Worked Solutions to Problems

Transcription

1 rd International Cheistry Olypiad Preparatory Probles Wored Solutions to Probles. Water A. Phase diagra a. he three phases of water oeist in equilibriu at a unique teperature and pressure (alled the triple point): tr 7.6. C P tr 6. bar b. If pressure dereases, boiling point dereases, but elting point inreases (slightly).. Beyond this point, there is no distintion between liquid and vapour phases of water. Put alternatively, it is possible to have liquid to vapour transition by a ontinuous path going around the ritial point. (In ontrast, solid-liquid transition is disontinuous.) d., P. bar : liquid phase 7, P. bar : solid phase e. Below P 6. bar, ie heated isobarially will subliate to vapour. f. If l and v are the ole frations of water in liquid and vapour phases, l v l l l v v v 4.6 l l v l l l ( ) l v Mubai, India, July 5

2 rd International Cheistry Olypiad Preparatory Probles B. Clausius Clapeyron equation a. dp d H H olar enthalpy hange in phase transition olar hange in volue in phase transition. For ie-liquid water transition : H > <, sine ie is less dense than water. dp d < Sine is not large, the P- urve for this transition is steep, with a negative slope. hus derease of pressure inreases the elting point slightly. For liquid water - vapour transition ûh > û < dp d > Derease of pressure dereases the boiling point. b. Clausius - Clapeyron equation for (solid) liquid - vapour transition is dp d P H R vap his equation follows fro the Clapeyron equation under the assuptions:. apour follows ideal gas law.. Molar volue of the ondensed phase is negligible opared to olar volue of vapour phase.. If further is assued to be onstant (no variation with ), the eq. Hvap is integrated to give P ln P H R vap 5 5 Mubai, India, July

3 rd International Cheistry Olypiad Preparatory Probles Here P. bar, R 8. J ol - P. bar Hvap 4.66 J ol he estiate is based on assuptions, and.. For ie - liquid water equilibriu, use Clapeyron equation At 7.5, P. bar H. Assue that for a sall hange in, is onstant. Integrating the Clapeyron equation above P P 7.95, H ûh (fusion) ln 68 Jol..97 P P 8.9 J.84 bar ol P. bar he estiate is based on assuption. C. Irreversible ondensation a. On the P- plane, this equilibriu state is a solid phase (ie). Water in liquid phase at this teperature and pressure is not an equilibriu state - it is a superooled state that does not lie on the given P- plane. b. reating the etastable state as equilibriu state, we an go fro the superooled liquid state to the solid state at the sae teperature and pressure by a sequene of reversible steps.. Superooled liquid at -. C to liquid at C q nuber of oles Cp (liquid water) hange of teperature Mubai, India, July 5

4 rd International Cheistry Olypiad Preparatory Probles 8.5g 76.J ol 8.5 g ol. 445 J. liquid at C to ie at C q 8.5 g (.5) J g 955 J. Ie at C to ie at. C q nuber of oles Cp (liquid water) hange of tep J ol 8.5 g ol (. ) 75. J q q q q 8765 J Sine all the steps are at the onstant pressure of. bar, q H But H is independent of the path, i.e., it depends only on the end points. hus for the irreversible ondensation of superooled liquid to ie q H 8765 J. he atual irreversible path between the two end states of the syste is replaed by the sequene of three reversible steps, as above. For eah reversible step, an be alulated. n C p d n C p ln 8.5 g 8.5 g ol 76. J ol 7.5 ln J Mubai, India, July

5 rd International Cheistry Olypiad Preparatory Probles H 4.79 J g 8.5 g ol 7.5 J ol ln J syste. J sur q sur sur J univ syste sur.54 J he entropy of the universe inreases in the irreversible proess, as epeted by the Seond aw of herodynais.. van der Waals gases a. For a van der Waals gas ý P n R b P R n a R n a b R he ratio of the agnitudes of the seond and third ters on the right side is : b b P R, taing P nr up to zeroth order. n a a he ratio of the agnitudes of the fourth and third ters on the right side is : nb bp R i. Fro the ratios above, it follows that at suffiiently high teperature for any given pressure, the seond ter doinates the third and fourth ters. herefore, b P R ý > For sall P, Z nearly equals unity. Mubai, India, July 55

6 rd International Cheistry Olypiad Preparatory Probles ii. At lower teperatures, the third ter an be greater (in agnitude) than the seond ter. It ay be greater (in agnitude) than the fourth ter also, provided P is not too large. Sine the third ter has a negative sign, this iplies that Z an be less than unity. iii. For a ý b P R whih shows that Z inreases linearly with P. b. Heliu has negligible value of a. Graph () orresponds to He and () orresponds to.. Above >, only one phase (the gaseous phase) eists, that is the ubi equation in has only one real root. hus isother () orresponds to <. d. At, the three roots oinide at his is an infleion point. dp d d P d he first ondition gives R ( nb) n a () he ( hese R seond ondition gives n b) n b equations give and n a For He, a 7b R () For, C 8 Sine, ( ) is greater than (He), is liquefied ore readily than He Mubai, India, July

