Advanced Placement Chemistry s3

Advanced Placement Chemistry 1988 Free Response Questions

1) At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas, as shown by the following equation.

SbCl5 <===> SbCl3 + Cl2 a) An 89.7-gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0-liter container at 182 °C.

(1) What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs? (2) What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs? b) If the SbCl5 is 29.2 percent decomposes when equilibrium is established at 182 °C, calculate the value for either equilibrium constant, Kp or Kc, for this decomposition reaction. Indicate whether you are calculating Kp or Kc. c) In order to produce some SbCl5, a 1.00-mole sample of SbCl3 is first placed in an empty 2.00- liter container maintained at a temperature different from 182 °C. At this temperature Kc equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?

2) Enthalpy of Combustion Absolute Entropy, S° Substance H° Kilojoules/mole Joules/mole-K C(s) - 393.5 5.740

H2(g) - 285.8 130.6

C2H5OH(l) - 1366.7 160.7

H2O(l) ------69.91 (a) Write a separate, balanced chemical equation for the combustion of each of the following: C(s), H2(g), and C2H5OH(l). Consider the only products to be CO2(g) and/or H2O(l). (b) In principle, ethanol can be prepared by the following reaction.

2 C(s) + 2 H2(g) + H2O(l) --> C2H5OH(l) Calculate the standard enthalpy change, H°, for the preparation of ethanol, as shown in the reaction above. (c) Calculate the standard entropy change, S°, for the reaction given in part (b) (d) Calculate the value of the equilibrium constant at 25 °C for the reaction represented by the equation in part (b). 3) An electrochemical cell consists of a tin electrode in an acidic solution of 1.00-molar Sn2+ connected by a salt bridge to a second compartment with a silver electrode in an acidic solution of 1.00-molar Ag+. (a) Write the equation for the half-cell reaction occurring at each electrode. Indicate which half- reaction occurs at the anode. (b) Write the balanced chemical equation for the overall spontaneous cell reaction that occurs when the circuit is complete. Calculate the standard voltage, E°, for this cell reaction. (c) Calculate the equilibrium constant for this cell reaction at 298 K. (d) A cell similar to the one described above is constructed with solutions that have initial concentrations of 1.00-molar Sn2+ and 0.0200-molar Ag+. Calculate the initial voltage, E, of this cell.

4) Use appropriate ionic and molecular formulas to show the reactants and the products for the following, each of which occurs in aqueous solution except as indicated. Omit formulas for any ionic or molecular species that remain unchanged after the reaction. You need not balance. In all cases a reaction occurs. (a) A solution of potassium iodide is added to an acidified solution of potassium dichromate. (b) A solution of sodium hydroxide is added to a solution of ammonium chloride. (c) A strip of magnesium is added to a solution of silver nitrate. (d) Solid potassium chlorate is heated in the presence of manganese dioxide as a catalyst. (e) Dilute hydrochloric acid is added to a solution of potassium carbonate. (f) Sulfur trioxide gas is added to excess water. (g) Dilute sulfuric acid is added to a solution of barium chloride. (h) A concentrated solution of ammonia is added to a solution of copper(II) chloride.

5) Using principles of chemical bonding and/or intermolecular forces, explain each of the following. (a) Xenon has a higher boiling point than neon has. (b) Solid copper is an excellent conductor of electricity, but solid copper chloride is not.

(c) SiO2 melts at a very high temperature, while CO2 is a gas at room temperature, event though Si and C are in the same chemical family.

(d) Molecules of NF3 are polar, but those of BF3 are not.

NH4HS(s) <===> NH3(g) + H2S(g) For this reaction, H° = + 93 kilojoules The equilibrium above is established by placing solid NH4HS in an evacuated container at 25 °C. At equilibrium, some solid NH4HS remains in the container. Predict and explain each of the following.

(a) The effect on the equilibrium partial pressure of NH3 gas when additional solid NH4HS is introduced into the container.

(b) The effect on the equilibrium partial pressure of NH3 gas when additional H2S gas is introduced into the container.

(c) The effect on the mass of solid NH4HS present when the volume of the container is decreased.

(d) The effect on the mass of solid NH4HS present when the temperature is increased.

A 30.00-milliliter sample of a weak monoprotic acid was titrated with a standardized solution of NaOH. A pH meter was used to measure the pH after each increment of NaOH was added, and the curve above was constructed. (a) Explain how this curve could be used to determine the molarity of the acid.

