Experiment 5- Chemical Equilibrium and Le Chatelier’s Principle PDF

Preview

Summary

LAB REPORT...

Description

Chem 132- 101 Experiment 5: Chemical Equilibrium and Le Chatelier’s Principle October 30, 2017

!2 Data and Calculations: 1. When red-colored Co(NO3)2×6H2O crystals are dissolved in 3mL of deionized water, the solution turns pink. To explain this, the equilibrium stress is on the product’s side (addition of water), so the solution shifts towards to reactants. Since the reactants are favored, the equilibrium produces more [Co(H2O)6]2+ ions, which results in a pink-colored solution. 2. When red-colored Co(NO3)2×6H2O crystals are dissolved in 3mL of 12M HCl, the solution turns blue. To explain this, the equilibrium stress is on the reactant’s side (addition of Cl-), so the solution shifts towards the products. Since the products are favored, the equilibrium produces more [CoCl4]2- ions, which results in a blue-colored solution.

Part B: Equilibrium: [Co(H2O)6]2+(aq) + 4Cl-(aq) [CoCl4]2- (aq) + 6H2O (L) 1. Table 1: Test Tube #1 Description of Condition

Prediction

Observation Before/ After

Predominant Species

Initial: 0.1 CoCl4×6H20

Light Pink

———

[Co(H2O)6]2+

3mL of 12M HCl

Blue

Blue

[CoCl4]2-

1mL of Deionized Water

Pink

Pink

[Co(H2O)6]2+

Hot Bath

Pink

Blue

[CoCl4]2-

Ice Bath

Blue

Pink

[Co(H2O)6]2+

• Adding water to the solution decreases the chlorine ion concentration, which causes the equilibrium to shift towards the reactants to create more [Co(H2O)6]2+. • Adding 3mL of 12M HCl puts stress on the reactants side (adding Cl-), which forces to equilibrium to shift towards products/right to create more [CoCl4]2-.

!3 • Adding deionized water puts a stress on H2O on the product side and ultimately decreases [CoCl4]2-, which will shift the equilibrium towards the reactants/left to create more [Co(H2O)6]2+. • Heating the solution shifts it towards products because it is an endothermic reaction; meaning, heat is on the reactant side. When heated, it will shift left and produce more [CoCl4]2-. • Cooling the solution shifts it towards reactants because since it is an endothermic reaction, more heat is required. Therefore it favors reactants and forms more [Co(H2O)6]2+. 2. Table 2: Test Tube #2 Description of Condition

Prediction

Observation Before/ After

Predominant Species

Initial: 0.1 CoCl4×6H20

Light Pink

———

[Co(H2O)6]2+

3mL of 6M HNO3

Pink

Pink

[Co(H2O)6]2+

1mL of Deionized Water

Blue

Pink

[Co(H2O)6]2+

Hot Bath

Pink

Pink

[Co(H2O)6]2+

Ice Bath

Blue

Pink

[Co(H2O)6]2+

• Adding water to the solution decreases the chlorine ion concentration, which causes the equilibrium to shift towards the reactants to create more [Co(H2O)6]2+. • Adding 3mL of puts a stress on the reactants side, which decreases the chloride ion; thus, shifting it to the left to create more [Co(H2O)6]2+. • Adding deionized water puts a stress on H2O on the product side and ultimately decreases [CoCl4]2-, which will shift the equilibrium towards the reactants/left to create more [Co(H2O)6]2+. • Heating the solution generally shifts the equilibrium towards the products side since it is an endothermic reaction. However, not enough energy/heat was produced in order to do so. • Cooling the solution shifts the equilibrium towards the reactants side which creates more [Co(H2O)6]2+.

!4 3. Table 3: Test Tube #3 Description of Condition

Prediction

Observation Before/ After

Predominant Species

Initial: 0.1 CoCl4×6H20

Light Pink

———

[Co(H2O)6]2+

Solid LiCl

Blue

Purple

[Co(H2O)6]2+ [CoCl4]2-

1mL of Deionized Water

Pink

Pink

[Co(H2O)6]2+

Hot Bath

Pink

Pink

[Co(H2O)6]2+

Ice Bath

Blue

Light Pink

[Co(H2O)6]2+

• Adding water to the solution decreases the chlorine ion concentration, which causes the equilibrium to shift towards the reactants to create more [Co(H2O)6]2+. • Adding solid LiCl puts a stress on the reactants side (increases Cl-), which will shift the equilibrium towards products to produce more [CoCl4]2-. • Adding deionized water puts a stress on H2O on the product side and ultimately decreases [CoCl4]2-, which will shift the equilibrium towards the reactants/left to create more [Co(H2O)6]2+. In this case, we may have put more deionized water than intended, which may have affected the shade of pink. • Heating the solution should have shifted the equilibrium to the products side because it is an endothermic reaction. However, results may have been affected because more deionized water was added than expected. • Cooling the solution shifts the equilibrium towards the reactants side which creates more [Co(H2O)6]2+. 4. Table 4: Test Tube #4 Description of Condition

