CHEMISTRY: ART, SCIENCE, FUN
THEORETICALEXAMINATION
SOLUTION andGRADING SCHEME
JULY 20, 2007MOSCOW, RUSSIA
Official English version
Problem 1. Proton tunneling
1.1.1The structures of propanedial and two of its isomersO=CH-CH2-CH=O1 markOHCHOCHCH1 markHOHHCCCHO1 mark3 marks maximum1.1.2
Acidic hydrogen atom is in CH2 (in enol forms acidic hydrogen is in OH).
1 mark
Acidity of СН2 group is caused by the stability of carbanion due to conjugation with two carbonylgroups. The first answer is correct.
2 marks
3 marks maximum1.2.1 The distance between two minima on the energy curve is 0.06 nm. In a purely aldehyde form
HH
OO
such distance between two possible positions of proton is impossible. Tunneling takes place only inenol Z-form:OHCHHOCHHOCCHOCHCH1 mark for each structure2 marks maximum1.3.1Expressions and plots of probability density122222ù(а)Y2(x,0)=Y+Y+Y-YYmark é(x)(x)(x)(x) = (x) 1 LRLRL?2?The probability density is concentrated in the left well:2
YL2-0,06-0,04L-0,020,000,02Distance, nmR0,040,060.5 marks(b) In the middle of the time interval?p?1é2ù1 markY2?x, ÷= YL(x)+Y2R(x)??2w2è?The probability density has a symmetric form, a proton is delocalized between two wells:(YL+YR)/222-0,06-0,04L-0,020,000,02Distance, nmR0,040,060.5 marks?p?1é22222mark(c)Y2?x, = YR(x) 1 ÷= ?YL(x)+YR(x)-YL(x)+YR(x)ù?èw?2The probability density is concentrated in the right well:YR2-0,06-0,04L-0,020,000,02Distance, nmR0,040,060.5 marks4.5 marks maximum1.3.2The probability of finding the proton in the left well is 1/2, because probability function issymmetric, and both wells are identical.
2 marks
3
1.3.3The time of transfer from one well to another ist =p /w.3.14t = 11= 4.85×10-12 s.2 marks6.48×10The proton velocity:0.06×10-9V = -12= 12 m/s.2 marks4.85×104 marks maximum1.3.4The uncertainty of proton position is approximately equal to half of the distance between minima,that is 0.03 nm (0.06 nm will be also accepted).1 markThe minimal uncertainty of velocity can be obtained from the uncertainty relation:1.055×10-34h3 marksDV ?1000 m/s.= = 0.0012mDx×0.03×10-92×236.02×10Comparing this uncertainty with the velocity 12 m/s we see that the notion of proton velocity duringtransfer from one well to another is senseless. Therefore, proton tunneling is a purely quantumphenomenon and cannot be described in classical terms.The second conclusion is correct.2 marks6 marks maximum4
Problem 2. Nanochemistry
2.1.1The Gibbs energy and the equilibrium constant of reaction (1)00DrG500(1) = DG0= -198.4+219.1 = 20.7kJ/mol 0.5 marksf,500(CoO,s)-DGf,500(H2O,g) K e= -0DrG500(1)RTe= -207008.314×500= 6.88×10-30.51 mark maximummarks2.1.2The standard Gibbs energy of the reaction (1) with the spherical cobalt nanoparticles of radiusra isDGo(1,r) G0= (CoO,s)+G0(H,g)-G0(HO,g)-G0(Co) =r500abulk,50050025002sphV(Co)?2s?0000G500= (CoO,s)+G500(H2,g)-G500(H2O,gas)-?G500(Co,s) +Co-gas÷=raè?V(Co)2so(=DrG500 1)-Co-gas;raMCo10-6×59.0m3-6V(Co) = = = 6.6×10;8.90molr(Co)for spherical particles withra = 10–8, 10–9 m one gets, respectively2sCo-gasV(Co) =210 and 2100 J/mol.raDGo(1,r) is equal to 20.5 (a), and 18.6 (b) kJ/mol, respectively.r500aThe equilibrium constant is calculated from the equation?DGo(1,r)?K(1,ra) = exp?-r500a÷;?÷RTè?-8-9K(1,ra) = 7.22′10-3; ra= 10 mK(1,ra) = 11.4′10-3; ra= 10 m2 marks maximum5
国际化学奥林匹克竞赛-39thIChOTheoreticalsolutions
