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1 roblem Set 4 Answers rofessor: C. E. Loader. At 5.0 ºC, the vapor pressure of ice is 3.03 mm and that of liquid water is 3.63 mm. Calculate G for the change of one mole of liquid water at 5.0 ºC to solid at 5.0 ºC. The system will not be in equilibrium, so G is not zero. We have to find a series of steps between the initial and final states. For S, we raised the temperture to 0 º but won't do that now because of the difficulty of calculating G for a change in T. The clue to the solution is given by the inclusion of vapor pressures in the problem. We can calculate G for a change in pressure at constant temperature. The alternative path for the process has three steps, all at 5 º; Step : Step 2: Step 3: Liquid water at 3.63 mm vapor at 3.63 mm. Vapor at 3.63 mm vapor at 3.03 mm. Vapor at 3.03 mm solid water at 3.03 mm. Steps () and (3) are equilibrium processes ( G 0), so the total G is the G of step (2). G nrt ln 2 (8.34)(268.2) ln 08.3 J Calculate the pressure at which the conversion of graphite to diamond would become thermodynamically possible. (a) (a) At K. (b) At 000 K. The densities of diamond and graphite are 3.53 g ml and g ml respectively, kj mol and J mol K r H V r S V. All of these values may be considered to be independent of temperature C (graphite) C (diamond) At constant temperature dg Vd or d( r G) ( r V)dp equation (i) V where r G K H V T r S J/mol (298.5 K)( J/mol) J mol age of 8 printed pages

2 and r V V(diamond) V(graphite) 2.0 g mol 3.53 g ml 2.0 g mol g ml ml mol x 0 3 L mol So, from equation (i), integrating with r V taken as constant (since densities are constant) d( r G) ( r V)d ( r V) d r G 2 r G 2 r V( 2 ) equation (ii) Set bar, Then G J mol r r G 0 2 will be the pressure required for r G 2 0 (thermodynamic possibility) J/mol x 0 3 L/mol ( 2 00 ka) 2.52 x 0 6 ka or.52 x 0 4 bar (b) at 000 K, V r G 000 K r H V T r S J/mol (000 K)( J/mol K) 550 J mol Again using equation (ii), J/mol x 0 3 L/mol ( 2 00 ka) x 0 6 ka or 2.77 x 0 4 bar A higher pressure is required at higher temperature. So, why do you think high temperatures are used to make artificial diamonds? 3. Calculate the freezing point of benzene at a pressure of 2000 atm. The normal freezing point is ºC and the densities of liquid and solid benzene are g cm 3 and g cm 3, respectively. H 0.58 kj mol FUS. Using the Clapeyron equation. d dt T M V FUS T T M V FUS or. where V FUS V(l) - V(s) V FUS 78. g mol g cm g mol g cm 3 2. cm 3 mol 2. x 0 6 m 3 mol T 0.58 x 0 3 J K - mol 2. x 0 6 m 3 mol x K.79 9 x 0 7 a K Now 2000 atm x 0 8 a T.26 K T M (2000 atm) ( ) K K x 08 a.799 x 0 7 a K age 2 of 8 printed pages

3 4. Using the data provided for benzene, calculate the vapor pressure of solid benzene at 20 ºC. Benzene: Boiling point 80.0 ºC Melting point 5.53 ºC H VA kj mol 0.59 kj mol We assume that the triple point occurs at the same temperature as the normal melting point. Then we can apply the Clausius-Clapeyron equation to the vaporization curve to find the vapour pressure of liquid benzene at the triple point (taken as 5.53 ºC). Remember that the vapour pressure at the normal boiling point is atm. Then ln t atm x 03 J\$mol ( 8.34 J\$ mol \$K K - ) K Hence t 6.07 x 0 2 atm. (vapour pressure at triple point) But at the triple point, solid is in equilibrium with the vapour at this same vapour pressure. We can now apply the Clausius-Clapeyron equation to the sublimation curve to determine the vapour pressure (p) at 20 ºC. For the sublimation curve we need H SUB H FUS + H VA 4.35 x 0 3 J mol atm 20 ºC TRILE OINT ressure SUBLIMATION CURVE VAORIZATION CURVE 5.53 ºC 80.0 ºC Temperature Then ln 6.07 x J\$ mol 8.34 J\$mol \$K ( K K ) Hence p.00 x 0 2 atm (Vapour pressure of solid benzene at 20 ºC) age 3 of 8 printed pages

4 5. The normal melting point of naphthalene (molar mass 28.2 g mol ) is 80.2 ºC and it rises by 2.32 ºC when the pressure increases by 00 atmospheres. It boils at 27.9 ºC and it's molar enthalpy of vaporization is 40.5 kj mol. The density of solid naphthalene is.5 g ml and that of the liquid is 0% less. Calculate. (a) The molar enthalpy of fusion of naphthalene g\$mol V(s).5 g\$ ml V(s).5 ml mol 28.2 g\$mol V(l).035 g\$ml V(l) 23.9 ml mol V FUS V FUS 2.4 ml mol L mol dt d T mp V FUS 2.32 K 00 x 0.3 ka ( K)(.024 L/mol) H 9.3 kj mol FUS (b) The vapor pressure of liquid naphthalene at the triple point (you may assume that this is the same temperature as the normal melting point). Apply the Clausius-Clapeyron equation to the vaporization curve, assuming that the triple point temperature is the same as the melting point, ln( ) H VA atm R T mpt T bpt ln() J\$ mol 8.34 J\$ mol \$ K ( K K ) ln() x 0 2 atm 2.2 ka (vapour pressure at the triple point) (c) The vapor pressure of solid naphthalene at 75.0 ºC. H SUB + H VA 9.3 kj mol kj mol age 4 of 8 printed pages

