Chemistry 531 Sring 2009 Problem Set 6 Solutions 1. In a thermochemical study of N 2, the following heat caacity data were found: t 0 C,m d 27.2Jmol 1 K 1 f t b f C,m d 23.4Jmol 1 K 1 C,m d 11.4Jmol 1 K 1 where t 35.61 K, the enthaly of transition is.229 kj mol 1 ; f 63.14 K, H melt,m.721 kjmol 1 ; b 77.321 K, H va,m 5.58 kj mol 1. Calculate the third law entroy of nitrogen vaor at the boiling oint. S t 0 C,m d + H f f + b f + H f t + t t C,m d + H va va C,m d 27.2J mol 1 K 1 +.229 103 J mol 1 + 23.4J mol 1 K 1 35.61K +.721 103 J mol 1 63.14K 2. A certain gas obeys the equation of state + 11.4J mol 1 K 1 + 5.58 103 J mol 1 152.02J mol 1 K 1 (V m b) R 77.32K 1
where b is a constant. Show that the artial molar enthaly of the gas is given by H H i b,v,n 1,n 2,...,V,n 1,n 2,... dh d +,{n i },V,{n j },j i d +,{n i } i dn i,,{n j }j i H + H i,{n i n } i,v,{n j }j i dh ds + V d + µ i dn i i H S + V,{n i },{n i } dg Sd + V d + i µ i dn i hen H i S,{n i },{n i } V V V V R + b V R R,{n i },V,{n j }j i,{n i } + b R b,{n i } b,v,{n j }j i 3. At -5. C the vaor ressure of ice is 3.012 mm of Hg. whereas the vaor ressure of of suer-cooled liquid water at -5. C is 3.163 mm Hg. Calculate G m for the transition water ice at -5. C. At equilibrium µ ice µ vaor µ + R ln ice 2
µ water µ vaor µ + R ln water G m µ ice µ water R ln ice water (8.3144J mol 1 K 1 )(268K) ln 3.012 3.163 109J 4. he density of lithium metal is.53 g ml 1. Assuming lithium to be incomressible, calculate G when the ressure on 5.0 g of lithium is raised isothermally from 1.0 to 100.0 atmosheres. dg Sd + V d G V G 2 1 V d V V 10 3 L 5.0g 0.00943 L 0.53g ( ) 8.3144J G (0.00943 L)(99 atm) 0.0821 L atm 94.6 J 5. he element hoshorus occurs in two solid hases often called red and white hoshorus. At 25. C and 1.0 bar ressure, the rocess P (red) P (white) has r G m-12134. J mol 1 and r Hm-17573. J mol 1. Assuming the enthaly change for the rocess to be temerature indeendent, at what temerature are the two hases in equilibrium at 1.0 bar ressure? G( 2 ) G( ( 1) 1 H 1 ) 2 1 2 1 At equilibrium G( 2 ) 0. hen 1 1 G( 1) 2 1 1 H 1 1 2 298K 12134 (17573)(298K) 2 963K 3
6. Consider the reaction between ideal gas molecules a A (g) + b B (g) e E (g) + d D (g) Define equilibrium constants and K e E d D a A b B K C Ce EC d D C a AC b B where C i is the concentration of secies i in moles er unit volume. (a) Show that where K C K (R ) ν ν e + d a b where hen i n ir V C i n i V C i R K e E d D a A b B where Ce ECD d (R ) ν CAC a B b ν e + d a b and the concentrations are exressed in moles er liter with R having units consistent with the defined standard state. hen K C K (R ) ν with K C Ce EC d D C a AC b B 4
(b) Show that d ln K C U d R 2 where U is the change in the internal energy for the reaction. For a gas-hase reaction d ln K C d d d [ln K + ln(r ) ν )] d ln K d ν d ln d ν rhm R 2 r Hm R ν R 2 r U m r H m (V ) hen 7. Consider the reaction r H m R ν d ln K C d ru m R 2 PCl 5(g) PCl 3(g) + Cl 2(g) (a) Show that if one starts with ure PCl 5 and a fraction, α, dissociates, the reaction quotient will be given by α 2 Q (1 α 2 ) 0 Let n the initial number of moles of PCl 5(g) hen at equilibrium n P Cl5 (1 α)n n P Cl3 n Cl2 αn n total 2αn + (1 α)n (1 + α)n P Cl5 χ P Cl5 1 α 1 + α P Cl3 Cl2 χ Cl2 5 α 1 + α
