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Chemical Equilibrium, Acid-Base Equilibria, and Diffusion, Exams of Chemistry

Various topics in chemistry including le chatelier's principle, endothermic reactions, chemical equilibrium, acid-base equilibria, and diffusion. It includes calculations and explanations related to these topics.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Practice Problems for CHGN 124 Final Exam, Spring 2008
1. Explain Le-Chatelier’s Principle, and predict how it will impact the following equilibria:
Fe3+ (aq) + Cl- (aq) FeCl2+(aq) endothermic reaction
Le Chatelier’s principle states that if a stress is applied to a reaction mixture at equilibrium,
reaction occurs in the direction that relieves the stress.
a. add Fe(NO3)3Adding reactant (Fe3+) causes reaction to go to right
b. add HCl Adding reactant (Cl-) causes reaction to go to right
c. add AgNO3Ag+ will precipitate with Cl-, causing reaction to go to left
d. add a catalyst Catalysts do not change equilibrium concentration
e. increase temperature Endothermic reaction will go to right when temp. is increased
2. The air pollutant NO can react with O2 by the following reaction:
2NO(g) + O2(g) 2NO2(g)
a. If we add 1 mole of NO and 2 moles of O2 to a 2 liter container and allow the mixture to
come to equilibrium, the final concentration of NO2 is 0.44M. What is the equilibrium
constant?
Initial concentrations: 0.5 M NO and 1.0 M O2
R 2NO + O2 2NO2
I 0.5 1.0 0
C -2x -x +2x
E 0.5-2x 1.0-x 2x
2x = 0.44M (final concentration of NO2), x = 0.22
[NO] = 0.5 – 2(.22) = 0.06M; [O2] = 1.0 – 0.22 = 0.78 M
Kc = [NO2]2 / [NO]2[O2] = (0.44)2/(0.06)2(0.78) = 69 = 70 (with one significant figure)
b. If we reduce the size of the container to 1 liter at equilibrium, will the concentrations of
reactants and products change? If so, in what direction will the reaction go (towards the
right or left)?
Reducing the volume of the container will increase the concentrations of all species.
Since there are three reactant gas molecules for every two product molecules, the reaction
will proceed to the right to reduce the overall number of gas molecules present.
3. Describe the major difference between strong and weak acids. Identify the conjugate acid of
each of the following bases, and rank then in order (from strongest to weakest base)
Strong acids completely dissociate when added to water. Weak acids only partially or
slightly dissociate in water (most of the acid molecules remain in their un-dissociated,
protonated form). The value of Ka for each acid indicates the degree to which the acid
dissociates. Similarly, the value of Kb for bases indicates the degree to which the base is
likely to accept protons.
a. CH3NH2CH3NH3+Kb = 3.7 x 10-4 2nd strongest
b. NaOH H2O Strong base Kb >>1 Strongest
c. H2OH3O+Kb ~ 1 x 10-14 Weakest
d. H2PO4-H3PO4Ka = 7.5 x 10-3, so Kb = 1.3 x 10-12 2nd weakest
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Practice Problems for CHGN 124 Final Exam, Spring 2008

  1. Explain Le-Chatelier’s Principle, and predict how it will impact the following equilibria: Fe 3+ (aq) + Cl - (aq) ↔ FeCl 2+ (aq) endothermic reaction Le Chatelier’s principle states that if a stress is applied to a reaction mixture at equilibrium, reaction occurs in the direction that relieves the stress. a. add Fe(NO 3 ) 3 Adding reactant (Fe3+) causes reaction to go to right b. add HCl Adding reactant (Cl

) causes reaction to go to right c. add AgNO 3 Ag

will precipitate with Cl

, causing reaction to go to left d. add a catalyst Catalysts do not change equilibrium concentration e. increase temperature Endothermic reaction will go to right when temp. is increased

