Le Chatleir's Principle
February 15, 2013
Purpose
The purpose of this lab is to subject a system to various stresses and observe the effects on the equilibrium, such as changes in concentration and changes in temperature. The equilibrium will be examined through the observation of cobalt chloride, hydrochloric acid, and silver nitrate in a system.
Procedure
1. Wear googles, gloves, and apron for safety.
2. Place beaker containing 100mL water on a hot plate and set it to a moderate setting
3. Label a well plate from left to right along the top with the numbers 1-6. Down the left side, label the rows of wells A-D. Place the well plate on a sheet of white paper.
4. Obtain 4 micropipets of CoCl2, HCl, H2O, and AgNO3 and place them upside down in an inverted petri dish cover
5. Place 5 drops of CoCl2 solution in in each of the 24 wells of the well plate.
6. Add two, four, six, eight, ten, and twelve drops of HCl to the CoCl2 solution in column 1, 2, 3, 4, 5, 6 of the well plate, respectively.
7. Mix the contents of each and record observations
8. In row B, add one drop of HCl to each well and stir; record observations.
9. In row C, add five drops of distilled water to each well and record observations.
10. In row D, add five drops of AgNO3 solution to each well and stir, then record observations.
11. Put 5mL of the cobalt solution in a test tube
12. Add enough HCl until the solution turns purple
13. Place the test tube in the beaker of hot water from step 2 and record observations
14. Place test tube in a 250 mL beaker containing ice water and record observations.
15. Dispose of all chemicals and clean up the work area.
2. Place beaker containing 100mL water on a hot plate and set it to a moderate setting
3. Label a well plate from left to right along the top with the numbers 1-6. Down the left side, label the rows of wells A-D. Place the well plate on a sheet of white paper.
4. Obtain 4 micropipets of CoCl2, HCl, H2O, and AgNO3 and place them upside down in an inverted petri dish cover
5. Place 5 drops of CoCl2 solution in in each of the 24 wells of the well plate.
6. Add two, four, six, eight, ten, and twelve drops of HCl to the CoCl2 solution in column 1, 2, 3, 4, 5, 6 of the well plate, respectively.
7. Mix the contents of each and record observations
8. In row B, add one drop of HCl to each well and stir; record observations.
9. In row C, add five drops of distilled water to each well and record observations.
10. In row D, add five drops of AgNO3 solution to each well and stir, then record observations.
11. Put 5mL of the cobalt solution in a test tube
12. Add enough HCl until the solution turns purple
13. Place the test tube in the beaker of hot water from step 2 and record observations
14. Place test tube in a 250 mL beaker containing ice water and record observations.
15. Dispose of all chemicals and clean up the work area.
Data table
|
|
CoCl2 at room temperature
CoCl2 in hot water CoCl2 in cold water |
purple/pink
dark purple/blue clear pink |
Conclusion
In conclusion, stress on the system due to changes in concentration caused the equilibrium to shift in the reaction, which changed the colors of he system, creating pinker or cloudier . When the temperature of the system was changed by adding heat with the hot water, the equilibrium was shifted to the right of the equation, causing the color to change to a dark purple/blue. When it was added to cold water, the equilibrium was shifted back left, which caused the color to go back to the original.
Discussion of Theory
This lab displays the effects of Le Chatleir's principle on the equilibrium of an equation when stress is added to certain factors of the reaction. In this particular lab, the stresses tested were concentrations of HCl, H2O and AgNO3, as well as increased and decreased temperature on the system. The imposed stresses were to cause the system to shift left or right to establish a new equilibrium condition in order to minimize the stress placed on it. Adding or removing a pure solid or liquid does not change the concentration, so it does not affect equilibrium. However, in this lab, the solutions added were aqueous, so they caused a shift. The addition of HCl was an increase in the products, so there was a shift to the right. The addition of water was an increase in the reactants, so there was a shift to the left. The addition of silver nitrate was an increase in the reactants, so there was a subsequent shift to the left for the equilibrium.
In addition to changes in concentration in this lab, changes in temperature were also tested and measured in the reaction. Increasing temperature in a reaction can be thought of as adding heat, or "H". When heat was added to the CoCl2 in the experiment through the hot water, the equilibrium of the reaction shifted right. When the test tube was placed into ice water, the equilibrium shifted left, back to where is started then beyond that point. This displays that the reaction that took place was endothermic. When written out with the reaction, the delta H would be a positive number, written on the left side of the equation with the products. Heat is a reactant in this equation, so adding any more would cause the shift right to minimize the stress placed onto that side of the equation. If the reaction were exothermic, all of these results would be opposite, causing the shift in equilibrium to be left when heat was added, and right when heat was taken away. In addition, changes in temperature also cause a change in the equilibrium constant, or k value. A shift right, when the heat was increased for this lab, favors the forward reaction which would cause the k to increase because the products over reactants ratio would be higher. A shift left, in this instance when heat was removed, would cause the k value to decrease because the reverse reaction would be favored.
In addition to changes in concentration in this lab, changes in temperature were also tested and measured in the reaction. Increasing temperature in a reaction can be thought of as adding heat, or "H". When heat was added to the CoCl2 in the experiment through the hot water, the equilibrium of the reaction shifted right. When the test tube was placed into ice water, the equilibrium shifted left, back to where is started then beyond that point. This displays that the reaction that took place was endothermic. When written out with the reaction, the delta H would be a positive number, written on the left side of the equation with the products. Heat is a reactant in this equation, so adding any more would cause the shift right to minimize the stress placed onto that side of the equation. If the reaction were exothermic, all of these results would be opposite, causing the shift in equilibrium to be left when heat was added, and right when heat was taken away. In addition, changes in temperature also cause a change in the equilibrium constant, or k value. A shift right, when the heat was increased for this lab, favors the forward reaction which would cause the k to increase because the products over reactants ratio would be higher. A shift left, in this instance when heat was removed, would cause the k value to decrease because the reverse reaction would be favored.
