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Department. of Physical Sciences Kingsborough Community College The City University of New York 2018 1
Experiment 1 Chemical Equilibrium and Le Châtelier’s Principle
Goals To become familiar with the law of mass action and Le Chatelier's Principle. Discussion Chemical equilibrium A system at chemical equilibrium is one in which the concentrations of all the components of the equilibrium are constant over time. For example, if 1 M dinitrogen tetraoxide gas (a heavy but invisible gas) is placed in a container and heated to 100°C it is found that nitrogen dioxide (a poisonous brown gas responsible for the smog seen in cities) will be formed according to the equation:
2 4 2 dinitrogen tetraoxide nitrogen dioxide
N O (g) (^) 2NO (g)
On the other hand, cooling the sample to a low temperature (–80°C) of will result in the formation of N 2 O 4. By changing the temperature we can make the system "shift" to one side of the equation or the other. When we say that the system "shifts" to the left we mean that the reactant (N 2 O 4 ) is produced in increasing amounts. When we say that the reaction "shifts" to the right we mean that the product (NO 2 ) is produced in increasing amounts.
The double arrow ( (^) ) in the equation is used to indicate that the system is reversible. In a reversible
reaction the system can react in the forward direction (→) or in the reverse direction (). At room temperature (25°C), it is found that, over time, a mixture of NO 2 and N 2 O 4 will result. The equilibrium concentrations of the two species will become:
species initial concentration at 25°C equilibrium concentration at 25°C NO 2 0.00 mol/L 0.067 mol/L N 2 O 4 1.00 mol/L 0.967 mol/L
If we keep the temperature constant then these equilibrium concentrations will remain constant over time. The system is said to be in chemical equilibrium. When a system is in chemical equilibrium the concentrations of the reactants and products are constant.
Stressing a system at equilibrium What happens if we take a system in chemical equilibrium and make a change to it? For example, suppose that we have a mixture of N 2 O 4 at 0.967 M and NO 2 at 0.067 M. We then add some NO 2 to the system so that the concentration of NO 2 is now 1.00 M. What happens?
What is observed is that the mixture responds to the change that we made. The response is one which counteracts the change we made. The nitrogen dioxide will begin to react more rapidly and form more dinitrogen tetraoxide. After some period of time, the system will again stabilize and the concentrations will stop changing. We can calculate the new concentrations. Once the system reaches the new equilibrium the concentration of NO 2 will be 2.16 M and N 2 O 4 will be 0.99 M. What happened?
The law of mass action
For the formation of N 2 O 4 from NO 2 it is found that at 25°C the ratio
2 2 2 4
(concentration of NO ) (concentration of N O )
is equal to about 0.0046 when the system is at equilibrium.
Sample calculation Consider the initial mixture we looked at earlier. It was stated that at equilibrium the concentration of NO 2 was 0.76 M and the concentration of N 2 O 4 was 0.12 M. Do these concentrations correspond to the
ratio
2 2 2 4
(concentration of NO ) (concentration of N O )
2
(0.12M)
(0.76M)
Since the ratio is equal to 0.21 we say that the system is at equilibrium because we find that once the ratio is equal to 0.21 the concentrations stop changing.
Now, what happens when the concentration of NO 2 is increased to 2.0 M as stated earlier? The system is no longer at equilibrium because NO 2 was added. Again, we can determine this by calculation. If the
ratio^2 2
(concentration of N O ) (concentration of NO )
is equal to 0.21 then the system is at equilibrium.
But for 2.0 M and 0.12 M the ratio is: 2
(0.12 M)
(2.0M)
= 0.03 so the system is not at equilibrium. In order for
the system to return to equilibrium it must form more N 2 O 4 and, in the process, consume NO 2. The concentration of N 2 O 4 will increase to and the concentration of NO 2 will decrease to.
nitrogen dioxide^ 2NO (g)^2 dinitrogen tetraoxideN O (g)^2
These observations can be summed up by Le Chatelier's Principle : When a system at equilibrium is subjected to a disturbance, it will respond such that the effect of the disturbance is minimized.
Le Chatelier's principle is commonly observed in chemical reactions. In this experiment we will use two separate methods to disturb systems at chemical equilibrium.
