Review Answers
2000 REVIEW #1
Kc & Kp
2H2S(g) ↔2H2(g) + S2(g)
When heated, hydrogen sulfide gas decomposes according to the equation above.
A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container.
The sealed container is heated to 483 K and 3.72x10-2 mol of S2(g) is present at
equilibrium.
a. Write the expression for the equilibrium constant, Kc, for the decomposition
reaction represented above.
b. Calculate the equilibrium concentration, in mol L1-, of the following gases in
the container at 483 K.
i. H2(g)
ii. H2S(g)
c. Calculate the value of the equilibrium constant, Kc, for the decomposition
reaction at 483 K.
d. Calculate the partial pressure of S2(g) in the container at 483 K.
e. For the reaction H2(g) + ½S2(g) H2S(g) at 483 K, calculate the value of the
equilibrium constant, Kc.
a.
- First find the molarity of the H2S and then recognize that the moles of S2 is the equilibrium moles of S2 and place both in an ICE table. The beginning moles of S2 and H2 is 0. Also be aware of the coefficients in the reaction.
Initial / 0.0800 / 0 / 0
Change / -0.0595 / +0.0595 / +0.0298
Equilibrium / 0.0205 / 0.0595 / 0.0298
Therefore: [H2]=0.0595M and [S2]=0.0298
- Simply plug into the equilibrium expression the equilibrium concentrations
- Use PV=nRT
- Recognize that this is ½ of the given reaction. So the value of K must be take to the ½ power. Note that a common misconception is to cut the value of K in half. But since this is a power function it must be take to the ½ power.
0.251(1/2)=0.500
2001 REVIEW #2
Ksp
Answer the following questions relating to the solubility of the chlorides of silver
and lead.
a. At 10°C , 8.9x10-5 g of AgCl(s) will dissolve in 100. mL of water.
i. Write the equation for the dissociation of AgCl(s) in water.
ii. Calculate the solubility, in mol L1-, of AgCl(s) in water at 10°C.
iii. Calculate the value of the solubility-product constant, Ksp, for
AgCl(s) at 10°C.
b. At 25°C, the value of Ksp for PbCl2(s) is 1.6x10-5 and the value of Ksp for
AgCl(s) is 1.8x10-10.
i. If 60.0 mL of 0.0400M NaCl(aq) is added to 60.0mL of 0.0300M
Pb(NO3)2(aq), will a precipitate form? Assume that volumes are
additive. Show calculations to support your answer.
ii. Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L of saturated
PbCl2 solution to which 0.250 moles of NaCl(s) have been added.
Assume that no volume change occurs.
iii. If 0.100M NaCl(aq) is added slowly to a beaker containing both
0.120M AgNO3(aq) and 0.150M Pb(NO3)2(aq) at 25°C, which will
precipitate first, AgCl(s) or PbCl2(s)? Show calculations to support
your answer.
- i. AgCl↔Ag+ + Cl-
- Ksp = [Ag+][Cl-]=(6.23 10-6)(6.23 10-6)=3.89 10-11
- i. Find Qsp and see if it is bigger than the Ksp.
[Cl-]=0.0200M, [Pb2+]=0.0150M (Note since the volume is doubled the concentrations are cut in half.
PbCl2↔ Pb2+ + 2Cl-, Qsp =[Pb2+][Cl-]2=[0.0150][0.0200]2 =6.00 10-6<1.6 10-5 therefore no ppt will form.
- Use common ion effect. You will know the [Cl-] and then solve for the [Pb2+] using the Ksp equation.
Ksp=1.6 10-5=[Pb2+][0.25M]2[Pb2+]=2.56 10-4 M
- Competitive precipitation: the hardest of all Ksp problems. Solve for the of the Cl-. The lowest answer wins.
PbCl2↔Pb2+ + 2Cl- / AgCl↔Ag+ + Cl-
Ksp=1.6 10-5=[0.150][2s]2 / Ksp= 1.8x10-10 = [0.120][s]
[Cl-]=0.0206M / [Cl-]=1.5 10-9 M
The AgCl will ppt first due to the lower concentration of the Cl-
1998 REVIEW #3
Ksp
Solve the following problem related to the solubility equilibria of some metal
hydroxides in aqueous solution.
a. The solubility of Cu(OH)2(s) is 1.72x10-6 gram per 100 milliliters of solution at
25°C.
i. Write the balanced chemical equation for the dissociation of
Cu(OH)2(s) at 25°C.
ii. Calculate the solubility (in moles per liter) of Cu(OH)2(s) at 25°C.
iii. Calculate the value of the solubility-product constant, Ksp, forCu(OH)2 at 25°C.
b. The value of the solubility-product constant, Ksp, for Zn(OH)2(s) is 7.7x10-17at 25°C.
i. Calculate the solubility (in moles per liter) of Zn(OH)2 at 25°C ina solution with a pH of 9.35.
ii. At 25°C, 50.0 milliliters of 0.100-molar Zn(NO3)2 is mixed with50.0 milliliters of 0.300-molar NaOH. Claculate the molarconcentration of Zn2+(aq) in the resulting solution onceequilibrium has been established. Assume that volumes are
additive.