5 H kj mol SUB The vapor pressure of solid naphthalene and liquid naphthalene are the same at the triple point. Apply the Clausius-Clapeyron equation to the sublimitation curve 2.2 ka ln 2.2 ka H SUB R ( K J\$ mol 8.34 J\$ mol \$ K ) K ln ( ) 2.2 ka ln ka ka (vapor pressure of solid at 75 ºC) (d) The free energy change when.00 mol of supercooled liquid naphthalene solidifies at 75.0 ºC. We already have the vapour pressure of solid naphthalene at 75 ºC. Use the extrapolated liquid vaporization curve to find the vapour pressure of supercooled liquid naphthalene at 75 ºC: ln 2.2 ka H VA R ( K K ) 2.2 ka ln ka ka (vapor pressure of (l) at 75 ºC) At 75 ºC (l) G? (s) G O G 3 O (g),.73 ka G 2 (g),.56 ka G G 2 nrt ln 2 x 8.34 x x ln(.56 G 293 J mol (spontaneous of course).73 ) age 5 of 8 printed pages

6 NOTE: That (l) naphthalene is in equilibrium with vapour at.73 ka at 75 ºC (see above), so G zero. Likewise, solid naphthalene is equilibrium with vapour at.56 ka at 75 ºC (part (c) above) and G 3 zero. This part of the problem is the same as problem # in this problem set g of hydrocarbon of the type; H(CH 2 ) n H, is dissolved in 90 g of ethylene bromide. The freezing point of the solution is 0.53 K lower than the pure solvent (K F 2.4 K m ). Calculate the value of n in the formula of the hydrocarbon. What assumption is it necessary to make about the hydrocarbon? Assume that the hydrocarbon is non-volatile. Then T F K F m T m F 0.53 K molal K F 2.4 K (molal) g ethylene bromide contains mol hydrocarbon 90 x g ethylene bromide contains mol hydrocarbon mol hydrocarbon But, from question, this hydrocarbon has a mass of 0.80 g 0.80 g Molar mass of the hydrocarbon mol 00 g mol For H(CH 2 ) n H, the molar mass 4n + 2 4n n n 7 4 The formula is H(CH 2 ) n H. 7. Bovine serum albumin (BSA) has a molar mass of g mol. Calculate the osmotic pressure of a solution of 2.0 g BSA in 0.00 L of water at 25.0 ºC. Assume V SOLN 0.00 L π π 2 g M 2 V g mol 0.00 L C x 0 4 mol L C 2 RT where C 2 in mol L, R in L atm, π in atm x 0 4 mol L x L atm K mol x 298 K 7.5 x 0 3 atm 8. The normal boiling point of nitrobenzene is 20.9 ºC. If the vapor pressure at 80 ºC is mm Hg, what is the vapor pressure at 200 ºC? age 6 of 8 printed pages

7 This problem has two steps. Since H, isn't given, you must calculate it. Remember that the normal boiling point is at atm. You have two pressures with corresponding temperatures, and, using the Clausius-Clapeyron equation H kj mol V (ln 760 ) 484.(453.2) In order to obtain the vapor pressure at a given temperature, use the Clausius-Clapeyron equation again and solve for the desired quantity. ln 2 ln + H V R T 2 T T 2 T Choosing mm as, x 0 ln 2 ln(353.2) ( 453.2) mm Hg 9. When 4.07 g of a new water soluble polymer was dissolved in 82.4 ml of water the measured osmotic pressure was 6.23 ka at 22.0 ºC. (a) Calculate the molar mass of the polymer a π C 2 RT so C x mol L J K mol x K 4.07 g molar mass.95 x 0 4 g mol x 0 3 mol L x 82.4 x 0 3 L (b) Calculate the freezing point of the solution. Assuming the density of the water is.0 g ml -, the molality is 4.07 g m.95 x 0 4 g mol 82.4 % 0 3 kg m T K f m T.86 K m % m K. So the freezing point of the solution is still 22.0 ºC to 3 significant digits - the change in fp is barely measurable. 0. henol (C 6 H 5 OH) is partially dimerized in the solvent bromoform. When 2.58 g of phenol is dissolved in 00 g of bromoform, the bromoform freezing point is lowered by 2.37 ºC. ure bromoform has a molal freezing-point-depression age 7 of 8 printed pages

8 constant of 4. ºC kg mol and freezes at 8.3 ºC. Calculate the equilibrium constant, K m. for the dimerization reaction of phenol in bromoform at 6 ºC, assuming an ideally dilute solution. 2 C 6 H 5 OH (C 6 H 5 OH) 2 T K f m a m a m a 0.68 mol kg (i) where m a is the apparent molality of the phenol. (i.e. based on actual # mole present at equilibrium) Also m p 2.58 g 94. g mol /0.00 kg m p mol kg (ii) where m p is the molality of the phenol if it it were all present as monomer (i.e. no dimerization). So, in molalities, I (all monomer) C E 2 C 6 H 5 OH (C 6 H 5 OH) x +x ( x) x (from (ii)) At equilibrium, actual # mole of solute per kilogram of solvent But from (i) above, this # of mole also x 0.68 x mol Finally, (m 2 ) eq x K m (m p ) 2 eq ( x) 2 ( x) + x x ( (0.060)) age 8 of 8 printed pages

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