  1. The air pollutant NO can react with O 2 by the following reaction: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) a. If we add 1 mole of NO and 2 moles of O 2 to a 2 liter container and allow the mixture to come to equilibrium, the final concentration of NO 2 is 0.44M. What is the equilibrium constant? Initial concentrations: 0.5 M NO and 1.0 M O 2 R 2NO + O 2 ↔ 2NO 2 I 0.5 1.0 0 C -2x -x +2x E 0.5-2x 1.0-x 2x 2x = 0.44M (final concentration of NO 2 ), x = 0. [NO] = 0.5 – 2(.22) = 0.06M; [O 2 ] = 1.0 – 0.22 = 0.78 M Kc = [NO 2 ] 2 / [NO] 2 [O 2 ] = (0.44) 2 /(0.06) 2 (0.78) = 69 = 70 (with one significant figure) b. If we reduce the size of the container to 1 liter at equilibrium, will the concentrations of reactants and products change? If so, in what direction will the reaction go (towards the right or left)? Reducing the volume of the container will increase the concentrations of all species. Since there are three reactant gas molecules for every two product molecules, the reaction will proceed to the right to reduce the overall number of gas molecules present.
  2. Describe the major difference between strong and weak acids. Identify the conjugate acid of each of the following bases, and rank then in order (from strongest to weakest base) Strong acids completely dissociate when added to water. Weak acids only partially or slightly dissociate in water (most of the acid molecules remain in their un-dissociated, protonated form). The value of Ka for each acid indicates the degree to which the acid dissociates. Similarly, the value of Kb for bases indicates the degree to which the base is likely to accept protons. a. CH 3 NH 2 CH 3 NH 3 +^ Kb = 3.7 x 10-4^2 nd^ strongest b. NaOH H 2 O Strong base Kb >>1 Strongest c. H 2 O H 3 O+^ Kb ~ 1 x 10-14^ Weakest d. H 2 PO 4 - H 3 PO 4 Ka = 7.5 x 10 - , so Kb = 1.3 x 10 - 2 nd weakest
  1. In lab, you are asked to titrate 500ml of 0.125 M H 2 SO 3 with a 0.237M solution of NaOH. a. What was the initial pH of your H 2 SO 3 solution before adding any NaOH? Before adding any NaOH, we can evaluate pH from equilibrium of H 2 SO 3 R H 2 SO 3 + H 2 O ↔ HSO 3 -^ + H 3 O+ I 0.125 M 0 0 C -x +x +x E 0.125 –x x x Ka1 = 1.5 x 10 - = x 2 /(0.125 – x); x = .036 M; pH = 1. b. What is the pH after the addition of 150 ml of NaOH? Amount H 2 SO 3 = (.125M)(.5L) = 0.0625 moles; amount of OH-^ = (0.237M)(0.15L) = 0.0356 moles N H 2 SO 3 + OH -  HSO 3 - + H 2 O I 0.0625 0.0356 0 -- C -0.0356 -0.0356 +0.0356 -- E 0.0269 0 0.0356 -- In M 0.0414M 0.0548M Equilibrium of remaining weak acid and base: R H 2 SO 3 + H 2 O  HSO 3 - + H 3 O + I 0.0414M -- 0.0548M 0 C -x +x +x E 0.0414-x 0.0548+x x Ka1 = 1.5 x 10 - = x(0.0548+x)/(0.0414 – x); x = 8.0 x 10 - M; pH = 2. c. How many ml of NaOH are required to reach the first equivalence? Need 0.0625moles OH - - (0.0625 moles/0.237M) = .264L = 264 ml d. What is the pH after the addition of 550 ml NaOH? After addition of 264 ml, all of H 2 SO 3 has converted to HSO 3 -. We have now added an additional 286 ml, so we have also passed the second endpoint and converted all of the HSO 3 -^ to SO 3 2-, with an excess of 22ml of NaOH (=5.21 x 10-3^ moles of OH-). So, our current concentration of OH-^ is 5.21 x 10-3^ moles/1.050L = 4.96 x 10-3M POH = 2.30, pH = 11.
  2. Arrange the following substances in order of increasing solubility in water and fully explain your reasoning .

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H H

A

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

HO OH

B

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H Cl

D

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H OH

C

d. Is the mechanism below consistent with the rate law you found in part a? 2 NO + H 2 ⇔ N 2 + 2 OH fast equilibrium H 2 + 2 OH → 2 H 2 O slow rate = d[H 2 O]/dt = 2k 2 [H 2 ][OH]^2 k 1 [NO]^2 [H 2 ] = k-1[N 2 ][OH]^2 so [OH]^2 = k 1 [NO]^2 [H 2 ]/ k-1[N 2 ] rate = 2k 2 [H 2 ][OH]^2 = 2k 2 [H 2 ]( k 1 [NO]^2 [H 2 ]/ k-1[N 2 ]) = = (2k 1 k 2 /k-1) [NO] 2 [H 2 ] 2 /[N 2 ] This is not consistent with part a as the reaction orders are different.

