Homework Assignment #4: Due at 500 pm Monday 8 July,. University of Washington Department of Chemistry Chemistry 45/456 Summer Quarter 0 ) o a very good approximation, ammonia obeys the Bertholet equation of state, whih reads: 9 nr V = nr + 8 where and are onstants and are alled the ritial pressure and temperature, respetively. For ammonia = 405.4 and =.5atm. a) Suppose we have 500 grams of ammonia under a pressure of =3.04 atm and at =33. Calulate the volume of ammonia aording to the Bertholet equation of state and ompare to the result predited by the ideal gas law. nr 9 nr 0.08 33 500g Latm mol V = + 6 = 8 7gmol 3.04atm 9 500g ( 0.08Latm mol )( 405.4 ) 405.4 + 6 8 7gmol.5atm 33 = 56.6L+ 0.67L 8.45 = 5.4L he ideal gas result is the first term V = 56.6L b) Assuming ammonia obeys the Bertholet equation of state obtain V expressions for the oeffiient of thermal expansion β = and the V V isothermal ompressibility κ =. Evaluate β and κ for 500 V grams of ammonia at =3.04 atm and at =33.
V nr 9 nr nr = + 6 = 8 ( 9.4mol)( 0.08Latm mol )( 33 ) ( 3.04atm) = = 84.4Latm V κ = = ( 84.4Latm ) = 0.336atm V 5.4L 3 V nr 9 nr nr 54 nr = + 6 = 8 8 3 nr 8 nr ( 9.4mol)( 0.08Latm mol ) = + = 8 3.04atm 3 7 ( 9.4mol)( 0.08Latm mol ) 405.4 + 3.5atm 33 0.794L + 0.083L.98 = 0.794L + 0.036L = 0.830L = V β = = ( 0.830L ) = 0.0033 V 5.4L U H ) Using your results from part b, alulate and for 500 grams V of ammonia at -3.04 atm and =33. U β 0.0033 = = ( 33) 3.04atm 0.3atm = V κ 0.336atm H = ( β + ) V = ( ( 33)( 0.0033 ) + )( 5.4L) = 6.6L ) Wind turbines are now reognized as useful devies for power generation. Wind turbines of various types are wide-spread in the U.S. and Europe and are bound to beome more ommon in the future. a) Calulate the power of wind moving through a irular area 6 meters in 3 diameter if the density of air at sea level is ρ =.3kgm and the wind veloity 0 ms -. 6m mv Svv Sv kgm ms 3 3 = = ρ = ρ = ( 0.5)(.3 )( 3.4) ( 0 ) 3 3 3 7 ( 0.5)(.3kgm )( 3.4)( 3969m )( 8000m s ) 6.3 Js wind = = 3
b) Consider an ideal wind turbine with a rotor diameter of 6m. What is the power generated by this turbine if the wind moves at 0 ms - into the turbine, exerts no drags on the rotor blades, et., and exits the turbine at ms -. What is the effiieny of this turbine? turbine = m ( v v) = ρsv( v v) = ρs ( v+ v)( v v) 3 6m = ( 0.5)(.3kgm )( 3.4) ( ms + 0ms )( 400m s 00m s ) 3 7 = 0.5.3kgm 3.4 3969m 30ms 300m s = 3.45 Js 7 turbine 3.45 Js ε = = = 0.56 7 6.3 Js wind ) he U.S. onsumes 0,680,000 barrels of rude oil per day. here are 59 L in a barrel of oil. Crude oil has an average density of 900 kg m -3 and an average heat of ombustion H omb =4.47x 4 kj/kg. Suppose we burn on average 90% of the oil that we onsume per day. How many wind turbines operating at the effiieny in part b would we require to replae the energy we obtain by burning oil? 3 3 9 m 0, 680, 000brlday 59Lbrl 0.00m L 900kgm =.96 kgday oil oil, total oil, burn 9 9 = ( 0.90)(.53 ) =.38 = 4.47 kjkg.96 kgday day / 86400s =.53 kjs 4 9 9 kjs kjs 4 wind = 3.45 kjs 9 oil, burn.38 kjs 4 = = 4 turbines 4 3.45 kjs wind Note: he United States is the world s largest produer of wind power with a total national wind power apaity of 35,000 MW=3.5x 7 kjs -. his is about.5% of our present energy output from oil. 3) Suppose a protein denatures from a strutured form N to an unstrutured form D at a melting temperature m =343 and at a pressure of m =atm. Suppose under these onditions the heat of denaturation is q m =638 kjmol -. Furthermore, when a protein 3 denatures it dereases in volume i.e. V = VD VN = 70m mol. a) Calulate H, U, S, w and G for the denaturation of this protein at m =343k and m =atm.