Prediction

Observation Before/ After

Predominant Species

Initial: 0.1 CoCl4×6H20

Light Pink

———

[Co(H2O)6]2+

NH4Cl

Pink

Darker Pink

[Co(H2O)6]2+ [CoCl4]2-

1mL of Deionized Water

Blue

Lighter Pink

[Co(H2O)6]2+

Hot Bath

Pink

Blue

[CoCl4]2-

Ice Bath

Blue

Pink

[Co(H2O)6]2+

• Adding water to the solution decreases the chlorine ion concentration, which causes the equilibrium to shift towards the reactants to create more [Co(H2O)6]2+.

!5 • Adding NH4Cl slightly puts a stress on the reactants side (adding more Cl- ions) but not enough is added to shift the equilibrium to the right which is why it remains a darker pink of [Co(H2O)6]2+. • Adding deionized water puts a stress on H2O on the product side and ultimately decreases [CoCl4]2-, which will shift the equilibrium towards the reactants/left to create more [Co(H2O)6]2+. • Heating the solution shifts it towards products because it is an endothermic reaction; meaning, heat is on the reactant side. When heated, it will shift left and produce more [CoCl4]2-. • Cooling the solution shifts the equilibrium towards the reactants side which creates more [Co(H2O)6]2+. 5. Table 5: Test Tube #5 Description of Condition

Prediction

Observation Before/ After

Predominant Species

Initial: 0.1 CoCl4×6H20

Pink

Pink

[Co(H2O)6]2+

———

————

————

[Co(H2O)6]2+

1mL of Deionized Water

Pink

Lighter Pink

[Co(H2O)6]2+

Hot Bath

Pink

Lighter Pink

[Co(H2O)6]2+

Ice Bath

Pink

Lighter Pink

[Co(H2O)6]2+

• Adding water to the solution decreases the chlorine ion concentration, which causes the equilibrium to shift towards the reactants to create more [Co(H2O)6]2+. • Adding deionized water puts a stress on H2O on the product side and ultimately decreases [CoCl4]2-, which will shift the equilibrium towards the reactants/left to create more [Co(H2O)6]2+. • Heating the solution generally shifts the equilibrium towards the products side since it is an endothermic reaction. However, not enough energy/heat was produced in order to do so. • Cooling the solution shifts the equilibrium towards the reactants side which creates more [Co(H2O)6]2+.

!6 Post-Lab Questions: 1. The equation referred to on page 36 is [Co(H2O)6]2+(aq) + 4Cl-(aq) [CoCl4]2- (aq) + 6H2O (L). This equation is an endothermic reaction; meaning, heat varies on the reactant’s side. For example, when temperature rises, the equilibrium tends to shift towards the products/right side. When this occurs, the products are favored and more [CoCl4]2- ions are formed, which produces the blue-colored effect. On the contrary, when temperature decreases, the equilibrium shifts towards the reactants/left side. When this occurs, the reactants are favored and more [Co(H2O)6]2+ ions are formed, which results in the pink-cored effect. 2. Based on the data collected, the equation referred to on page 36 is an endothermic reaction. The equation with the addition of heat is as follows: [Co(H2O)6]2+(aq) + 4Cl-(aq) +heat [CoCl4]2- (aq) + 6H2O (L)

3. The value of the equilibrium constant (K) for this reaction increases as temperature becomes warmer because the the solution/equation is an endothermic reaction, which requires heat to produce products. For example, if there was a significant amount of heat added, more products will be formed and the equilibrium will shift towards the products/right side. Therefore, because the equilibrium expression can be written as K= [C]c[D]d/[A]a[B]b, the more products that are formed (i.e. more heat), the higher the equilibrium constant. 4. The value of the equilibrium constant (K) for this reaction increase as the concentration of chloride increases because Cl- is on the reactants side. Because of this, the equilibrium stress is on the reactants side (i.e. addition of Cl-), which will cause a shift towards the products/ right side. Therefore, because the equilibrium expression can be written as K= [C]c[D]d/ [A]a[B]b, the more products that are formed, the higher the equilibrium constant (because reactants are in the denominator)....