Lab - Disturbing An Equilibrium System

OBJECTIVE: To study factors which can disturb an equilibrium system
MATERIALS:
Spatula, several test tubes, beakers, foil, test tube rack, pipette, test tube clamp
PROCEDURE:
General Procedure: 1. Obtain a test tube rack and six test tubes (clean!). 2. Prepare a hot water bath (one per table) using a 250 mL beaker, tap water and hot plate. 3. Prepare an ice bath (one per table) using a 250 mL beaker, crushed ice and tap water.
Reaction #1 – Iron (III) and thiocyanate 1. Half-fill the first test tube in your rack with distilled water. 2. Add 2 drops on 0.1 M Fe(NOɜ)ɜ and 2 drops of 0.1 M KSCN to this tube. Mix the contents thoroughly. 3. If the contents of the tube are not red-Orange, repeat Step 2 until the solution is red-orange. 4. Divide the red-orange solution in the first tube among the other tubes so each tube contains the same volume.

Fe ³ (aq) + SCN‾ (aq) ↔ FeSCN ² (aq) + heat

5. Leave tube #1 undisturbed; use it as the control. 6. Use a clean, dry spatula to add a small crystal or two of solid iron (III) nitrate, Fe(NOɜ)ɜ, to tube #2. Mix. Record what you added to change the system and what you observed. 7. Use a clean, dry spatula to add one or two small crystals of solid potassium thiocyanate, KSCN, to tube #3. Mix and record info. 8. Add 5 drops of 0.1 M sodium hydroxide, NaOH, to tube #4. Mix, record info. 9. Use a test tube clamp to hold tube #5 in the hot water bath. Record info. 10. Use a test tube clamp to hold #6 in the cold water bath. Record info. 11. Discard all test tube contents in waste container (NOT in the sink). Rinse tubes completely, first with tap water and then with distilled water.

Reaction #2 – Complex Ions of Copper (II) (Cu ²) 1. Half-fill the first test tube with 1.5 M copper (II) chloride, CuCl , solution. 2. Divide so that five tubes contain equal volumes. Equilibrium has already been established in the solution.

CuCl ‾² (aq) + 4H O (l) ↔ Cu(H O) ² + 4Cl‾ (aq) + heat

3. Add a small quantity (the size of a rice grain) of solid calcium chloride, CaCl, to tube #2. Mix to dissolve the solid. Repeat the addition until no more solid will dissolve. Record. 4. Add enough ethyl alcohol,, CHOH, to triple the volume of the solution in tube #3. Mix and record. 5. Using a clamp, place tube #4 in the hot water bath. Record. 6. Using a clamp, place tube #5 in the cold water bath. Record. 7. Dispose of waster and clean tubes as before.

Reaction #3 – complex Ions of Cobalt (II)(Co) 1. Half-fill a test tube with 1.5 M cobalt (II) chloride, CoCl . 2. Divide so that five tubes contain equal volumes. Equilibrium has been established in the solution.

Heat + Co(H O) ² (aq) + 4Cl‾ (aq) ↔ CoCl ‾²(aq) + 6H O (l)

3. Add a small quantity (the size of a rice grain) of solid calcium chloride, CaCl , to tube #2 Mix to dissolve the solid. Repeat the addition until no more solid will dissolve. Record. 4. Add enough acetone, CHɜCOCHɜ, to double the volume of the solution of tube #3. Mix and record. 5. Using a clamp, place tube #4 in 6he hot water bath. Record. 6. Using clamp, place tube #5 in the cold water bath. Record. 7. Clean up as before. Be sure to wash your hands thoroughly!!

DATA/RESULTS Reaction #1
|Tube # |Disturbance ( what you did) |Observed Change |Directions of Shift |
|2 |FeNOɜ |Red |Right |
|3 |KSCN |Dark Red |Right |
|4 |NaOH |Lighter in color |Left |
|5 |Hot Water |Light Yellow |Left |
|6 |Cold Water |Darker Yellow |Right |

Reaction #2
|Tube # |Disturbance (what you did) |Observed Change |Direction of Shift |
|2 |CaCl2 |Light green |Left |
|3 |Alcohol |Yellow Green |Left |
|4 |Hot Water |Green |Left |
|5 |Cold Water |Blue |Right |

Reaction #3
|Tube # |Disturbance (what you did) |Observed Change |Direction of Shift |
|2 |CaCl2 |Dark Purple |Right |
|3 |Acetone |Blue then Light Pink |Right |
|4 |Hot Water |Magenta/Dark Red |Right |
|5 |Cold Water |No Change |Left |

DISCUSSION QUESTIONS: 1. Explain why no visible change can be observed when a system is at equilibrium. Express your answer in terms of rates of the forward and reverse reactions. Both the forward and reverse reactions are exactly the same meaning that once they reaction equilibrium the reaction is still occurring just occurring at the same rate which causes no visible change. 2. Are the rates of the forward and reverse reactions still equal immediately after an equilibrium system is disturbed? No 3. What effect would each of the following have on the rate of a reaction? a. increasing the concentration of a reactant Right Shift b. decreasing the concentration of a reactant Left Shift c. increasing the temperature of a system Shifts to opposite side of heat d. decreasing the temperature of the system Shifts to side of heat

CONCLUSION: Describe how equilibrium can be affected when changing the conditions of the reactants or products. Use examples from your lab.

A decrease in the concentration of either a reactant or product will cause an equilibrium shift to the side in which the decrease has occurred. The same will occur if the concentration was to increase, which would ultimately cause a shift to the increased side. During this lab a simple mislabeling of a reactant or product could cause an error in data.