- Change the concentration of one or more species in the chemical equilibrium
- Change the temperature of the system
Acid-base indicators Acid-base indicators are materials that change color to reflect changes is solution acidity. For example, phenolphthalein is colorless in water when the solution is acidic (the concentration of H+^ ions is greater than the concentration of – OH ions). But when the solution becomes basic (the concentration of – OH ions is greater than the concentration of H+^ ions) phenolphthalein solutions are pink. The phenolphthalein is reflecting the acidity of the solution it is in.
O
O
O H
O H
O
O
O
O
colorless in acidic solution pink in basic solution
- Add 6 M sodium hydroxide solution to the test tube. Follow the same procedure as in step 2. Drops of NaOH solution Color 0
What can you conclude from your observations? What is the effect of adding the sodium hydroxide solution to the crystal violet solution?
Methyl Red: Methyl red is another common acid-base indicator. In solution it ionizes according to the equation :
red yellow
HMR (aq) (^) H (aq) +MR (aq)
- Using a graduated cylinder add 5 mL of distilled or deionized water to a clean test tube. Add 2- drops of methyl red indicator.
What is the color of the solution?
Which species has a higher concentration in the solution, HMR or MR–?
Is the ratio
(concentration of MR )- (concentration of HMR)
large or small?
- Add either HCl (6 M) or NaOH (6 M) to the solution dropwise to change the color of the solution. Which one, HCl or NaOH, should you add?
Did the solution change color as you predicted?
Department of Physical Sciences Kingsborough Community College The City University of New York
Part 2. Chemical equilibrium in cobalt complexes Some ionic compounds exist as hydrates. They form weak bonds to water molecules. The attachment of these water molecules can affect the electronic structure of the compound and affect its color. An example is cobalt(II) chloride. Without the attached water molecules cobalt(II) chloride is a blue solid. When exposed to humid air, however, the salt forms a hydrate and turns a dark red. The compound is called cobalt(II) chloride hexahydrate and its formula is: CoCl 2 ∙6H 2 O. This process can be represented by equation 1:
equation 1 2 2 2 2 blue fromair dark red
CoCl (s) + 6H O(g) CoCl 6H O(s)
By heating the hydrate water can be driven off:
equation 2
heat CoCl (^2) dark red 6H O(s) 2 coolCoCl (s) + 6H O(g)blue 2 2
When dissolved in water the cobalt(II) chloride salt decomposes, resulting in the formation of the Co(H 2 O) 6 2+^ ion and a deep red solution. This process is represented by equation 3:
equation 3 (^2 2) H O(l) 2 2 62 - dark red crystals redsolution
CoCl 6H O(s) Co(H O) (aq) + 2Cl (aq)
Alternatively, solutions high in chloride concentration can form the dark blue aqueous CoCl 4 2-^ ion (equation 4):
equation 4 2 6 2+^ - 4 2- 2 red blue
Co(H O) (aq) + 4Cl (aq) CoCl (aq) +6H O(l)
In this part of the experiment we will test LeChatelier's Principle using the cobalt(II) chloride salt.
Procedure
- Place a few crystals of CoCl 2 ·6H 2 O into each of three test tubes.
- Add 2 mL of water to the first test tube. Label the test tube (H 2 O). Stir the crystals to dissolve them into solution.
What is the color of this solution?
What is the dominant species of cobalt in this solution?
Explain your observations equation 3.
- Add 2 mL of 12 M HCl to the second test tube. Label the test tube (HCl). Stir the crystals to dissolve them into solution.
What is the color of this solution?
What is the dominant species of cobalt in this solution?
- Slowly add distilled or deionized water drop by drop with stirring, until no further color change occurs.
Department of Physical Sciences Kingsborough Community College The City University of New York
Part 3: Solubility and Complex Ion Equilibria of Zinc and Magnesium Ions
Both zinc (II) ions and magnesium (II) ions form insoluble hydroxide precipitates. However, these hydroxide precipitates are quite different in their properties, as this part of the experiment will show. For example, Zn(OH) 2 forms [Zn(OH) 4 ]^2 –^ with excess OH–^ and [Zn(NH 3 ) 4 ]2+^ upon addition of NH 3. All hydroxide precipitates dissolve in acid. Hydroxide precipitates that react with bases are called amphoteric hydroxides.
