxide, CO 2 and water, H 2 O. 7 When a slice of potato is placed into a solution of hydrogen peroxide, bubbles of oxygen form on the potato surface and rises to the top of the solution. Potatoes contain catalase, an enzyme that breaks down hydrogen peroxide into water and oxygen. 2. Obtain a skinless rectangle of raw potato 7 (~ 1cm x 1 cm x 3 cm), dice it into tiny bits then crush with a mortar and pestle. 3. Add the crushed potato to tube #3 of H 2 O 2 and stopper. Over the next few minutes observe any reaction. 4. Record your observations for tubes 1, 2, and Where gas bubbles are produced how might you go about verifying the identity of the gas? 8 Potato Catalase In H 2O 2 8 You could use a match to light the end of a wooden splint. Allow the fire to burn a few seconds, then blow out the flames, leaving glowing embers. A glowing splint will flicker into a flame in the presence of oxygen, O 2, gas. CH105 Lab 11: Rates & Equilibrium (F15) 5

6 II. IRON (III) THIOCYANATE EQUILIBRIUM: Fe 3+ (aq) + 2SCN 1- (aq) Fe(SCN) 1+ (aq) Iron(III) ion thiocyanate ion Iron(III) thiocyanate complex (pale yellow) (colorless) (deep red) 1. Obtain: A. 1 ml of 0.1M Iron(III) chloride (FeCl 3 ) solution. Note the color on the report sheet. B. 1 ml of 0.1 M Ammonium Thiocyanate (NH 4 SCN) solution. Note the color on the report sheet. C. 60 ml of deionized water in a clean 150 ml beaker. 2. Add both the 1 ml of 0.1 M FeCl 3 solution and the 1 ml of 0.1 M NH 4 SCN solution into the 60 ml of deionized water. Stir to mix. Note the color of this prepared stock equilibrium mixture on the report sheet as your control solution. 3. Set up 6 clean test tubes in a test tube rack and label each tube A through F. 8 Divide your stock equilibrium solution evenly 9 into the 6 prepared tubes. 4. Prepare a hot water bath by filling a 250 ml beaker about ½ full of water and heating on a hot plate. Do not heat to boiling. Save this bath for the testing of tube E. 5. Prepare an ice water bath by filling a 250 ml beaker about ½ full of ice and water. Save this bath for the testing of tube F. 6. Apply the following stresses to your equilibrium mixture. Record your observations and conclusions about any equilibrium shifts on the report sheet. Tube A = Control for color comparisons. 10 Tube B = Increase [Fe 3+ ]. Add 2 ml of 0.1 M FeCl 3 and compare the color to that of test tube A. Tube C = Increase [SCN 1- ]. Add 2 ml of 0.1 M NH 4 SCN and compare the color to that of test tube A. Tube D = Increase [Cl 1- ]. Add 2 ml of 0.1 M NH 4 Cl and compare the color to that of test tube A. 11 Tube E = Increase Temperature. Place tube E in the hot (but not boiling) water bath 12. After 10 minutes, compare the color to that of test tube A. Tube F = Decrease Temperature. Place tube F in the ice water bath. After 10 minutes, compare the color to that of test tube A. NOTES: 8 Tubes can be labeled by Marking with a grease pencil. Writing on the white patch with pencil. Writing on stickers and attaching to each tube. 9 The amount of solution does not need to be exact but you want to have approximately the same amount in each tube. By dividing evenly you will have samples of about 10 mls each. 10 You will not make any changes to the mixture in tube A so that you can compare its color to the colors that form in the other test tubes. A B C D E F 11 When FeCl 3 reacts with excess chloride ion, Cl 1-, the colorless complex, 1- FeCl 4 is formed, thus removing the yellow Fe 3+ ions from solution. FeCl 3 + Cl 1-1- FeCl 4 12 Turn off the heat source to prevent the water from boiling. Boiling away water will concentrate the solution and thus intensify its color. 7. (Optional: After a time switch tubes E and F and see if the results are reversible.) E Hot Water Bath F Ice Bath 6 CH105 Lab 11: Rates & Equilibrium (F15)

