CHEMISTRY 51, INSERT FOR SECOND MIDTERM EXAM

SOLUTIONS TO THE EXAMINATION

CHEMISTRY 51, INSERT FOR THE SECOND MIDTERM EXAM

A) USEFUL CONSTANTS AND CONVERSION FACTORS

1) N A = 6.022 x 1023 molecules/mole

2) R = 8.31451 J/K-mole = 0.082058 L-atm/K-mole

3) 1 L-atm = 101.325 J

4) 0.00?C = 273.15 K

5) F = 96485.31 C/mole

6) RTln(10)/F = 0.0592 V at 298 K and 0.0602 V at 303 K

B) DATA RELEVANT TO PROBLEM ONE

The following thermodynamic data were measured at 25.00?C (298.15 K) Species CH3OH(l) O2(g) CO2(g) H2O(l)

ΔH f? (kJ/mole) -238.66 -393.509 -285.830

S? (J/K-mole) 126.8 205.138 213.74 69.91

C) DATA RELEVANT TO PROBLEM FOUR

Values of Equilibrium Constants of Selected Reactions at 25.00?C

Reaction K

Ni+2(aq) + 6 NH3(aq) → Ni(NH3)6+2(aq) 4.1 x 108

NH4+(aq) → NH3(aq) + H+(aq) 5.6 x 10-10

D) DATA RELEVANT TO PROBLEM FIVE

Standard Entropies of Solvation at 25.00?C and Ionic Radii, r M+, of Group 1 Cations ΔS? is provided for the process M+(g) → M+(aq).

Species ΔS? (J/K-mole) r M+ (?)

Li+

Na+(aq) 15.3 0.95

K+(aq) 52.2 1.33

Rb+(aq) 61.5 1.48

Cs+(aq) 67.5 1.69

NAME:____________________________________ Chem 51, Fall, 2005, 2nd. midterm

The enclosed insert contains useful information and data needed to answer the questions. You may use the back for scratch work but enter all work to be graded in the space provided with each question. Show your work in order to receive credit.

1) (50 points) Professor Harry Gray at Cal Tech suggested at last year's Robbins Lectures that liquid methanol, CH3OH, might be the fuel of the future if it could be produced photochemically via the following reaction:

2 CO2(g) + 4 H2O(l) → 2 CH3OH(l) +

3 O2(g)

a) Using the data found on the insert, calculate the equilibrium constant of the

above reaction at 30.00?C.

In order to solve the problem, we shall assume that the enthalpy and entropy

changes for the reaction are relatively independent of temperature.

ΔH? = 3(0) + 2(-238.66) - 2(-393.509) - 4(-285.830) = 1453.018 kJ = 1.4540 x106 J ΔS? = 3(205.138) + 2(126.8) - 2(213.74) - 4(69.91) = 161.89 J/K

ΔG? (1454000) - (303.15)(161.89) = 1.4039 x 106 J

K = exp(-ΔG?/RT) = exp{(-1.4039 x 106)/[(8.314)(303.15)]} = exp(-557.00) =

10-241.90 = 1.2 x 10-242. N.B. 241.90 = 557.00/[ln (10)]

b) Suggest a convenient but accurate experimental method of obtaining the

standard enthalpy change for the above reaction. In your essay, note what

quantities are measured and succinctly indicate how the experimental data are

converted into the required enthalpy change.

K for the forward reaction is too small. One instead examines the reverse

reaction knowing that the desired ΔH(forward) equals -ΔH(reverse). The reverse reaction is combustion. Accurate measurements require a closed system and

hence a bomb, i.e. constant volume, calorimeter. One charges the bomb with

liquid methanol, a few mL of liquid water, and oxygen. The calorimeter yields

ΔE = qσ = -qθ = -CθTθ. One obtains a value of ΔH(reverse) by noting that

ΔH(reverse) = ΔE(reverse) + ΔpV and ΔpV ≈ (2-3)RT = RT.

2) (40 points) Methanol can be burned in an internal combustion engine or reacted with oxygen in a fuel cell. A fuel cell is simply a battery in which the reagents can be easily supplied. Consider a fuel cell based on the reverse of the reaction in Problem One.

a) Draw a sketch of the electrochemical cell and clearly label the components.

Provide the half reaction occurring at the anode and the cathode as well as the

polarity of the anode and cathode. The two cells are connected by a salt bridge.

the two electrodes are connected to a resistive load which much be large if the cell runs reversibly.