(b) Explain how this curve could be used to determine the acid dissociation constant Ka of the weak monoprotic acid. (c) If you were to repeat the titration using an indicator in the acid to signal the endpoint, which of the following indicators should you select. Give the reason for your choice.

5 Methyl red Ka = 1 x 10¯ 8 Cresol red Ka = 1 x 10¯ 11 Alizarin yellow Ka = 1 x 10¯ (d) Sketch the titration curve that would result if the weak monoprotic acid were replaced by a strong monoprotic acid, such as HCl of the same molarity. Identify differences between this titration curve and the curve shown above.

8) The normal boiling and freezing points of argon are 87.3 K and 84.0 K, respectively. The triple point is at 82.7 K and 0.68 atmosphere. (a) Use the data above to draw a phase diagram for argon. Label the axes and label the regions in which the solid, liquid, and gas phases are stable. On the phase diagram, show the position of the normal boiling point. (b) Describe any changes that can be observed in a sample of solid argon when the temperature is increases from 40 K to 160 K at a constant pressure of 0.50 atmosphere. (c) Describe any changes that can be observed in a sample of liquid argon the pressure is reduced from 10 atmospheres to 1 atmosphere at a constant temperature of 100 K, which is well below the critical temperature. (d) Does the liquid phase of argon have a density greater than, equal to, or less than the density of the solid phase? Explain your answer, using information given in the introduction to this question.

9) An experiment is to be performed to determine the standard molar enthalpy of neutralization of a strong acid by a strong base. Standard school laboratory equipment and a supply of standardized 1.00-molar HCl and standardized 1.00-molar NaOH are available. (a) What equipment would be needed? (b) What measurements should be taken? (c) Without performing calculations, describe how the resulting data should be used to obtain the standard molar enthalpy of neutralization. (d) When a class of students performed this experiment, the average of the result was -55.0 kilojoules per mole. The accepted value for the standard molar enthalpy of neutralization of a strong acid by a strong base is -57.7 kilojoules per mole. Propose two likely sources of experimental error that could account for the result obtained by the class. Advanced Placement Chemistry 1988 Free Response Answers Notes  [delta] and [sigma] are used to indicate the capital Greek letters.  [square root] applies to the numbers enclosed in parenthesis immediately following  All simplifying assumptions are justified within 5%.  One point deduction for a significant figure or math error, applied only once per problem.  No credit earned for numerical answer without justification.

1) Average score = 4.95 a) four points

(1) 89.7 g SbCl5 / 299.0 g mole = 0.300 mole SbCl5

[SbCl5] initial = 0.300 mole / 15.0 liter = 0.200 M (2) T = 182 °C + 273 = 455 K P = nRT / V = [(0.300 mole) (0.0821 L atm) (455 K)] / [(15.0 L) (mole K)] = 0.747 atm OR (15.0 L / 0.300 mol) x (273 K / 455 K) = 30.0 L / mol (at std. temperature) 1 atm x [(22.4 L / mol) ÷ (30.0 L / mol)] = 0.747 atm b) three points Equilibrium concentrations: 3 [SbCl3] = [Cl2] = (0.0200 mol / L) x 0.292 = 5.84 x 10¯ M 2 [SbCl5] = (0.0200 mol L) x 0.708 L = 1.42 x 10¯ M 3 2 2 3 Kc = ([SbCl3][Cl2]) ÷ [SbCl5] = (5.84 x 10¯ ) ÷ (1.42 x 10¯ ) = 2.41 x 10¯ OR Equilibrium pressures:

PSbCl3 = PCl2 = 0.747 atm x 0.292 = 0.218 atm

PSbCl5 = 0.747 atm x 0.708 = 0.529 atm 2 2 Kp = (PSbCl3 x PCl2) ÷ PSbCl5 = (0.218) ÷ 0.529 = 8.98 x 10¯ c) two points

K = ([SbCl3][Cl2]) ¯ [SbCl5] = 0.117 Equilibrium concentrations:

[SbCl5] = (1.00 - 0.70) mol / 2.00 L = 0.15 M

[SbCl3] = 0.700 mole / 2.00L = 0.350 M

[Cl2] = x

Kc = [(0.350) (x)] ÷ (0.15) = 0.117 x = [Cl2] = 0.50 M

Moles Cl2 at equilibrium = 0.050 mol L x 2.00 L = 0.10 mol

Moles Cl2 needed to make 0.300 mol SbCl3 into SbCl5 = 0.30 mol

Moles Cl2 that must be added = 0.40 mol

2) Average score = 3.25 a) two points

C + O2 --> CO2

2 H2 + O2 --> 2 H2O

C2H5OH --> 2CO2 + 3 H2O

(Half credit for recognition that combustion is reaction with O2) b) three points