7 LAB 11: RATES & EQUILIBRIUM PRE LAB EXERCISES: NAME DATE 1. If concentrated acids or bases contact your skin you should A. remove any clothing covering the skin affected. D. wash with soap. B. flush with copious amounts of water. E. all of these. C. inform your instructor 2. The word copious means A. a very little bit. B. a large quantity. C. hot. D. ice cold. 3. The rate of a chemical reaction may be increased by A. increasing the frequency of molecular collisions B. increasing the energy of molecular collisions C. lowering the activation energy of the reaction D. all of these 4. Write the complete and balanced equation for the equilibrium reaction between Iron (III) chloride and Ammonium thiocyanate: 5. Write the equilibrium expression (K eq ) for the net ionic equation of this reaction: Fe 3+ (aq) + 2SCN 1- (aq) Fe(SCN) 2 1+ (aq) 6. Given the following equilibrium A + B C + D, what will be the effect on the equilibrium if each of the following takes place? Will equilibrium Shift or or none (N)? A. Increase [C] A and B B. Decrease [A] C and D C. Increase [B] C and D D. Decrease [D] A and B Will the concentration of these substances increase ( ) or decrease ( ) or be unaffected (U)? CH105 Lab 11: Rates & Equilibrium (F15) 7

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9 LAB 11: RATES & EQUILIBRIUM REPORT: NAME PARTNER DATE I. REACTION RATES: A. Concentration Effects: Complete and Balance: Equation = Mg (s) + HCl (aq) Equation = H 2(g) + O 2(g) 1.0 M HCl Time (sec) for complete reaction Reaction Rate (#1 fastest ) Exothermic (Exo) or Endothermic (Endo) Observations: Include verification for identity of gas 2.0 M HCl 3.0 M HCl Summary and Conclusions: 1. According to our experimental data, increasing the concentration of a reactant the rate of a reaction. A. increases B. decreases C. has no effect on D. has variable effects on Explanation/Analysis: Give an explanation for why the rate results were the way they were. Explain any anomalies. B. Temperature Effects: Complete and Balance: Account for any gas produced. Equation= HC 2 H 3 O 2(aq) + NaHCO 3(s) Test Tube Temperature Reaction Rate (#1 fastest ) Observations H: Hot HC 2 H 3 O 2 C: Cold HC 2 H 3 O 2 Summary and Conclusions: 1. According to our experimental data, increasing the temperature of a reactant the rate of a reaction. A. increases B. decreases C. has no effect on D. has variable effects on Explanation/Analysis: Give an explanation for why the temperature results were the way they were. Explain any anomalies. CH105 Lab 11: Rates & Equilibrium (F15) 9

10 C. Catalyst Effects: Complete and Balance: Equation= H 2 O 2(aq) Test Tube: Reaction Rate (#1 fastest ) Observations Where bubbles form how might you go about confirming the identity of the gas? Explain what you d do and what you d expect to see. 1. Hydrogen Peroxide 3% H 2 O 2 (control) 2. Hydrogen Peroxide 3% H 2 O 2 (warm) 3. Hydrogen Peroxide H 2 O 2 w/ potato catalase Summary and Conclusions: 1. According to our experimental data, adding a catalyst the rate of a reaction. A. increases B. decreases C. has no effect on D. has variable effects on 2. The gas given off in the decomposition of hydrogen peroxide is. Explanation/Analysis: Give an explanation for why the catalyst results were the way they were. Explain any anomalies. 3. Commercial hydrogen peroxide, H 2 O 2, is stored in dark bottles in which a trace of the inhibitor acetanilide has been added. Both the dark bottle and the acetanilide help prevent the decomposition of the hydrogen peroxide. An inhibitor acts the opposite of a catalyst and so increases the activation energy of the reaction. Assuming the decomposition of H 2 O 2 is slightly exothermic complete the given potential energy diagram using a solid line ( ) to represent the decomposition of hydrogen peroxide, H 2 O 2. dotted line (.) to represent the decomposition of H 2 O 2 in the presence of raw potato containing the enzyme catalase. starred line (*****) to represent the decomposition of H 2 O 2 in the presence of acetanilide. Energy Reaction Progress 4. The purpose of storing commercial H 2 O 2 in a dark bottle is A. To keep the hydrogen peroxide cool. B. To increase the E act and so prevent hydrogen peroxide from decomposing. C. to prevent light from entering the bottle because light can act as a catalyst and so lower the E act enabling hydrogen peroxide to decompose at room temperature. D. To prevent light from entering the bottle because light can provide the E act needed to cause hydrogen peroxide to decompose. 10 CH105 Lab 11: Rates & Equilibrium (F15)

