Exam 2 Solutions. for a gas obeying the equation of state. Z = PV m RT = 1 + BP + CP 2,

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1 Chemistry 360 Dr. Jean M. Standard Fall 016 Name KEY 1.) (14 points) Determine # H & % ( $ ' Exam Solutions for a gas obeying the equation of state Z = V m R = 1 + B + C, where B and C are constants. Since a partial derivative of H is requested, it is helpful to start with the fundamental equation for H, dh = ds + V d. ake the partial derivative with respect to V (at constant ) of both sides of the equation, H $ = S $ + V. he partial derivative on the right side of the equation cannot be determined from the equation of state. However, one of the Maxwell relations can be used, S $ = V $. " % Substituting yields H $ = V $ " % + V. Now, the partial derivative on the right side can be calculated from the equation of state. Since the equation of state is Z = V m R = 1 + B + C, solving for V m gives V m R = 1 + B + C V m = R ( 1 + B + C) or V m = R + BR + CR. Now, converting to V instead of V m leads to the expression V = nr + nbr + ncr.

2 1. Continued aking the partial derivative, V $ " % = nr + nbr + ncr. Substituting into the expression for the partial derivative of H gives the result, H $ H $ = V $ + V " % p = nr $ # + nbr + ncr& + V " % = nr $ # + nbr + ncr& + V " % = V + V = 0.

3 .) (14 points) he standard molar enthalpy of formation of Fe O 3 (s) is ΔH f = 84. kj/mol, and the standard molar enthalpy of formation of SO (g) is ΔH f = 96.8 kj/mol (both at 98 K). Use this information, along with the standard molar enthalpy change of the following reaction at 98 K, 3 FeS (s) + 11 O (g) Fe O 3 (s) + 4 SO (g) = 1655 kj/mol, to determine the standard molar enthalpy change of the reaction shown below at 98 K: Fe (s) + S (s) FeS (s). [Note: he standard state of iron is Fe(s) and the standard state of sulfur is S(s) at 98 K.] he formation reaction for Fe O 3 is Fe( s) + 3 O ( g ) Fe O 3 ( s). he formation reaction for SO is S( s) + O ( g) SO ( g). he desired reaction can be constructed from the reactions given in the problem in the following way: 1 [ ( ) + O ( g) SO ( g) ] ( ) + 3 O ( g ) Fe O 3 ( s) S s 1 Fe s [ ] 11 [ Fe O 3 ( s) + 4 SO (g) FeS (s) + O (g) ] Fe( s) + S( s) FeS ( s) From the combination of reactions above, the enthalpy of reaction can be calculated from Hess' Law, ΔH f ( FeS ) = ΔH f ( SO ) + 1 ΔH f ( Fe O 3 ) 1 ΔH r, where ΔH r corresponds to the standard molar enthalpy of the reaction between FeS and O given in the problem. Substituting, ( ) = ΔH f ( SO ) + 1 ΔH f ( Fe O 3 ) 1 ΔH r, = ( 96.8 kj/mol) kj/mol ( ) 1 ( 1655kJ/mol ) ΔH f FeS ΔH f ( FeS ) = 178. kj/mol.

4 3.) (14 points) Explain why we are able to obtain absolute molar entropies of substances; that is, what is the theoretical foundation that allows the determination of absolute molar entropies? Discuss any trends that may be observed in the magnitudes of the absolute molar entropies of the following substances and the reasoning behind your answer: He (g), O (g), CH 4 (g), C 6 H 6 (g). 4 he opportunity for an absolute entropy scale is provided by the hird Law of hermodynamics, which sets the absolute entropy of a perfect crystalline substance as 0 at 0 K. he absolute entropies tabulated above are all for gases of varying structural complexity. As the substances vary from a monatomic gas (He) to a diatomic gas (O ) to polyatomic gases (CH 4 and C 6 H 6 ), the entropy is expected to increase. his is a result of the increase in the number of degrees of freedom of the substances. As we saw in our discussion of the Equipartition heorem, the number of degrees of freedom increases as the complexity of the molecule increases. For the complex molecules with more degrees of freedom, there are more ways of distributing the available energy among the degrees of freedom, which increases the disorder in the system. hus, more complex substances have higher entropies than less complex substances.