7 rd International Cheistry Olypiad Preparatory Probles e. W P d R b a d b R ln b 56.7 bar ol a. Rates and reation ehaniss a. Mehanis : d [HI] Sine the first step is fast, there is a pre - equilibriu : [I d[hi] [I] Mehanis : ] [I] [H ] [I ] [H ] [H ] [I ] d[hi] ' d[hi] [I ] [I ] d [I ] d [H ] ' [I ][H ] [H ][I ] Both ehaniss are onsistent with the observed rate law. b. i. E With E a a A e the Ea R R ln given 7 J ol nuerial values, Mubai, India, July 57

8 rd International Cheistry Olypiad Preparatory Probles ii. he ativation energy is greater than the bond dissoiation energy of I. Hene the seond step is rate deterining in both the ehaniss.. he ativation energy E a for the reverse reation is E ' a E a û J ol d. i. d [I] '' d [I ] [IAr] [I] [IAr][Ar] [I][Ar] '' [I] [I] [Ar] [Ar] ii. A possible reason why this is negative is that Ea is positive and less in agnitude than H, while H o is negative. A '' e we now û G A E a R e û6 R e e û þ û* R (E û þ ) a R û S he ativation energy for the overall reationis E a û þ 4. Enzye atalysis a. i. he differential rate equations for the Mihaelis-Menten ehanis are d [ES] [E] d[p] [ES] - ' [ES] [ES] () () Mubai, India, July

9 ' rd International Cheistry Olypiad Preparatory Probles In the steady-state approiation, d[es] () [E] Eq. () then gives [ES] ' (4) ow [E] [E] [ES] (5) where [E] is the total enzye onentration. Eqs. (4) and (5) gives [E] [ ES] where is the Mihaelis-Menten onstant. Fro eq. (), d[p] [E] (7) (6) Sine the baward rate is ignored, our analysis applies to the initial rate of foration of P and not lose to equilibriu. Further, sine the enzye onentration is generally uh saller than the substrate onentration, is nearly equal to in the initial stage of the reation. hus, aording to the Mihaelis-Menten ehanis, the initial rate versus substrate onentration is desribed by eq. (7), where is replaed by. For <<, Initialrate [E] i.e., initial rate varies linearly with. o (8) For >>, Initial rate [E] (9) i.e., for large substrate onentration, initial rate approahes a onstant value [E]. hus the indiated features of the graph are onsistent with Mihaelis-Menten ehanis. Mubai, India, July 59

10 rd International Cheistry Olypiad Preparatory Probles ii. he asyptoti value of initial rate is [E] Fro the graph, [E]. -6 M s With [E].5 9 M we get. s iii. Fro eq. (7), for, the initial rate is half the asyptoti value. Fro the graph, therefore, 5. 5 M For. 4 M, using eq. (7) again, 9 [. s ] [.5 M] [. Initial rate 5 4 [5. ] M [. ] M 4 ]M. 6 M s iv. We have 5. 5 M he enzye equilibrates with the substrate quily, that is the first step of equilibration between E, S and [ES] is very fast. his eans that is uh greater than. herefore, negleting above, 5 5. M he equilibriu onstant for the foration of ES fro E and S is, 5. M b. Fro the graph at the new teperature, [E] 6. 6 M s i.e., 6. M s.5 M 6 4. s Mubai, India, July

11 rd International Cheistry Olypiad Preparatory Probles Using Arrhenius relation for teperature dependene of rate onstant : A e E a R () where E a is the olar ativation energy. ( ) ( ) E a R e ( ) ln i.e. ( ) E () a R () ow., R 8. J ol (85) E a.4 J ol. i. he fration of the enzye that binds with the substrate is, fro eq. (6): [ES] () [E] where is nearly equal to in the initial stage of the reation. ow. 6 ol. M and 5. 5 M [ES] [E] (5.. 5 M. )M.98 early the whole of the enzye is bound with the substrate. ii. Fro eq. (7), Integrating the equation gives, d [E] Mubai, India, July 6

12 rd International Cheistry Olypiad Preparatory Probles [E] t ln () If at t, /, [E] ln. ol Here [E]... s, 5. 5 M, 9 M (4). M Substituting these values in eq. (4) gives 84 s hus 5% of the antibioti dose is inativated in 84 s. d. i. he differential rate equations for the situation are : d [ES] [E] [ES] [ES] (5) d [E] [I] [EI] [EI] (6) d [P] [ES] (7) where and are the forward and baward rate onstants for the enzye-inhibitor reation. Applying steady-state approiation to [ES] and [EI], [ES] [E] (8) [E] [I] and [EI] (9) ow [E] [E] [ES] [EI] () 6 6 Mubai, India, July