1)Cu(OH)2↔ Cu2+ + 2OH-
2)
3)Ksp =[Cu2+][OH-]2 =[s][2s]2=4s3=4(1.76 10-7)3=2.18 10-20
4)pOH = 14-9.35 = 4.6
5)10-pOH=10-4.65 = 2.24 10-5
6)Ksp = 7.7 10-17 = [Zn2+][2.24 10-5]2
7)[Zn2+]=1.53 10-7 M
- i. Ksp = 7.7 10-17 = [Zn2+][OH-]2 =4s3 s= molar solubility = 2.68 10-6 M
Zn(NO3)2 / + 2NaOH / / Zn(OH)2 / +2 NaNO3
5 mmol / 15mmol / 0
-5 / -10 / +10
0 / 5 mmol / 10mmol / Not important
[OH-]=
Ksp = 7.7 10-17 = [Zn2+][OH-]2 = [Zn2+][0.050M]2
[Zn2+]=3.08 10-14
1998 REVIEW #4
Le Châtelier’s Principle
C(s) + H2O(g) ↔CO(g) + H2(g) ΔH°= +131 kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), and H2(g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement.
a. Additional H2(g) is added to the equilibrium mixture at constant volume.
b. The temperature of the equilibrium mixture is increased at constant volume.
c. The volume of the container is decreased at constant temperature.
d. The graphite pellets are pulverized.
This is a essay question, but don’t think of it like an essay question: think rather of a short answer.
- moles of CO will go down due to the reaction shifting to the left
- Since this reaction is endothermic (H is positive): this causes the reaction to shift to the right making more CO
- Volume decreased makes the pressure go up. When pressure goes up the reaction shifts to the side with the least moles of gas. Threfore this will shift to the left since the reactants have only one mole of gas and the products have 2 moles of gas.
- No effect. Solids do not affect the position of equilibrium.
1996 REVIEW #5
Acid-Base
Ka & Kb
HOCl OCl1- + H1+
Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. Theacid-dissociation constant, Ka, for the reaction represented above is 3.2 x 10-8.
a. Calculate the [H1+] of a 0.14-molar solution of HOCl.
b. Write the correctly balanced net ionic equation for the reaction that occurs when NaOCl is dissolved in water and calculate the numerical value of theequilibrium constant for the reaction.
c. Calculate the pH of a solution made by combining 40.0 milliliters of 0.14- molar HOCl and 10.0 milliliters of 0.56-molar NaOH.
d. How many millimoles of solid NaOH must be added to 50.0 milliliters of0.20-molar HOCl to obtain a buffer solution that has a pH of 7.49? Assumethat the addition of the solid NaOH results in a negligible change in volume.
e. Household bleach is made by dissolving chlorine gas in water, as represented below.
Cl2(g) + H2O(l) → H1+ + Cl1- + HOCl(aq)
Calculate the pH of such a solution if the concentration of HOCl in thesolution is 0.065-molar.
a.
HOCl / ↔ / H+ / + / OCl-0.14 / 0 / 0
-x / +x / +x
0.14-x / X / X
b) OCl- + HOH HOCl + OH-
c)Start with stoich
OH- / + HOCl / / HOH / + OCl-5.6 mmol / 5.6 mmol / 0
-5.6 / -5.6 / +5.6
0 / 0 / 5.6
At equivalence point
[OCl-] = 5.6mmol/50mL = 0.112 M
OCl- / + HOH / ↔ / HOCl / + OH-0.112 / 0 / 0
-x / +x / +x
0.112-x / X / x
This is the conjugate base so I must find Kb for this substance
Kw / Ka =1 10-14/3.2 10-8 =3.12 x10-7
d) First put in and ICE table
OH- / + HOCl / / HOH / + OCl-x / 10 mmol / 0
-x / -x / +x
0 / 10-x / x
Now use Henderson-Hasselbach Equation
Find pKa = -log(3.2 x 10-8)=7.49.