When sodium hydroxide is added to a solution of zinc nitrate, Zn(NO 3 ) 2 , the following reaction occurs.
Zn(NO 3 ) 2 (aq) + 2NaOH(aq) Zn(OH) 2 (s) + 2NaNO 3 (aq)
Write the complete ionic equation and the net ionic equation for the above reaction:
Complete ionic equation:
Net ionic equation:
A magnesium ion solution has a similar reaction:
Mg(NO 3 ) 2 (aq) + 2NaOH(aq) Mg(OH) 2 (s) + 2NaNO 3 (aq)
Write the complete ionic equation and the net ionic equation for the above reaction:
Complete ionic equation:
Net ionic equation:
Step 1: To each of three test tubes, add about 2 mL of 0.1 M Zn(NO 3 ) 2. To each test tube add one drop of 6 M NaOH and stir. Record what you see below.
Observation: _______________________________________________________
Step 2: To the first test tube from Step 1 above, add 6 M HCl drop by drop with stirring. To the second test tube add 6 M NaOH, again drop by drop with stirring. To the third test tube, add 6 M NH 3 one drop at a time with stirring. Record your observations below:
Addition of HCl __________________________________________________________
Addition of excess NaOH _________________________________________________
Addition of NH 3 __________________________________________________________
Repeat steps 1 and 2 using a solution of 0.1 M Mg(NO 3 )2. Record what you observe below.
Addition of 1 drop NaOH _________________________________________________
Addition of HCl __________________________________________________________
Addition of excess NaOH _________________________________________________
Addition of NH 3 __________________________________________________________
Below write the equations for each of the steps above. (Your instructor will write these on the chalkboard. Copy them carefully).
For zinc hydroxide:
The addition of HCl ____________________________________________________
The addition of excess NaOH ____________________________________________
The addition of NH 3 ____________________________________________________
For magnesium hydroxide:
The addition of HCl ____________________________________________________
The addition of excess NaOH ____________________________________________
The addition of NH 3 _____________________________________________________
From your observations above, how is Mg(OH) 2 similar to Zn(OH) 2?
From your observations above, how is Mg(OH) 2 different from Zn(OH) 2? (Note that some cations form many complex ions and others do not.) Which metal hydroxide is an amphoteric hydroxide?
Complete ionic equation for the above reaction:
Net ionic equation for the above reaction:
Step 6: Now add 6 M NH 3 drop by drop to the test tube in Step 5 above. Record what happens below including any color changes.
Addition of excess NH 3 ____________________________________________________
Molecular equation for the above reaction (only for the dissolving of Cu(OH) 2 ):
From your observations above, discuss how solutions of copper(II) ions differ from those of Zn(II) and Mg(II) ions as regarding their reactions with sodium hydroxide and ammonia. Is Cu(OH) 2 amphoteric?
An Aside: Copper(II) sulfate in crystalline form actually has water of hydration, just as cobalt(II) chloride does. The formula is correctly written as CuSO 4 5H 2 O and the name is copper(II) sulfate pentahydrate. Your instructor will show you some of the crystals and then heat them in a large test tube to drive off the water. A few drops of water will then be added to the dried solid. Record the color changes below. Note: In describing a solid whose water of hydration has been removed, the word “anhydrous” is sometimes added to the name. “Anhydrous” means “without water.”
Color of crystalline copper(II) sulfate pentahydrate ___________________________
Color of the heated powder, copper(II) sulfate anhydrous _____________________ [The equation for the reaction is CuSO 4 5H 2 O(s) CuSO 4 (s) + 5H 2 O(g)]
Color of the heated powder after water has been added _______________________ [Adding water turns the copper(II) sulfate anhydrous back into copper(II) sulfate pentahydrate, but now it is probably in aqueous solution. The equation is CuSO 4 (s) + 5H 2 O(l) CuSO 4 5H 2 O(aq)]
Hydroxide Properties Summary Table
Metal hydroxide Dissolves in excess OH–^ (yes or no)
Forms complexes with NH 3 (yes or no)
Amphoteric? (yes or no)
Zn(OH) 2
Mg(OH) 2
Cu(OH) 2