13 rd International Cheistry Olypiad Preparatory Probles Eliinating [E] and [EI] fro eqs. (8) to () gives : [ES] [E] [I] I (M) () d[p] [E] [I] I (M) () Here, I(M) to E and I. is the equilibriu onstant for the dissoiation of EI he degree of inhibition is i r r [I] I (M) Using eq. (), i () [I] I (M) For fied [I], i dereases with inrease in (opetitive inhibition). and for large, i, i.e., the inhibitor eases to play any role. ii. For sall i [I] (M) I [I] If r r, i 4 4 i.e., [I] I (M).5 4 M he inhibitor onentration required to redue the rate of inativation by a fator of 4 is.5 4 M; i.e.,.5 µol in a volue of.. Mubai, India, July 6

14 rd International Cheistry Olypiad Preparatory Probles 5. Shrödinger equation a. i. One-diensional Shrödinger equation for a free partile of ass :! d % E%! d h Œ where E stands for the energy of the partile and ψ its wave funtion. ii. he boundary onditions are : ψ () ψ() Only % n () sin n Œ satisfies the required boundary onditions. Other funtions are not possible wave funtions of the eletron in a one-diensional rigid bo. iii.! d d sin n Œ! Œ n sin nœ En! Œ n h n 8 iv. Ground state (n ) Œ % () sin First eited state (n ) % () Œ sin Seond eited state (n ) % () Œ sin uber of nodes in points. % n ψ () ψ () ψ () n, apart fro the nodes at the end / / / Mubai, India, July

15 rd International Cheistry Olypiad Preparatory Probles v. % % () () b. In the eaple % ( ) Œ[ sin sin sin Œ[ Œ[ he first three energy levels are: d d ( is hosen os to be real Œ ) d E h. 8 9 J E 4 E J E 9 E.98 9 J In the ground state, the four eletrons will oupy the levels E and E, eah with two eletrons. E E E E E E Ground state owest eited state Mubai, India, July 65

16 rd International Cheistry Olypiad Preparatory Probles he lowest eitation energy E E 6. 9 J. he ondition that ψ(φ) is single valued deands that (φ) (φ π) % % e iλφ e iλ(φπ) e i πλ i.e. λ, where, ±, ±, ±,. his shows that angular oentu projetion ( z ) annot be an arbitrary real nuber but an have only disrete values:!, where is a positive or negative integer (inluding zero). 6. Atoi and oleular orbitals A. Atoi orbitals a. i. % s e r a o % s dv 4 Œ a o 4 Œ [ ŒD o ] a o 4 Œ a o ( hosen to be real) % s ao [ Œ a ] e o r ii. Probability of finding an eletron between r and r dr 4Œ r r a [ ŒD ] e dr his is a aiu at r r a, given by d dr r e r a r ra Mubai, India, July

17 rd International Cheistry Olypiad Preparatory Probles his gives r a a he s eletron is ost liely to be found in the neighborhood of r a. b. % s at r a odal surfae is a sphere of radius a at % p z Œ odal surfae is the y plane. % d z at os, i.e., os ± odal surfaes are ones with these values of half-angle, one above the y plane and the other below it. (ote: all three wave funtions vanish as r. At r, ψls does not vanish, but the other two wave funtions vanish.). Eah eletron in n shell of heliu ato has energy Z eff.6 e Heliu ground state energy Z eff 7. e Energy of He ground state e Ionization energy ( 54.4 Z eff 7.) e 4.46 e his gives Z eff.7 B. Moleular orbitals a. % and % are bonding orbitals % ~ and are antibonding orbitals % ~ Bonding orbital o nodal surfae between the nulei. Eletroni energy has a iniu at a ertain internulear distane. Qualitative reason: eletron has onsiderable probability of being between the nulei and thus has attrative potential energy due to both the nulei. Mubai, India, July 67

18 rd International Cheistry Olypiad Preparatory Probles Antibonding orbital odal surfae between the nulei. Eletroni energy dereases onotonially with internulear distane. Hene bound state is not possible. b. R e. D.6 (5.6).76 e. It will dissoiate to a hydrogen ato in s state and a bare hydrogen nuleus (proton). d. he two eletrons oupy the sae oleular orbital with the lowest energy. By Pauli s priniple, their spins ust be antiparallel. Hene the total eletroni spin is zero. e. In the first eited state of H, one eletron is in ψ (bonding orbital) and the other in ψ (antibonding orbital). It will dissoiate into two hydrogen atos. f. Using the aufbau priniple, in the ground state two eletrons of He are in ψ (bonding orbital) and two in ψ (antibonding orbital). he bond order is ½ ( ) herefore, bound He is unstable and diffiult to detet. However, if one or ore eletrons are elevated fro the antibonding orbital to (higher energy) bonding orbitals, the bond order beoes greater than zero. his is why it is possible to observe He in eited states. 7. Fission a. 5 9 U n 94 8 Sr 4 54 Xe n 5 9 U n 4 56 Ba 9 6 r n b. he net nulear reation is 5 9 U n 94 4 Zr 4 58 Ce n 6e (Q) he energy released is Q [ ( 5 U) ( 94 Zr) ( 4 Ce) n 6 e ] Mubai, India, July