Chapter Twenty
Thinking It Through
T20.1 The work is calculated by the quantity PDV, where P is the external pressure against which the expansion takes place. We need to determine DV from the given information.
PDV = 1 atm ´ 5.00 L = 5.00 L atm
Next, we convert from units L atm to units J by remembering that the unit Pa is equal to a Newton per meter squared, or 1 Pa = 1 N/m2:
Finally, we convert liters to the units m3 and multiply in order to arrive at a final answer in units joules = N m.
T20.2 This type of calculation is done using the equation from the text: DH = DE + DngasRT. Here the value for Dngas is equal to 3 – 2 = +1. We convert the mass of material that is to react to a number of moles since we are given the value of DH for a 2 mole reaction.
This answer must then be multiplied by the number of moles we calculated above. The second calculation is the same, but the temperature that is used is 498 K.
T20.3 Values of DS are strongly influenced by changes in state or by changes in the number of moles of high entropy substances, such as gases. Here the reaction produces three moles of gaseous products starting with only two moles of gaseous reactants. We conclude that entropy increases, meaning that DS has a positive value.
T20.4 The spontaneity of the reaction depends on DG (=DH – TDS). We therefore need to know a value for DH and the temperature, in addition to a value for DS.
T20.5 The sign of DS should be negative if more order is found among the products. In this case, two homonuclear diatomic molecules possess more order than the two heteronuclear diatomic molecules of the reactants. The sign of DS should therefore be negative.
T20.6 Every substance possesses an entropy, given the symbol S°, and listed in tables such as Table 20.1. Evidently, what is called for here is the entropy change for the reaction in which one mole of HI is formed from the elements H2 and I2. This would be termed the DS° for the reaction. Its value could be calculated using equation 20.6, once values of S° for each reactant and product are known. Appendix E is also a good source for this type of information.
T20.7 The value for DG° is given in the problem, and values for the free energy of formation of all other reactants and products can be obtained in Appendix E.
T20.8 First, it is necessary to determine values for DH° and DS° using the data of Appendix E. Once these are determined, we use the temperature T = 500 K.
T20.9 The maximum work that can be obtained from the combustion of one mole under standard conditions is given by the value of DG° for this reaction. First, it is necessary to calculate the number of moles of butane that are actually present (using the ideal gas equation, where P = 2.00 atm, V = 10.0 L, T = 293 K). Next, the balanced equation is used to calculate the value for DG°, using equation 20.7 and the data of appendix E. Last, we multiply the value of DG° (in units kJ/mol) by the number of moles of gas that are present, in order to arrive at the number of kJ to be expected from the combustion of the sample.
T20.10 We must rearrange these equations so that they can be added together, giving a net reaction in which one mole of N2O5 is formed from elements only. When this has been accomplished, the value for the net DG° will correspond to the free energy of formation of N2O5. Remember to adjust the individual values for DG°, as was done for enthalpy when applying Hess's Law in Chapter 7.
2HNO3(l) ® N2O5(g) + H2O(l) DG° = 37.6 kJ
N2(g) + 3O2(g) + H2(g) ® 2HNO3(l) DG° = –159.82 kJ
H2O(l) ® H2(g) + 1/2O2(g) DG° = 237.2 kJ
N2(g) + 5/2O2(g) ® N2O5(g) DG° = sum of the other equations
T20.11 Add the energy of each of the bonds.
T20.12 Sum the heats of formation of the gaseous elemental atoms and subtract the sum of the bond energies in the HCN molecule.
T20.13 If we multiply P in units N/m2 (i.e., Pascals), by volume in units m3, the result has units N ´ m, which is the correct set of units for a joule.
T20.14 For an ideal gas, there are no interactions among the various gas molecules. This means that no inelastic collisions can take place, and consequently, no energy transfer is possible for a system that has an unchanging temperature.
T20.15 DE = 0, so q = PDV = (2.0 atm)(6.0 L – 12.0 L) = –12.0 L atm
T20.16 It is necessary to use the equation for DG (=DH – TDS) to answer this question. Since DS is negative, the term –TDS will be positive. We expect that, as the temperature rises, the value of DG will become increasingly less negative. Eventually the term –TDS will become sufficiently positive to offset the exothermic nature of the reaction, causing the value for DG to become positive. Above this temperature, the reaction is nonspontaneous.
T20.17 We know that DG = –RT ln(K). We can calculate the value of the equilibrium constant because we can determine the pressure of each of the gases from the data provided if, in addition, we know the initial amount of ClNO.
T20.18 Using DH and DS we can determine DG and from that K. We can then set up the resulting equilibrium problem to determine the equilibrium concentration of each component of the reaction.
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