CHEMISTRY 51, FALL, 2002

INSERT FOR SECOND MIDTERM EXAM

CONSTANTS, CONVERSION FACTORS

AND USEFUL DATA

I) CONSTANTS AND CONVERSION FACTORS

1 liter-atm = 101.325 J

R = 8.31451 J/mole-K = 0.082058 liter-atm/mole-K

0 K = 273.15°K

k_B = Boltzmann’s constant = 1.38066 x 10^-23 J/molecule-K

N_A = 6.02205 x 10²³ molecules/mole

RT/F = 0.0592 V at 25.00°C

F = 96484.55 C/mole

II) THERMODYNAMIC DATA AT 25.00 °C USED IN PROBLEM ONE

C₂H₄O₂ = methyl formate

species C₂H₄O(g) H₂(g) CO(g)

DH_f°(kJ/mole) -350.2 -110.525

S°(J/mole-K) 285.39 130.684 197.674

III) THERMODYNAMIC DATA AT 25.00 °C USED IN PROBLEM TWO

CH₃COOH = acetic acid

species CH₃COOH(aq) H⁺(aq) CH₃COO^-(aq)

DH_f°(kJ/mole) -485.76 0.00 -486.01

S°(J/mole-K) 178.7 0.00 86.6

IV) THERMODYNAMIC DATA AT 25.00 °C USED IN PROBLEM THREE

species CH₃OCH₃(l) CH₃OCH₃(g)

DH_f°(kJ/mole) -203.50 -184.00

S°(J/mole-K) 187.60 267.34

V) THERMODYNAMIC DATA AT 25.00 °C USED IN PROBLEM FOUR

DH°(kJ/mol) reaction

-181.7 CH₃OH(aq) + 0.5 O₂(g) ® CH₂O(g) + H₂O(l)

33.2 CH₂O(aq) ® CH₂O(g)

-283.6 CH₂O(aq) + 0.5 O₂(g) ® HCOOH(aq)

NAME:____________________________________________ FALL,2002, SECOND MIDTERM EXAM

This examination has 5 questions on 3 pages. Data required for the solutions as well as useful constants are provided on the insert. You may use the back of the insert for scratch work but enter ALL work to be graded in the space provided with each question. Show your work in order to receive any credit in problems involving computation.

1) (45 points) Synthesis gas, a mixture of carbon monoxide and hydrogen that is produced from coal and water, is used by Eastman Chemicals to produce gaseous methyl formate (C₂H₄O₂) according to the following equation:

2 H₂(g) + 2 CO(g) ® C₂H₄O₂(g)

a) Using the thermodynamic data at 25.00°C provided on the insert, calculate the value of the equilibrium constant for the reaction at 20.00 °C.

DH° = DH_f°[C₂H₄O(g)] - 2DH_f°[CO(g)] - 2DH_f°[H₂(g)]

= -350.2 kJ -2(-110.525 kJ) – 2(0) = -129.15 kJ (don’t round off yet)

DS° = S°[C₂H₄O(g)] – 2S°[CO(g)] – 2S°[H₂(g)]

= 285.39 J/K -2(197.674 J/K) – 2(130.684 J/K) = - 371.326 J/K

Assume that H and S are independent of temperature to calculate DG° at 20.00 °C.

T = 273.15 K + 20.00 = 293.15 K.

DG°[293.15 K] = DH°[298.15 K] – (293.15 K)DS°[298.15 K]

= -129150 J – (293.15 K)(-371.326 J/K) = -20,296 J.

K = exp(-DG°/RT) = exp[-(-20296 J)/(8.31451 J/K)(293.15 K)]

= exp(8.327) = 4.13 x 10^3.

b) Discuss the effect on the equilibrium constant K if liquid rather than gaseous methyl formate is the product. Will K increase or decrease? The result may be indeterminate.

S° of liquid methyl formate is less than that of gaseous methyul formate so DS° of the reaction will be less. This factor will lead to a reduction in K. On the other hand, liquid methyl formate will be stabler than the gaseous ester so DH° of the reaction will also be less. This second factor leads to an increase in K. The two effects work at cross purposes and the final result is indeterminate unless more information is provided.

One student suggested an informative alternate route to the answer. One can combine the following reactions to yield the production of liquid methyl formate.

2 H₂(g) + 2 CO(g) = C₂H₄O₂(g)

C₂H₄O₂(g) = C₂H₄O₂(l).

The effect on the equilbrium by producing liquid rather than gaseous product is determined by the equilibrium constant for the second condensation reaction. This equilibrium constant is the inverse of the vapor pressure of methyl formate in atm. However, the state of methyl formate at ambient temperature and standard conditions was not provided. In fact, it is a liquid so K for the second process is greater than unity (the vapor pressure is less than one atm)

2) (20 points) Abundant experimental data clearly show that acetic acid is a weak acid in water. Why? Use the data at 25.00 °C on the insert and discuss the molecular basis for this experimental fact.