2 C + 2 O2 ---> 2 CO2 [delta]H° = 2 (- 393.5) = - 787.0 kJ

2 H2 + O2 ---> 2 H2O [delta]H° = 2(- 285.8) = - 571.6 kJ

2 CO2 + 3 O2 ---> C2H5OH + 3 O2 [delta]H° = - (- 1366.7) = 1.366.7 kJ Sum of three equations above

2 C + 2 H2 + H2O ---> C2H5OH [delta]H° = - 8.1 kJ (1 point for correct [delta]H° for each of the first 3 reations) OR

[delta]H° combustion C(s) = [delta]Hf° CO2(g)

[delta]H° combustion H2(g) = [delta]Hf° H2O(l)

C2H5OH + 3 O2 ---> 2 CO2 + 3 H2O [delta]H° = -1,366.7kJ

[delta]H° reaction = [sigma] [delta]Hf° (products) - [sigma] [delta]Hf° (reactants)

= [2( -393.5) + 3(- 258.8) - [delta]Hf°C2H5OH - 0] kJ = - 277.7 kJ 2 C + 2 H2 + H2O ---> C2H5OH [delta]H° reaction = [sigma] [delta]H° (products) - [sigma] [delta]H° (reactants) = [- 277.7 - 0 - 0 - (- 285.8)] kJ = + 8.1 kJ c) one point [delta]S° = [sigma] S° (products) - [sigma] S° (reactants) = (160.7 - 11.5 - 261.2 - 69.9) J / mol K = - 181.9 J / mol K d) three points (1) [delta]G° = [delta]H° - T [delta]S° = 8,100 J - (298) (- 181.9) J = 8,100 J + 54,200 J = 62,300 J (2) [delta]G° = - RTlnK lnK = - [delta]G / RT = - 62.300J ÷ [(8.31J/mol-K) (298K)] = - 25.2 K = 1.1 x 10¯11 OR [delta]G° = - 2.303 RT logK logK = - [delta]G / 2.303 RT = - 62.000J ÷ [(2.303) (8.31) (298)] = - 10.9 K = 1.3 x 10¯11 If the terms are not rounded, K = 1.2 x 10¯11 correct substitutions in (1) and (2) earns one point

3) Average score = 5.02 a) two points Sn ---> Sn2+ + 2e¯ Ag+ + e¯ ---> Ag b) two points 2 Ag+ + Sn ---> 2 Ag + Sn2+ E° = [0.80 - (- 0.14)] V = 0.94 V c) two points E = (0.0591 ÷ n) log K (or - nFE = - RT ln K) log K = (0.94 x 2) ÷ 0.0591 = 31.8 K = 6 x 1031 d) two points E = E° - (0.0591 ÷ n) log [Sn2+] / [Ag+]2 OR E = E° - (RT / nF) ln Q Q = [Sn2+] / [Ag+]2 E = 0.94 = (0.0591 ÷ 2) log (1 ÷ (0.022)) E = 0.94 - 0.10 = 0.84 V The final point is for completion of calculations in (c) and (d).

4) Average score = 5.66 3 points for each reaction; 1 pt. for reactants and 2 for products. Two points (of the three) for correct response in inappropriate form, i.e. molecular when ionic is correct. One point penalty for spurious product, e.g. redox product in an acid-base reaction. One point penalty for inclusion of spectator ions. + + 2 3+ a) H (or H3O ) + I¯ + Cr2O7 ¯ ---> I2 + Cr H2O (Any reasonable I¯ oxidation product accepted.) + b) OH¯ + NH4 ---> H2O + NH3 (or NH4OH) c) Mg + Ag+ ---> Mg2+ + Ag d) KClO3 ---> KCl + O2

(MnO2 is also shown over the reaction arrow in the original.) + 2 e) H + CO3 ¯ ---> CO2 + H2O (or H2CO3 or HCO3¯) + + 2 f) SO3 + H2O ---> H + HSO4¯ (or H + SO4 ¯) 2+ 2 g) Ba + SO4 ¯ ---> BaSO4 2+ + (or Ba + HSO4¯ ---> BaSO4 + H ) 2+ 2+ h) Cu + NH3 ---> Cu(NH3)4

(Partial credit for other logical Cu(II) ammonia complexes and for Cu(OH)2)