pH = pKa + log(base/acid)
e. don’t overthink this question. Simply realize that the: [HOCl]=[H+]=0.065M:: pH = -log(o.o65M)=1.19
1999 REVIEW #6
Acid-BaseNH3 + H2O ↔ NH4+ + OH-
Ka & Kb
In aqueous solution, ammonia reacts as represented above. In 0.0180M NH3(aq) at 25°C, the hydroxide ion concentration, [OH1-], is 5.60 x 10-4M. In answering the following, assume that the temperature is constant at 25°C and that the volumes are additive.
a. Write the equilibrium-constant expression for the reaction represented above.
b. Determine the pH of 0.0180M NH3(aq).
c. Determine the value of the base ionization constant, Kb, for NH3(aq).
d. Determine the percent ionization of NH3 in 0.0180M NH3(aq).
e. In an experiment, a 20.0 mL sample of 0.0180M NH3(aq) was placed in a flask and titrated to the equivalence point and beyond using 0.0120M HCl(aq).
i. Determine the volume of 0.0120M HCl(aq) that was added to reach the equivalence point.
ii. Determine the pH of the solution in the flask after a total of 15.0 mL of 0.0120M HCl(aq) was added.
iii. Determine the pH of the solution in the flask after a total of 40.0 mL of 0.0120M HCl(aq) was added.
a.
b)[OH1-], is 5.60 x 10-4M
pOH=-log(5.6 10-4)=3.25
pH=14-pOH=14-3.25=10.75
c)Use ICE table and solve for x
NH3 / + H2O / ↔ / NH4+ / + OH-0.0180M / 0 / 0
-5.6 10-4 / +5.6 10-4 / +5.6 10-4
0.01744 / 5.6 10-4 / 5.6 10-4
d)Percent Ionization
e)Reaction
NH3 / + H+ / / NH41+Convert everything to mmol
NH3 / + H+ / / NH41+36 / 18 / 0
-18 / -18 / +18
18 / 0 / 18
This is ½ the equivalence point so pH=pKa
pKb =-log(1.8E-5) = 4.74
pKb + pKa = 14: pKb=9.26=pH
NH3 / + H+ / / NH41+36 / 48 / 0
-36 / -36 / +36
0 / 12 / 36
Excess acid: [H+]=12mmol/60mL=0.20M: -log(0.20)=0.70=pH
1998 REVIEW #37
Lab
An approximately 0.1-molar solution of NaOH is to be standardized by titration. Assume that the following materials are available.
- • Clean, dry 50 mL buret • Analytical balance
- • 250 mL Erlenmeyer flask • Phenolphthalein indicator solution
- • Wash bottle filled with distilled water • Potassium hydrogen phthalate, KHP, a pure solid monoprotic acid (to be used as the primary standard)
(a) Briefly describe the steps you would take, using materials listed above, to standardize the NaOH solution.
- Mass some amount of KHP (maybe 0.50g) and dissolve. Also add a few drops of phenolphthlein
- Place 0.1 M NaOH in buret
- Add NaOH unitl the solution turns pink
- Do Stoich to determine the concentration of the NaOH. Stoich would look something like this:
(b) Describe (i.e., set up) the calculations necessary to determine theconcentration of the NaOH solution.
Done above
(c) After the NaOH solutions has been standardized, it is used to titrate a weak monoprotic acid, HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. In the space provided at the right, sketch the titration curve, showing the pH changes that occur as the volume of NaOH solution added increases from 0.0 to 35.0 mL. Clearly label the equivalence point on the curve.
(d) Describe how the value of the acid-dissociation constant, Ka, for the weak acid HX could be determined from the titration curve in part (c).
At ½ the equivalence point (12.5mL) the pH=pKa.. Then use the equation 10-pKa= Ka
(e) The graph to the right shows the results obtained by titrating a different weak acid, H2Y, with the standardized NaOH
solution. Identify the negative ion that is present in the highest concentration at the point in the titration represented by the letter A on the curve.
H2Y + OH- HY- + HOH
HY- + OH- Y2- + HOH
Answer: Y2-
1997 REVIEW #12
Thermodynamics
For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased.
PCl5(g) ⇌PCl3(g) + Cl2(g)
a. What is the sign of ΔS° for the reaction? Explain.
Since you are going from one mole of a gas to two moles of a gas this system is getting more disordered and therefore entropy is increased. ΔS° is positive
b. What change, if any, will occur in ΔG° for the reaction as the temperature is increased? Explain your reasoning in terms of thermodynamics principles.
Since when Temp is increased there are more molecules of PCl3 and Cl2, this reaction is endothermic and ∆H is positive.