H = qm = 638kJmol H 638kJmol S = = =.86kJ mol 343 w = V = atm 0.070 Lmol.kJL atm = 0.007kJmol U = qm V 638kJmol G = 0 b) When the protein denatures its heat apaity inreases. Assume the heat apaity D N hange is given by C C C 8.37kJ = = mol. Calulate H, S, and G for the protein at =3 and = atm. Assume C is onstant between =343 and =3. Is the N form of the protein more stable or less stable than the D form under these onditions. Explain. H 3 = H 343 + C = 638kJmol + 8.37kJ mol 3 343 = 36.8kJmol 3 S ( 3 ) = S ( 343 ) + C ln =.86kJ mol + ( 8.37kJ mol ) ln =.0kJ mol m 343 G = H S = 36.8kJmol 3.0kJ mol = 48.7kJmol G>0 so N form is more stable. ) Suppose the pressure on the protein solution is inreased to =00atm at m =343. Calulate Dg for the onversion of N to D at =00atm and m=343. Whih form of the protein is more stable under these onditions? Explain. Assume V is a onstant in the pressure range under onsideration. Set up the yle: N N m = atm m = 343 = 00atm = 343 D D hen: G = 343, = atm = 0 = G + G = 343, = 00atm + G ( m m ) N ( m ) D G ( 343, 00atm) G G V ( 00atm atm) V ( atm 00atm) m = = = D N = N + D = V V atm 00atm = 0.07L 999atm = 69.9Latm JL atm = 7.06kJmol ( D N) G<0 so the D form is favored beause it has a smaller volume. 4) Many biologial maromoleules undergo a transition alled denaturation. Denaturation is a proess whereby a strutured, biologial ative moleule, alled the native form, unfolds or beomes unstrutured and biologially inative. he equilibrium is
unfolds native denatured folds For the enzyme hymotrypsin at ph= the enthalpy hange assoiated with denaturation 0 is H = 48kJ mol 0 and the entropy hange is S =.3kJ mol a) Calulate the Gibbs energy hange for the denaturation of hymotrypsin at ph= and =303. Assume the enthalpy and entropy are temperature independent between 98 and 303. G den = H den S den = 48 3 Jmol 303 = 48kJmol 400kJmol = 8.0kJmol ( 30J mol ) b) Calulate the equilibrium onstant for the denaturation of hymotrypsin at ph and =303. 303 = exp G den R = exp 8000Jmol = 7.86 ( 8.3J mol 4 )303 ) Based on your answer for parts a and b, is hymotrypsin struturally stable at ph and =303. fd = so at 303 fd fn. he protein is struturally stable. f N d) he melting temperature is defined as the temperature at whih the equilibrium onstant for denaturation =. Assuming that the enthalpy of denaturation is temperature independent, alulate the melting temperature of hymotrypsin at ph. G = H S so 0 = Hden m Sden. herefore den den den m = H den S den = 48 3 Jmol 30J mol = 37 5) he tehnology to build small mirosopi-sized motors (e.g. moleular motors) exists. In priniple, if a mirosopi motor an be built, whose ation is based on flutuations in the environment, the limit imposed on the effiieny of an engine by the Seond Law an be irumvented! Or an it? Consider the following mirosopi engine, held at onstant temperature:
Assume the motor is so small that a gas moleule (random motion a flutuation) hits the paddle wheel ausing it to turn lokwise or ounterlokwise. When the paddle turns the axle turns lifting a one ell organism whih is tethered to the axle in some way. Now beause the motion of gas moleules is random, as many moleules hit the paddle ausing it to turn lokwise, as hit it in the opposite diretion ausing it to turn ounterlokwise, so the ell is randomly raised up and down with no net work aomplished. But it is proposed that a rathet, held down with a spring against a gear, an prevent ounterlokwise motions of the paddle. hen the ell will be raised by lokwise motions of the paddle. Net work is done (mass of the ell times the hange in gravitational potential) and beause the fore ausing the work originates in flutuations, the effiieny of this engine is not governed by Seond Law limits. a) Will the motor work? Hint: onsider and disuss the proposed ation of the rathet in terms of random motions imposed by flutuations. he engine does not work. he reason it does not work is that the temperature is assumed uniform over this tiny mahine. herefore the same number of moleules that bombard a unit area of the paddle wheel per unit time also bombard the rathet per unit time. Insofar as the rathet and the wheel are both tiny, they are both perturbed by moleular ollisions. he rathet annot work as advertised unless there is some way to abate the moleular ollisions that fall upon it. b) If the motor will not work as shown, an you think of a simple hange that might make it work? One way to abate moleular ollisions against the rathet would be to plae it in an enlosure muh older than the area around the wheel. But then we would have reated a temperature gradient, and this mahine would then just operate as any other mahine. We have to provide fuel to pump heat out of the enlosure around the rathet, so the mahine requires fuel and will have a finite effiieny.