The reverse reaction has a large, positive value of ΔG and will run spontaneously

so the cathode will be positive with respect to the anode and the electrons will

flow from anode to cathode. The reactants and products are not conductors so

employ an inert electrode, e.g. platinum, in both cells.

In the anode cell, insert a Pt electrode in the nitric acid. Bubble carbon dioxide

over the Pt. Probably coat the Pt electrode with finely divided catalytic Pt. The

half reaction is CH3OH(l) + H2O(l) → CO2(g) + 6 H+(aq) + 6e-.

In the cathode cell, insert a Pt electrode, probably also coated with platinum black, into an aqueous solution of methanol and nitric acid. Buybble oxygen over the

electrode. The half reaction is 4 e- + O2(g) + 4 H+(aq) → 2 H2O(l).

b) Calculate the standard, reversible cell potential at 30?C. You will need the

value of ΔG? which you calculated in answering Problem One. If you were

unable to answer part (a) of Problem One, use a value of -250 kJ.

E? = -ΔG?/(n F) = -(-1403900 J)/[(12 mol)(96485 C/mol)] = 1.213 V

c) Calculate the reversible potential of a cell running at 30?C with the following

concentrations: O2, 0.20 atm; CO2, 0.050 atm; CH3OH, 0.40 M. The electrolyte

in each cell is 2.0 M nitric acid.

E = E? - [RTln(10)/F][1/n]log10Q

= 1.213 V - (0.0602/12)log10{(0.050)2/[(0.40)2(0.20)3]}

= 1.213 V - 0.0015 V = 1.212 V

3) (25 points) A piston was filled with a mixture of air and ethane and was

immersed in a large water bath. The heat capacity of the water bath plus piston

was 2.00 x 105 J/K. The piston was slowly compressed at a constant external

pressure of 8.0 atm. When the volume of the piston reached 1.00L, the mixture

ignited and the volume of the piston quickly increased to 10.00 L. The expansion occurred against a constant external pressure of 8.0 atm and the temperature of the water bath increased by 0.40 K.

Treating the contents of the piston as the system, calculate the following

system quantities for the expansion described above: work, heat, ΔE and ΔH.

qθ = CθTθ = (200,000 J/K)(0.40 K) = 8.0 x 104 J

qσ = -qθ = -8.0 x 104 J

wσ = -p extΔV = -(8.0 atm)(10 L - 1 L) = -72 L-atm = -7.3 x 103 J

ΔEσ = qσ + wσ = -8.7 x 104 J

ΔHσ = qσ = -8.0 x 104 J (Pressure is constant; there is on p-V work.)

4) (15 points) Using the insert, calculate at 25.00?C the value of ΔG? for the reaction:

Ni(NH3)6+2(aq) + 6 H+(aq) → Ni+2(aq) + 6 NH4+(aq)

The net reaction can be broken down to reactions A and B. For each of these,

ΔG? = -RTln(K). Both reactions A and B are the reverse of reactions given on the insert so their equilibrium constants are the inverse of those on the insert.

reaction K ΔG?(kJ/mole)

A Ni(NH3)6+2(aq) → Ni+2(aq) + 6 NH3(aq) 1/(4.1 x 108) = 2.4 x 10-9 49.16

B NH3(aq) + H+(aq) → NH4+(aq) 1/(5.6 x 10-10) = 1.8 x 109 -52.81

Since the net reaction is A + 6)B), ΔG?(net) = 49.16 + 6(-52.81) = -268.0 kJ.

5) (20 points) Ionic radii and the standard entropy of solvat ion, i.e. ΔS? for the process

M+(g) → M+(aq), are provided in the insert for Group 1 cations. Provide a molecular interpretation for the entropy data.

Focus on the trend in the data, a monotonic dependence of ΔS? on ionic radius . The motion of the ion, random in the gas phase and nearly random in the aqueous phase, leads to a dependence of the standard entropy on molecular weight. The contribution of this dependence is largely cancelled out by considering the entropy change. The increase in the standard entropy change as one goes down the column reflects the solvation of water molecules around the central cation. With a small ion, the electric field at the ion's surface binds the waters more tightly and the waters are less flexible. If one considers all the solvated ions in a solution (recall that thermodynamics is the study of large numbers of molecules), there will be fewer solvated structures than in the case of less tightly bound and hence more flexible water molecules. Fewer structures or microstates results in a lower entropy.