∆G = ∆H-T∆S: = + - T(+). At higher temperatures the T∆S term will be larger and make the value of ∆G become more negative, thus making the reaction more spontaneous at higher temperatures.
c. If He gas is added to the original reaction mixture at constant volume and temperature, what will happen to the partial pressure of Cl2? Explain.
He gas is an inert gas. It will not increase the partial pressures of any of the gasses therefore there will be no change in the equilibrium mixture.
d)If the volume of the reaction mixture is decreased at constant temperature to half the original volume, what will happen to the number of moles of Cl2 in the reaction vessel? Explain.
When volume is decreased the pressure goes up. This favors the formation of the reactant (less moles of gas). Therefore the number of moles of Cl2 will go down.
1999 REVIEW #13
Thermodynamics
Answer the following questions in terms of thermodynamic principles and concepts of kinetic molecular theory.
a. Consider the reaction represented below, which is spontaneous at 298 K.
CO2(g) + 2NH3(g) → CO(NH2)2(s) + H2O(l) ΔH°298 = -134 kJ
- For the reaction, indicate whether the standard entropy change, ΔS°298, is positive, or negative, or zero. Justifyyour answer.
This system is going from 3 moles of gas on the reactant side to 1 mole of solid and one mole of liquid. The system is becoming more ordered, therefore ΔS is negative.
- ii. Which factor, the change in enthalpy, ΔH°298, or the change in entropy, ΔS°298, provide the principal drivingforce for reaction at 298 K. Explain
the enthalpy will be the principle driving force. Since ΔS is negative this does not favor a spontaneous change. The value of ΔH is negative and this does contribute to the value of ΔG being negative and will drive the reaction to be spontaneous.
- iii. For the reaction, how is the value of the standard freeenergy change, ΔG°, affected by an increase in temperature? Explain.
Since ΔG=ΔH-TΔS: and we have a (-) value of ΔH and a (-) value of ΔS
ΔG = (-) –T(-): Higher values of T will cause the reaction to be less spontaneous, more positive.
b. Some reactions that are predicted by their signs of ΔG° to bespontaneous at room temperature do not proceed at a measurable rate at room temperature.
- Account for this apparent contradiction.
This is a separate question of reaction kinetics. Just because ΔG is negative only tells the chemist that the reaction will proceed, not how fast it will proceed. Diamonds graphite is a spontaneous process, but the kinetics of the reaction cause it to happen very very slowly.
- A suitable catalyst increases the rate of such a reaction. What effect does the catalyst have on ΔG° for the reaction? Explain.
No effect. The kinetics do not have an effect on the thermodynamics of a reaction. The catalyst will speed up the process, but not make it more or less spontaneous.
1996 REVIEW #14
Thermodynamics
C2H2(g) + 2H2(g) → C2H6(g)
Information about the substances involved in the reaction represented above is summarized in the following tables.
Substance / S° ( J mol⋅K ) / (kJ/mol)C2H2 (g) / 200.9 / 226.7
H2(g) / 130.7 / 0
C2H6 / ----- / -84.7
Bond / Bond energy (kJ/mol)
C-C / 347
C=C / 611
C-H / 414
H-H / 436
a. If the value of the standard entropy change, ΔS°, for the reaction is - 232.7 joules per mole ▪ Kelvin, calculate the standard molar entropy, S°, of C2H6 gas.
The trick on this one is to realize that the value -232.7 joules/mol K is the overall value for the entire reaction. Here we want to find the individual amount for the C2H6.
b. Calculate the value of the standard free-energy change, ΔG°, for the reaction. What does the sign of ΔG° indicate about the reaction above?
ΔG=ΔH-TΔS: First though we must find ΔH. Note that when you then covert it that the value of S must be converted to kJ.
c. Calculate the value of the equilibrium constant, K, for the reaction at 298K?
G=-RTlnK
d. Calculate the value of the C ≡ C bond energy in C2H2 in kilojoules per
mole.
Bonds Broken-bonds Formed = ΔH
C≡C 1(x) / C=C 1(611)H-H 2(436) / C-H 6(414)
C-H 2(414)
Total / 1700+x / 3095
1700+x -3095 = -379
X=1016kJ/mol=Bond energy of C≡C
1998 REVIEW #15
Thermodynamics
C6H5OH(s) + 7O2(g) → 6CO2(g) + 3H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burnedaccording to the equation above, 64.98 kilojoules of heat is released. Use the information in the table below to answer the questions that follow.