5 4.) (15 points) rue/false, short answer, multiple choice. 5 a.) rue or False : In the statistical definition of entropy, S = k B lnw, the quantity W corresponds to the energy. b.) rue or False : he Debye equation gives the form of the constant pressure molar heat capacity of a solid at low temperature and has the form C p,m ( ) = a. c.) Short answer Adiabatic (or bomb) calorimetry is carried out at constant volume, whereas Solution calorimetry is carried out at constant pressure. d.) Short answer For two chemical reactions that combine to give a third reaction (the overall reaction), the enthalpy change of the overall reaction (3) is given as ΔH 3 = ΔH 1 + ΔH, which is an example of the application of Hess' Law. e.) Multiple Choice: Which of the following corresponds to the correct relationship? 1) = G $. " ) = U % $ '. # V & " 3) = A % $ '. # V & 4) = H $. " S % S S S

6 5.) (14 points) Using the values in the table below reported at 98 K, determine hydrogenation of acetylene to ethane, C H ( g) + H g ( ) C H 6 ( g). Assume that C p,m is independent of temperature for all species in the reaction. at 700 K for the 6 ΔH f (kj/mol) C p,m (Jmol 1 K 1 ) C H (g) C H 6 (g) H (g) he first step is to determine the enthalpy of reaction at 98 K: = ΔH f ( C H 6 ) ΔH f ( C H ) ΔH f ( H ) ( ) ( 0) = kj/mol 6.73 kj/mol = kj/mol. he temperature dependence of the molar enthalpy of reaction is given by ( ) = ( 1 ) + ΔC p,m ( 1 ). his equation assumes that the molar heat capacities are independent of temperature. In the equation, ΔC p,m is the difference in molar heat capacities between products and reactants with the appropriate signed stoichiometric coefficients included, ΔC p,m = ν i C p,m ( i). i For this reaction, the heat capacity difference is ( ) ΔC p,m = ν i C p,m i i = C p,m ( C H 6 ) C p,m ( C H ) C p,m ( O ) = ( 8.8) Jmol 1 K 1 ΔC p,m = Jmol 1 K 1. Substituting, the molar enthalpy of reaction at 700 K is ( 700 K) = ( 98 K) + ΔC p,m ( K) = kj/mol + ( Jmol 1 K 1 )( K) = kj/mol ( 700 K) = kj/mol. 1 kj 1000 J

7 6.) (14 points) Calculate the entropy change ΔS for the following transformation involving one mole of H O, 7 H O (s, 00 K) H O (l, 300 K). You may need to use the following data: = 73 K ; = 6.01 kj mol 1 vap = 373 K; ΔH vap = kj mol 1 C p,m s ( ) = 36. J mol 1 K 1 ; C p,m ( l) = 75.3 J mol 1 K 1 ; C p,m ( g) = 33.6 J mol 1 K 1 he process may be broken up into three steps: (1) heating the solid from 00 to 73 K; () the S-L phase transition at 73 K; and (3) heating the liquid from 73 to 300 K. he molar entropy change can then be written as a sum of these three steps, = C p,m ( s) fus d C p,m ( l) d. Assuming that the heat capacities are independent of temperature (they are listed in the problem as constants), the expression becomes = C p,m ( s) d C p,m ( l) d = C p,m ( s) 1 fus d + 1 = C p,m ( s) ln fus C p,m l + C p,m l ( ) ( ) ln 1 d. Substituting, = C p,m ( s) ln fus + 1 ( ) ln 73K = 36. J mol 1 K J mol 1 K 1 + C p,m l + 00 K 73K ( ) ln 300K = J mol 1 K 1 = 40.4 J mol 1 K 1. ( ) ln ( 6010 J/mol) 73K So, for one mole of H O, the entropy change is ΔS = 40.4 J K 1.

8 7.) (15 points) rue/false, short answer, multiple choice. 8 a.) rue or False: For the reaction CO (g) + O (g) CO (g), the entropy change is expected to be negative. b.) rue or False : he standard state is defined to be the pure form of an element at 1 bar and 5 C. c.) Short answer he Second Law of hermodynamics states that entropy increases for a spontaneous process in an isolated system. d.) Short answer he Helmholtz Energy A is defined as U S. e.) Multiple Choice: At 98 K, ΔG R = 158 kj/mol for a certain chemical reaction. In addition, this reaction has > 0. Which of the following is a reasonable expectation for ΔG R at 500 K? 1) ΔG R = 195 kj/mol. ) ΔG R = 158 kj/mol. 3) ΔG R = 141 kj/mol. 4) ΔG R = +158 kj/mol.

Physical Chemistry I Exam points

Physical Chemistry I Exam points Chemistry 360 Fall 2018 Dr. Jean M. tandard October 17, 2018 Name Physical Chemistry I Exam 2 100 points Note: You must show your work on problems in order to receive full credit for any answers. You must