Sources of Error
A source of error in this lab was that there was no distilled water for use in the experiment. Instead, tap water had to be used which contains impurities that would skew the results that were observed throughout the lab. In addition, different sized pipettes were used to drop the contents into the wells of the well plate. The built in droppers produced larger drops than that of the micropipettes that were used for the HCl and water. This would cause the amounts added to the wells to be slightly uneven, producing slightly different results that expected. In well C-5 of the well plate, an excess amount of water was accidentally added. This would cause the equilibrium to shift more than it should, and the reaction and the subsequent color change would be different. Also, the test tube may not have been left in the cold or hot water as long as it should have. This would result in the equilibrium shift and subsequent color change to not be as dramatic as they should be.
Pre-Lab Questions
1. Le Chatleir's principle states that when an equilibrium system is subjected to a stress, the system responds by attaining a new equilibrium condition that minimizes the imposed stress.
2. When equilibrium is reached, the rate of the forward and reverse reactions are equal.
3. Changes in concentration of water, HCl, and CoCl2; and changes in temperature are the stresses that will be studied in this experiment.
4. A compound that has water as part of its crystal structure is called a hydrate.
5. Goggles and lab apron and gloves should be worn at all times during the lab. Hydrochloric acid is extremely corrosive and will irritate skin. Silver nitrate is toxic and stains.
6. a) When HCl is added, the equilibrium would shift right.
b) When H2O is added, the equilibrium would shift left.
c) When NaOH is added, the equilibrium would shift left.
2. When equilibrium is reached, the rate of the forward and reverse reactions are equal.
3. Changes in concentration of water, HCl, and CoCl2; and changes in temperature are the stresses that will be studied in this experiment.
4. A compound that has water as part of its crystal structure is called a hydrate.
5. Goggles and lab apron and gloves should be worn at all times during the lab. Hydrochloric acid is extremely corrosive and will irritate skin. Silver nitrate is toxic and stains.
6. a) When HCl is added, the equilibrium would shift right.
b) When H2O is added, the equilibrium would shift left.
c) When NaOH is added, the equilibrium would shift left.
Post-Lab Questions
1. a) In this case, the equilibrium would shift to the right.
b) In this case, the equilibrium would shift to the left.
c) In this case, the equilibrium would shift to the left.
d) In this case, the equilibrium would shift to the right.
e) In this case, the equilibrium would shift to the left.
2. In 1a, the equilibrium would shift right because the addition of HCl would cause the moles of Cl- to increase, which would cause a need for the equilibrium to move to the right of the reaction. In 1b, the reaction would shift left because the addition of water would cause the number of moles of H2O to increase, which would cause a need for the equilibrium to move left to even the reaction out.
3. When AgNO3 was added to row D in the well plate, the pink solution turned very cloudy and milky whitish. This is because a precipitate formed within the solution.
4. This reaction is endothermic because the delta H of the equation is a positive number. When there is more energy at the start of the reaction with the reactants than at the end of the reaction with the products, the reaction is known to be endothermic.
5. k= [CoCl4][H20]^6/[Co(H2O)6][Cl-]^4
b) In this case, the equilibrium would shift to the left.
c) In this case, the equilibrium would shift to the left.
d) In this case, the equilibrium would shift to the right.
e) In this case, the equilibrium would shift to the left.
2. In 1a, the equilibrium would shift right because the addition of HCl would cause the moles of Cl- to increase, which would cause a need for the equilibrium to move to the right of the reaction. In 1b, the reaction would shift left because the addition of water would cause the number of moles of H2O to increase, which would cause a need for the equilibrium to move left to even the reaction out.
3. When AgNO3 was added to row D in the well plate, the pink solution turned very cloudy and milky whitish. This is because a precipitate formed within the solution.
4. This reaction is endothermic because the delta H of the equation is a positive number. When there is more energy at the start of the reaction with the reactants than at the end of the reaction with the products, the reaction is known to be endothermic.
5. k= [CoCl4][H20]^6/[Co(H2O)6][Cl-]^4
Critical Thinking
1. If NaCl were added, the equilibrium would likely shift right because it would add to the moles of the Cl-.
2. Co(H2O)6^(2+) + 4Cl- + 50 kJ/mol <--> CoCl4^(2-) + 6H2O
3. There would likely be more solid silver chloride than silver and chloride ions. This is because the k value is quite high, insinuating that the top half of the fraction for calculating the number is larger than the bottom half. Because the top half of the fraction is 1 due to the fact that solids are not taken into account in calculating the equilibrium constant, it is evident that the amount of reactants is much lower than that of the product.
2. Co(H2O)6^(2+) + 4Cl- + 50 kJ/mol <--> CoCl4^(2-) + 6H2O
3. There would likely be more solid silver chloride than silver and chloride ions. This is because the k value is quite high, insinuating that the top half of the fraction for calculating the number is larger than the bottom half. Because the top half of the fraction is 1 due to the fact that solids are not taken into account in calculating the equilibrium constant, it is evident that the amount of reactants is much lower than that of the product.