  1. Consider the following representation showing the dissociation of an acid, HA, in water. a. Fully label the parts of this graph and explain at an atomic level what is going on at various points along the curve Identify those points and regions the correspond to: Bond making, Bond breaking, activation energy, Delta E, Energy of the hydrated ions, b. Explain how this graph will vary between a strong and weak acid A strong acid will have a greater Delta E. This will be caused when the hydrogen to conjugate bond is weaker or if the the dissociation products are stabilized (moved down in energy) c. Explain how the hydration spheres forming around the conjugate base will change the graph The more bonds between the conjugate base and the water molecules the more stable are dissociation products. This will strengthen an acid.
  2. Put the following lists of acids in order of increasing strength. Reaction Coordinate

E

a. ClOH, ClO 2 H, ClO 3 H and ClO 4 H ClOH, < ClO 2 H, < ClO 3 H < ClO 4 H b. AsH 3 , HI, HBr, H 2 O AsH 3 , < H 2 O<HBr<HI, c. Fully explain why the structures of each acid affect the energy of the A- to H+ bond and how the size of the hydration sphere around the conjugate base influences acid strength For (a) the more O atoms will remove more charge from the O-H bond—weakening it, and destabilizing the reactants. A larger hydration sphere will stabilize the reactants.

  1. State the Second Law of Thermodynamics. Briefly explain how this law is related to tables of ∆G°(formation) for compounds and tables of E° for half reactions. The second law of thermodynamics asserts that things can only happen if the entropy of the universe increases. The entropy of the universe is sum of two parts the entropy of the system and its surroundings. We can construct a single quantity called the free energy which also measures the total entropy of the universe. For every process there must be a change in the free energy called ∆G. If this is negative, then the process is allowed by the second law of thermodynamics. If it is positive, then the process is not allowed. Now we can take certain standard reactions, such as the formation of a compound from its elements and determine ∆G for all of these reactions. It then becomes an easy matter to use these standard ∆Gs also called ∆G 0 and combine them to find the change in free energy for any reaction. From this information we can determine if a reaction can happen.
  2. For each of the following mixtures, write a balanced chemical equation (net ionic) for the most likely redox reaction that will occur. Write "no reaction" if nothing should occur. a. Br 2 (l) is mixed with Fe(NO 3 ) 2 (aq) Br 2 (l) + 2 Fe+2^  2 Br-^ + 2Fe+ b. Ag(s) is dropped into HNO 3 (aq) 3 Ag(s) + NO 3 - (aq) + 4 H +  NO(g) + 3 Ag + + 2 H 2 O(l) c. Cr(s) is dipped into Cu(ClO 4 ) 2 (aq) Cr(s) + Cu+2^  Cu(s) + Cr2+
  3. Draw the following cell Fe(s) | Fe
    • (aq, 1.30 M) || Ag + (aq, 0.0024 M) | Ag(s) a. In your drawing indicate the directions of electron and ion flow. Electrons are flowing from the iron electrode to the silver electrode. Cations are moving from the salt bridge into the silver half-cell.

b. Write an expression giving the equilibrium constant associated with this

defect reaction.

c. Derive an expression that relates the diffusivity of O

2-

ions through

calcium-stabilized zirconia to the concentration of added CaO.

By stoichiometry,

substituting into the above equilibrium expression gives, rearranging we have,

16. The power of a lead-acid battery is specified at two temperatures. The so-

called “cold cranking power” refers to the number of amps produced by the

battery at 0

o

F while the “cranking power” refers to the number of amps

produced at 32

o

F.

a. Explain why the number of amps delivered by the battery changes with

temperature and at which temperature the number of amps delivered is

the greater.

The current is proportional to the rate of the chemical reaction producing the power. The rate-determining step for this reaction in the lead acid battery is the diffusion of ions through the electrolyte. As with all reactions, temperature will have an affects. At lower temperatures diffusion will occur more slowly and hence the number of amps produced will also decrease

b. The specifications for Sears’ Die Hard battery gives cold cranking power

of 875 amps and cranking power of 1000 amps. Use these specifications

to calculate the activation energy for diffusion of HSO 4

ions through the

battery electrolyte.

0 o^ F = 255.37 K, 32o^ F = 273.15 K

Giving

c. What would be the cranking power of this Die Hard battery on a winter

day in Anchorage Alaska with a temperature of -

o

F?

-40o^ F = 233.15 K At - o F the battery will deliver 72% of the cranking power as at 32 o F or 720 amps.

17 Using what you know about diffusion and electrochemistry explain, with

all else being equal, why it is that batteries with liquid electrolytes can

deliver more power than those with solid-state electrolytes. Give an

order of magnitude estimate of the difference.

The current is ultimately controlled by the diffusion of ions. In the solid-state the mechanism providing the largest diffusivities is interstitial diffusion. Even for very small ions, as for example Li+ ions, the activation energy for diffusion is ~ 15 kJ/mole as compared to the number we just calculated for liquid diffusion of roughly 5 kJ/mole. We can use these numbers to estimate the relative current carrying capacity of solid state versus liquid electrolyte batteries. We note,