Question

21C Transition-state theory 899 from which it follows that the Arrhenius factor A can be identi fied as harpoon mechanism, the product yield is expected to be opti- mal if the second pulse is applied when the wavepacket is at a point where the Xe-distance is just right for electron transfer from Xe to I, to occur. This type of control of the 12+ Xe reac tion has been demonstrated. Are Be Transition state theory Afactor (210.15b) 210.2 Thermodynamic aspects The entropy of activation is negative because throughout the system reactant species are combining to form reactive pairs. However, if there is a reduction in entropy below what would be expected for the simple encounter of A and B. then the Arrhenius factor A will be smaller than that expected on the basis of simple collision theory. Indeed, we can identify that additional reduction in entropy. Atceries as the origin of the steric factor of collision theory, and write The statistical thermodynamic version of transition state the- ory rapidly runs into difficulties because only in some cases is anything known about the structure of the activated complex However, the concepts that it introduces, principally that of an equilibrium between the reactants and the activated complex, have motivated a more general, empirical approach in which the activation process is expressed in terms of thermodynamic functions. Transition state theory actor (210.15c) Thus, the more complex the steric requirements of the encoun- ter, the more negative the value of Asset and the smaller the value of P. (a) Activation parameters If we accept that is an equilibrium constant (despite one mode of Chaving been discarded), we can express it in terms of a Gibbs energy of activation, A'G, through the definition A'GE-RTIR Definition Gbbs energy of activation (2011) All the A'X in this section are standard thermodynamic quan tities, A'x, but we shall omit the standard state sign to avoid overburdening the notation. Then the rate constant becomes Brief illustration 21CA Activation parameters The reaction of propylxanthate ion in acetic acid buffer solu tions can be represented by the equation A+HP. Near 30°C, A=2.05 x 10 dm.mol-'s'. To evaluate the entropy of activation at 30°C we first note that because the reaction is in solution the e' of eqn 21.15 should be replaced by e (see foot note 1), and then use eqn 21C.Isb in the form KT RT hp =1.592x10 dm.mol's KT RT k= 121C.121 Therefore, 2.05x10'dm'mol's" ASR in ex(1.592x10'dm'mol =-25.41K mol =Rin 0.047... Because G=H-TS, the Gibbs energy of activation can be divided into an entropy of activation, AS, and an enthalpy of activation, A'H. by writing Entropy and AG=A'H-TA'S Definition enthalpy of (210.13) activation When eqn 21C.13 is used in eqn 21C.12 and is absorbed into the entropy term, we obtain KT RT Besi-MRT (210.14 Self-test 210.5 The reaction A+H-Pin solution has A=6.92 x 102 dm'mols. Evaluate the entropy of activation at 25°C Answer:-1.1 kmol The formal definition of activation energy (en 200.2 of Topic 20D, E =RT-a Ink,/OT)), then gives E =A'H+2RT, so! Gibbs energies, enthalpies, and entropies of activation (and volumes and heat capacities of activation) are widely used to report experimental reaction rates, especially for organic reactions in solution. They are encountered when relation ships between equilibrium constants and rates of reaction are explored using correlation analysis, in which In K (which is equal to -4,G/RT) is plotted against Ink, (which is propor tional to -AG/RT). In many cases the correlation is linear, sig- nifying that as the reaction becomes thermodynamically more kweBee (210.15al For reactions of the type A+BP in the gas phase. F.- S'H2RT. For such reactions in solution - AHRT 900 21 Reaction dynamics

21C.1(a) The reaction of propylxanthate ion in acetic acid buffer solutions
has the mechanism A− +H+→P. Near 30 °C the rate constant is given by the
empirical expression kr=(2.05×1013) e−(8681K)/T dm3mol−1 s−1. Evaluate the
energy and entropy of activation at 30 °C.
21C.1(b) The reaction A− +H+→P has a rate constant given by the empirical
expression kr=(6.92×1012)e−(5925K)/T dm3mol−1 s−1. Evaluate the energy and
entropy of activation at 25 °C.

Please explain why they calculated H=E-RT like that like why is it 8.3145 × (5925 K – 298 K) = 46.9... Like there's only multiplication there. There is supposed to be minus H=E-RT

Please also show how they calculated for A factor.

Answer 1

م ہے 2 The Archeniul equation i kr = A 7 ELRT compare it with given equation Ky = 6.92X.1012 e 5925k/t 소 - Eq = 5925 K R A = 6.92 x 1012 EQ = 5925K XR Now, substitute it in the given equation OH = EQ-RT 5925 RxR - RT Take "R" common out stH = R(5925k-T) T=25°c = 273+25K = 298 K ATH = R(5925K - 298k) R is gal constant R = 8.3145 J K mot! s*H = 8.3145 I C'mot! (5925k-2981)

The comparison of equation @ and @ gives A = 6.928 10 12 dm3 moi's e Ats/R Ky = Be -o*HIRT Ath = ER-RT . subukitute it above equation 26#SIR (EX-RT)/RT kr = B Beats/R e e Asie Falet Relof = B e ats/R ē Galerie B e -EaIRT e Beatsir e where A=eBeatsir AFSIR -Ea/RT Ky = де Take In on both sidel InA = IneBeats/R) = 'metlmB+ line = 1+ ImB + otsir me = It In B + Atsir (1) AtsiR = INA-IMB-1 sts = R [ın (All) -1 in (678) -1