11 II. EQUILIBRIUM Stress Applied Observation (resulting color) Fe 3+ (aq) + 2SCN 1- (aq) 1+ Fe(SCN) 2 (aq) iron(iii) ion thiocyanate ion iron(iii) thiocyanate complex (pale yellow) (colorless) (deep red) [Fe 3+ (aq)] [SCN 1- (aq)] or Original color or Original color Shift or or N [Fe(SCN) 2 1+ (aq) ] or A. Control Original color of mix B. Add [Fe 3+ ] Increase C. Add [SCN 1- ] Increase D. Add [Cl 1- ] E. Add heat (Increase Temp) F. Remove heat (Decrease Temp) Write the equation for the reaction that occurred upon addition of NH 4 Cl. Give an explanation (reason) for why adding NH 4 Cl causes the effect observed on the equilibrium. Give an explanation (reason) for the effect of heat and cold on the equilibrium. Be specific. Is heat a reactant or a product? 1. According to our experimental data, the reaction Fe 3+ (aq) + 2SCN 1- (aq) Fe(SCN) 2 1+ (aq) is. A. endothermic B. exothermic C. neither endo nor exothermic Explanation/Analysis: Explain why the overall results were the way they were. Explain any anomalies. RELATED EXERCISES: 1. The fact that chewed food digests faster than unchewed food can best be explained by A. increased surface area in chewed food than in unchewed. B. increased temperature of chewed food than of unchewed. C. greater concentration of food in chewed than in unchewed. D. more catalyst present in chewed food than in unchewed. CH105 Lab 11: Rates & Equilibrium (F15) 11

12 2. Hospitals take special precautions when oxygen is administered to a patient because A. oxygen is flammable. B. increased concentrations of oxygen will increase rates of combustion. C. oxygen acts as a catalyst in combustion reactions. D. all of these. 3. Increasing the concentration of a reactant A. increases the frequency of molecular collisions B. increases the energy of molecular collisions C. increases both the frequency and energy of molecular collisions D. lowers the activation energy of the reaction E. increases both the frequency and energy of molecular collisions and also lowers the activation energy of the reaction. 4. Increasing the temperature of a reactant A. increases the frequency of molecular collisions B. increases the energy of molecular collisions C. increases both the frequency and energy of molecular collisions D. lowers the activation energy of the reaction E. increases both the frequency and energy of molecular collisions and also lowers the activation energy of the reaction. 5. Increasing the surface area of a solid reactant A. increases the frequency of molecular collisions B. increases the energy of molecular collisions C. increases both the frequency and energy of molecular collisions D. lowers the activation energy of the reaction E. increases both the frequency and energy of molecular collisions and also lowers the activation energy of the reaction. 6. Unlike today s blimps that are filled with nonflammable helium, the famous Hindenburg dirigible was filled with 190,000,000 liters of hydrogen. After making 10 crossing of the Atlantic Ocean, it was destroyed by fire while landing in New Jersey in1937, killing 36 of the 92 people on board. The cause of the fire was never determined, but it is likely that static electricity ignited leaking hydrogen. When ignited, hydrogen reacts with oxygen in the air to produce water vapor. The burning of hydrogen is. A. endothermic B. exothermic C. neither endo nor exothermic 7. The cause of the Hindenburg fire was never determined, but it is likely that static electricity ignited leaking hydrogen. In the ignition of the hydrogen static electricity A. is a catalyst and so lowers the activation energy enabling hydrogen to combust at atmospheric temperature. B. is a catalyst and so increases the activation energy enabling the hydrogen to combust at atmospheric temperature. C. is not a catalyst but is simply providing the activation energy needed to cause hydrogen and oxygen to react. D. caused the equilibrium to shift to product and thus consume the hydrogen gas. 8. The rate of a reaction usually doubles for every 10 o C increase in temperature. Is heat a catalyst? A. Yes B. No Explain your answer 12 CH105 Lab 11: Rates & Equilibrium (F15)