hen Q ( P Cl 3 / )( Cl2 / ) P Cl5 / [(α/(1 + α))(/ )] 2 [(1 α)/(1 + α)(/ )] α2 1 α 2 (b) At 250. C the equilibrium constant for the reaction is K 1.78. Calculate α for 0.1 bar and 1.0 bar at equilibrium. K α2 1 α 2 (1 α 2 )K α 2 ) α 2 ( + K K ( ) 1/2 K α K + / 1.78 1/2 α(.1) 0.973.1 + 1.78 1.78 1/2 α(1.) 0.800 2.78 (c) Calculate r G m for an equimolar mixture at 1 bar ressure and 250. C. r G m r G m + R ln Q R ln K + R ln Q R ln K + R ln ( P Cl 3 / )( Cl2 / ) ( P Cl5 / ) (1/3)(1/3) R ln K + R ln 1/3 ( 8.3144J mol 1 K 1 )(523K)[ln 1.78 ln(1/3)] 7285J (d) Calculate r G m for 20 er cent dissociation at 250. C and 1 bar ressure. Q.22 1.2 2 0.0417 r G m ( 8.3144J mol 1 K 1 )(523K)[ln 1.78 ln 0.0417] 16323J 6
8. he following results were obtained for the degree of dissociation, α, of CO 2 in the reaction CO 2(g) CO (g) + 1 2 O 2(g) at a ressure of 1.0 bar: ( K) α 1000. 2. x 10 7 1400. 1.27 x 10 4 2000. 1.55 x 10 2 Calculate r S m for the reaction at 1400. K. Let n initial number of moles of CO 2(g) n CO2 (1 α)n n CO αn n O2 α 2 n n total (1 α)n + αn + 1 2 αn (1 + α 2 )n hen hen O2 2(1 α) CO2 2 + α CO 2α 2 + α (1/2)α 1 + (1/2)α α 2 + α K ( CO/ )( O2 / ) 1/2 CO2 / ( (2α/[2 + α])(α/[2 + α])1/2 2(1 α)/(2 + α) α 3/2 1/2 (1 α)(2 + α) 1/2 K (1000K) 6.32 10 11 ) 1/2 7
K (1400K) 1.01 10 6 K (2000K) 1.38 10 3 r G m(1400k) R ln K (8.3144J mol 1 K)(1400K) ln(1.01 10 6 ) 160.7kJ mol 1 ln ln K (1400K) K (1000K) rhm R 1.01 10 6 6.32 10 11 r H m 8.3144J mol 1 K ( 1 1000K 1 r H m 281.7kJ mol 1 r S m rh m r G m (281700 160700)J mol 1 1400K 9. Hydrogen iodide gas dissociates according to the reaction HI (g) 1 2 H 2(g) + 1 2 I 2(g) 1400K ) ( 1 1000K 1 1400K 86.43J mol 1 K 1 At 0. C and 1.0 bar ressure, the degree of dissociation is α.40, and at 100. C, α.88. Assuming r H m to be indeendent of temerature, calculate r H m, r G m and r S m for the reaction at 0. C. Let ninitial number of moles of HI n HI (1 α)n n H2 n I2 αn 2 n total αn + (1 α)n n HI (1 α) H2 I2 α 2 K [(α/2)1/2 ] 2 1 α K (273K) 0.333 K (373K) 3.67 α 2(1 α) r G m(273k) R ln K (273K) (8.3144J mol 1 K 1 )(273K) ln 0.333 2496J mol 1 8 )
ln K ( 2 ) K ( 1 ) rhm ( 1 1 ) R 1 2 ln 3.67 0.333 r H m 8.3144J mol 1 K 1 r H m 20318J mol 1 r S m rh m r G m (20318 2496)J mol 1 273K 10. Hydrogen bromide dissociates according to the reaction HBr (g) 1 2 H 2(g) + 1 2 Br 2(g) ( 1 273K 1 ) 373K 65.28J mol 1 K 1 Use ables 4.2 and 7.2 of the text to find the degree of dissociation of HBr at 35. C and 2.0 bars ressure. [ ] 1 r G m 2 (3.110) ( 53.45) kj mol 1 55.005kJ mol 1 As in roblem 9 or r H m [ 1 2 (30.907) ( 36.40) ] kj mol 1 51.854kJ mol 1 K (298K) ex[ r G m/r ] ex[ 55005J mol 1 /(8.3144J mol 1 K 1 )(298K)] 2.28 10 10 ln K ( (308K) 51854J mol 1 1 2.28 10 10 8.3144J mol 1 K 1 298K 1 ) 308K K (308K) 4.50 10 10 α K α 2(1 α) 2K 1 + 2K 9.0 10 10 9