Show your work in all problems involving a calculation or derivation in order to receive any credit. Useful constants and a table of standard reduction potentials are printed on the insert. You may use the back of the insert for scratchwork but enter all work to be graded in the space provided with each question. Please note that the last question is worth 25%! An essay written in proper English is expected.
1) (38 points) Thermodynamic data at 25.00 ºC are provided below. Calculate the following quantities for the reaction C2H4(g) + 2 H2(g) = 2 CH4(g):
species C2H4(g) H2(g) CH4(g)
DHfº (kJ/mol) 52.26 -74.81
Sº (J/K-mol) 219.56 130.684 186.264
a) the value of DGº at 30.00 ºC,
DHº = 2(-74.81) - 2(0) -1(52.26) = -201.88 kJ = -201880 J.
DSº = 2(186.264) - 2(130.684) - (219.56) = -108.4 J/K
DGº = -201880 - (273.15+30.00)(-108.4) = -169,018 J = -169.0 kJ.
b) the value of K, the thermodynamic equilibrium constant, at 30.00 ºC, and
K = exp[--169018/(273.15+30.00)(8.31451)] = exp(67.056) 1.33 x 1029.
c) the value of the standard cell potential at 30.00 ºC of the cell based on the reaction.
Eº = --169018/(4)(96485.4) = 0.4379 V.
2) (37 points) Consider the reaction Al(c) + 3 Fe+3(aq) = Al+3(aq) + 3 Fe+2(aq).
a) Calculate the standard cell potential for the reaction at 25.00 ºC. Use the data provided on the insert.
e- + Fe+3 = Fe+2 for which Eº(red) = 0.77 V.
Al = Al+3 + 3 e- for which Eº(ox) = - Eº(red) = --1.66 V.
Eº(cell) = 0.77 + 1.66 = 2.43 V
b) A chemist wishes to characterize the reaction's thermodynamics using electrochemical measurements. Provide a sketch of an electrochemical cell that could be used in the study. Label the electrodes (anode or cathode). Also provide the half reaction that occurs at each electrode and the sign of each electrode.
Set up two beakers connected by a salt bridge. One beaker should contain aluminum nitrate solution and an aluminum electrode which is the anode. The half reaction is Al(c) = Al+3(aq) + 3 e-. The reaction occurs spontaneously under standard conditions and the anode is negative with respect to the cathode. The other beaker should contain an aqueous solution of both iron(II) nitrate and iron(III) nitrate. Use an inert electrode such as platinum; this is the cathode. The half reaction is e- + Fe+3(aq) = Fe+2(aq).
c) A cell is constructed in which one electrolyte is 0.200 M Al(NO3)3 and the other electrolyte is 0.100 M Fe(NO3)3 and 0.050 M Fe(NO3)2. Calculate the reversible potential of the cell at 25.00ºC.
E = 2.43 - (0.059/3)log[(0.05)3(0.20)/(0.10)3] = 2.43 - (0.059/3)log[0.025] = 2.43 + 0.03 = 2.46 V.
3) (25 points) Propose an experiment that would yield the standard enthalpy of formation of gaseous sulfur dioxide or a quantity from which the enthalpy change could be readily calculated. Include in your discussion a treatment of what would be measured and how you would calculate the enthalpy change from the measurements. Also specify any special experimental conditions.
Measure the heat of combustion of elemental sulfur in a bomb calorimeter for which the volume is constant and there is no other work term. The relevant reaction is S(c) + O2(g) = SO2(g) so DH(reaction) = DHfº(SO2(g)). Since Dn(gas) = 0, (E, the quantity measured in the calorimeter, equals the desired enthalpy of formation. In addition, DE(system) = q(system) = -q(surroundings) and q(surroundings) equals the product of the temperature change of the calorimeter and the heat capacity of the calorimeter.
4) (13 points) Another thermodynamic state function is the Helmholz free energy, A, where A is defined
as A = E - TS.
a) Obtain an expression for DA when the temperature T is constant.
DA = DE - TDS
b) Use the result from part (a) and the First and Second Laws of Thermodynamics as presented in class and show that DA of a process run reversibly and isothermally (i.e. at constant temperature) equals the total work. This is a straightforward algebraic exercise if you use fundamental principles and simple algebra.
DA = DE - TDS (eqn. 1,part a)
Substitute the First Law, (E = q + w, into equation 1 and obtain DA = q + w - TDS(eqn. 2).
The Second Law declare that under reversible conditions DS = q/T or TDS = q (eqn. 3).
Substituting equation 3 into equation 2, one obtains DA = TDS + w - TDS = w.
5) (37 points) Although fluorine is the most electronegative element, HF is unique among the hydrogen halides in that it is a moderately weak acid in water. Use the following thermodynamic data and provide an explanation for this anomaly.
Species HF(aq) HCl(aq) H+(aq) F-(aq) Cl-(aq) F-(g) Cl-(g)
DHfº (kJ/mol -320.08 not available 0.000 -332.63 -167.15 -255.39 -167.16
Sº (J/K-mol) 88.7 not available 0.00 -13.8 56.5 147.22 153.6
For the dissociation of HF in water, the standard enthalpy change is -12.55 kJ. The sign of this favorable term is consistent with the electronegative character of fluorine. The HF would be expected to have significant ionic character so the enthalpy term would suggest that HF would be a strong acid. In contrast, the standard entropy change is an unfavorable -102.55 J/K. Furthermore the value of -TDS is large compared with the enthalpy change and the reaction is entropy driven. The standard entropies of gaseous fluoride and chloride anions are very similar so the striking difference in the entropy of solvated fluoride and chloride cannot be attributed to intrinsic properties of the anions. The solvent, water, consequently plays a major role. The negative value of the net entropy change indicates that the favorably positive contributions to the entropy change from the cleavage of the HF bond and the breakup of the water structure are dwarfed by the negative entropy changes associated with the solvation of fluoride and hydrogen ions. Fluoride is a small ion and the electrostatic attraction between the anion and the water dipoles is particularly strong and the water molecules surrounding the anion are highly ordered. The unexpected high order in the aqueous phase is the source of the weakness of HF as an acid.
last revised, 15 October 1999