1. Consider two identical four-atom solids, one having 4 quanta of energy and the other having 10.
(a) What observable property differs between these two solids?
The two solids have different amounts of thermal energy and will be at different temperatures. The four-atom
solid with 10 energy quanta is at a higher temperature than the four-atom solid with 4 energy quanta.
(b) Allow 3 quanta of energy to be transferred from the solid with 10 quanta to the solid with 4 quanta of energy. Use
data from class on July 7th (the values of W) to determine whether this energy transfer is likely, explaining your
reasoning.
W = 35 for n =4
W = 286 for n =10
W = 120 for n = 7
Winitial = W4 x W10 = 35 x 286 = 10,010
Wfinal = W7 x W7 = 120 x 1206 = 14,400
This thermal energy transfer is likely since the final arrangement is more probable than the initial arrangement
(Wfinal > Winitial).
2. For each pair of substances, choose the one that you expect to have the higher standard molar entropy S° at 25°C.
Explain your reasoning.
(a) CH3CH2OH(l); CH3CH2OH(g)
CH3CH2OH(g) is expected to have the higher standard molar entropy since molecules in the gas phase have greater freedom of movement than molecules in the liquid phase.
(b) NO2(g); NO(g)
NO2(g) is expected to have the higher standard molar entropy because it is a more complex molecule (3 atoms vs. 2 atoms).
(c) Kr(g); HBr(g)
HBr(g) is expected to have the higher standard molar entropy because it is a more complex particle (2 atoms vs. 1 atom) even though it has a slightly lower molar mass than Kr.
3. Without using a table of standard molar entropies, rank each of the following in order of increasing (smallest to
largest) standard molar entropy S°. Justify your reasoning.
(a) N2O4(g); N2O5(g); and N2O(g)
S° [N2O(g)] < S° [N2O4(g)] < S° [N2O5(g)]
The value of the standard entropy (S°) increases with molecular complexity; in this instance increasing molecular complexity means an increase in the number of atoms present in the molecule and an increase in molar mass.
(b) H2O(l) at 0°C; H2O(s) at -10°C; and H2O(s) at 0°C
S° [H2O(s) at -10°C] < S° [H2O(s) at 0°C] < S° [H2O(l) at 0°C]
The value of the standard entropy is higher for H2O(l) at 0°C than that for H2O(s) at 0°C due to an increase in the freedom of movement of the water molecules in the liquid state (translational, rotational, and vibrational motion present) compared to the solid state (only vibrational motion present). The value of the standard entropy for H2O(s) at 0°C is higher than that for H2O(s) at -10°C because ice at the higher temperature has more thermal energy available to be arranged among the various energy levels. The increase in thermal energy leads to an increase in the number of ways energy can be arranged among the energy levels which results in a higher value for S°.
(c) H2O2(g); Ne(g); HCN(g); CH3CH2OH(g); and He(g)
This set contains both atoms and molecules; atoms have lower standard molar entropy values than molecules.
S° [He(g)] < S° [Ne(g)] < S° [HCN(g)] < S° [H2O2(g)] < S° [CH3CH2OH(g)]
From He to Ne there is an increase in mass, beyond that, the molecules increase in complexity.
4. Without doing any calculations, determine the sign of DSsys for each of the following processes. Justify your reasoning.
(a) O2(g) + O(g) ® O3(g)
The entropy change for this chemical reaction is predicted to be negative. There is a decrease in the number of moles of gas from the reactants to the product (2 moles ® 1 mole). The decrease in the number of independent particles means that the entropy of the product (the final state) is less than the entropy of the reactants (the initial state). Since DS°system = S°f – S°i the value of DS°system is negative.
(b) I2(s) ® I2(g)
The entropy change for this phase change is predicted to be positive. The I2 molecules experience a large increase in the freedom of movement when going from the physical state with the least freedom of movement (solid – only vibrational motion possible) to the physical state with the most freedom of movement (gas or vapor – translational, rotational, and vibrational motion present). This increase in freedom of movement means that I2 molecules in the vapor phase have a higher value of S° than I2 molecules in the solid phase. This results in a positive value for DS°system.
(c) CH3OH(l) ® CH3OH(s)
The entropy change for this phase change is predicted to be negative. The available freedom of motion for molecules decreases going from the liquid phase to the solid phase.
5. Without doing any calculations, determine the sign of DSsys and DSsurr for each of the chemical reactions below. Explain
your reasoning.
Remember that DSsurr = – DHsys/T
(a) 2 CO(g) + O2(g) ® 2 CO2(g) DH°rxn = -566.0 kJ
DSsys < 0; decreasing the number of gas molecules
DSsurr > 0; exothermic reaction
(b) 2 NO2(g) ® 2 NO(g) + O2(g) DH°rxn = +113.1 kJ
DSsys > 0; increasing the number of gas molecules
DSsurr < 0; endothermic reaction
6. Why is the enthalpy of vaporization of methanol (CH3OH) greater at room temperature than it is at its normal boiling
point?
Methanol molecules have a lower kinetic (or thermal) energy at room temperature than at the normal boiling point. The enthalpy of vaporization is the energy difference between molecules in the liquid phase and the gas phase. Since the energy of the liquid is lower at room temperature, the energy difference that must be overcome to become a vapor is greater which results in a greater enthalpy of vaporization.