6) A sample onsisting of.50 moles of an ideal gas at 98 is expanded from an initial volume of.0l to a final volume of 50.0L. Calulate G and A for: a) an isothermal reversible path; b) an isothermal expansion against a onstant external pressure of 0.750 bar. Explain why G and A do not differ. 7) In the last two problem sets we have alulated H and S at =00 for the 3 reation of nitrogen rwith hydrogen to form ammonia: N( g) + H( g) NH3( g). Using the standard heats of formation at =98 and the heat apaities at onstant pressure as a funtion of temperature, alulate G for this reation at =500 and alulate the equilibrium onstant at =500. Based on your alulation, does the formation of ammonia beome more or less favorable as temperature is inreased? Explain. 00 f f 98 ( 00 ) ( 98 ) H = H + C d 3 from HW : C ( ) = 3.+ 30.88 5.895 C d d 30.88 5.895 = 3.( 500 98) + ( 500 98 ) ( 500 98 ) J mol 3 500 500 3 = 3.+ 30.88 5.895 98 98 3 3 3
( 3.( 0) 5.44 (.6 ).965 ( 9.85 )) = + = 6304 + 486 94 Jmol = 40Jmol 3 5 7 00 f f 98 Jmol H 00 = H 98 + C d = 46.kJmol 4.0kJmol = 50.kJmol ( 500 ) ( 98 ) From last hom ework : 500 S S d = + 3 98 3 500 3. 3 S 500 S 98 30.88 5.895 d 99J 3.ln 9 98 C S 98 = S NH S N S H = 9.8J 95.8J 96J = 99.0J = + + = + 500 5.895 + 30.88 500 98 500 98 8 = 99J + 6.5 + 6.4 0.48 J =09.4J J 3 (,500 ) (,500 ) (,500 ) G NH = H NH S NH rxn 3 rxn 3 rxn 3 = 50.kJ 500 9.4J = 4.58kJ (,98 ) (,98 ) (,980 ) G NH = H NH S NH rxn 3 rxn 3 rxn 3 = 46.kJ 98 99J =6.6kJ herefore the reation is thermodynamially less favored at higher temperature. ( J mol ) S H f f,9 C J mol (kj mol - ) NH3 ( g ) 9.8-46. 3 5.89 + 3.58 3.046 N ( g ) 9.6 3 6.98 + 5.9 0.3376 H ( g ) 30.7 3 9.07 0.837 +.0 8) he van t Hoff equation relates the temperature dependene of the equilibrium onstant to the enthalpy for a reation : 0 Hrxn dln = d R 0 0
For the reation: CuO( s) Cu ( s) + O ( g ), the following data are available for =98: CuO(s) Cu(s) O (g) H ( f kjmol ) -57 G kjmol -30 C 4.3 4.4 9.4 f ( J mol ) a. Integrate the van t Hoff equation given above assuming H rxn is independent of temperature. Calulate for the reation at =0. 0 0 0 Hrxn ln = R 98 98 d d 0 0 0 Hrxn d H rxn ln ( 0) ln ( 98) = R = R 98 3 G = G O + G Cu G CuO = 30 Jmol f f f 3 ( 30 Jmol ) G ln ( 98) = = =05 R ( 8.3J mol )( 98 ) 0 H rxn ln ( 0) = ln ( 98) R 3 ( 57 Jmol ) =05 8.3J mol 0 98 =05 37786 0.00083 0.0034 = 7.89 b. Assume H rxn is temperature-dependent and given by: Hrxn = Hrxn ( 0) + C ( 0) Integrate the van t Hoff equation assuming this temperature dependene for H and alulate. How is the equilibrium affeted assuming rxn H rxn is temperature dependent?
H ln ( 0) = ln ( 98) R 0 98 C 0 C ( 98 ) + ln + R 98 R 0 98 ( 6.4)( 98) 6.4 = 7.89 (.39) + 0.005 = 7.89.07 0.57 = 9.53 8.3 8.3 he equilibrium onstant is diminished by the temperature dependene of the enthalpy.