For the dissociation process, HA(aq) = H+(aq) + A-(aq), one quickly calculates at 25.00 °C that DH° = -0.25 kJ, DS° = -92.1 J/K, and -TDS° = +27.5 kJ. Surprisingly, the net energetics makes a very small contribution to DG° and the reaction is dominated by the unfavorable entropy change. The value of DS° is at first glance counterintuitive as dissociation should lead to an increase in entropy. However, the contribution to DS° from dissociation is more than counterbalanced by the organization of water molecules around that ions.

3) (30 points) The system in this problem is 2.000 mole of liquid dimethyl ether (CH₃OCH₃) whose normal boiling point is –24.8 °C. The liquid is inserted into a frictionless piston that is immersed in a large thermal bath at 25.00 °C and the liquid quickly evaporates.

a) Using the data at 25.00 °C in the insert, calculate the following quantities for the vaporization process: the heat (q), the work (w), and the change in energy (DE).

It was stated during the examination that a constant external pressure of one atmosphere was applied on the piston during the evaporation process. The process occurs at constant pressure and only pressure-volume work occurs. Hence, the system heat equals the standard enthalpy change.

q_s = DH° = n{DH_f°[CH₃OCH₃(g)] -D°H_f[CH₃OCH₃(l)]}

= (2.000 mole)[(-184.0 kJ/mole) – (-203.50 kJ/mole)] = 39.0 kJ.

w_s = - p_extDV = -p[V(g) – V(l)] @ -pV(g) = -RT

= -(2.000 mole)(8.31451 J/K-mole)(298.15 K) = -4.958 kJ.

DE_s = q_s + w_s = (39.0 kJ) + (-4.96 kJ) = 34.0 kJ.

b) Also calculate DS at 25.00 °C and compare this result with q/T. Comment on the result.

DS° = n[S°(g) - S°(l)] = (2.000 mole)(267.34 J/K-mole – 187.60 J/K-mole) = 158.5 J/K.

q/T = (39,000 J)/(298.15 K) = 130.8 J/K.

Note that DS > q/T. This result comes as no surprise. The rapid evaporation of the methyl ether is clearly irreversible and the direction of the inequality is required by the Second Law. The two terms would be equal at the boiling point.

4) (20 points) Using the thermodynamic data in the insert, calculate the standard enthalpy change (D°H) at 25.00 °C for the following biochemical transformation that occurs when a poor devil ingests methanol:

CH₃OH(aq) + O₂(g) ® HCOOH(aq) + H₂O(l)

Consider a three-step thermodynamic cycle:

reaction I: CH₃OH(aq) + 0.5 O₂(g) ® CH₂O(g) + H₂O(l)

reaction II: CH₂O(g) ® CH₂O(aq)

reaction III: CH₂O(aq) + 0.5 O₂(g) ® HCOOH(aq)

These add up to yield the net reaction.

net reaction: CH₃OH(aq) + O₂(g) ® HCOOH(aq) + H₂O(l)

Hence: DH(net) = D(I) + DH(II) + DH(III) = (-181.7 kJ) + (-1 x 33.2 kJ) + (-283.6 kJ) = -498.5 kJ.

Note that reaction II is the reverse of the vaporization reaction on the insert; hence, the minus sign in the middle. term.

5) (35 points) A technician plans to set up an electrochemical cell that will yield thermodynamic data for the reaction in Problem 4. The value of DG° for the reaction is –434.1 kJ at 25.00 °C.

a) Sketch the cell. Your sketch should show the anode and cathode and the half reactions occurring at each. Also provide the sign of each electrode.

Formic acid (HCCOH) and methanol are diffusible species so a salt bridge is required.

ANODE VESSEL:

Dip a platinum electrode in an acidified solution containing formic acid and methanol. The value of DG° is large and negative; the process occurs spontaneously and hence the electrode will be negative with respect to the cathode. The half reaction for the oxidation process is H₂0(l) + CH₃OH(aq) = HCOOH(aq) + 4 H⁺(aq) + 4 e^- .

ANODE VESSEL:

Dip a platinum electrode, preferably coated with finely divided catalytic platinum, in an acidified solution. Bubble a stream of oxygen over the electrode. The anode is positive. The half reaction for the reduction process is 4 e^- + 4 H⁺(aq) + O₂(g) = 2 H₂O(l).

Connect the two electrodes to a resistor with a high resistance. Also connect the two vessels with a salt bridge.

b) Calculate the Voltage of the cell when it is run under standard conditions at 25.00 °C. Indicate any experimental conditions required for your calculation to be valid.

The cell must be run reversibly. From part (a), n = 4.

DG° = -nFE° so E° = -DG°/nF = -(-434,100 J)/(4 mole)(96484.55 C/mole) = 1.125 V.