5) Average score = 2.99 a) two points Xe and Ne are monatomic elements held together by London dispersion (van der Waals) forces. The magnitude of such forces is determined by the number of electrons in the atom. A Xe atom has more electrons than a neon atom has. (Size of the atom was accepted but mass was not.) b) two points The electrical conductivity of copper metal is based on mobile valence electrons (partially filled bands). Copper chloride is a rigid ionic solid with the valence electrons of copper localized in individual copper(II) ions. c) two points

SiO2 is a covalent network solid. There are strong bonds many of which must be broken simultaneously to volatize SiO2. CO2 is composed of discrete, nonpolar CO2 molecules so that the only forces holding the molecules together are the weak London dispersion (van der Waals) forces. d) two points A lone pair of electrons on the central atom results in a pyramidal shape. The dipoles don't cancel, thus the molecule is polar. There is no lone pair on the central atom so the molecule has a trigonal planar shape in which the dipoles cancel, thus the molecule is nonpolar.

6) Average score = 4.31 a) two points

The equilibrium pressure of NH3 gas would be unaffected Kp = (PNH3) (PH2S). Thus the amount of solid NH4HS present does not affect the equilibrium. b) two points

The equilibrium pressure of NH3 gas would decrease. In order for the pressure equilibrium constant, Kp, to remain constant, the equilibrium pressure of NH3 must decrease when the pressure of H2S is increased.

Kp = (PNH3) (PH2S) (A complete explanation based on Le Chatelier's principle is also acceptable.) c) two points

The mass of NH4HS increases. A decrease in volume causes the pressure of each gas to increase. To maintain the value of the pressure equilibrium constant, Kp, the pressure of each of the gases must decrease. That decrease realized by the formation of more solid NH4HS.

Kp = (PNH3) (PH2S) (A complete explanation based on Le Chatelier's principle is also acceptable.) d) two points The mass of NH4HS decreases because the endothermic reaction absorbs heat and goes nearer to completion (to the right) as the temperature increases. (One point was assigned for each correct prediction and one point for each correct explanation.)

7) Average score = 1.76 a) two points The sharp vertical rise in the pH on the pH-volume curve appears at the equivalence point (about 23 mL). Because the acid is monoprotic, the number of moles of acid equals the number of moles of NaOH. That number is the product of the exact volume and the molarity of the NaOH. The molarity of the acid is the number of moles of the acid divided by 0.030 L, the volume of the acid. b) two points At the half-equivalence point (where the volume of the base added is exactly half its volume at the equivalence point), the concentration [HX] of the weak acid equals the concentration [X¯] of + its anion. Thus, in the equilibrium expression, [H ] [X¯] / [HX] = Ka. Therefore, pH at the half- equivalence point equals pKa c) one point

Cresol red is the best indicator because its pKa (about 8) appears midway in the steep equivalence region. This insures that at the equivalence point the maximum color change for the minimal change in the volume of NaOH added is observed. d) three points 8) Average score = 3.82 a) four points

One point each for: correct identification of axes diagram of correct shape correct labeling of regions correct position of normal boiling point b) one point The argon sublimes. c) one point The argon vaporizes. d) two points The liquid phase is less dense than the solid phase. Since the freezing point of argon is higher than the triple point temperature, the solid-liquid equilibrium line slopes to the right with increasing pressure. Thus, if a sample of liquid argon is compressed (pressure increased) at constant temperature, the liquid becomes a solid. Because increasing pressure favors the denser phase, solid argon must be the denser phase.

9) Average score = 3.31 a) two points Equipment needed includes a thermometer, and a container for the reaction, preferably a container that serves as a calorimeter, and volumetric glassware (graduated cylinder, pipet....) b) two points Measurements include the difference in temperatures between just before the start of the reaction and the completion of the reaction, and amounts (volume, moles) of the acid and the base. c) two points Determination of heat (evolved or absorbed): The sum of the volumes (or masses) of the two solutions, the change in temperature and the specific heat of water are multiplied together to determine the heat of solution for the sample used, (q = m x Cp x [delta]T). For full credit, two of the three factors must be identified. Division of the calculated heat of neutralization by moles of water produced, or moles of H+, or moles of OH¯, or moles of limiting reagent. d) two points Experimental errors: heat loss to the calorimeter wall, to air, to the thermometer; incomplete transfer of acid base from graduated cylinder; spattering of some of the acid or base so that incomplete mixing occurred....Experimenter errors: dirty glassware, spilled solution, misread volume or temperature....(One point for each of two experimental errors: one point for an experimenter error if only one experimental error is given.)