Substance / Standard Heat of Formation at 25 C (kJ/mol) / Absolute Entropy at ° C (J/mol K)C(graphite) / 0.00 / 5.69
CO2(g) / -393.5 / 213.6
H2(g) / 0.00 / 130.6
H2O(l) / -385.85 / 69.91
O2(g) / 0.00 / 205.0
C6H5OH(s) / ? / 144.0
a. Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C.
b. Calculate the standard heat of formation, of , of phenol in kilojoules permole at 25°C.
c. Calculate the value of the standard free-energy change, ΔG°, for the combustion of phenol at 25°C.
Frist find ∆S:
Now find ∆G using: ∆G=∆H-T∆S
=-3054-298(-0.08767)=-3028kJ/mol
- If the volume of the combustion container is 10.0 liters, calculate the final pressure in the container when the temperature is changed to 110. °C. (Assume no oxygen remains unreacted and that all products are gaseous.)
Here I will use PV=nRT. The trick will be finding the moles of gaseous product. So I will do stoich to determine the moles: Volume and temperature are given in the problem above:
1996 REVIEW #16
Electrochemistry
Sr(s) + Mg2+⇌Sr2+ + Mg(s)
Consider the reaction represented above that occurs at 25°C. All reactants and products are in their standard states. The value of the equilibrium constant, Keq, for the reaction is 4.2 x 1017 at 25°C.
a. Predict the sign of the standard cell potential, E°, for a cell based on the reaction. Explain your prediction.
Since Keq is very large, the reaction favors the products. This means that the reaction is spontaneous and therefore the value of E° is positive.
b. Identify the oxidizing agent for the spontaneous reaction.
Oxidizing agent is the substance that reduces. Mg2+ is losing charge and therefore is the oxidizing agent
c. If the reaction were carried out at 60°C instead of 25°C, how would the cell potential change? Justify your answer.
This can be determined by using the Nernst Equation:
Ecell = E°cell –RT/nF (lnQ)
Since E°cell is (+) and the value of ln(Q) is (+) then a higher T will make the –RT/nF (lnQ) term become larger which will lower the value of Ecel.
d. How would the cell potential change if the reaction were carried out at 25°C with a 1.0-molar solution of Mg(NO3)2 and a 0.10-molar solution of Sr(NO3)2? Explain.
Again we would use the Nernst Equation but now the Q term becomes important:
Ln(0.1)=a negative number so this will cause the –RT/nF (lnQ) term to be negative. And when you subtract a negative number you get a (+) number. So this will increase the value of cell potential
- When the cell reaction in (d) reaches equilibrium, what is the cell potential?
This akin to a battery dying. So at equilibrium the value of the cell potential is =0. This can also be proven mathematically by realizing that at equilibrium ∆G=0. And using the equation ∆G=-nFE, E must be 0 if ∆G=0
1998 REVIEW #17
Electrochemistry
Answer the following questions regarding the electrochemical cell shown to the right.
a. Write the net-ionic equation for the spontaneous reaction that occurs as the cell operates, and determine the cell voltage.
Ag+ +1e- Ag0.80 V
Cd Cd2+ + 2e-+0.40 V
2Ag+ + Cd Cd2+ +2Ag1.20V
b. In which direction do anions flow in the salt bridge as the cell operates? Justify your answer.
Since the electrons are leaving the Cd side(Anode) that is making the silver side(cathode) more negative. This means that the cations are flowing toward the Silver side and the anions are flowing through the salt bridge towards the Cd side.
c. If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right, what will happen to the cell voltage? Explain.
Adding additional AgNO3 is adding Ag+ to the reaction. This follows LeChatlier’s principle and adding more silver ions (a reactant) will drive the reaction to the right and cause the products to be favored and the voltage will go UP.
d. If 1.0 gram of solid NaCl is added to each half-cell, what will happen to the cell voltage? Explain.
NaCl will react with the Ag+ (Ag+ + Cl- AgCl). This will take away Ag+ ions which will lower the [Ag+], thus driving the reaction to the right, favoring the reactants. This will lower the value of the voltage. Cl- does not react with the Cd2+.
Note: this could also be explained using the Nernst equation and what happens to Q with lower amounts of Ag+
- If 20.0 mL of distilled water is added to both half-cells, the cell voltage decreases. Explain.
This would lower the concentrations of each. This must be understood by looking at the Nernst Equation.
The trick on this one has to do with the fact that the [Ag+] is squared. This will cause the value of Q to be larger since you are now dividing by a smaller number squared. This will make the value of ln(Q) to be smaller and make the value of Ecell to be larger.
2001 REVIEW #18
Electrochemistry